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## Notes of Jingyin Huang’s Cambridge lecture 11-01-2017

Groups quasi-isometric to right angled Artin groups

Joint with Bruce Kleiner.

1. Result

Counterexamples by Behrstock-Neumann, Ruiz-Kazachkov-Zakharov

Theorem 1 (Huang 2014) Let ${G}$ and ${H}$ be quasi-isometric right angled Artin groups. If ${Out(G)}$ is finite, then ${G}$ and ${H}$ are commensurable.

Theorem 2 (Huang 2014) Let ${G}$ and ${H}$ be quasi-isometric f.g. groups. Assume ${G}$ is a right angled Artin group, on a star-rigid graph (any automorphism fixing the star of a vertex pointwise is the identity) without induced 4-cycle. If ${Out(G)}$ is finite, then ${G}$ and ${H}$ are commensurable.

Conversely, if ${\Gamma}$ is a graph which contains an induced 4-cycle, this fails: there exists a f.g. generated group quasi-isometric but not commensurable to ${G(\Gamma)}$.

2. Proof

2.1. Analysis of quasi-isometries

The Salvetti complex ${X}$ of a graph ${\Gamma}$ consists of gluing tori, one for each clique of ${\Gamma}$. It is NPC (in fact, the cubical completion of the Cayley graph), it is a classifying sace for ${G:=G(\Gamma)}$.

We show that any quasi-isometry of ${X}$ permutes maximal flats. Conversely, from a morphism of the space of flats of ${X}$, we are able to reconstruct a map ${G\rightarrow G}$. Furthermore, provided ${Out(G)}$ is finite, every quasi-isometry is within finite distance of a flat preserving bijection (BKS 2007, Huang 2013).

2.2. Construction of an adapted space

Assume ${H}$ is quasi-isometric to ${G}$. Then ${H}$ acts by flat-preserving bijections. They may change the order of flats along standard geodesics. Add little branches to ${X}$ in order separate flats from standard geodesics. One obtains a new ${CAT(0)}$ space ${Y}$ on which ${H}$ acts isometrically. ${Y}$ has a labelling and an orientation. If ${\Gamma}$ is star-rigid, ${H}$ preserves the labelling, but need not preserve the orientation. It does not suffice to go to a finite index subgroup.

2.3. Specialness

Say a NPC cube complex is special if it admits a local isometric embedding into some Salvetti complex. ${Y/H}$ is a graph of spaces, where vertex spaces are again of the form ${Y'/H'}$. We prove a combination theorem for virtually special NPC cube complexes, inspired from Haglund-Wise 2009. There is a malnormality assumption, that Wise removed in his celebrated 2011 paper. In Haglund and Wise’s theory, hyperbolicity is essential. What saves us is that Salvetti complexes are better behaved than general NPC complexes: one may collapse bad tubes, get a hyperbolic cube complex and apply Haglund-Wise, and go back to the original space. Also, failure of malnormality is limited if no induced 4-cycles. Nevertheless, we need to modify Haglund-Wise’s canonical completion.

2.4. Counterexamples

If there are induced 4-cycles, the proof collapses. The construction of counterexamples involve replacing reducible lattices by irreducible lattices in products of trees.

Agol’s theorem (hyperbolic ${CAT(0)}$-cube complexes are virtually special) is more general than Wise’s. No assumption that the complex be a graph of spaces. Gives me hope.


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# Fibonacci numbers as numbers of strings over alphabets

i am asked to show that for $n \geq 3$ $F_{n} =$ the number of strings of size $n - 2$ over the alphabet ${a,b}$ with no consecutive $b$s.

Base case is obvious. $F_{n+1} = F_{n} + F_{n-1}$ so we use strong induction and get that its the sum of the number of strings of size $n-2$ and $n-3$ with no consecutive $b$s. So i figure we should count how many such strings there are:

$2^{n-2} + 2^{n-3} - (n-3)(n-4)! - (n-4)(n-5)!,$ where the subtractions come from the fact that, for example, in the case of $n-2$ long strings we have $n-3$ consecutive pairs and the rest can be permuted. I am quite sure this is wrong, because what if say two of the symbols are the same, so that the permutation leaves the string unchanged, and i am assuming in this case it shouldn't count as a new string.

In any case, i am stuck here, so any help would be appreciated. Thanks

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Suppose that we want to construct the sequence of $a$, $b$ with length $n+1$.

If the last term is $a$, other $n$ terms can be any sequence which satisfies the condition, so we have $F_n$ sequences.

If the last term is $b$, $n$th term has to be $a$, since $n$th and $n+1$th terms cannot be both $b$. Now, other $n-1$ terms can be any sequence which satisfies the condition, so we have $F_{n-1}$ sequences.

Thus we have: $$F_{n+1}=F_n+F_{n-1}.$$

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$F_{n}$ is the number of sequences of length $n-2.$ If the last term is $a,$ you are left with $n$ terms – user168593 Aug 8 '14 at 3:15
@user168593 Oh, it seems to be. Shifting subscripts might be needed. – The Great Seo Aug 8 '14 at 3:40

Let $a_n$ and $b_n$ be the number of strings of size $n-2$ that ends with $a$ and $b$, respectively. Then you have the following recursion. $$a_n = a_{n-1}+b_{n-1}$$ and $$b_n = a_{n-1}$$, with the obvious identity $F_n = a_n+b_n$. Now you can easily prove the rest using induction.

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I am sorry, but why is $F_{n} = a_{n} + b_{n}$ so obvious? – user168593 Aug 8 '14 at 3:20
Because a string can end with either $a$ or $b$, according to your formulation of the problem. – dezdichado Aug 8 '14 at 16:51

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# Tag Archives: trigonometry

## Deep Unit Circle Understanding

Chris Bolognese started a new Honors Trigonometry unit a few days ago and posted a new problem set, welcoming feedback.  Clicking on the document image gives you access to Chris’ problem set.

This post is my feedback.  I wrote it as a conversation with Chris.  it

FEEDBACK…

Question 2 suggests you have already discussed special angles (or your students recalled them from last year’s algebra class).  What I love about this problem set is how you shift the early focus away from these memorized special angles and onto the deep symmetry underlying the unit circle (and all of trigonometry).  Understanding deep structure grants always far more understanding than rote memorizations ever does!

HINTS OF IDENTITIES TO COME…

The problem set’s initial exploration of trig symmetry starts in parts a-d of Question 1, asking students to justify statements like $sin(x+\pi )=sin(x)$ and $cos(2\pi -x)=cos(x)$.  The symmetry is nice, and I hope you refer back to this problem set when your class turns to its formal exploration of trig identities, helping them see that not all proofs of identities require algebraic justifications.

CONSIDERATION 1:  Put some creative power in your students’ hands.  Challenge them to discover additional algebraic/transformational identities like Questions 1a-1d.  For example, what is the relationship between $sin \left( \frac{\pi}{2} + x \right)$ and $sin \left( \frac{\pi}{2} - x \right)$ ?  There are LOTS of symmetry statements they could write.

CONSIDERATION 2:  I would shift Question 1e to Question 4, as 1e is the basis of the Pythagorean identities you explore in the later group.

CONSIDERATION 3:  I suggest dropping Question 4d and asking students to discover another equation showing another Pythagorean relationship between trig functions.  That cotangent and cosecant have not yet been used is hopefully a loud, silent hint.

CONSIDERATION 4:  They’re not ready for this one until they see sinusoidal graphs, but another trig pre-identity I love is graphing $y=cos^2 (x)$ and asking students to write an equivalent equation using only translations and dilations.

Unlike the symmetric relationships in 1a-1d, I don’t think you can actually KNOW it is true without algebraic relationships.  I use problems like this to set up and justify the transition to identities.

VERY CLEVER SUMMATIONS …

I don’t recall ever seeing something like your Question 3 before.  In my opinion, is the gold nugget in the assignment, especially with your students’ prior exposure to special angles.

CONSIDERATION 5:  In some ways the large number of addends in  gives away that there must be a simpler approach than the problem suggests on its surface.  What about something that suggests a direct solution if you don’t invoke the symmetry, like ?

CONSIDERATION 6:  Closely related to this, why not shamelessly take advantage of the memorized values to see if students notice the symmetry to notice a simpler approach?  I suggest , with sine or cosine.

CONSIDERATION 7:  My prior two examples tweaked your initial  problem.  Like Consideration 1, why not challenge your students to develop their own summations?  I bet they can develop some clever alternatives.

DO YOU UNDERSTAND RADIANS???

In Question 3, you had some nice explorations using degrees.  Unfortunately, I’m not aware of many equally clever early questions involving radians that aren’t degree-oriented in disguise.  Here are my final suggestions to address this gap.

CONSIDERATION 8:  Without using any technology, rank

sin(1), cos(2), tan(3), cot(4), sec(5), csc(6)

in ascending order.  Note that all angles are expressed in radian measures.  One pair of expressions is very difficult.  (A softer version of this question ranks only sin(1), cos(2), & tan(3).)

CONSIDERATION 9:  One pair of expressions in Consideration 8 is very difficult to rank without technology.  Which pair is this and why is it so difficult to rank?  Exchange angles or functions to change this question in a way that makes it easier to rank.

Thanks for the fun and thoughtful problem set.

## Quadratics + Tangent = ???

Here’s a very pretty problem I encountered on Twitter from Mike Lawler 1.5 months ago.

I’m late to the game replying to Mike’s post, but this problem is the most lovely combination of features of quadratic and trigonometric functions I’ve ever encountered in a single question, so I couldn’t resist.  This one is well worth the time for you to explore on your own before reading further.

My full thoughts and explorations follow.  I have landed on some nice insights and what I believe is an elegant solution (in Insight #5 below).  Leading up to that, I share the chronology of my investigations and thought processes.  As always, all feedback is welcome.

WARNING:  HINTS AND SOLUTIONS FOLLOW

Investigation  #1:

My first thoughts were influenced by spoilers posted as quick replies to Mike’s post.  The coefficients of the underlying quadratic, $A^2-9A+1=0$, say that the solutions to the quadratic sum to 9 and multiply to 1.  The product of 1 turned out to be critical, but I didn’t see just how central it was until I had explored further.  I didn’t immediately recognize the 9 as a red herring.

Basic trig experience (and a response spoiler) suggested the angle values for the tangent embedded in the quadratic weren’t common angles, so I jumped to Desmos first.  I knew the graph of the overall given equation would be ugly, so I initially solved the equation by graphing the quadratic, computing arctangents, and adding.

Insight #1:  A Curious Sum

The sum of the arctangent solutions was about 1.57…, a decimal form suspiciously suggesting a sum of $\pi/2$.  I wasn’t yet worried about all solutions in the required $[0,2\pi ]$ interval, but for whatever strange angles were determined by this equation, their sum was strangely pretty and succinct.  If this worked for a seemingly random sum of 9 for the tangent solutions, perhaps it would work for others.

Unfortunately, Desmos is not a CAS, so I turned to GeoGebra for more power.

Investigation #2:

In GeoGebra, I created a sketch to vary the linear coefficient of the quadratic and to dynamically calculate angle sums.  My procedure is noted at the end of this post.  You can play with my GeoGebra sketch here.

The x-coordinate of point G is the sum of the angles of the first two solutions of the tangent solutions.

Likewise, the x-coordinate of point H is the sum of the angles of all four angles of the tangent solutions required by the problem.

Insight #2:  The Angles are Irrelevant

By dragging the slider for the linear coefficient, the parabola’s intercepts changed, but as predicted in Insights #1, the angle sums (x-coordinates of points G & H) remained invariant under all Real values of points A & B.  The angle sum of points C & D seemed to be $\pi/2$ (point G), confirming Insight #1, while the angle sum of all four solutions in $[0,2\pi]$ remained $3\pi$ (point H), answering Mike’s question.

The invariance of the angle sums even while varying the underlying individual angles seemed compelling evidence that that this problem was richer than the posed version.

Insight #3:  But the Angles are bounded

The parabola didn’t always have Real solutions.  In fact, Real x-intercepts (and thereby Real angle solutions) happened iff the discriminant was non-negative:  $B^2-4AC=b^2-4*1*1 \ge 0$.  In other words, the sum of the first two positive angles solutions for $y=(tan(x))^2-b*tan(x)+1=0$ is $\pi/2$ iff $\left| b \right| \ge 2$, and the sum of the first four solutions is $3\pi$ under the same condition.  These results extend to the equalities at the endpoints iff the double solutions there are counted twice in the sums.  I am not convinced these facts extend to the complex angles resulting when $-2.

I knew the answer to the now extended problem, but I didn’t know why.  Even so, these solutions and the problem’s request for a SUM of angles provided the insights needed to understand WHY this worked; it was time to fully consider the product of the angles.

Insight #4:  Finally a proof

It was now clear that for $\left| b \right| \ge 2$ there were two Quadrant I angles whose tangents were equal to the x-intercepts of the quadratic.  If $x_1$ and $x_2$ are the quadratic zeros, then I needed to find the sum A+B where $tan(A)=x_1$ and $tan(B)=x_2$.

From the coefficients of the given quadratic, I knew $x_1+x_2=tan(A)+tan(B)=9$ and $x_1*x_2=tan(A)*tan(B)=1$.

Employing the tangent sum identity gave

$\displaystyle tan(A+B) = \frac{tan(A)+tan(B)}{1-tan(A)tan(B)} = \frac{9}{1-1}$

and this fraction is undefined, independent of the value of $x_1+x_2=tan(A)+tan(B)$ as suggested by Insight #2.  Because tan(A+B) is first undefined at $\pi/2$, the first solutions are $\displaystyle A+B=\frac{\pi}{2}$.

Insight #5:  Cofunctions reveal essence

The tangent identity was a cute touch, but I wanted something deeper, not just an interpretation of an algebraic result.  (I know this is uncharacteristic for my typically algebraic tendencies.)  The final key was in the implications of $tan(A)*tan(B)=1$.

This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving

$\displaystyle tan(A) = \frac{1}{tan(B)} = cot(B)$.

But cotangent is also the co-function–or complement function–of tangent which gave me

$tan(A) = cot(B) = tan \left( \frac{\pi}{2} - B \right)$.

Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \frac{\pi}{2} - B$, or $A+B = \frac{\pi}{2}$.  Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation,

$(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\pi ]$ and any $\left| b \right| \ge 2$,

is always $3\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions.  QED

Insight #6:  Generalizing the Domain

The posed problem can be generalized further by recognizing the period of tangent: $\pi$.  That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\pi$ each, as shown in the GeoGebra construction above.

Insights 4 & 5 proved the sum of the angles at points C & D was $\pi/2$.  Employing the periodicity of tangent,  the x-coordinate of $E = C+\pi$ and $F = D+\pi$, so the sum of the angles at points E & F is $\frac{\pi}{2} + 2 \pi$.

Extending the problem domain to $[0,3\pi ]$ would add $\frac{\pi}{2} + 4\pi$ more to the solution, and a domain of $[0,4\pi ]$ would add an additional $\frac{\pi}{2} + 6\pi$.  Pushing the domain to $[0,k\pi ]$ would give total sum

$\displaystyle \left( \frac{\pi}{2} \right) + \left( \frac{\pi}{2} +2\pi \right) + \left( \frac{\pi}{2} +4\pi \right) + \left( \frac{\pi}{2} +6\pi \right) + ... + \left( \frac{\pi}{2} +2(k-1)\pi \right)$

Combining terms gives a general formula for the sum of solutions for a problem domain of $[0,k\pi ]$

$\displaystyle k * \frac{\pi}{2} + \left( 2+4+6+...+2(k-1) \right) * \pi =$

$\displaystyle = k * \frac{\pi}{2} + (k)(k-1) \pi =$

$\displaystyle = \frac{\pi}{2} * k * (2k-1)$

For the first solutions in Quadrant I, $[0,\pi]$ means k=1, and the sum is $\displaystyle \frac{\pi}{2}*1*(2*1-1) = \frac{\pi}{2}$.

For the solutions in the problem Mike originally posed, $[0,2\pi]$ means k=2, and the sum is $\displaystyle \frac{\pi}{2}*2*(2*2-1) = 3\pi$.

I think that’s enough for one problem.

APPENDIX

My GeoGebra procedure for Investigation #2:

• Graph the quadratic with a slider for the linear coefficient, $y=x^2-b*x+1$.
• Label the x-intercepts A & B.
• The x-values of A & B are the outputs for tangent, so I reflected these over y=x to the y-axis to construct A’ and B’.
• Graph y=tan(x) and construct perpendiculars at A’ and B’ to determine the points of intersection with tangent–Points C, D, E, and F in the image below
• The x-intercepts of C, D, E, and F are the angles required by the problem.
• Since these can be points or vectors in Geogebra, I created point G by G=C+D.  The x-intercept of G is the angle sum of C & D.
• Likewise, the x-intercept of point H=C+D+E+F is the required angle sum.

## Envelope Curves

My precalculus class recently returned to graphs of sinusoidal functions with an eye toward understanding them dynamically via envelope curves:  Functions that bound the extreme values of the curves. What follows are a series of curves we’ve explored over the past few weeks.  Near the end is a really cool Desmos link showing an infinite progression of periodic envelopes to a single curve–totally worth the read all by itself.

GETTING STARTED

As a simple example, my students earlier had seen the graph of $f(x)=5+2sin(x)$ as $y=sin(x)$ vertically stretched by a magnitude of 2 and then translated upward 5 units.  In their return, I encouraged them to envision the function behavior dynamically instead of statically.  I wanted them to see the curve (and the types of phenomena it could represent) as representing dynamic motion rather than a rigid transformation of a static curve.  In that sense, the graph of f oscillated 2 units (the coefficient of sine in f‘s equation) above and below the line $y=5$ (the addend in the equation for f).  The curves $y=5+2=7$ and $y=5-2=3$ define the “Envelope Curves” for $y=f(x)$.

When you graph $y=f(x)$ and its two envelope curves, you can picture the sinusoid “bouncing” between its envelopes.  We called these ceiling and floor functions for f.  Ceilings happen whenever the sinusoid term reaches its maximum value (+1), and floors when the sinusoidal term is at its minimum (-1).

Those envelope functions would be just more busy work if it stopped there, though.  The great insights were that anything you added to a sinusoid could act as a midline with the coefficient, AND anything multiplied by the sinusoid is its amplitude–the distance the curve moves above and below its midline.  The fun comes when you start to allow variable expressions for the midline and/or amplitudes.

VARIABLE MIDLINES AND ENVELOPES

For a first example, consider $y= \frac{x}{2} + sin(x)$.  By the reasoning above, $y= \frac{x}{2}$ is the midline.  The amplitude, 1, is the coefficient of sine, so the envelope curves are $y= \frac{x}{2}+1$ (ceiling) and $y= \frac{x}{2}-1$ (floor).

That got their attention!  Notice how easy it is to visualize the sine curve oscillating between its envelope curves.

For a variable amplitude, consider $y=2+1.2^{-x}*sin(x)$.  The midline is $y=2$, with an “amplitude” of $1.2^{-x}$.  That made a ceiling of $y=2+1.2^{-x}$ and a floor of $y=2-1.2^{-x}$, basically exponential decay curves converging on an end behavior asymptote defined by the midline.

SINUSOIDAL MIDLINES AND ENVELOPES

Now for even more fun.  Convinced that both midlines and amplitudes could be variably defined, I asked what would happen if the midline was another sinusoid?  For $y=cos(x)+sin(x)$, we could think of $y=cos(x)$ as the midline, and with the coefficient of sine being 1, the envelopes are $y=cos(x)+1$ and $y=cos(x)-1$.

Since cosine is a sinusoid, you could get the same curve by considering $y=sin(x)$ as the midline with envelopes $y=sin(x)+1$ and $y=sin(x)-1$.  Only the envelope curves are different!

The curve $y=cos(x)+sin(x)$ raised two interesting questions:

1. Was the addition of two sinusoids always another sinusoid?
2. What transformations of sinusoidal curves could be defined by more than one pair of envelope curves?

For the first question, they theorized that if two sinusoids had the same period, their sum was another sinusoid of the same period, but with a different amplitude and a horizontal shift.  Mathematically, that means

$A*cos(\theta ) + B*sin(\theta ) = C*cos(\theta -D)$

where A & B are the original sinusoids’ amplitudes, C is the new sinusoid’s amplitude, and D is the horizontal shift.  Use the cosine difference identity to derive

$A^2 + B^2 = C^2$  and $\displaystyle tan(D) = \frac{B}{A}$.

For $y = cos(x) + sin(x)$, this means

$\displaystyle y = cos(x) + sin(x) = \sqrt{2}*cos \left( x-\frac{\pi}{4} \right)$,

and the new coefficient means $y= \pm \sqrt{2}$ is a third pair of envelopes for the curve.

Very cool.  We explored several more sums and differences with identical periods.

WHAT HAPPENS WHEN THE PERIODS DIFFER?

Try a graph of $g(x)=cos(x)+cos(3x)$.

Using the earlier concept that any function added to a sinusoid could be considered the midline of the sinusoid, we can picture the graph of g as the graph of $y=cos(3x)$ oscillating around an oscillating midline, $y=cos(x)$:

IF you can’t see the oscillations yet, the coefficient of the $cos(3x)$ term is 1, making the envelope curves $y=cos(x) \pm 1$.  The next graph clear shows $y=cos(3x)$ bouncing off its ceiling and floor as defined by its envelope curves.

Alternatively, the base sinusoid could have been $y=cos(x)$ with envelope curves $y=cos(3x) \pm 1$.

Similar to the last section when we added two sinusoids with the same period, the sum of two sinusoids with different periods (but the same amplitude) can be rewritten using an identity.

$cos(A) + cos(B) = 2*cos \left( \frac{A+B}{2} \right) * cos \left( \frac{A-B}{2} \right)$

This can be proved in the present form, but is lots easier to prove from an equivalent form:

$cos(x+y) + cos(x-y) = 2*cos(x) * cos(y)$.

For the current function, this means $y = cos(x) + cos(3x) = 2*cos(x)*cos(2x)$.

Now that the sum has been rewritten as a product, we can now use the coefficient as the amplitude, defining two other pairs of envelope curves.  If $y=cos(2x)$ is the sinusoid, then $y= \pm 2cos(x)$ are envelopes of the original curve, and if $y=cos(x)$ is the sinusoid, then $y= \pm 2cos(2x)$ are envelopes.

In general, I think it’s easier to see the envelope effect with the larger period function.  A particularly nice application connection of adding sinusoids with identical amplitudes and different periods are the beats musicians hear from the constructive and destructive sound wave interference from two instruments close to, but not quite in tune.  The points where the envelopes cross on the x-axis are the quiet points in the beats.

A STUDENT WANTED MORE

In class last Friday, my students were reviewing envelope curves in advance of our final exam when one made the next logical leap and asked what would happen if both the coefficients and periods were different.  When I mentioned that the exam wouldn’t go that far, she uttered a teacher’s dream proclamation:  She didn’t care.  She wanted to learn anyway.  Making up some coefficients on the spot, we decided to explore $f(x)=2sin(x)+5cos(2x)$.

Assuming for now that the cos(2x) term is the primary sinusoid, the envelope curves are $y=2sin(x) \pm 5$.

That was certainly cool, but at this point, we were no longer satisfied with just one answer.  If we assumed sin(x) was the primary sinusoid, the envelopes are $y=5cos(2x) \pm 2$.

Personally, I found the first set of envelopes more satisfying, but it was nice that we could so easily identify another.

With the different periods, even though the  coefficients are different, we decided to split the original function in a way that allowed us to use the $cos(A)+cos(B)$ identity introduced earlier.  Rewriting,

$f(x)=2sin(x)+5cos(2x) = 2cos \left( x - \frac{ \pi }{2} \right) + 2cos(2x) + 3cos(2x)$ .

After factoring out the common coefficient 2, the first two terms now fit the $cos(A) + cos(B)$ identity with $A = x - \frac{ \pi }{2}$ and $B=2x$, allowing the equation to be rewritten as

$f(x)= 2 \left( 2*cos \left( \frac{x - \frac{ \pi }{2} + 2x }{2} \right) * cos \left( \frac{x - \frac{ \pi }{2} - 2x }{2} \right) \right) + 3cos(2x)$

$\displaystyle = 4* cos \left( \frac{3}{2} x - \frac{ \pi }{4} \right) * cos \left( - \frac{1}{2} x - \frac{ \pi }{4} \right) + 3cos(2x)$.

With the expression now containing three sinusoidal expressions, there are three more pairs of envelope curves!

Arguably, the simplest approach from this form assumes $cos(2x)$ from the $latex$3cos(2x)\$ term as the sinusoid, giving $y=2sin(x)+2cos(2x) \pm 3$ (the pre-identity form three equations earlier in this post) as envelopes.

We didn’t go there, but recognizing that new envelopes can be found simply by rewriting sums creates an infinite number of additional envelopes.  Defining these different sums with a slider lets you see an infinite spectrum of envelopes.  The image below shows one.  Here is the Desmos Calculator page that lets you play with these envelopes directly.

If the $cos \left( \frac{3}{3} x - \frac{ \pi}{4} \right)$term was the sinusoid, the envelopes would be $y=3cos(2x) \pm 4cos \left( - \frac{1}{2} x - \frac{ \pi }{4} \right)$.  If you look closely, you will notice that this is a different type of envelope pair with the ceiling and floor curves crossing and trading places at $x= \frac{\pi}{2}$ and every $2\pi$ units before and after.  The third form creates another curious type of crossing envelopes.

CONCLUSION:

In all, it was fun to explore with my students the many possibilities for bounding sinusoidal curves.  It was refreshing to have one student excited by just playing with the curves to see what else we could find for no other reason than just to enjoy the beauty of these periodic curves.  As I reflected on the overall process, I was even more delighted to discover the infinite spectrum of envelopes modeled above on Desmos.

I hope you’ve found something cool here for yourself.

## Squares and Octagons, A compilation

My last post detailed my much-too-long trigonometric proof of why the octagon formed by connecting the midpoints and vertices of the edges of a square into an 8-pointed star is always 1/6 of the area of the original square.

My proof used trigonometry, and responses to the post on Twitter  and on my ‘blog showed many cool variations.  Dave Radcliffe thought it would be cool to have a compilation of all of the different approaches.  I offer that here in the order they were shared with me.

Method 1:  My use of trigonometry in a square.  See my original post.

Method 2:  Using medians in a rectangle from Tatiana Yudovina, a colleague at Hawken School.

Below, the Area(axb rectangle) = ab = 16 blue triangles, and
Area(octagon) = 4 blue triangles – 2 red deltas..

Now look at the two green, similar triangles.  They are similar with ratio 1/2, making

Area(red delta) = $\displaystyle \frac{b}{4} \cdot \frac{a}{6} = \frac{ab}{24}$, and

Area(blue triangle) = $\displaystyle \frac{1}{16} ab$

So, Area(octagon) = $\displaystyle 2 \frac{ab}{24}-4\frac {ab}{16}=\frac{1}{6}ab$.

QED

Method 3:  Using differences in triangle areas in a square (but easily extended to rectangles)from @Five_Triangles (‘blog here).

Full solution here.

Method 4:  Very clever shorter solution using triangle area similarity in a square also from @Five_Triangles (‘blog here).

Full second solution here.

Method 5:  Great option Using dilated kitesfrom Dave Radcliffe posting as @daveinstpaul.

Full pdf and proof here.

Method 6:  Use fact that triangle medians trisect each other from Mike Lawler posting as @mikeandallie.

Tweet of solution here.

Method 7:  Use a coordinate proof on a specific square from Steve Ingrassia, a colleague at Hawken School.  Not a quick proof like some of the geometric solutions, but it’s definitely different than the others.

If students know the formula for finding the area of any polygon using its coordinates, then they can prove this result very simply with nothing more than simple algebra 1 techniques.   No trig is required.

The area of polygon with vertices (in either clockwise or counterclockwise order, starting at any vertex) of $(x_1, y_1)$, $(x_2, y_2)$, …, $(x_n, y_n)$ is

$\displaystyle Area = \left| \frac{(x_1y_2-x_2y_1)+(x_2y_3-x_3y_2)+...+(x_{n-1}y_n-x_ny_{n-1})}{2} \right|$

Use a 2×2 square situated with vertices at (0,0), (0,2), (2,2), and (2,0).  Construct segments connecting each vertex with the midpoints of the sides of the square, and find the equations of the associated lines.

• L1 (connecting (0,0) and (2,1):    y = x/2
• L2 (connecting (0,0) and (1,2):   y=2x
• L3 (connecting (0,1) and (2,0):  y= -x/2 + 1
• L4 (connecting (0,1) and (2,2):  y= x/2 + 1
• L5 (connecting (0,2) and (1,0):  y = -2x + 2
• L6 (connecting (0,2) and (2,1):  y= -x/2 + 2
• L7 (connecting (1,2) and (2,0):  y = -2x + 4
• L8 (connecting (2,2) and (1,0):  y = 2x – 2

The 8 vertices of the octagon come at pairwise intersections of some of these lines, which can be found with simple substitution:

• Vertex 1 is at the intersection of L1 and L3:   (1, 1/2)
• Vertex 2 is at the intersection of L3 and L5:  (2/3, 2/3)
• Vertex 3 is at the intersection of L2 and L5:  (1/2, 1)
• Vertex 4 is at the intersection of L2 and L4:  (2/3, 4/3)
• Vertex 5 is at the intersection of L4 and L6:  (1, 3/2)
• Vertex 6 is at the intersection of L6 and L7:  (4/3, 4/3)
• Vertex 7 is at the intersection of L7 and L8:  (3/2, 1)
• Vertex 8 is at the intersection of L1 and L8:  (4/3, 2/3)

Using the coordinates of these 8 vertices in the formula for the area of the octagon, gives

$\displaystyle \frac{ \left| 1/3 +1/3+0+(-1/3)+(-2/3)+(-1/3)+0 \right|}{2} = \frac{2}{3}$

Since the area of the original square was 4, the area of the octagon is exactly 1/6th of the area of the square.

Thanks, everyone, for your contributions.

## Old school integral

This isn’t going to be one of my typical posts, but I just cracked a challenging indefinite integral and wanted to share.

I made a mistake solving a calculus problem a few weeks ago and ended up at an integral that looked pretty simple.  I tried several approaches and found many dead ends before finally getting a breakthrough.  Rather than just giving a proof, I thought I’d share my thought process in hopes that some students just learning integration techniques might see some different ways to attack a problem and learn to persevere through difficult times.

In my opinion, most students taking a calculus class would never encounter this problem.  The work that follows is clear evidence why everyone doing math should have access to CAS (or tables of integrals when CAS aren’t available).

Here’s the problem:

Integrate $\int \left( x^2 \cdot \sqrt{1+x^2} \right) dx$.

For convenience, I’m going to ignore in this post the random constant that appears with indefinite integrals.

While there’s no single algebraic technique that will work for all integrals, sometimes there are clues to suggest productive approaches.  In this case, the square root of a binomial involving a constant and a squared variable term suggests a trig substitution.

From trig identities, I knew $tan^2 \theta + 1 = sec^2 \theta$, so my first attempt was to let $x=tan \theta$, which gives $dx=sec^2 \theta d\theta$.  Substituting these leads to

$(tan \theta)'=sec^2 \theta$, claiming two secants for the differential in a reversed chain rule, but left a single secant in the expression, so I couldn’t make the trig identities work because odd numbers of trigs don’t convert easily using Pythagorean identities.  Then I tried using $(sec \theta)'=sec \theta \cdot tan \theta$, leaving a single tangent after accounting for the potential differential–the same problem as before.  A straightforward trig identity wasn’t going to do the trick.

Then I recognized that the derivative of the root’s interior is $2x$.  It was not the exterior $x^2$, but perhaps integration by parts would work.  I tried $u=x \longrightarrow u'=dx$ and $v'=x\sqrt{1+x^2} dx \longrightarrow v=\frac{1}{2} \left( 1+x^2 \right)^{3/2} \cdot \frac{2}{3}$.  Rewriting the original integral gave

The remaining integral still suggested a trig substitution, so I again tried $x =tan \theta$ to get

but the odd number of secants led me to the same dead end from trigonometric identities that stopped my original attempt.  I tried a few other variations on these themes, but nothing seemed to work.  That’s when I wondered if the integral even had a closed form solution.  Lots of simple looking integrals don’t work out nicely; perhaps this was one of them.  Plugging the integral into my Nspire CAS gave the following.

OK, now I was frustrated.  The solution wasn’t particularly pretty, but a closed form definitely existed.  The logarithm was curious, but I was heartened by the middle term I had seen with a different coefficient in my integration by parts approach.  I had other things to do, so I employed another good problem solving strategy:  I quit working on it for a while.  Sometimes you need to allow your sub-conscious to chew on an idea for a spell.  I made a note about the integral on my To Do list and walked away.

As often happens to me on more challenging problems, I woke this morning with a new idea.  I was still convinced that trig substitutions should work in some way, but my years of teaching AP Calculus and its curricular restrictions had blinded me to other possibilities.  Why not try a hyperbolic trig substitution? In many ways, hyperbolic trig is easier to manipulate than circular trig.  I knew

$\frac{d}{dt}cosh(t)=sinh(t)$ and $\frac{d}{dt}sinh(t)=cosh(t)$,

and the hyperbolic identity

$cosh^2t - sinh^2t=1 \longrightarrow cosh^2t=1+sinh^2t$.

(In case you haven’t worked with hyperbolic trig functions before, you can prove these for yourself using the definitions of hyperbolic sine and cosine:  $cosh(x)=\frac{1}{2}\left( e^x + e^{-x} \right)$ and $sinh(x)=\frac{1}{2}\left( e^x - e^{-x} \right)$.)

So, $x=sinh(A) \longrightarrow dx=cosh(A) dA$, and substitution gives

Jackpot!  I was down to an even number of (hyperbolic) trig functions, so Pythagorean identities should help me revise my latest expression into some workable form.

To accomplish this, I employed a few more hyperbolic trig identities:

1. $sinh(2A)=2sinh(A)cosh(A)$
2. $cosh(2A)=cosh^2(A)+sinh^2(A)$
3. $cosh^2(A) = \frac{1}{2}(cosh(2A)+1)$
4. $sinh^2(A) = \frac{1}{2}(cosh(2A)-1)$

(All of these can be proven using the definitions of sinh and cosh above.  I encourage you to do so if you haven’t worked much with hyperbolic trig before.  I’ve always liked the close parallels between the forms of circular and hyperbolic trig relationships and identities.)

If you want to evaluate $\int x^2 \sqrt{x^2+1} dx$ yourself, do so before reading any further.

Using equations 3 & 4, expanding, and then equation 3 again turns the integral into something that can be integrated directly.

The integral was finally solved!  I then used equations 1 & 2 to rewrite the expression back into hyperbolic functions of A only.

The integral was solved using the substitution $x=sinhA \longrightarrow A=sinh^{-1}x$ and (using $cosh^2A-sinh^2A=1$), $coshA=\sqrt{x^2+1}$.  Substituting back gave:

but that didn’t match what my CAS had given.  I could have walked away, but I had to know if I had made an error someplace or just had found a different expression for the same quantity.  I knew the inverse sinh could be replaced with a logarithm via a quadratic expression in $e^x$.

Well, that explained the presence of the logarithm in the CAS solution, but I was still worried by the cubic in my second term and the fact that my first two terms were a sum whereas the CAS’s solution’s comparable terms were a difference.  But as a former student once said, “If you take care of the math, the math will take care of you.”  These expressions had to be the same, so I needed to complete one more identity–algebraic this time.  Factoring, rewriting, and re-expanding did the trick.

What a fun problem (for me) this turned out to be.  It’s absolutely not worth the effort to do this every time when a CAS or integral table can drop the solution so much more quickly, but it’s also deeply satisfying to me to know why the form of the solution is what it is.  It’s also nice to know that I found not one, but three different forms of the solution.

Morals:  Never give up.  Trust your instincts. Never give up. Try lots of variations on your instincts. And never give up!

## A Student’s Powerful Polar Exploration

I posted last summer on a surprising discovery of a polar function that appeared to be a horizontal translation of another polar function.  Translations happen all the time, but not really in polar coordinates.  The polar coordinate system just isn’t constructed in a way that makes translations appear in any clear way.

That’s why I was so surprised when I first saw a graph of $\displaystyle r=cos \left( \frac{\theta}{3} \right)$.

It looks just like a 0.5 left translation of $r=\frac{1}{2} +cos( \theta )$ .

But that’s not supposed to happen so cleanly in polar coordinates.  AND, the equation forms don’t suggest at all that a translation is happening.  So is it real or is it a graphical illusion?

I proved in my earlier post that the effect was real.  In my approach, I dealt with the different periods of the two equations and converted into parametric equations to establish the proof.  Because I was working in parametrics, I had to solve two different identities to establish the individual equalities of the parametric version of the Cartesian x- and y-coordinates.

As a challenge to my precalculus students this year, I pitched the problem to see what they could discover. What follows is a solution from about a month ago by one of my juniors, S.  I paraphrase her solution, but the basic gist is that S managed her proof while avoiding the differing periods and parametric equations I had employed, and she did so by leveraging the power of CAS.  The result was that S’s solution was briefer and far more elegant than mine, in my opinion.

S’s Proof:

Multiply both sides of $r = \frac{1}{2} + cos(\theta )$ by r and translate to Cartesian.

$r^2 = \frac{1}{2} r+r\cdot cos(\theta )$
$x^2 + y^2 = \frac{1}{2} \sqrt{x^2+y^2} +x$
$\left( 2\left( x^2 + y^2 -x \right) \right) ^2= \sqrt{x^2+y^2} ^2$

At this point, S employed some CAS power.

[Full disclosure: That final CAS step is actually mine, but it dovetails so nicely with S’s brilliant approach. I am always delightfully surprised when my students return using a tool (technological or mental) I have been promoting but hadn’t seen to apply in a particular situation.]

S had used her CAS to accomplish the translation in a more convenient coordinate system before moving the equation back into polar.

Clearly, $r \ne 0$, so

$4r^3 - 3r = cos(\theta )$ .

In an attachment (included below), S proved an identity she had never seen, $\displaystyle cos(\theta) = 4cos^3 \left( \frac{\theta }{3} \right) - 3cos \left( \frac{\theta }{3} \right)$ , which she now applied to her CAS result.

$\displaystyle 4r^3 - 3r = 4cos^3 \left( \frac{\theta }{3} \right) - 3cos \left( \frac{\theta }{3} \right)$

So, $\displaystyle r = cos \left( \frac{\theta }{3} \right)$

Therefore, $\displaystyle r = cos \left( \frac{\theta }{3} \right)$ is the image of $\displaystyle r = \frac{1}{2} + cos(\theta )$ after translating $\displaystyle \frac{1}{2}$ unit left.  QED

Simple. Beautiful.

Obviously, this could have been accomplished using lots of by-hand manipulations.  But, in my opinion, that would have been a horrible, potentially error-prone waste of time for a problem that wasn’t concerned at all about whether one knew some Algebra I arithmetic skills.  Great job, S!

S’s proof of her identity, $\displaystyle cos(\theta) = 4cos^3 \left( \frac{\theta }{3} \right) - 3cos \left( \frac{\theta }{3} \right)$ :

## Transformations II and a Pythagorean Surprise

In my last post, I showed how to determine an unknown matrix for most transformations in the xy-plane and suggested that they held even more information.

Given a pre-image set of points which can be connected to enclose one or more areas with either clockwise or counterclockwise orientation.  If a transformation T represented by matrix $[T]= \left[ \begin{array}{cc} A & C \\ B & D \end{array}\right]$ is applied to the pre-image points, then the determinant of $[T]$, $det[T]=AD-BC$, tells you two things about the image points.

1. The area enclosed by similarly connecting the image points is $\left| det[T] \right|$ times the area enclosed by the pre-image points, and
2. The orientation of the image points is identical to that of the pre-image if $det[T]>0$, but is reversed if $det[T]<0$.  If $det[T]=0$, then the image area is 0 by the first property, and any question about orientation is moot.

In other words, $det[T]$ is the area scaling factor from the pre-image to the image (addressing the second half of CCSSM Standard NV-M 12 on page 61 here), and the sign of $det[T]$ indicates whether the pre-image and image have the same or opposite orientation, a property beyond the stated scope of the CCSSM.

Example 1: Interpret $det[T]$ for the matrix representing a reflection over the x-axis, $[T]=\left[ r_{x-axis} \right] =\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]$.

From here, $det[T]=-1$.  The magnitude of this is 1, indicating that the area of an image of an object reflected over the line $y=x$ is 1 times the area of the pre-image—an obviously true fact because reflections preserve area.

Also, $det \left[ r_{x-axis} \right]<0$ indicating that the orientation of the reflection image is reversed from that of its pre-image.  This, too, must be true because reflections reverse orientation.

Example 2: Interpret $det[T]$ for the matrix representing a scale change that doubles x-coordinates and triples y-coordinates, $[T]=\left[ S_{2,3} \right] =\left[ \begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array} \right]$.

For this matrix, $det[T]=+6$, indicating that the image’s area is 6 times that of its pre-image area, while both the image and pre-image have the same orientation.  Both of these facts seem reasonable if you imagine a rectangle as a pre-image.  Doubling the base and tripling the height create a new rectangle whose area is six times larger.  As no flipping is done, orientation should remain the same.

Example 3 & a Pythagorean Surprise:  What should be true about  $det[T]$ for the transformation matrix representing a generic rotation of $\theta$ units around the origin,  $[T]=\left[ R_\theta \right] = \left[ \begin{array}{cc} cos( \theta ) & -sin( \theta ) \\ sin( \theta ) & cos( \theta ) \end{array} \right]$ ?

Rotations preserve area without reversing orientation, so $det\left[ R_\theta \right]$ should be +1.  Using this fact and computing the determinant gives

$det \left[ R_\theta \right] = cos^2(\theta ) + sin^2(\theta )=+1$ .

In a generic right triangle with hypotenuse C, leg A adjacent to acute angle $\theta$, and another leg B, this equation is equivalent to $\left( \frac{A}{C} \right) ^2 + \left( \frac{B}{C} \right) ^2 = 1$, or $A^2+B^2=C^2$, the Pythagorean Theorem.  There are literally hundreds of proofs of this theorem, and I suspect this proof has been given sometime before, but I think this is a lovely derivation of that mathematical hallmark.

Conclusion:  While it seems that these two properties about the determinants of transformation matrices are indeed true for the examples shown, mathematicians hold out for a higher standard.   I’ll offer a proof of both properties in my next post.


text:
# Some new sufficient conditions for $2p$-Hamilton-biconnectedness of graphs

@article{Chen2017SomeNS,
title={Some new sufficient conditions for \$2p\$-Hamilton-biconnectedness of graphs},
author={Ming-Zhu Chen and Xiaodong Zhang},
journal={arXiv: Combinatorics},
year={2017}
}
• Published 2017
• Mathematics
• arXiv: Combinatorics
A balanced bipartite graph $G$ is said to be $2p$-Hamilton-biconnected if for any balanced subset $W$ of size $2p$ of $V(G)$, the subgraph induced by $V(G)\backslash W$ is Hamilton-biconnected. In this paper, we prove that "Let $p\geq0$ and $G$ be a balanced bipartite graph of order $2n$ with minimum degree $\delta(G)\geq k$, where $n\geq 2k-p+2$ and $k\geq p$. If the number of edges $e(G)>n(n-k+p-1)+(k+2)(k-p+1),$ then $G$ is $2p$-Hamilton-biconnected except some exceptions." Furthermore… Expand

#### References

SHOWING 1-10 OF 16 REFERENCES
A method in graph theory
• Computer Science, Mathematics
• Discret. Math.
• 1976
Biclosure and bistability in a balanced bipartite graph
• Mathematics, Computer Science
• J. Graph Theory
• 1995
On the First Eigenvalue of Bipartite Graphs
• Mathematics, Computer Science
• Electron. J. Comb.
• 2008


text:
# 1/3x=4/27

## Simple and best practice solution for 1/3x=4/27 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.

If it's not what You are looking for type in the equation solver your own equation and let us solve it.

## Solution for 1/3x=4/27 equation:

1/3x=4/27
We move all terms to the left:
1/3x-(4/27)=0

Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R

We add all the numbers together, and all the variables
1/3x-(+4/27)=0
We get rid of parentheses
1/3x-4/27=0
We calculate fractions
There is no solution for this equation

`


text:
# The Unapologetic Mathematician

## Engel’s Theorem

When we say that a Lie algebra $L$ is nilpotent, another way of putting it is that for any sufficiently long sequence $\{x_i\}$ of elements of $L$ the nested adjoint

$\mathrm{ad}(x_n)\left[\dots\mathrm{ad}(x_2)\left[\mathrm{ad}(x_1)[y]\right]\right]$

is zero for all $y\in L$. In particular, applying $\mathrm{ad}(x)$ enough times will eventually kill any element of $L$. That is, each $x\in L$ is ad-nilpotent. It turns out that the converse is also true, which is the content of Engel’s theorem.

But first we prove this lemma: if $L\subseteq\mathfrak{gl}(V)$ is a linear Lie algebra on a finite-dimensional, nonzero vector space $V$ that consists of nilpotent endomorphisms, then there is some nonzero $v\in V$ for which $l(v)=0$ for all $l\in L$.

If $\dim(L)=1$ then $L$ is spanned by a single nilpotent endomorphism, which has only the eigenvalue zero, and must have an eigenvector $v$, proving the lemma in this case.

If $K$ is any nontrivial subalgebra of $L$ then $\mathrm{ad}(k)\in\mathrm{End}(L)$ is nilpotent for all $k\in K$. We also get an everywhere-nilpotent action on the quotient vector space $L/K$. But since $\dim(K)<\dim(L)$, the induction hypothesis gives us a nonzero vector $x+K\in L/K$ that gets killed by every $k\in K$. But this means that $[k,x]\in K$ for all $k\in K$, while $x\notin K$. That is, $K$ is strictly contained in the normalizer $N_L(K)$.

Now instead of just taking any subalgebra, let $K$ be a maximal proper subalgebra in $L$. Since $K$ is properly contained in $N_L(K)$, we must have $N_L(K)=L$, and thus $K$ is actually an ideal of $L$. If $\dim(L/K)>1$ then we could find an even larger subalgebra of $L$ containing $K$, in contradiction to our assumption, so as vector spaces we can write $L\cong K+\mathbb{F}z$ for any $z\in L\setminus K$.

Finally, let $W\subseteq V$ consist of those vectors killed by all $\in K$, which the inductive hypothesis tells us is a nonempty collection. Since $K$ is an ideal, $L$ sends $W$ back into itself: $k(l(w))=l(k(w))-[l,k](w)=0$. Picking a $z\in L\setminus K$ as above, its action on $W$ is nilpotent, so it must have an eigenvector $w$ with $z(w)=0$. Thus $l(w)=0$ for all $l\in L=K+\mathbb{F}z$.

So, now, to Engel’s theorem. We take a Lie algebra $L$ consisting of ad-nilpotent elements. Thus the algebra $\mathrm{ad}(L)\subseteq\mathfrak{gl}(L)$ consists of nilpotent endomorphisms on the vector space $L$, and there is thus some nonzero $z\in L$ for which $[L,z]=0$. That is, $L$ has a nontrivial center — $z\in Z(L)$.

The quotient $L/Z(L)$ thus has a lower dimension than $L$, and it also consists of ad-nilpotent elements. By induction on the dimension of $L$ we assume that $L/Z(L)$ is actually nilpotent, which proves that $L$ itself is nilpotent.

August 22, 2012 - Posted by | Algebra, Lie Algebras


text:
Soccer impulse problem

1. Oct 26, 2005

Punchlinegirl

Recent studies have raised concern about heading' in youth soccer (i.e., hitting the ball with the head). A soccer player heads' a size 3 ball deflecting it by 54.0°, and keeps its speed of 10.30 m/s constant. A size 3 ball has a mass of approximately 2.000 kg. What is the magnitude of the impulse which the player must impart to the ball?
I started by finding the change in velocity.
In the x-direction:
the final velocity= cos 54 * 10.30 * 2.000 = 12.1 m/s
the inital velocity = 10.30 * 2.000 = 8.3 m/s
Change in velocity in x= 3.8 m/s
In the y-direction:
the final velocity= sin 54 * 10.30 * 2.000= 16.6
the initial velocity= 0 m/s
change in velocity in y= 16.6 m/s
Then i used the pythagorean theorem.
$$sq rt 16.6^2 + 3.8^2$$ = 17.0 m/s
I multiplied this by the mass 2.000 to get the impulse= 34.0 kg*m/s
This wasn't right.. can someone tell me what I'm doing wrong?thanks

2. Oct 26, 2005

daniel_i_l

The impulse is the change in momentum. So first you have to find the new speed vector (the new x and y speeds, multiply (the sqrt of them squared as you did) by mass to get the momentum vector (then get the new x and y momentum components with trig), and then subtract the initial momentum vector from the knew one (x-axis minus x-axis and y=axis minus y-axis), then find the size of the new vector to get the impulse.
-Also, when you calculated the velocities, why did you multiply by 2.000?
you found the new momentum, not the impulse.

3. Oct 26, 2005

Punchlinegirl

I'm sorry but I really don't follow you. I think you're saying what I did was right but it's momentum not impulse. I'm not sure how to get the initial momentum.

4. Oct 26, 2005

Staff: Mentor

The speed doesn't change, just the direction. Find the change in momentum by subtracting the initial momentum vector from the final momentum vector. Then find the magnitude of that change, which is all you need. (Hint: Pretend the initial momentum is in the +x direction.)

5. Oct 26, 2005

daniel_i_l

You get the initial momentum by simply multiplying the speed by the mass, as Doc AI said, all of the initial speed is along the x-axis.

6. Oct 26, 2005

Punchlinegirl

I see what I was doing wrong. Thanks


text:
Math Help - fourier analysis

1. fourier analysis

hello......
i am completely unaware of fourier analysis. please suggest me some elementary text to grab the topic.........and please look into the question........

2. there is a formula that involves integration to find the fourier series of a function. it is a series that represents signals by an infinite sum of sines and cosines (it can be finite, but it is still an infinite sum with amplitudes equal to zero). The sines represent the odd parts of a function (sensible considering sine is odd) and cosines represents the even parts of a function (again, sensible). A third conceptual rule is that the non-sinusoidal component, often called the DC-component, represents the average value of the signal over the range that the fourier series represents (and is a cosine with a frequency of zero + is even)

You can eliminate many possibilities by using these three rules. First off, this function is even. Therefore, (b), since it has a sinewave, is wrong. Also, since this function never goes negative, it has an average value. This makes (c) wrong.

you can find the equation here. Fourier Sine and Cosine Series

I assume there may be another little trick you can use to eliminate one of the last two possibilities. However, I do not know it. You may just have to suffer the integration.


text:
Prealgebra (7th Edition)

Percent increase is found by dividing the difference by the original amount. Find the difference. $26460-25200=1260$ Find the percentage. $1260\div 25200=0.05=5$%


text:
Date

Introduction

I have been using the OSH Park's 4 layer process a lot on my own projects. It has FR408 substrate that has better controlled permittivity and lower losses than ordinary FR-4 that other low cost PCB manufacturers use. In my opinion currently it is the best low cost process for making RF PCBs.

My previous boards have worked pretty well, but I decided to make a test board that I can use to characterize the process better.

Test board for TRL calibration.

In the above picture is the test board that I made. It has two 50 ohm microstrip lines of different length, one open microstrip line, one microstrip line terminated with 50 ohm resistor and line with 0402 footprint that I populated with a 1 nF capacitor. I'm using this same type of capacitor as a general DC blocking capacitor in my VNA so I'm interested in finding out how it performs at high frequencies.

Block diagram of the board.

Using this board and TRL calibration, effect of the connectors can be calibrated out resulting in accurate measurements of the transmission line parameters. Accurate measurement of the capacitor is also obtained, which is useful in a system design. Other 0402 footprint passives can also be populated in place of capacitor to measure them at high frequencies.

If you are not familiar with TRL calibration it is a different method for calibrating a VNA. The normal SOLT calibration uses Short, Open, Load and Thru calibration standard measurements and sets the measurement reference plane to the end of the coaxial connector. Calibration standards need to be known accurately or otherwise there will be errors in the measurements.

TRL calibration uses at least two lines of different lengths and one reflect. No short, open or load standards are needed and unlike standards used with SOLT calibration, lines or reflect don't need to be fully known. Line lengths need to be known, but their electrical behaviour doesn't need to be known accurately. Phase of the reflection coefficient of reflect standard needs to be known within +- 90 degree accuracy. Real reflection coefficient and propagation constant of the transmission lines will be solved during the calibration.

After the TRL calibration reference plane is placed after the SMA connectors on the board. The exact reference plane location is in the middle of the shortest line. For example the capacitor test structure on the board would include the SMA connectors if measured using the SOLT calibration, but with TRL calibration the reference planes are right next to the capacitor pads.

There are few limitations on the TRL calibration:

• Length difference between the lines should ideally be a quarter wavelength. If the measured phase is a multiple of 180 degrees, the resulting system of equation is singular and calibration fails. This results in a limited bandwidth for the calibration but it can be extended by using multiple lines with different lengths.
• Reference impedance of the calibration is the characteristic impedance of the transmission lines and not 50 or 75 ohms as is usual. The measured S-parameters are referenced to this possibly frequency dependent and complex impedance. If the characteristic impedance is not known accurately, the measured results might not be very useful.

Propagation constant

Propagation constant of the transmission line characterizes how phase and amplitude of a wave traveling in a transmission line varies as a function of distance. During the TRL calibration propagation constant is solved and it is one of the outputs of the calibration. Propagation constant varies significantly as a function of frequency and thus can't be really called a constant but that is the established term.

S21 coefficient of a transmission line with length $$l$$ can be calculated using propagation constant as: $$S_{21} = e^{-\gamma l}$$, where $$\gamma = \alpha + j \beta$$ is the propagation constant.

Real part $$\alpha$$ of the propagation constant causes loss and imaginary part $$\beta$$ causes phase shift. Units of $$\alpha$$ are in Nepers/meter, which can be converter to more practical units of dB/m with equation $$20\log_{10}(e)\alpha$$. $$\beta$$ includes the frequency dependence and looks like a very steep straight line when plotted as is. It can be normalized by dividing by $$2\pi f$$, but I think that effective permittivity of the transmission line is a more useful quantity.

Effective permittivity of the transmission line is related to the speed of electromagnetic wave on the transmission line via equation: $$v = \frac{c}{\sqrt{\epsilon_{\text{eff}}}}$$, where $$c$$ is the speed of light. Wavelength of the electromagnetic wave in a transmission line can be calculated using effective permittivity with: $$\lambda = \frac{v}{f} = \frac{c}{f\sqrt{\epsilon_{\text{eff}}}}$$. Since many RF circuits need transmission lines with accurately known length in wavelengths, knowing the effective permittivity accurately is very useful.

Effective permittivity of the transmission line depends on the permittivities of the surrounding materials. With stripline or coaxial cable where the signal conductor is completely surrounded by one material the effective permittivity of the transmission line is equal to the permittivity of the surrounding material. If the TRL calibration would be done for a stripline, the effective permittivity calculated from the solved propagation constant would directly give the permittivity of the material. With microstrip some of the field travels also on the air and the effective permittivity is somewhere between permittivities of substrate material and air. The exact value depends on the line geometry.

Effective permittivity can be calculated from the propagation constant with:

$$\epsilon_{\text{eff}} = - \left(\frac{\gamma}{2\pi f/c}\right)^2$$

Measurements

All of the following measurements are done using the second version of my homemade VNA.

Unknown thru

Before doing the TRL calibration let's first do a regular calibration to the SMA connectors to see what the S-parameters look like with the SMA connectors.

I'm using scikit-rf Python library for doing the calibration. In the library there are several possible calibration methods to choose from. There is ordinary SOLT that most of the commercial VNAs have builtin, but also several more advanced calibration routines. This time I'm using UnknownThru calibration, because I know the short, open and match standards I'm using well because I have measured them with a commercial VNA, but I don't have a measurement of the through standard. Unknown thru calibration doesn't requires knowledge of the thru standard S-parameters so it is a good calibration to use in this case.

Below is the code for doing UnknownThru calibration:

import sys
import skrf
#Use skrf plot style
skrf.stylely()

#Load the measured uncalibrated S-parameter files
oo = skrf.Network('open_open.s2p')
through = skrf.Network('through.s2p')

sw_terms = skrf.Network('sw_terms.s2p')
switch_terms = (sw_terms.s11, sw_terms.s22)

#Network to plot
dut = skrf.Network(sys.argv[1])

#Calibration kit measurements
o_i = skrf.Network('../cal_kit/open.s1p')
s_i = skrf.Network('../cal_kit/short.s1p')

frequency = through.frequency

#Interpolate the calibration standard S-parameters
o_i = o_i.interpolate(frequency)
s_i = s_i.interpolate(frequency)
l_i = l_i.interpolate(frequency)

#Make two-port networks from two one-ports
oo_i = skrf.two_port_reflect(o_i, o_i)
ls_i = skrf.two_port_reflect(l_i, s_i)
sl_i = skrf.two_port_reflect(s_i, l_i)

#Make ideal through Network
through_s = [[[0,1],[1,0]] for f in through.f]
through_i = skrf.Network(s=through_s, frequency=frequency)

cal = skrf.UnknownThru(
measured = [oo, sl, ls, through],
ideals = [oo_i, sl_i, ls_i, through_i],
n_thrus = 1,
isolation= ll,
switch_terms = switch_terms
)

cal.apply_cal(dut).plot_s_db()

The measured networks follow the std1-std2 naming rule, where std1 is the standard connected to port 1 and std2 on port 2. Measurements of open, short and load are needed on the both ports. Standards are usually measured as one port on both ports, but in this case standards are connected to both ports and they are measured as a two port. Through also needs to be measured, but with this calibration real through S-parameters don't need to be known. Through can even be a line on the board being measured or any other transmissive circuit that has S21 equal to S12.

Switch terms are needed to account for reflection coefficient of the port switch inside the VNA that causes different termination depending on the measurement direction. They are measured with the ports connected together using any transmissive standard. They can't be calculated from the S-parameters and instead raw receiver measurements are required. It's not an issue with my homemade VNA since accessing the raw receiver outputs is easy, but it can be tricky with some older VNAs. They can also be extracted from SOLT measurements if the receiver measurements are not possible.

Short line calibrated to SMA connectors.

Close up to S21 of short line.

Above is the shortest 5mm long line on the board calibrated to the SMA connectors. It is matched quite well which suggest that the SMA connector matching is also good and that the line is not at least very far off from 50 ohms. Loss is smaller than 0.5 dB even at 6 GHz.

Solved SMA through S-parameters.

Close up to S21 of SMA through.

Above are the solved S-parameters of the SMA through (right side on this picture) that I used for the calibration. Loss is okay, but matching is not spectacular. I guess the through is not exactly 50 ohms. Same results are obtained if the line on the board is used as a through in the calibration.

Long line calibrated to SMA connectors.

Close up to S21 of the long line.

Longer 18mm line however isn't performing as well as the shorter one. There are resonances at 4.2 and 5.8 GHz causing increased losses. I measured also another identical board and it has the same resonances. I'm not exactly sure why it happens and I haven't investigated it much yet. Maybe the bottom ground plane works as an antenna? Board has 4 layers and the bottom three layers have a ground plane. The two bottom ground planes are connected to the microstrip ground plane only near the SMA connectors and it is possible that the bottom ground plane is resonating.

Series capacitor test structure calibrated to SMA connectors.

Close up to S21 of the capacitor.

Capacitor test structure doesn't have resonance at 4.2 GHz, but maybe there is a small resonance at 5.8 GHz? Compared to the through line, capacitor test structure has little higher losses as is expected.

TRL

There are three different TRL calibrations in the scikit-rf: Ordinary TRL calibration (TRL), Least squares multiline TRL (MultilineTRL) and NIST Multiline TRL (NISTMultilineTRL).

MultilineTRL is just an alias of TRL, but difference between the TRL and NISTMultilineTRL calibrations is that they behave differently when using multiple lines. Ordinary least squares TRL uses the same system of equations with multiple lines as with just one, but solves it with least squares. It is simple to implement but doesn't avoid singularities if any pair of lines has phase difference that is multiple of 180 degrees. NIST multiline TRL combines the lines more intelligently avoiding the singularities as long as there is one pair of lines that has non-singular phase difference. It allows very wideband calibration even with just a few lines if their lengths are chosen correctly.

In this simple case with one through and one line there shouldn't be difference in the accuracy but I'm using NISTMultilineTRL because it has few features that the TRL class doesn't have.

Below is the code for calibrating the uncorrected measurements:

import skrf
import matplotlib.pyplot as plt
#Use the skrf plot style
skrf.stylely()

trl_thru = skrf.Network('trl_through.s2p')
trl_line = skrf.Network('trl_line.s2p')
#Assemble the two reflection measurements into same network
trl_open = skrf.two_port_reflect(trl_open1.s11, trl_open2.s22)
trl_dut = skrf.Network('trl_cap.s2p')

sw_terms = skrf.Network('sw_terms.s2p')
switch_terms = (sw_terms.s11, sw_terms.s22)

#Measurement with load on both ports for isolation calibration

cal_trl = skrf.NISTMultilineTRL(
measured = [trl_thru, trl_open, trl_line],
Grefls = [1], #Estimate of the reflection coefficient
er_est = 3, #Estimated effective permittivity
l = [0, 13e-3], #Line lengths
gamma_root_choice = 'real', #Assumes that lines are lossy
switch_terms = switch_terms,
isolation = ll
)

#Plot the calibrated capacitor S-parameters
cal_trl.apply_cal(trl_dut).plot_s_db()

13 mm long line TRL calibration.

The 13 mm long line still has the resonance after the calibration. The resonance will also be visible in all the other measurements since the line with resonance is used in calculating the calibration error coefficients. S11 and S22 are perfect because reference impedance of the TRL calibration is equal to the characteristic impedance of the lines.

Series capacitor test structure calibrated with TRL.

Series capacitor S21 detail.

Above is the capacitor test structure this time calibrated with TRL. With the unknown thru calibration S21 of the capacitor was -0.4 dB at 4 GHz, but with TRL calibration the S21 is only -0.1 dB as the connector losses are calibrated out.

Self resonance frequency of this capacitor is 380 MHz. Equivalent series resistance (ESR) of the capacitor and pads can be determined from the loss at this frequency. Since at self resonance frequency inductive and capacitive reactances cancel out, the capacitor looks like a small valued resistor. S21 of a series resistor can be solved as: $$S_{21} = 2Z_0/(2Z_0+ R)$$. S21 of the capacitor is 0.022 dB dB at 380 MHz, giving a series resistance of 0.26 Ω assuming 50 ohm reference impedance. But the reference impedance in this measurement is really the characteristic impedance of the lines. It is probably close to 50 ohms, but characteristic impedance needs to be determined more accurately if a more accurate estimate of ESR is needed.

The solved effective permittivity and line attenuation can be plotted with:

plt.figure()
plt.title('Effective permittivity')
plt.xlabel('Frequency [Hz]')
plt.plot(freqs, np.real(cal_trl.er_eff))

plt.figure()
plt.title('Attenuation dB/m')
plt.xlabel('Frequency [Hz]')
plt.plot(freqs, 20*np.log10(np.exp(1))*cal_trl.gamma.real)

Solved effective permittivity.

Solved line attenuation.

The resonance at 4.2 GHz affects both effective permittivity and attenuation, but below 4 GHz both should be correct. Effective permittivity is quite noisy, but it appears to be around 2.84 at high frequencies and more at lower frequencies.

Line attenuation is around 4.3 dB/m at 1 GHz and 11.6 dB/m at 3 GHz. Using a calculator the theoretical values are 4.9 dB/m at 1 GHz and 11.9 dB/m at 3 GHz.

Accuracy of the calibration can be estimated using the normalized standard deviation plot. It is normalized such that when one through and one line calibration is the most accurate, the phase difference is 90 degrees, the normalized standard deviation is one. When the phase difference goes further from 90 degrees accuracy decreases and the normalized standard deviation increases.

plt.figure()
plt.title('Normalized standard deviation')
plt.xlabel('Frequency [Hz]')
plt.plot(freqs, cal_trl.nstd)

Normalized standard deviation.

As a rule of thumb phase difference between through and line should be between 20 and 160 degrees for an accurate calibration. In the normalized standard deviation plot this corresponds to about three. From the above plot it is clear that at around below 500 MHz the accuracy is much worse than this. To improve the accuracy a longer line that is 90 degrees at low frequencies should be added.

Characteristic impedance

Transmission line lumped element model. Source: Wikipedia

Transmission line can be modeled using sections of lumped elements. Each section represents a short section of transmission line. One section has series resistor and inductor and parallel capacitor and conductance.

Using the lumped section model propagation constant of the transmission line can be defined as:

$$\gamma = \sqrt{(R + j\omega L)(G + j \omega C)}$$

and characteristic impedance as:

$$Z_0 = \sqrt{\frac{R + j \omega L}{G + j \omega C}}$$

or written using the propagation constant:

$$Z_0 = \frac{\gamma}{G + j \omega C} \approx \frac{\gamma}{j\omega C}$$

Typically conductance (G) of the transmission line is very close to zero if the substrate is not conductive and it can be approximated to be zero. Because propagation constant is solved during the TRL calibration characteristic impedance of the transmission line can be determined if the capacitance per length of the transmission line is known.

C could be measured directly, but it is hard to measure accurately since capacitance of a typical transmission line is very low. Here is a good paper of the possible methods for measuring it.

Measuring characteristic impedance

There are several ways to measure the characteristic impedance of the microstrip lines. The first most simple way would probably be trying to curve fit to the measured lines, but the problem is that we can't fit to the TRL calibrated measurements because TRL calibration's reference impedance is the characteristic impedance of the lines that we want to measure. Fitting to the measurements corrected to the SMA connectors is possible, but accuracy is limited by the matching of the connectors.

Better method without extra measurements is to first calibrate measurements to the SMA connector for example with unknown thru calibration and then do TRL calibration as a second level calibration. Now the TRL calibration error terms represent the S-parameters of the circuit between the reference planes of the calibrations. Unknown thru reference plane was at the SMA connectors and TRL calibrations reference plane is after the SMA connector on the PCB, thus the error terms of the TRL calibration are the S-parameters of the connector. Port impedances of the error terms are 50 ohm from the unknown thru and characteristic impedance of the line on the TRL side. If the SMA connector is short and well matched then the error terms only include the impedance change. Modeling the error terms as an ideal impedance transformer the TRL reference impedance can be solved. There are even enough degrees of freedoms to add a shunt capacitance to the model and still solve the characteristic impedance. For overview of this method see this paper.

We can get the error parameters of the calibration as follows from the calibration:

s21 = np.sqrt(cal_trl.coefs['forward reflection tracking'])
connector_s = np.array(\
[[cal_trl.coefs['forward directivity'], s21],\
[s21, cal_trl.coefs['forward source match']]] \
).transpose().reshape(-1,2,2)
connector = skrf.Network(s=connector_s, frequency=cal_trl.frequency, name='SMA')

connector.plot_s_db()

VNA calibration can't solve for both S21 and S12 of the error network and instead only the product S12*S21 is solved, which is called reflection tracking. For passive circuit S12 = S21 and we can take a square root of the reflection tracking to solve for S12 and S21. Taking square root leaves 180 degree phase ambiquity, but that is not important when only looking a the magnitudes. Directivity is S11 and source match is the S22.

SMA connector S-parameters from the error terms.

Above is the plot of the S-parameters of the connector from the error terms. Port 1 is the connector side and port 2 is the board side. Port 1 reference impedance is the reference impedance of the unknown thru calibration which is 50 ohms in this case. Port 2 reference impedance is the TRL calibration reference impedance which is the characteristic impedance of the lines that we don't know. At low frequencies we would expect the connector to be matched well, but S11 is only -22 dB suggesting that characteristic impedance of the line is not 50 ohms.

Using scikit-rf we can solve the characteristic impedance with the following code:

coefs = cal_trl.coefs
r = np.sqrt(coefs['forward reflection tracking'])
s1 = np.array([[coefs['forward directivity'],r],
[r, coefs['forward source match']]]).transpose()
t = skrf.s2t(s1)
g = np.sqrt( (t[:,0,1] + t[:,1,0])**2/(4 + (t[:,0,1] + t[:,1,0])**2))
z = 50*(1+g)/(1-g)

t11 = (t[:,0,0] - 1/np.sqrt(1-g**2))*2/(50*np.sqrt(z/50.))
t22 = (t[:,1,1] - 1/np.sqrt(1-g**2))*2/(50*np.sqrt(z/50.))

#Plot the characteristic impedance
plt.figure()
plt.title('Z0 from error terms')
plt.plot(freqs, z)

Characteristic impedance solved from the error terms.

The method is only going to be accurate at low frequencies and in the above plot we can see that the estimated characteristic impedance of the lines is around 57 ohms. Accuracy is not very good and issue with low frequency measurements is that characteristic impedance increases as the frequency decreases. Reason for the increase can be found from the lumped element model characteristic impedance definition:

$$Z_0 = \sqrt{\frac{R + j \omega L}{G + j \omega C}}$$

As $$\omega \rightarrow 0$$, $$Z_0 = \sqrt{R/G}$$. $$G$$ is usually much smaller than $$R$$ and $$Z_0$$ approaches a large value. At high enough frequencies $$Z_0 \approx \sqrt{L/C}$$, which is constant.

Better estimate for the characteristic impedance can be obtained by solving for the capacitance per unit length and using the solved propagation constant from the TRL calibration to estimate the characteristic impedance. This assumes that $$G \approx 0$$, which should be true when substrate is not conductive.

c = np.real(-1j*cal_trl.gamma/(2*np.pi*freqs*z))

plt.figure()
plt.title('C0 from error terms')
plt.ylabel('Capacitance [F/m]')
plt.xlabel('Frequency [Hz]')
plt.plot(freqs, c)

Solved capacitance per unit length.

At low frequencies the capacitance per unit length is about 99 pF/m, but the obtained value is not very accurate because accuracy of the calibration is poor at low frequencies.

Assuming that the capacitance per unit length doesn't vary as a function of frequency, we can use the capacitance solved at low frequencies for solving the characteristic impedance over the whole frequency range. This is usually good approximation to make as long as the relative permittivity of the substrate doesn't vary much as a function of frequency.

z0 = cal_trl.gamma/(1j*2*np.pi*freqs*99e-12)

plt.figure()
plt.title('Z0 from C0 and $\gamma$')
plt.ylabel('Z0 [$\Omega$]')
plt.xlabel('Frequency [Hz]')
plt.plot(freqs, z0)

Characteristic impedance from C0 and propagation constant.

The trace is quite noisy and the resonance around 4.2 GHz in the line affects the propagation constant solution causing also the characteristic impedance to be solved incorrectly, but below 4 GHz the solution should be correct. At very low frequencies it can be seen that the characteristic impedance rises as is expected and after 1 GHz it is practically constant. From the plot characteristic impedance of the lines can be estimated to be about 56.7 Ω.

Second method

The previous method assumed that the SMA connector was well matched. A better method that doesn't assume can be used if line there is a terminated line on the board.

At low frequencies parasitics of the termination can be assumed to be low and impedance of the termination is the same as its DC resistance that can be measured very accurately with a multimeter. When the termination is measured with a VNA the expected reflection coefficient is:

$$\Gamma = \frac{R_{\text{load}} - Z_0}{R_{\text{load}} + Z_0}$$

This equation can be solved for $$Z_0$$, but it is again better to instead solve for capacitance per unit length because $$Z_0$$ is not constant at low frequencies.

#Calibrated termination
plt.figure()

#Solve for Z0
r_load = 50.2 #Measure DC resistance of the termination

plt.figure()
plt.title('Z0 from termination')
plt.xlabel('Frequency [Hz]')
plt.plot(f_ghz, np.real(z0))

#Solve for C0

plt.figure()
plt.title('C0 from termination')
plt.xlabel('Frequency [Hz]')
plt.plot(freqs, c)

Measured S11 of the termination

Above are the TRL calibrated S-parameters of the termination. At low frequencies the S11 should be very low if $$Z_0=R_{\text{load}}$$, but the measured S11 is only -22 dB.

Solved Z0 using termination.

Solved capacitance using termination.

Solved Z0 seems to be around 57 Ω agreeing with the first method. Solved capacitance per unit length is about 98.8 pF/m, when previous method gave 99 pF/m. Now that we have gotten the same characteristic impedance using two methods we should be pretty confident in the result.

Renormalized TRL calibration

Finally the reference impedance of the TRL calibration be renormalized to 50 ohms by giving the capacitance per unit length to the calibration routine:

cal_trl = skrf.NISTMultilineTRL(
measured = [trl_thru, trl_open, trl_line],
Grefls = [1], #Estimate of the reflection coefficient
er_est = 3, #Estimated effective permittivity
l = [0, 13e-3], #Line lengths
gamma_root_choice = 'real', #Assumes that lines are lossy
switch_terms = switch_terms,
isolation = ll,
c0 = 98.8e-12,
z0_ref = 50 #New reference impedance
)

#Plot the smoothed characteristic impedance
#http://scipy.github.io/old-wiki/pages/Cookbook/SavitzkyGolay

plt.figure()
plt.title('Z0')
plt.ylabel('Z0 [$\Omega$]')
plt.xlabel('Frequency [Hz]')
plt.plot(f_ghz[1:], savitzky_golay(np.real(cal_trl.z0[1:]), 31, 2))

Characteristic impedance of the transmission lines. Smoothed using Savitzky Golay filter.

The resonance at 4.2 GHz gives incorrect results around that frequency, but the characteristic impedance can be expected to be constant at high frequencies. Reading from the plot the characteristic impedance seems to be around 56.7 Ω.

The line was supposed to be 50 ohms, but characteristic impedance seems to be over 10% bigger than that, so where's the problems?

Photo of the line with calibers for scale.

The first thing to check is that the manufactured dimensions of the line match the designed dimensions. In the picture is the microstrip line with calibers set to 1.00 mm for scale. In the image distance between the caliber jaws is 118 pixels, while line is 32 pixels wide. Converting this to mm gives line width of about 0.27 mm. The designed width was 0.34 mm so the manufactured line is thinner, which raises the characteristic impedance. Inputting the measured width to microstrip characteristic impedance calculator with nominal 35 µm trace thickness, 170 µm substrate height and substrate dielectric constant of 3.66 give characteristic impedance of 57.0 ohms.

The observed increase in the characteristic impedance can be fully explained by overetching of the lines. Overetching is about 35 µm, which is not bad considering the price of the PCBs. I have seen bigger overetching on much more expensive PCBs.

Termination S11.

Above is S11 of the termination with renormalized calibration. Now the low frequency reflection coefficient is small as expected. At high frequencies the reflection coefficient is not so good, but this is an issue with the termination and not measurement.

SMA connector S-parameters.

Plotting the solved SMA connector S-parameters from the error terms now gives the true 50 ohm normalized S-parameters. This time the connector matching looks much better. S11 and S22 are below -25 dB up to 4 GHz.

SMA connector S21.

Loss of the SMA connector is also very small.

Capacitor S-parameters.

Capacitor S-parameters don't look that different from before.

Capacitor S21.

At self-resonance frequency S21 is about 0.025 dB, when previously it was 0.022 dB. This gives ESR of 0.29 Ω, when without renormalizing it was estimated to be 0.26 Ω. Error is not very big in this case.

Part number of this capacitor is GRM155R71H102KA01J and Murata makes its S-parameter measurements available on their website.

Capacitor S-parameters from manufacturer.

My measurements include the PCB pads which cause some extra capacitance at high frequencies, but S-parameters measured by the manufacturer look very similar to what I measured. The manufacturer has measured the ESR to be 0.277 Ω which agrees really well with my measurement of 0.29 Ω.

Conclusion

Even though my VNA is much less accurate than commercial VNAs it is accurate enough for many useful measurements. I tried the same measurement also with the first version of the VNA and due to the much worse isolation the results were unusable.

The board had unfortunate resonance at 4.2 GHz, which I think comes from the poor grounding of the multiple ground planes. I think I'll need to order a new PCB with better grounding.


text:
## Problem B. Binary Operation ≡

 Author: ACM ICPC NEERC 2010 Jury Time limit: 2 sec Input file: binary.in Memory limit: 256 Mb Output file: binary.out

### Statement

Consider a binary operation op defined on digits 0 to 9, op: digs × digs ↦ digs, such that 0op 0 = 0.

A binary operation opd is a generalization of op to the set of non-negative integers, opd: nneg × nneg ↦ nneg. The result of aopd b is defined in the following way: if one of the numbers a and b has fewer digits than the other in decimal notation, then append leading zeroes to it, so that the numbers are of the same length; then apply the operation op digit-wise to the corresponding digits of a and b.

Let us define opd to be left-associative, that is, aopd bopd c is to be interpreted as (aopd b)opd c.

Given a binary operation op and two non-negative integers a and b, calculate the value of res.

### Input file format

The first ten lines of the input file contain the description of the binary operation op. The i-th line of the input file contains a space-separated list of ten digits — the j-th digit in this list is equal to (i − 1)op (j − 1).

The first digit in the first line is always 0.

The eleventh line of the input file contains two non-negative integers a and b (0 ≤ a ≤ b ≤ 1018).

### Output file format

Output a single number — the value of res without extra leading zeroes.

### Sample tests

No. Input file (binary.in) Output file (binary.out)
1
0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 0
2 3 4 5 6 7 8 9 0 1
3 4 5 6 7 8 9 0 1 2
4 5 6 7 8 9 0 1 2 3
5 6 7 8 9 0 1 2 3 4
6 7 8 9 0 1 2 3 4 5
7 8 9 0 1 2 3 4 5 6
8 9 0 1 2 3 4 5 6 7
9 0 1 2 3 4 5 6 7 8
0 10
15

0.128s 0.016s 15


text:
# Curve Fitting (Polynomial) (G Dataflow)

Version:

Finds the set of polynomial fit coefficients that best represents an input signal or input data set using a specific fitting method.

## coefficient constraint

Constraints on the polynomial coefficients of a certain order.

Use this input if you know the exact values of certain polynomial coefficients.

This input is available only if you wire one of the following data types to signal or y:

• Waveform
• 1D array of waveforms
• 1D array of double-precision, floating-point numbers

### order

Constrained order.

Default: 0

### coefficient

Coefficient of the specific order.

Default: 0

## polynomial order

Order of the polynomial.

polynomial order must be greater than or equal to 0. If polynomial order is less than zero, this node sets polynomial coefficients to an empty array and returns an error. In real applications, polynomial order is less than 10. If polynomial order is greater than 25, the node sets polynomial coefficients to zero and returns a warning.

Default: 2

## reset

A Boolean that specifies whether to reset the internal state of the node.

 True Resets the internal state of the node. False Does not reset the internal state of the node.

This input is available only if you wire a double-precision, floating-point number to y or signal.

Default: False

## signal

Input signal.

This input accepts a waveform or a 1D array of waveforms.

This input changes to y when the data type is a double-precision, floating-point number or a 1D array of double-precision, floating-point numbers.

## y

Dependent values representing the y-values of the data set.

This input accepts a double-precision, floating-point number or a 1D array of double-precision, floating-point numbers.

This input changes to signal when the data type is a waveform or a 1D array of waveforms.

## x

Independent values representing the x-values of the data set.

This input accepts a double-precision, floating-point number or a 1D array of double-precision, floating-point numbers.

This input is available only if you wire a double-precision, floating-point number or a 1D array of double-precision, floating-point numbers to y or signal.

## weight

An array of weights for the data set.

This input is available only if you wire one of the following data types to signal or y:

• Waveform
• 1D array of waveforms
• 1D array of double-precision, floating-point numbers

## tolerance

Value that determines when to stop the iterative adjustment of coefficients when you use the Least Absolute Residual or Bisquare methods.

If tolerance is less than or equal to 0, this node sets tolerance to 0.0001.

This input is available only if you wire one of the following data types to signal or y.

• Waveform
• 1D array of waveforms
• 1D array of double-precision, floating-point numbers

How tolerance Affects the Outputs with Different Fitting Methods

For the Least Absolute Residual method, if the relative difference of the weighted mean error of the polynomial fit in two successive iterations is less than tolerance, this node returns resulting polynomial coefficients. For the Bisquare method, if any relative difference between polynomial coefficients in two successive iterations is less than tolerance, this node returns the resulting polynomial coefficients.

Default: 0.0001

## block size

Length of each set of data. The node performs computation for each set of data.

When you set block size to zero, the node calculates a cumulative solution for the input data from the time that you called or initialized the node. When block size is greater than zero, the node calculates the solution for only the newest set of input data.

This input is available only if you wire a double-precision, floating-point number to signal or y.

Default: 100

## error in

Error conditions that occur before this node runs.

The node responds to this input according to standard error behavior.

Standard Error Behavior

Many nodes provide an error in input and an error out output so that the node can respond to and communicate errors that occur while code is running. The value of error in specifies whether an error occurred before the node runs. Most nodes respond to values of error in in a standard, predictable way.

error in does not contain an error error in contains an error
If no error occurred before the node runs, the node begins execution normally.

If no error occurs while the node runs, it returns no error. If an error does occur while the node runs, it returns that error information as error out.

If an error occurred before the node runs, the node does not execute. Instead, it returns the error in value as error out.

Default: No error

## algorithm

Algorithm this node uses to compute the polynomial curve that best fits the input values.

Name Value Description
SVD 0 Uses the SVD algorithm.
Givens 1 Uses the Givens algorithm.
Givens2 2 Uses the Givens2 algorithm.
Householder 3 Uses the Householder algorithm.
LU Decomposition 4 Uses the LU Decomposition algorithm.
Cholesky 5 Uses the Cholesky algorithm.
SVD for Rank Deficient H 6 Uses the SVD for Rank Deficient H algorithm.

Default: SVD

## method

Method of fitting data to a polynomial curve.

Name Value Description
Least Square 0 Uses the least square method.
Least Absolute Residual 1 Uses the least absolute residual method.
Bisquare 2 Uses the bisquare method.

Algorithm for the Least Square Method

The least square method finds the polynomial coefficients of the polynomial model by minimizing the residue according to the following equation:

$\frac{1}{N}\underset{i=0}{\overset{N-1}{\sum }}{w}_{i}{\left({f}_{i}-{y}_{i}\right)}^{2}$

where

• N is the length of y or the number of data values in a waveform
• wi is the ith element of weight
• fi is the ith element of best polynomial fit
• yi is the ith element of y or the ith data value in a waveform

Algorithm for the Least Absolute Residual Method

The least absolute residual method finds the polynomial coefficients of the polynomial model by minimizing the residue according to the following equation:

$\frac{1}{N}\underset{i=0}{\overset{N-1}{\sum }}{w}_{i}|{f}_{i}-{y}_{i}|$

where

• N is the length of y or the number of data values in a waveform
• wi is the ith element of weight
• fi is the ith element of best polynomial fit
• yi is the ith element of y or the ith data value in a waveform

Algorithm for the Bisquare Method

The bisquare method finds the polynomial coefficients using an iterative process, as shown in the following illustration.

The node calculates residue according to the following equation:

$\frac{1}{N}\underset{i=0}{\overset{N-1}{\sum }}{w}_{i}{\left({f}_{i}-{y}_{i}\right)}^{2}$

where

• N is the length of y or the number of data values in a waveform
• wi is the ith element of weight
• fi is the ith element of best polynomial fit
• yi is the ith element of y or the ith data value in a waveform

Default: Least Square

## best polynomial fit

Polynomial curve that best fits the input signal.

This output can return the following data types:

• Waveform
• 1D array of waveforms
• 1D array of double-precision, floating-point numbers

## polynomial coefficients

Coefficients of the fitted model in ascending order of power. The total number of elements in polynomial coefficients is m + 1, where m is polynomial order.

## error out

Error information.

The node produces this output according to standard error behavior.

Standard Error Behavior

Many nodes provide an error in input and an error out output so that the node can respond to and communicate errors that occur while code is running. The value of error in specifies whether an error occurred before the node runs. Most nodes respond to values of error in in a standard, predictable way.

error in does not contain an error error in contains an error
If no error occurred before the node runs, the node begins execution normally.

If no error occurs while the node runs, it returns no error. If an error does occur while the node runs, it returns that error information as error out.

If an error occurred before the node runs, the node does not execute. Instead, it returns the error in value as error out.

## residue

Weighted mean error of the fitted model.

This output can return a double-precision, floating-point number or a 1D array of double-precision, floating-point numbers.

Algorithm for Calculating residue When the Input Signal is a Double-Precision, Floating-Point Number

When the input signal is a double-precision, floating-point number, this node calculates residue according to the following equation:

$\mathrm{residue}=\text{\hspace{0.17em}}\frac{1}{N}\underset{i=0}{\overset{N-1}{\sum }}{\left({f}_{i}-{y}_{i}\right)}^{2}$

where

• N is the number of elements in the data set
• fi is the ith element of best linear fit
• yi is the y component of the ith input data point

## Algorithm for Calculating best polynomial fit

This node fits data to a polynomial function of the general form described by the following equation:

${f}_{i}=\underset{j=0}{\overset{m}{\sum }}{a}_{j}{x}_{i}^{j}$

where

• f is the output sequence best polynomial fit
• x is the input sequence calculated from the dt component of the input signal
• a is polynomial coefficients
• m is polynomial order

This node finds the value of a that best fits the observations (X, Y). When the input signal is an array of double-precision, floating-point numbers, X is the x component of the input signal and Y is y component of the input signal. When the input signal is a waveform or an array of waveforms, X is the input sequence calculated from the start time of the waveform and Y is the data values in the waveform.

The following equation describes the polynomial curve resulting from the general polynomial fit algorithm:

$y\left[i\right]=\underset{j=0}{\overset{m}{\sum }}{a}_{j}{\left(x\left[i\right]\right)}^{j}$

The following illustration shows a general polynomial fit result using this node:

Where This Node Can Run:

Desktop OS: Windows

FPGA: DAQExpress does not support FPGA devices

Web Server: Not supported in VIs that run in a web application


text:
Is it possible to unload a LaTeX package?

I have a preamble file which I use for most documents. Now suppose I want to use a preamble file preamble.tex and explicitly don't want to use one of the packages loaded in it. For example in this file there is a line:

\usepackage{mypackage}

and then I want to have a file test.tex which contains something like this:

\input{preamble}
\usepackage{myotherpackage} %myotherpackage can only be loaded when mypackage is not present
-
–  Torbjørn T. Dec 27 '11 at 10:25
no, it is a special problem ... –  Herbert Dec 27 '11 at 10:56
No, you can't unload any packages. TeX does not have namespaces so it isn't possible to know what macros got defined by which package. –  Martin Scharrer Dec 27 '11 at 11:27

You can fool LaTeX into believing that it has already loaded package foo with options bar and baz by defining a command with name "ver@foo.sty" and one with name "opt@foo.sty" in a suitable way; so

\makeatletter
\newcommand{\dontusepackage}[2][]{%
\@namedef{ver@#2.sty}{9999/12/31}%
\@namedef{opt@#2.sty}{#1}}
\makeatother

will prevent both loading the package and also keep LaTeX happy with the options possibly passed to it.

However, \dontusepackage[bar,baz]{foo} must be issued before \input{preamble}.

I'm in general contrary to "do-it-once-and-for-always" preambles, which raise problems such as this one. There's no such thing as a universal preamble and later developments make some packages obsolete or at least superseded by new ones.

Just copy preamble.tex into the working directory and change it for what's needed for the current document you're writing.

-
@egreg I think it is useful to emphasize that, to avoid option clash later, bar and baz must be the only options for package foo. This will require knowing all the options of foo. This also assumes that foo wasn't loaded before \dontusepackage. –  Ahmed Musa Dec 27 '11 at 19:25
I've chosen to pass also the options used in preamble.tex in order to raise an error if some other package tries to load it with different options. However it's not that important. –  egreg Dec 27 '11 at 20:21


text:
# Photons and matter waves: The photon of quantum light: I need help

## Homework Statement

How fast must an electron move to have a Kinetic Energy equal to the photon energy of sodium light at wavelength 590 nm.

## Homework Equations

photon energ E =hf

h = 6.63 * 10 ^ -34 J * s = 4.14 = 10^-15 eV *s

f = c/ lamda

c = 3 * 10^ 8 m/s

mass of an electron is 9.11 * 10 ^ -31 (kg) or 511 keV

## The Attempt at a Solution

Ok I solved for the energy of a photon for sodium light

E = 6.63 * 10 ^ -34 J *s * ( (3* 10 ^ 8 m/s) / 590 * 10 ^ -9 meters) = 1.989 * 10 ^ -25 Joules

Were solving for Velocity. How do I do this problem it's confusing.

Last edited:

Pengwuino
Gold Member
Well you're looking for the kinetic energy (and thus velocity) right? Well $$E_{kinetic} = \frac{{mv^2 }}{2}$$ and you know the energy of the photon so equate them and find the velocity.

Sorry about that I'll make my information much clearer next time. So that's the equation I use.

Last edited:
Would it be like this

$$v = \frac{{\sqrt{{2E }} \div {m} }$$

Last edited:
Pengwuino
Gold Member
Yes, where E = hf = hc/lamda

So is that E of photon energy of soidum light the same as the kinetic energy

Check your math again. I get a different value for energy using the same numbers you gave as input.

so then it would be

It would be v = sqrt { 2 (3.37 * 10 ^ -19 joules / 8.19 * 10^ -14 joules}

m of electron in joules is 8.19 * 10 ^ -14 joules

Last edited:
Oh yea the Enery of a photon for sodium light is actually
$$E = 3.371186441 \times 10^-19$$

That's what I got.

I got 2.86 * 10 ^ -3 m/s

That doesn't seem correct. Your equation for v doesn't look right, but I don't know if you are just typing in the tex code wrong. Did you really intend for the mass, m, to be in the numerator of that equation?

Also, I'm not sure where you are getting the mass of the electron in joules. If you are using SI units of joules for energy, then the electron mass should be in kg.

Pengwuino
Gold Member
Wait yes, your equation for the velocity in terms of the energy changed... its divided by m.

Well my last word on this...it looks like you threw another factor of c in there somewhere. If I divide the answer I get in meters/sec by 3*10^8 m/s, then I get the answer you posted above. So your answer would be correct if it was in units of "c".