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# 9.6: Supplementary and Complementary Angles

Difficulty Level: At Grade Created by: CK-12

Paul is fixing a house. The wood frames have already been placed and he is almost done with one of the rooms. He looks around the room and sees that along each wall, the boards are placed at a 90 degree angle. There are pairs of 90 degree angles next to each other. He knows that there is a special name for the angle pair, but he can't remember the name. What type of angle pair is formed by two 90 degree angles?

In this concept, you will learn about complementary and supplementary angles.

### Guidance

Sometimes, two angles are a part of each other or are connected to each other. These angles are called angle pairs.

Let's look at two special types of angle pairs, supplementary angles or complementary angles.

Supplementary angles are two angles whose sum is equal to 180\begin{align*}180^\circ\end{align*}. In other words when you add the measure of one angle in the pair with the other angle in the pair, they equal 180 degrees.

These two angles are supplementary because together they form a straight line. You can also tell that they are supplementary because when you add their angle measures the sum is equal to 180 degrees.

120+60=180\begin{align*}120 + 60 = 180^\circ\end{align*}

Complementary angles are a pair of angles whose sum is 90\begin{align*}90^\circ\end{align*}. Here is an example of complementary angles.

If you add up the two angle measures, the sum is equal to 90 degrees. Therefore, the two angles are complementary.

You can find missing angle measures by using information about supplementary and complementary angles.

Let's look at an example.

Find the measure of x\begin{align*}x\end{align*}.

First, identify that these two angles are supplementary. They form a straight line. The total number of degrees in a straight line is 180. Therefore, you can write the following equation to solve.

130+xx=180=50

The missing angle is equal to 50\begin{align*}50^\circ\end{align*}.

Here the two angles are complementary. Therefore the sum of the two angles is equal to 90 degrees. Write an equation and solve for the missing angle measure.

75+xx=90=15

The missing angle measure is equal to 50\begin{align*}50^\circ\end{align*}.

### Guided Practice

Identify the angle pair as complementary or supplementary.

First, add the two angles to calculate their sum.

75 + 105 = 180

Then, identify which angle pair adds up to 180 degrees.

Supplementary angles

### Examples

Write whether each pair is complementary or supplementary.

#### Example 1

If the sum of the angles is equal to 180 degrees.

First, identify which angle pair adds up to 180 degrees.

Supplementary angles

#### Example 2

If one angle is 60 degrees and the other angle is 120 degrees.

First, add the two angles together to find the sum.

60 + 120 = 180

Then, identify which angle pair adds up to 180 degrees.

Supplementary angles

#### Example 3

First, add the two angles to calculate their sum.

30 + 60 = 90

Then, identify which angle pair adds up to 90 degrees.

Complementary angles

The answer is that the pair are complementary angles.

Remember Paul and the house? He has an angle pair that is formed by two 90 degree angles. What type of angle pair is formed when two 90 degree angles are combined?

First,add the two angles to calculate their sum.

90 + 90 = 180

Then, identify which angle pair adds up to 90 degrees.

Supplementary angles

The answer is that the pair are supplementary angles.

### Explore More

Identify each angle pair as supplementary or complementary angles.

1.

2.

3.

4.

Use what you have learned about complementary and supplementary angles to answer the following questions.

5. If two angles are complementary, then their sum is equal to _________ degrees.

6. If two angles are supplementary, then their sum is equal to ________ degrees.

7. True or false. If one angle is 120\begin{align*}120^\circ\end{align*}, then the second angle must be equal to 90\begin{align*}90^\circ\end{align*} for the angles to be supplementary.

8. True or false. If the angles are supplementary and one angle is equal to 100\begin{align*}100^\circ\end{align*}, then the other angle must be equal to 80\begin{align*}80^\circ\end{align*}.

9. True or false. The sum of complementary angles is 180\begin{align*}180^\circ\end{align*}.

10. True or false. The sum of supplementary angles is 90\begin{align*}90^\circ\end{align*}.

Identify whether the angles are supplementary, complementary or neither based on the angle measures.

11. One angle is 50 degrees. The other angle is 130 degrees.

12. One angle is 30 degrees. The other angle is 60 degrees.

13. One angle is 112 degrees. The other angle is 70 degrees.

14. One angle is 110 degrees. The other angle is 50 degrees.

15. One angle is 35 degrees. The other angle is 55 degrees.

2. [2]^ License: CC BY-NC 3.0
3. [3]^ License: CC BY-NC 3.0

## Date Created:

Sep 24, 2015

Sep 24, 2015
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# Compute covariance matrix via rank-1 update to $M^\top M$

I have a large, sparse matrix $M\in\mathbb{R}^{n\times p}$. Centering $M$ to compute the covariance matrix $\Sigma$ would, in general, destroy the "zeros aren't stored" property of sparse matrices. Instead, we can center $\Sigma$ after computing the second moment. There are any number of ways to do this, but I want one that is cheap and numerically stable.

The goal of estimating $\Sigma$ is to compute many evaluations of $f(x) = x^\top \Sigma^{-1} x$.

My first thought was to compute $M^\top M$ and apply the Bunch-Nielsen-Sorensen formula for a rank-1 update to a spectral decomposition, which allows one to compute the eigenvalues and eigenvectors to $A$ for $D$ diagonal and $\rho\in\mathbb{R}$ $$A = D + \rho zz^\top$$ (Some algebra extends this to updating $\Sigma$.) However, BNS requires that $z_i>0 ~\forall ~ i$, and that does not hold for the mean vector $\mu$ in general.

As a work-around, I could apply two successive BNS updates, one for a $\tilde{\mu}$ which shifts the nonpositive elements of $\mu$ to be positive, and a second which offsets that shift. This is (probably) more efficient than two spectral decompositions.

Another way would form a pseudoinverse and apply the Sherman-Morrison formula, but we all know that one should never explicitly compute a matrix inverse, hence my interest in factorizations.

• How many of those quadratics are you planning to compute? If it's not many (relative to the size of your $S$), then solving $\Sigma^{-1}x$ each time may be cheaper than computing a pseudoinverse and storing it for repeated use. For the few-quadratics case, there are solvers that rely on repeated multiplications of the form $\frac{1}{n-1}(M^TM-n\mu\mu^T)v$. These multiplications can be done cheaply without forming $\frac{1}{n-1}(M^TM-n\mu\mu^T)$ explicitly. Where to find a software package that does this, though, I'm not sure. – eric_kernfeld Oct 6 '17 at 16:36
• $S$ is approximately $10^5 \times 10^5$ and I'll be evaluating $f$ on the order of $10^7$ times. I'm not afraid of coding stuff up myself, as long as I have a good strategy to solving the problem. :-) – Sycorax Oct 6 '17 at 16:42
• I'm still sorting through this mentally, and I realized that "using versus not using decompositions" is a slightly different decision than "explicitly or not explicitly forming an inverse". I will focus on fast, stable solutions by any means. Or any covariances. – eric_kernfeld Oct 6 '17 at 17:16
• @eric_kernfeld It might be that the best approach is the simplest -- compute the Cholesky factorization $\Sigma = LL^\top$ and forwardsolve the linear system to evaluate $f$. This can be done directly, or indirectly if there exists a cheaper way to apply a rank-1 update to $L$. – Sycorax Oct 6 '17 at 18:38
• That's what I'm thinking of. There may be a bonus in there -- if $A^TA=B$, the $R$ from the QR decomposition of $A$ is the same as the $R$ from the Cholesky factorization of $B$. If you use Givens rotators to compute the QR of $A$, you can take advantage of the sparsity by only rotating the nonzero elements. I know that my A-B scenario is not identical to your M-mu-sigma problem, but maybe some tweak would allow this trick to apply. – eric_kernfeld Oct 6 '17 at 18:43

The canonical way to compute $f(x)$ is to compute the Cholesky factor of $\Sigma=\tilde{R}^\top \tilde{R}$ where $\tilde{R}$ is and upper-triangular. Then we can write $$f(x)=x \tilde{R}^{\top}\tilde{R}^{-\top} x = y^\top y$$ and solve the triangular system for $y$. So we should focus on how to get to a Cholesky factor of $\Sigma$.
Using the equation $$\Sigma=\frac{1}{n-1}M^\top M - \frac{n}{n-1}\mu\mu^\top$$ I observe that this is just a rank-one down-date to $M^\top M$. Or, in terms of Cholesky factors, we can use the factorization $M=QR$ for orthonormal $Q$ and upper triangular $R$ to write $$M^\top M = R^\top Q^\top QR = R^\top R$$
We can update $R$ to be Cholesky factors of $\Sigma$ using hyperbolic rotations, as outlined in Golub & Van Loan, Matrix Computations 4th edition, p. 339-341. The procedure covers three pages of text, so it would be impractical to reproduce here. The point is that it is a down-dating framework similar to Givens rotations. (To efficiently compute $M=QR$ for sparse $M$ is what Golub & Van Loan call the "The Sparse QR Challenge" and it is discussed in pp. 606-608.)

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# $K[X]$-modules are $K$-vector spaces with a linear transformation

In the Atiyah's book there is this example about $A$-modules.

Let $A=\mathbb{K}[x]$, where $\mathbb{K}$ is a field. An $A$-module is a $\mathbb{K}$-vector space with a linear transformation.

Can anyone explain me this claim?

Let $$\;V\;$$ be a $$\;\Bbb K\,-$$ vector space, and let $$\;T:V\to V\;$$ be any linear operator. Then, $$\;V\;$$ gets a structure of $$\;\Bbb K[x]\,-$$ module if we define
$$\forall\,v\in V\;,\;\;x\cdot v:=Tv$$
$$\sum_{j=0}^m k_jx^j\cdot v:=\sum_{j=0}^mk_jT^jv\;,\;\;k_j\in\Bbb K$$
• "$T:A\to A$ is any linear operator$" , it think it has to be$V$not$A$. Please correct me if I'm wrong Aug 7 '20 at 2:30 • @LordShadow Yep, thanks. Aug 7 '20 at 9:31 Say$M$is the$k[x]$module in question which is also a$k$vector space by means of the inclusion$k \hookrightarrow k[x]$. The action of$x \in k[x]$is that of a linear transformation i.e.$x(am) = a x(m)$and$x(m+m') = x(m)+x(m')$. Conversely, given any linear$L: M \to M$we can let$x(m) = L(m)$to define a$k[x]$-module structure on the vector space$M$. Note this while the action of$k$is faithful because it is a field, the action of$k[x]$may not be. For instance the linear transformation could be nilpotent or something. Working with$K$-vector spaces and (unital)$K$-algebras.... • Every$K[x]$-module is a$K$-vector space, because$K \subseteq K[x]$. • The$K[x]$-module structures on a vector space$V$are in bijection with$K$-algebra homomorphisms$K[x] \to \mathrm{End}(V)$. •$K$-algebra homomorphism$K[x] \to \mathrm{End}(V)$are in bijection with elements of$\mathrm{End}(V)\$.


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# Learn Electronics/Resistors

Real Resistors

## Voltage Drop

• Think of a bucket full of water; nothing happens yet, if there is no hole at the bottom of the bucket. The height of the water in the bucket represents the electrical voltage.
• If there is a hole at the bottom of the bucket, then water will flow out of it. That water flow represents electric current flow.
• If the source of the voltage is a battery, then, in time, the battery's voltage will be reduced until there is no voltage left, just as the level of the water in the bucket will get lower and lower until no water is left in the bucket, with zero level.
• If the source of the voltage is the socket in the wall, then the voltage will remain about the same; normally there will be only slight changes of that kind of voltage.

The voltage is the source of the current, and the current has some purpose, it flows through what is called a "load", which may be one or more resistors. No current flows through a resistor or equivalent, unless there is a voltage acrosss the two terminals of the resistor. If there are at least 2 resistors, or equivalent, in series (like a river flowing via 3 points) then the voltage across each resistor is called voltage drop. Think of a river passing via points A, B, and C, where A is highest and C is lowest. There is a difference in the elevations/levels, with B minus A representing a voltage drop, and C minus B being like another voltage drop. The sum of all of the voltage drops in a circuit always equals the supply voltage, such as the battery voltage.

## Ohms Law

Resistors: Resistors "resist" the flow of electricity, reducing the amount of the current (measured in units called amps). Without a resistor in the circuit the current would be too high, you would ruin parts of the circuit, use up the battery too fast, or even blow the battery. Since resistors are too small to print numbers on them they are labeled with color coded bands that have to be interpreted. Resistors have 3 or 4 colorbands on them, but only the first 3 matter for now. The fourth one is either silver or gold. Silver means it is 10% above or below the value shown by the first 3 bands, while gold means 5% above or below. The first 2 bands represent a one decimal number, the third represents the number of zeros that follow.

Resistors follow Ohm's Law: $V=IR$ (V=Voltage, I=Current (in amps), R=Resistance (in ohms))

Ohm's law is important to remember since most electronic components have a maximum and minimum voltage and amp rating: if they're too low it wouldn't work, and if they're too high it will destroy the component. As long as you have at least two values you can solve for the last one.

The Standard EIA Color Code Table per EIA-RS-279 is as follows:

Colour 1st band 2nd band 3rd band (multiplier) 4th band (tolerance) Temp. Coefficient
Black 0 0 ×1
Brown 1 1 ×10
Red 2 2 ×100 (10^2)
Orange 3 3 ×1000 (10^3)
Yellow 4 4 ×10000 (10^4)
Green 5 5 ×100000 (10^5)
Blue 6 6 ×1000000 (10^6) Highest Multiplier band normally seen
Violet 7 7
Gray 8 8
White 9 9
Gold ×0.1 ±5% (J)
Silver ×0.01 ±10% (K)
None ±20% (M)

A trick to remember is that the decimal number of a color band is equal to the amount of zeros added to the end if it's the third band.

## Series & Parallel

(need circuit diagram of resistors in parallel and in series)

Definition of series: In electronics when something is connected in series like a resistor it means that you have resistors that are in a "line". The resistors are connected end to end.

Definition of Parallel: The same wire is connected to one end of all the parts in parallel and another wire is connected to the other end of all the parts in parallel.

(Math for resistors in series and parallel):

Resistors in series: Add their values (in Ohms) to obtain the total resistance of their combination.

Resistors in parallel: The reciprocal (means: one divided by) of the total resistance of the combination of two resistors R1 and R2 is

          1/R = 1/R1 + 1/R2,    also l/R = 1/R1 + 1/R2 + 1/R3  if there are 3, and so on.

Resistors are very useful in electronic design.

Previous chapter Basics | Next chapter: Capacitors | Go back to Index


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# Simplify ((b^(-5/2))/(c^(-2/3)))^2(b^(-1/2)c^(-1/3))^-1

(b-52c-23)2(b-12c-13)-1
Move b-52 to the denominator using the negative exponent rule b-n=1bn.
(1c-23b52)2(b-12c-13)-1
Move c-23 to the numerator using the negative exponent rule 1b-n=bn.
(c23b52)2(b-12c-13)-1
Apply the product rule to c23b52.
(c23)2(b52)2(b-12c-13)-1
Multiply the exponents in (c23)2.
Apply the power rule and multiply exponents, (am)n=amn.
c23⋅2(b52)2(b-12c-13)-1
Multiply 23⋅2.
Combine 23 and 2.
c2⋅23(b52)2(b-12c-13)-1
Multiply 2 by 2.
c43(b52)2(b-12c-13)-1
c43(b52)2(b-12c-13)-1
c43(b52)2(b-12c-13)-1
Multiply the exponents in (b52)2.
Apply the power rule and multiply exponents, (am)n=amn.
c43b52⋅2(b-12c-13)-1
Cancel the common factor of 2.
Cancel the common factor.
c43b52⋅2(b-12c-13)-1
Rewrite the expression.
c43b5(b-12c-13)-1
c43b5(b-12c-13)-1
c43b5(b-12c-13)-1
Rewrite the expression using the negative exponent rule b-n=1bn.
c43b5(1b12c-13)-1
Rewrite the expression using the negative exponent rule b-n=1bn.
c43b5(1b12⋅1c13)-1
Multiply 1b12 and 1c13.
c43b5(1b12c13)-1
Change the sign of the exponent by rewriting the base as its reciprocal.
c43b5(b12c13)
Cancel the common factor of b12.
Factor b12 out of b5.
c43b12b92(b12c13)
Factor b12 out of b12c13.
c43b12b92(b12(c13))
Cancel the common factor.
c43b12b92(b12c13)
Rewrite the expression.
c43b92c13
c43b92c13
Combine c43b92 and c13.
c43c13b92
Multiply c43 by c13 by adding the exponents.
Use the power rule aman=am+n to combine exponents.
c43+13b92
Combine the numerators over the common denominator.
c4+13b92
c53b92
c53b92
Simplify ((b^(-5/2))/(c^(-2/3)))^2(b^(-1/2)c^(-1/3))^-1

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Bitcoin Forum
April 29, 2017, 11:49:13 AM
 News: Latest stable version of Bitcoin Core: 0.14.1  [Torrent]. (New!)
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 Pages: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 [38] 39 40 41 42 43 44 45 46 47 48 49 50
 Author Topic: ICBIT Derivatives Market (USD/BTC futures trading) - LIVE  (Read 91659 times)
StevenPine
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 April 11, 2013, 10:51:17 PM

thanks for the forum edit clarification.

Fireball, please, what are the current margin conditions and when will you be performing the next clearance?

Let's get 2 factor authentication
https://bitcointalk.org/index.php?topic=178568.0
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killerstorm
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 April 11, 2013, 10:55:47 PM

It looks like I was right that 'traditional' futures do not work so well with Bitcoin.

I gotta build my own exchange with blackjack and hookers...

Err, I mean with capped forward contracts and no margin calls.

colored coins proof-of-concept: private currencies, stock/bond p2p exchange

Tips and donations: 16v13Fa9cPmfFzpm9mmbWwAkXY4gyY6uh4
StevenPine
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 April 11, 2013, 11:29:58 PM

We've had large long positions closed without being fully compensated because the margin was not increased. And yet now he is increasing the margin when it isn't in our favor, this is a clear case of bias on the part of the admin/fireball

Let's get 2 factor authentication
https://bitcointalk.org/index.php?topic=178568.0
zebedee
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 April 11, 2013, 11:51:42 PM

We've had large long positions closed without being fully compensated because the margin was not increased. And yet now he is increasing the margin when it isn't in our favor, this is a clear case of bias on the part of the admin/fireball

huh? I'd be very happy right now they were closed out. I'm sure they'd be worth a lot less now.
zebedee
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 April 12, 2013, 12:13:07 AM

It's hard to justify treating someone differently just because they left BTC on the exchange that you can seize.

Moral of the story : if there's any chance you might want out, don't leave excess coins on icbit.
picobit
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Decor in numeris

 April 12, 2013, 07:30:26 AM

We've had large long positions closed without being fully compensated because the margin was not increased. And yet now he is increasing the margin when it isn't in our favor, this is a clear case of bias on the part of the admin/fireball
Yes, we were lucky - at least I count myself lucky that my large, long position was largely closed a few days before the crash.  The profit I lost was less than the loss I would have incurred now (I would have been wiped out).

Of course it is not good at all that Fireball changes the rules during the game.  On the other hand, it is probably better than ICBIT just closing with all our funds being lost.  Remember, we all get angry when our profitable position is closed or we are margin called, but how many of us would leave a negative account alone (and perhaps even open a new one), leaving Fireball with the loss. It is likely that our combined wallets at ICBIT exceeds his private funds, and he also has to make an effort at staying solvent.  We have chosen to gamble our bitcoins on an anonymous, unregulated website.  We have to factor that risk into our risk management.  But that being said, Fireball should be more specific in his descriptions about what he is allowed to do in a crisis.

Disclaimer: I am not in any way associated with ICBIT, but I have made significant (by my modest standards at least) profit on the site, which probably gives me a positive bias.  I currently have money on the site that I would withdraw if I could, but I understand and (reluctantly) accept that they are frozen until the dust settles.
Fireball
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 April 12, 2013, 01:29:29 PM

Hardware maintenance, sorry for downtime.

Bytheway, we are enlarging our infrastructure to include more servers and ultimately provide interruption-less service.

Margin trading platform OrderBook.net (ICBIT): https://orderbook.net
Fireball
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 April 14, 2013, 07:50:07 PM

20:00 UTC clearing will take a bit longer than usual due to contracts expiration.

Margin trading platform OrderBook.net (ICBIT): https://orderbook.net
Fireball
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 April 14, 2013, 08:59:29 PM

Final settlement report for three contracts:

BTC/USD-4.13 (BUJ3)
Settlement price as of 24h volume weighted average price from MtGox: \$98.94082
Total volume of the contract: \$754550
Open interest by the moment of settlement: 5005

GOLD-4.13 (GDJ3)
Settlement price \$1507.0 / \$98.94082 = 15.23 BTC
Total volume of the contract: 119 contracts
Open interest by the moment of settlement: 17

OIL-4.13 (CLJ3)
Settlement price \$90.72 / \$98.94082 = 0.9169 BTC
Total volume of the contract: 187 contracts
Open interest by the moment of settlement: 17

Everyone got paid in full during the final settlement. As the OIL and GOLD contracts show rather low volume, we are not introducing new contracts until more interest in trading them shows up.

Thank you for trading with us and sorry for all the turbulence during the crazy rate rise/fall days.

Margin trading platform OrderBook.net (ICBIT): https://orderbook.net
Ichthyo
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 April 14, 2013, 09:03:48 PM

20:00 UTC clearing will take a bit longer than usual due to contracts expiration.

...which brings up again the topic of gold and oil futures.

In the last time those futures were pretty much dead in the water. In case you did hold a position, there was basically just the option to hold until settlement. But we must concede that this situation became similar unfortunate for the BTC/USD futures. Right now, the only market that works is S&P.

On the other hand, the first batch of Gold and especially Oil was really interesting. They were always a bit illiquid, but they had their own specific behaviour and reacted to the movement of Gold and Oil, instead of just reflecting the BTC/USD rate.

Thus my proposal would be maybe to wait some time, until the BTC/USD development is back in a sane region, so that trading any BTC based future makes sense for both parties involved. And then to reconsider adding Gold and Oil again.
papamoi
Sr. Member

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 April 14, 2013, 09:06:14 PM

Final settlement report for three contracts:

BTC/USD-4.13 (BUJ3)
Settlement price as of 24h volume weighted average price from MtGox: \$98.94082
Total volume of the contract: \$754550
Open interest by the moment of settlement: 5005

GOLD-4.13 (GDJ3)
Settlement price \$1507.0 / \$98.94082 = 15.23 BTC
Total volume of the contract: 119 contracts
Open interest by the moment of settlement: 17

OIL-4.13 (CLJ3)
Settlement price \$90.72 / \$98.94082 = 0.9169 BTC
Total volume of the contract: 187 contracts
Open interest by the moment of settlement: 17

Everyone got paid in full during the final settlement. As the OIL and GOLD contracts show rather low volume, we are not introducing new contracts until more interest in trading them shows up.

Thank you for trading with us and sorry for all the turbulence during the crazy rate rise/fall days.

hi fireball
i ve pmed you

let me know if you can help

thanks
Stephen Gornick
Legendary

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Activity: 2142

 April 14, 2013, 11:19:31 PM

Total volume of the contract: \$754550

The exact same volume for the last two BTCUSDs?

Total volume for this contract is \$754550.

Here's a summary of the settled contracts:

Dec 15, 2012: BTCUSD-12.12 (BUZ2) Total volume: \$500,000, Final settlement: \$13.4870
Mar 15, 2013: BTCUSD-3.13 (BUH3) Total volume: \$754,550, Final settlement: \$47.0764
Apr 14, 2013: BTC/USD-4.13 (BUJ3) Total volume: \$754,550, Final settlement: \$98.94082

Feb 15, 2013: GOLD-02.13 (GDG3) Total volume: 268 contracts, Final settlement: 65.0314 BTC (\$1634.30 USD)
Apr 14, 2013: GOLD-04.13 (GDJ3) Total volume: 119 contracts, Final settlement: 15.23 BTC (\$1507.0 USD)

Feb 15, 2013: OIL-02.13 (CLG3) Total volume: 1513 contracts, Final settlement: 3.8709 BTC (\$97.28 USD).
Apr 14, 2013: OIL-04.13 (CLJ3) Total volume: 187 contracts, Final settlement: 0.9169 BTC (\$90.72)

BUZ2 volume obtained from news release: https://plus.google.com/u/0/117990284188664727142/posts

picobit
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Decor in numeris

 April 15, 2013, 06:44:13 AM

Thus my proposal would be maybe to wait some time, until the BTC/USD development is back in a sane region, so that trading any BTC based future makes sense for both parties involved. And then to reconsider adding Gold and Oil again.

I tend to agree.  The first round of Gold futures were essentially gold futures.  There was a contango in Gold at the same time as there was a contango in BTC versus USD.  The second round were BTC futures traded in the currency "gold", a bit more complicated to figure out, but essentially just the same as BTCUSD.  There was constantly a backwardation corresponding pretty much to the contango in BTC/USD.  Gold futures probably only make sense in times where the BTC is pretty stable, and even then there would be a tendency to treat them as another BTC futures.
Fireball
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 April 15, 2013, 08:42:49 AM

Total volume of the contract: \$754550
The exact same volume for the last two BTCUSDs?

Total volume for this contract is \$754550.

Wow, what a coincidence!

Margin trading platform OrderBook.net (ICBIT): https://orderbook.net
Fireball
Hero Member

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Activity: 673

 April 18, 2013, 09:34:44 AM

Proposed changes to the futures trading aimed to improve the trading process:

• Additional clearing happens automatically if the price hits the limit for N minutes (N may be 5, 10 or more minutes - to be discussed).
• Margin calls are executed automatically against the market according to the existing 75% rule.
• Margin requirements (both initial and maintenance) would be different for long and short positions. The formula for margin requirements calculation will include contango/backwardation into account, so that if some traders drive the price away from spot price, they are guaranteed to have enough money reserved in case contango is reduced.
• Daily settlement price is going to be calculated as the last hour average
• Contract specifications will be united into one (with the only difference of settlement dates), so that e.g. BUM3, BUU3 would be the same spec, with different settlement dates. The contract specification may be amended, and a list of all amendments must be maintained along with the recent edition of the contract spec. Amendments may come into effect at least in the next trading session after announcement, but usually for big changes the waiting interval after announcement may be a few days.

Thanks to alpet and potential market makers who proposed one or more of these changes. Your help is really appreciated!

Margin trading platform OrderBook.net (ICBIT): https://orderbook.net
Super T
Full Member

Offline

Activity: 124

 April 18, 2013, 12:01:08 PM

Fireball,

It's great to be getting the opportunity to discuss this stuff, my feedback below.

1. Additional clearing happens automatically if the price hits the limit for N minutes (N may be 5, 10 or more minutes - to be discussed).

There are two scenarios in which limits are typically locked:
1) During a spot price reversal, spot moves away from futures range and the range is trending TOWARDS the spot but is not able to move quickly enough.
2) During a bull/bear run, where the range is moving AWAY from the spot and speculation results in contango/backwardation reaching range thresholds.

Is the objective to solve one or both of these issues?

As for the proposed solution, it is not clear what "Price" is being referred to here, do you mean Mt Gox spot price, ICBIT "last" price, or some kind of VWAP?

If basing on Mt Gox prices, this might be effective in times of trend reversal, but will not prevent limit locks during contango/backwardation stretching.

If basing on ICBIT prices, I would suggest using modified VWAP rather than last price to reduce the scope for manipulation, and I would also suggest any trigger times are measured in hours rather than minutes, again to reduce the scope for abuse.

Whilst limit-locked ranges are frustrating at times, the range ceilings/floors do provide a buffer against intraday volatility which is both useful and relevant in a futures market.  Nobody will benefit from changes which make 100% daily range movement possible if the primary outcome of this is increased counterparty default and contract liquidation.

2. Margin calls are executed automatically against the market according to the existing 75% rule.

Does this not currently happen?  I agree this should happen, if a holders margin is exhausted contracts should be auto-offered to market and either sold/bought at the best available price, or placed as an offer at the minimum level which will guarantee 0 counterparty loss.  If no-one picks up the distressed contracts, the corresponding contracts should be liquidated at some specified point (e.g. the subsequent clearing session).

3. Margin requirements (both initial and maintenance) would be different for long and short positions. The formula for margin requirements calculation will include contango/backwardation into account, so that if some traders drive the price away from spot price, they are guaranteed to have enough money reserved in case contango is reduced.

This needs to be thought through very carefully.  A blanket rule to automatically raise margin requirements because of increased backwardation/contango could have the effect of moving innocent contract holders into negative margin and possible liquidation, even if they are on the gaining side of a contract (!).  This could potentially be mitigated if such a margin adjustment were programmed to occur at clearing after unrealised P/L has been transferred to balance (and then used to refil margin), but I suspect that is not what is being proposed.

4. Daily settlement price is going to be calculated as the last hour average

This is far better than using last trade price. However, if there has been no trade in the past hour the "last second single trade manipulation" would appear to remain possible.  To prevent this could you ensure the last hour average somehow apportions appropriate "weight" to periods in which no trades have occurred (e.g. because ranges are limit locked).

5. Contract specifications will be united into one (with the only difference of settlement dates), so that e.g. BUM3, BUU3 would be the same spec, with different settlement dates. The contract specification may be amended, and a list of all amendments must be maintained along with the recent edition of the contract spec. Amendments may come into effect at least in the next trading session after announcement, but usually for big changes the waiting interval after announcement may be a few days.

Cool.
picobit
Hero Member

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Decor in numeris

 April 18, 2013, 12:01:49 PM

I think these are all good ideas - except for the first point.

On one hand, it was a problem during the rally (and probably also during the crash) that ICBIT had difficulties following the movements of the market - but two daily clearances must have helped.  On the other hand, during the many "mini-crashes" leading up to the big crash, the market moved far down for an hour or so, and then moved back up again.  In these cases, the lack of clearing gave considerable stability to ICBIT compared to the CFD that 1broker was having at the time, where most long positions were liquidated in those situations.

Also, I would fear that market manipulation would be easier for big players: they can almost trigger a clearing "on demand".  Open a huge position at the end of the trading range during a "sleepy" period, and force a lot of people out of the market.

random_cat
Jr. Member

Offline

Activity: 30

 April 18, 2013, 06:19:56 PM

Proposed changes to the futures trading aimed to improve the trading process:

• Additional clearing happens automatically if the price hits the limit for N minutes (N may be 5, 10 or more minutes - to be discussed).

This makes no sense as proposed.  Are there range limits or not?  Pick one.

Given the thin nature of the market I can understand a desire to prevent a certain sort of manipulation via range limits.   If the desire is to
retain limits, then perhaps the limits (as well as margin requirements) should be based on historical volatility of spot.  Unfortunately the
options market is now thin with extremely wide spreads.

More sessions ending at previously scheduled times is a better choice (and allows for scheduled market down time).

If/when there is sufficient maker, why have limits at all?

In any event, note that a lower bound on the probability of an arbitrageur missing a margin call in any given session of a fast market is given by
$u=http://latex.codecogs.com/gif.latex?\sum_{c=0}^{R-1}e^{-0.1N}(0.1N)^{c}/c!&t=576&c=v1zTCQqeczHeIw$
where R = required number of confirmations for a deposit (R>0).   Please choose R and N carefully to avoid dysfunction.

Quote
• Margin requirements (both initial and maintenance) would be different for long and short positions. The formula for margin requirements calculation will include contango/backwardation into account, so that if some traders drive the price away from spot price, they are guaranteed to have enough money reserved in case contango is reduced.

I hope this is published for trader review.   Something fixed with a varying (as often as per session) table of margin factors would be excellent.

Ichthyo
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 April 19, 2013, 02:01:53 AM

I too share the concerns regarding additional clearings.
What exactly to you want to achieve with those?

As long as the limits are not moved, all positions are covered (not entirely true, but roughly speaking). The market might go dead, but no harm is done. But whenever the limits are moved, the probability of forced liquidations is increased. Thus, why should we strive at increasing the speed of possible adjustments to the trading limits?

We have now two clearings per day (correct?).
Even in case of a crash of the spot market, this gives a moderate adjustment rate.

If there is really the need for additional adjustments, then these should be triggered by the spot rate's movement solely (never by the futures). And there should be a relation to the absolute distance between the trading range and the current spot rate.

In the past, we had several times a stable market situation, where the spot rate was outside the futures trading range, due to contango or backwardation. Only when the spot moves significantly and stays at the new position for more than one trading session, plus when additionally the futures market is dysfunctional (e.g. only bids or only asks, stacked to the limits) -- only then a rule for additional clearings seems indicated.
Ichthyo
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 April 19, 2013, 02:33:15 AM

on a second thought....

why not using the shape of the futures market to a greater extent?

The rule would be: whenever the market becomes totally loopsided, i.e one side of the orderbook is empty within the trading range, and the other side is stacked with orders up to the very limit of the range.

When this condition is met, then the market needs help, e.g. one additional clearing per day.
But if this condition is not met, then there is no necessity for any intervention.

The whole point of futures is that they represent a bet on a future situation.
Thus, if the BTC spot crashes to \$30 right now, of what relevance is this for a future due next september? If market participants still feel fine trading (both buying and selling) them at \$150, then everything should be fine. Only if everyone wants to get out of their positions, but can't due to the trading range, only then an intervention seems adequate.
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text:
Corpus ID: 126350882

Optimal Transport on the Probability Simplex with Logarithmic Cost

@article{Khan2018OptimalTO,
title={Optimal Transport on the Probability Simplex with Logarithmic Cost},
author={Gabriel Khan and Jun Zhang},
journal={arXiv: Optimization and Control},
year={2018}
}
• Published 30 November 2018
• Mathematics, Economics
• arXiv: Optimization and Control
Motivated by the financial problem of building financial portfolios which outperform the market, Pal and Wong considered optimal transport on the probability simplex $\triangle^n$ where the cost function is induced by the free energy. We study the regularity of this problem and find that the associated $MTW$ tensor is non-negative definite and in fact constant on $\triangle^n \times \triangle^n$. We further find that relative $c$-convexity corresponds to the standard notion of convexity in the… Expand
3 Citations
On the difference between entropic cost and the optimal transport cost.
Consider the Monge-Kantorovich problem of transporting densities $\rho_0$ to $\rho_1$ on $\mathbb{R}^d$ with a strictly convex cost function. A popular relaxation of the problem is the one-parameterExpand
Logarithmic Divergences: Geometry and Interpretation of Curvature
• Mathematics, Computer Science
• GSI
• 2019
The logarithmic divergence is shown to be equivalent to a conformal transformation of the Bregman divergence, and, via an explicit affine immersion, is equivalent to Kurose's geometric divergence. Expand
Optimal transport and information geometry
• Mathematics
• 2019
Optimal transport and information geometry both study geometric structures on spaces of probability distribution, and their connections have attracted more and more attention. In this paper we showExpand

References

SHOWING 1-10 OF 54 REFERENCES
Logarithmic divergences from optimal transport and Rényi geometry
It is shown that if a statistical manifold is dually projectively flat with constant curvature, then it is locally induced by an L(∓α)-divergence, and a generalized Pythagorean theorem holds true. Expand
Hessian Curvature and Optimal Transport
• Computer Science, Mathematics
• GSI
• 2019
A new complex geometric interpretation of the optimal transport problem is introduced by considering an induced Sasaki metric on the tangent bundle of the domain of \(\varPsi) and the Ma-Trudinger-Wang tensor is proportional to the orthogonal bisectional curvature. Expand
On the regularity of solutions of optimal transportation problems
We give a necessary and sufficient condition on the cost function so that the map solution of Monge’s optimal transportation problem is continuous for arbitrary smooth positive data. This conditionExpand
Multiplicative Schrödinger problem and the Dirichlet transport
• Mathematics
• 2018
We consider an optimal transport problem on the unit simplex whose solutions are given by gradients of exponentially concave functions and prove two main results. First, we show that the optimalExpand
About the analogy between optimal transport and minimal entropy
• Mathematics
• 2015
We describe some analogy between optimal transport and the Schr\"odinger problem where the transport cost is replaced by an entropic cost with a reference path measure. A dual Kantorovich typeExpand
Hölder Continuity and Injectivity of Optimal Maps
• Mathematics
• 2011
Consider transportation of one distribution of mass onto another, chosen to optimize the total expected cost, where cost per unit mass transported from x to y is given by a smooth function c(x, y).Expand
Continuity of optimal transport maps and convexity of injectivity domains on small deformations of 2
• Mathematics
• 2009
Given a compact Riemannian manifold, we study the regularity of the optimal transport map between two probability measures with cost given by the squared Riemannian distance. Our strategy is toExpand
A Jacobian Inequality for Gradient Maps on the Sphere and Its Application to Directional Statistics
In the field of optimal transport theory, an optimal map is known to be a gradient map of a potential function satisfying cost-convexity. In this article, the Jacobian determinant of a gradient mapExpand
From the Schr\"odinger problem to the Monge-Kantorovich problem
The aim of this article is to show that the Monge-Kantorovich problem is the limit of a sequence of entropy minimization problems when a fluctuation parameter tends down to zero. We prove theExpand
Optimal Entropy-Transport problems and a new Hellinger–Kantorovich distance between positive measures
• Mathematics
• 2015
We develop a full theory for the new class of Optimal Entropy-Transport problems between nonnegative and finite Radon measures in general topological spaces. These problems arise quite naturally byExpand


text:
# Retrieving values from InterpolatingFunction

I have code for finding the flight of a golf ball and I would like to retrieve the values of horizontal displacement x[t] and at the time when the golf ball hits the ground (y[t]=0, when t=tmax)

(* Question 1. (B) *)
(* Experimenting with different life and drag coefficiants cl and cd, respectively, and different starting angles \[Theta] to study the behaviour of the motion of the golf ball *)

Clear["Global*"]

(* Defining the variables density of air at 20C \[Rho], the cross-sectional area of the golf ball A, acceleration due to gravity g, initial velocity u and the mass of the golf ball m *)

\[Rho]=1.2014; A=(0.02^2*Pi); g=9.81; u=60; m=0.05;

(* Defining the proportionality factor \[Alpha] by half the density of air \[Rho] multiplied with the cross-sectional area A *)

\[Alpha]=1/2*\[Rho]*A;

(* Setting the solution of the differential equations to be sol which can be called with the number of the solution n, the drag coefficiant cd, the lift coefficiant cl and the initial angle \[Theta]. Using the Block\[Rule]RecursionLimit funtion to set the limit of recursions to infinity *)

sol[n_,cd_,cl_,\[Theta]_]:=Block[{\$RecursionLimit=Infinity},

(* Using NDSolve function with all the previous differential equations and initial conditions to calculate the numerical solution to the differential equations *)

NDSolve[{
x''[t]==-\[Alpha]*(1/m)*v[t]*x'[t]*(cl*Sin[P[t]]+cd*Cos[P[t]]),
y''[t]==-g-\[Alpha]*(1/m)*v[t]*y'[t]*(cd*Sin[P[t]]-cl*Cos[P[t]]),
v[t]==Sqrt[x'[t]^2+y'[t]^2],
P[t]==ArcTan[y'[t]/x'[t]],P[0]==\[Theta],
y'[0]==u*Sin[P[0]],x'[0]==u*Cos[P[0]],
y[0]==x[0]==0,

(* Using the WhenEvent function to set a new variable tmax that indicates when the solutions dips below the x-axis, which would be when the golf ball hits the ground *)

WhenEvent[y[t]==0,tmax[n]=t]},

(* Telling the NDSolve function to calculate the vertical distance y[t] and the horizontal distance x[t] as a function of time *)

{x[t],y[t]},

(* Setting the time limit to a large number of t=200 as to make sure the *)

{t,0,200}]];

As can be seen, I would like to retrieve the x[tmax[n]] value from the interpolating function returned from NDsolve function. I have tried numerous things, but i can't seem to find a solution.

• have a look at this mathematica.stackexchange.com/a/137787/2079 Oct 3 '17 at 11:53
• I don't understand. Aren't you storing the time in tmax[n]? If I run sol[1, 1, 1, 1] and then evaluate tmax[1], it returns 6.54454, and when I do ParametricPlot[{x[t], y[t]} /. sol[1, 1, 1, 1] // Evaluate, {t, 0, tmax[1]}], I get a nice plot showing the trajectory from launch to when it hits the ground. Finally, just do x[t] /. First@sol[1, 1, 1, 1] /. t -> tmax[1]`. Oct 3 '17 at 15:57


text:
# How to connect a Dask cluster (in Docker) to Amazon S3

I am going to run all of my code on the local machine, but when you have Dask workers configured you can deploy them in a Mesos/Kubernetes cluster using the same configuration (just change the connection strings).

# Docker Compose

First, I’m going to define my Docker Compose file (I will call it: docker-compose.yaml). The file starts with the version, so in the first line I put:

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version: "3.1"

## Scheduler

The first service I must run is the Dask scheduler, so I add the following code to my docker-compose.yaml file:

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services:
scheduler:
ports:
- "8786:8786"
- "8787:8787"

We see that it will deploy the Dask scheduler as a dask-scheduler host and expose the endpoints used by the workers.

## Workers

In the next step, I have to deploy the workers. In this example, I will configure only one worker, but in a real-life scenerio, you will need many of them (if one worker is enough, you don’t need Dask at all).

Note that the worker configuration needs the address of the Dask scheduler and we must mount the directory containing the AWS credentials. I’m going to assume that you pass the credentials as a Docker volume, so the sensitive data is not displayed in any configuration console containing environment variables.

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worker:
volumes:
- \$HOME/.aws/credentials:/home/app/.aws/credentials:ro

## Workers (with environment variables)

If for some reason, you need to specify the AWS credentials as environment variables, you can configure the worker in the following way:

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worker:
environment:
- AWS_ACCESS_KEY_ID=your_access_key
- AWS_SECRET_ACCESS_KEY=your_secred
- AWS_DEFAULT_REGION=the_aws_region

Please remember, that it is not a recommented way because the environment variables (so also the keys and passwords) are usually visible in an administration console, so everyone who can manage your deployed services has access to the keys.

## Notebook

Finally, when I have the scheduler and workers configured, I can add the notebook to the configuration.

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notebook:
hostname: notebook
ports:
- "8888:8888"
environment:

Now, when I run the docker-compose --file docker-compose.yaml up command, all of the services will start, and I can open the localhost:8888 address in my browser to access the notebook.

Obviously, when you deploy the configuration using Kubernetes or Mesos, the command is going to be different. Check the documentation of the service you are using for deployment.

To create a DataFrame which contains data from an S3 bucket, I have to run the following code in the notebook:

1
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text:
Db questions

1. Nov 18, 2005

dervast

Hi i want to ask sth for the db
We have denoted db as log10(P/Po)
I want to ask sth if a^x=theta=>x=log(a)theta
then the db is 10*a^x=10*log(a)theta Is that right?
I want to know why we multiply the log(P/Po) by 10? Why we didnt leave it alone?

2. Nov 18, 2005

SGT

Because $$log\frac{P}{P_0}$$, called Bel is too large a unit to be useful. So we use the deciBel, that is more manageable.

3. Nov 18, 2005

dervast

If bel is too large to be manageable then why we multiplt by 10?

4. Nov 18, 2005

Averagesupernova

Dervast, it is the same thing as microfarads vs farads. Most of the time we deal in signal ratios that make it more suitable to use decibels instead of bels. Same thing with farads and microfarads. However, if you choose NOT to multiply by 10 and work with bels it is up to you. It will work just fine.

5. Nov 18, 2005

Ouabache

I think you are looking for an analytical reason.
Try applying dimensional analysis.
log (Pout/Pin) = y (Bel) equ (i)
how to convert to dB? Since a dB is 1/10 of a Bel, then 10db = 1 Bel
To convert equat (i) to db, multiply both sides by (10db/1Bel)
10 log (Pout/Pin) = y (Bel)(10db/1Bel) <------dimensions Bel/Bel = 1 leaving just db
10 log (Pout/Pin) = 10y (db) (ii)
right side of equation (ii) means we multiply the solution y in Bels by 10 to obtain a value in dB.


text:
Question

# The  correct increasing order for second ionization energy of  $$He, Li, Be, B, C, N, O, F, Ne$$  is :

A
Be<C<B<N<O<F<Ne<Li<He
B
Be<B<C<N<O<F<Ne<He<Li
C
Be<C<B<N<F<O<Ne<He<Li
D
Be<C<B<O<N<F<Ne<He<Li

Solution

## The correct option is B $$Be\, <\, C\, <\, B\, <\, N\, <\, F\, <\, O\, <\, Ne\,<\, He\,<\,Li$$$$He^+- 1s^1$$$$Li^+- [He]$$$$Be^+- [He]2s^1$$$$B^+- [He]2s^2$$$$C^+- [He]2s^22p^1$$$$N^+- [He]2s^22p^2$$$$O^+- [He]2s^22p^3$$$$F^+- [He]2s^22p^4$$$$Ne^+- [He]2s^22p^5$$$$Li^+$$ is most stable and then comes $$He^+$$. Half filled orbitals are more stable than partially filled.Hence the correct order will be $$Be\, <\, C\, <\, B\, <\, N\, <\, F\, <\, O\, <\, Ne\,<\, He\,<\,Li$$Chemistry

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text:
# Why do the geosynchronous TDRS satellites have this distribution of inclinations?

From the Tracking and Data Relay Satellite System (TDRSS) option on Celestrak's NORAD Two-Line Element Sets Current Data TLE page I've compiled current TLE inclinations for "on line" TDRS spacecraft:

COSPAR       NORAD    Launched       Spacecraft   current inc(degs)
1988-091B    19548    1988-09-29     TDRS 3          14.4405
1991-054B    21639    1991-08-02     TDRS 5          14.5306
1993-003B    22314    1993-01-13     TDRS 6          14.0861
1995-035B    23613    1995-07-13     TDRS 7          15.0545
2000-034A    26388    2000-06-30     TDRS 8           7.9573
2002-055A    27389    2002-03-08     TDRS 9           5.8274
2002-055A    27566    2002-12-05     TDRS 10          5.5187
2013-004A    39070    2013-01-31     TDRS 11          5.0141
2014-004A    39504    2014-01-24     TDRS 12          5.6613

While not included (yet) on that page, according to this answer TDRS-13 has entered into service, and can still be found in Celestrak:

2017-047A    42915    2017-08-18     TDRS 13          6.7494

While these satellites are in geosynchronous orbits, they would not be considered geostationary orbits; their large inclinations result in analemma-shaped ground tracks, and the antennas of TDRS ground-segment stations would need to trace their daily, nearly North-South movement in order to remain in contact.

This is probably not a big deal, considering the attention and resources already necessary to maintain this critical segment of so many ongoing missions.

But I am curious about the inclinations themselves. There is a cluster (TDRS-3 through TDRS-7) between 14 and 15 degrees, and another (TDRS-9 through TDRS-13) between 5 and 6 degrees. The newest member TDRS-13 is and has been close to 7 degrees for a while now, and TDRS-8 is near 8 degrees.

Question: Are these inclinations specified, station-kept and optimal in some way, or do they simply reflect non-station-kept "inclination creep" that is a natural phenomenon for circular orbits at this distance, or is there another way to uderstand this distribution in TDRS inclinations?

note: I understand that without station keeping, inclination will increase. But I'm looking for more of an answer than "It might be...". Considering that TDRS-13 started at almost 7 degrees suggests that it might be more than just an intrinsic lack of North-South station-keeping ability.

above: Screen shot of TDRS-13 analemma-shaped ground track from N2Y0.

below: Inclination and mean motion (rev/day) for TDRS satellites from the first TLE found in the first week of each calendar year. Dot for TDRS13

• I can't offer more than "it might be" but here goes a comment -- It might be to increase the chances of seeing one of the TDRS birds. If some mission parameter or vehicle obstruction blocks the angle towards the equator, it's OK because some satellites are above and below it. – Saiboogu Mar 28 '18 at 18:49
• @Saiboogu that sounds like a great hypothesis. – uhoh Mar 28 '18 at 22:57
• @Saiboogu take a look at the angles between each pair of TDRS-5, -10, and -11 which are located near each other in longitude, as shown in this answer. It certainly looks (a little bit) like 3-phase AC to me! I think you should pursue this further. – uhoh Mar 29 '18 at 9:49
• I've got some sit-and-wait errands coming up, I'll see if I can do some reading and find some more clues. – Saiboogu Mar 29 '18 at 14:07
• @Saiboogu finished with your errands yet? I've added a bounty. – uhoh Apr 15 '18 at 11:24

Please have a look at the "Onboard Orbit Computations" chapter on the "ACS Without an Attitude" book by Harold L. Hallock Gary Walter David G. Simpson Christopher Rouff.

## Part of section 9.2 - HST Onboard Orbit Models

...In fact, only five parameters are needed to specify the motion of the center of the box in GCI (which we’ll call TDRS from now on for simplicity), namely the TDRS inclination (ideally zero), the TDRS longitude (assigned by the Network Control Center (NCC)), the distance from the Earth center to TDRS (nominally 42,164km), the TDRS rotation rate in the GCI frame (nominally the Earth’s rotation rate), and the Greenwich Hour Angle at the epoch time (i.e., the right ascension of the Greenwich Meridian at the epoch time). The TDRS inclination actually does require calculation because the TDRSS inclinations are allowed to drift up to $14^o$ to conserve fuel and because of the time dependent nature of the GCI reference frame arising from the precession and wobble behavior of the Earth’s spin axis relative to the “fixed stars”. Operationally, exploiting the Flight Dynamics Facility’s (FDF’s) inclined-center- of-box (ICOB) modeling, the advantage of using this so-called “TDRS on-a-stick” model is that the number of uplink parameters (including the epoch time) is reduced from 15 to 6 (the 5 specified above plus the epoch time), and the frequency for parameter updates is reduced from once a week to once a year, or less

Another source of information can be found here on page 18-19.

TDRSs have the same orbit period as a geostationary satellite but are not North/South Station-kept and therefore their orbits may be highly inclined.

• They move in a figure-8 pattern as viewed from earth.

• Inclination evolves between 0 to 15 degrees.

• High inclination provides visibility to South Pole.

• Inclination is not managed as part of stationkeeping.

I think that inclination have a minor effect on the constellation performance. By not controlling inclination the satellites reduce the required amount of propellant dramatically and increase the constellation lifetime.

• Excellent, thanks! I've added a more detailed plot of the inclinations versus time to the question. While conservation of propellant is certainly much of the answer TDRS 8, 9, 10, 11, 12, and 13 were all started somewhere between 6 and 8 degrees of inclination. Curiouser and curiouser! en.wikipedia.org/wiki/Geostationary_orbit#Orbital_stability – uhoh Apr 17 '18 at 13:43
• If your system already accepts drift up to 14° for conservation of propellant, it seems to make sense to accept the same precision on deployment targets as well, especially since initial inclinations begin higher and you're aiming to reduce them as close as necessary to zero. Necessary is <14° evidently. – Saiboogu Apr 17 '18 at 14:32
• @Saiboogu the answer will turn out to be much more interesting than that. – uhoh Apr 17 '18 at 14:43
• Thanks for the additional source. It confirms there is no North-South station-keeping, but doesn't directly address the reason "why" (e.g. nothing about fuel). However it does provide the statement that TDRS can be used by stations at the South Pole, and I'd say that that's more than a "minor effect on the constellation performance". It seems to be a major effect on performance, as standard GEO orbit is invisible from the poles! It's not yet clear if that addresses the "Why?" but it certainly is a positive impact! – uhoh Apr 22 '18 at 1:34


text:
# Platinum as a Metallic Material

## Example of Application: Glass for Flat Panel Displays

Feeder systems of platinum-rhodium alloys are used, for example, in the manufacture of glass for flat panel displays. The system is resistively heated. The blue flanges are the water-cooled current connectors. A feeder system of this type can be up to 3$m$ long, the individual cells have a diameter up to 400$mm$. The glass temperature can exceed 1,600 $°C$. The high mechanical loading at the operating temperature and the severe thermal expansion during heating and cooling demand a material with high strength and good ductility, primarily PtRh90/10 DPH. The further figures show the stirrer for the cell on the left and the plunger for the cell on the right.

Fig.1
Feeder system
Fig.2
Stirrer

Stirrer for the left hand cell of the feeder system.

Fig.3
Plunger

Plunger for the right hand cell of the feeder system.

Page 17 of 20


text:
Corpus ID: 119364135

# Potentials for light moduli in supergravity and string theory

@article{Alwis2006PotentialsFL,
title={Potentials for light moduli in supergravity and string theory},
author={S. Alwis},
journal={arXiv: High Energy Physics - Theory},
year={2006}
}
• S. Alwis
• Published 2006
• Physics
• arXiv: High Energy Physics - Theory
• This is an account of lectures that were given at TASI 2005, the Shanghai Summer School in M-theory 2005 and the Perimeter Institute. I review 1) the derivation of the potential for chiral scalar fields in ${\cal N}$=1 supergravity 2) the relation between F and D terms for chiral scalars, Weyl anomalies and the generation of non-perturbative terms in the superpotential and 3) the derivation of effective potentials for light moduli in type IIB string theory.


text:
• Corpus ID: 118533992

# Comparing mesons and W_L W_L TeV-resonances

@article{Dobado2015ComparingMA,
title={Comparing mesons and W\_L W\_L TeV-resonances},
journal={arXiv: High Energy Physics - Phenomenology},
year={2015}
}`
• Published 13 October 2015
• Physics
• arXiv: High Energy Physics - Phenomenology
Tantalizing LHC hints suggest that resonances of the Electroweak Symmetry Breaking Sector might exist at the TeV scale. We recall a few key meson-meson resonances in the GeV region that could have high-energy analogues which we compare, as well as the corresponding unitarized effective theories describing them. While detailed dynamics may be different, the constraints of unitarity, causality and global-symmetry breaking, incorporated in the Inverse Amplitude Method, allow to carry some…
2 Citations

## Figures from this paper

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• Physics
• 2014
The apparent finding of a 125-GeV light Higgs boson closes unitarity of the minimal Standard Model (SM), that is weakly interacting: this is an exceptional feature not generally true if new physics
ATLAS diboson excess from Stueckelberg mechanism
• Physics
• 2015
A bstractWe discuss the diboson excess seen by the ATLAS collaboration around 2 TeV in the LHC run I at s=8$$\sqrt{s}=8$$ TeV. We explore the possibility that such an excess can arise from a Z′

## References

SHOWING 1-10 OF 13 REFERENCES
Interpreting a 2 TeV resonance in WW scattering
• Physics
• 2016
A diboson excess has been observed ---albeit with very limited statistical significance--- in $WW$, $WZ$ and $ZZ$ final states at the LHC experiments using the accumulated 8 TeV data. Assuming that
Production cross section estimates for strongly-interacting Electroweak Symmetry Breaking Sector resonances at particle colliders
• Physics
• 2015
We are exploring a generic strongly-interacting Electroweak Symmetry Breaking Sector (EWSBS) with the low-energy effectie field theory for the four experimentally known particles ($W_L^\pm$, $Z_L$,
Diboson excess in the Higgs singlet and vectorlike quark models
• Physics
• 2015
Diboson resonance with mass of $1.8-2$ TeV is reported successively by CMS and ATLAS experiments in proton-proton collisions at $\sqrt{s}=8$ TeV. We investigate the potentiality of Higgs singlet as
Review of Scalar Mesons
The formulae needed to parametrise σ, κ, a0(980) and f0(980) are reviewed. For σ and κ, the Adler zero due to Chiral Symmetry Breaking plays a crucial role. The f0(980) and a0(980) are locked to the
Search for high-mass diboson resonances with boson-tagged jets in proton-proton collisions at s=8$$\sqrt{s}=8$$ TeV with the ATLAS detector
• Physics
• 2015
A bstractA search is performed for narrow resonances decaying into WW, WZ, or ZZ boson pairs using 20.3 fb−1 of proton-proton collision data at a centre-of-mass energy of s=8$$\sqrt{s}=8$$ TeV
Diffractive pion production at COMPASS: First results on 3$\pi$ final states - neutral mode
The COMPASS experiment at CERN is designed for light hadron spectroscopy with emphasis on the detection of new states, in particular the search for exotic states and glue‐balls. After a short pilot
Augmenting the diboson excess for the LHC Run II.
• Physics
• 2015
The ATLAS collaboration recently reported an excess of events in the high invariant mass tail of reconstructed diboson events. We investigate their analysis and point to possible subtleties and
Phys
• Rev. Lett. 114, 221803
• 2015


text:
### 统计代写|数据可视化代写Data visualization代考|INFS6023

statistics-lab™ 为您的留学生涯保驾护航 在代写数据可视化Data visualization方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数据可视化Data visualization代写方面经验极为丰富，各种代写数据可视化Data visualization相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|数据可视化代写Data visualization代考|Using Ggplot

Recall that long is a dataset with ten participants, two groups (control and target), three tests, and test scores. A natural question to ask is whether the scores in both groups are different. ${ }^{21}$ For that, we could create a bar plot with scores on the $\gamma$-axis and the two groups on the $x$-axis. We want bars (which represent the mean for each group) as well as error bars (for standard errors)-see $₫ 1.3 .4$. An example is shown in Fig. 2.3.

You should look at any plot in $R$ as a collection of layers that are “stitched” together with a “+” sign. Each subsequent layer is automatically indented by RStudio and can add more information to a figure. The very first thing we need to do when using ggplot2 is to tell the package what data you need to plot. You can do that with the function ggplot(). Inside the function, we will also tell gaplot2 what we want to have on our axes. Let’s carefully go over the code that generates Fig. 2.3, shown in code block 9 .

In line 1 , we source our dataPrep. $R$ script (which itself will source other scripts). A month from now, we would simply open our R Project, click on our eda. $R$ script and, by running line 1 in code block 9 , all the tasks discussed earlier would be performed in the background. $R$ would import your data, load the necessary packages, and prepare the data, and we’d be ready to go. This automates the whole process of analyzing our data by splitting the task into separate scripts/components (which we created earlier). Chances are we won’t even remember what the previous tasks are a month from now, but we can always reopen those scripts and check them out.

As with anything we do in $\mathrm{R}$, there are different ways to save your plot. However, before saving it, we should create a folder for it in our current directory (basics) -let’s call it figures. ${ }^{22}$ One way to save plots created with ggplot2 is to use the function ggsave() right after you run the code that generates your plot. Inside ggsave(), we specify the file name (and extension) that we wish to use (file), and we can also specify the scale of the figure as well as the DPI (dots per inch) for our figure (dpi). Thus, if you wanted to save the plot generated in code block 9 to the figures folder, you’d add a line of code after line 12 : ggsave (file = “figures/plot.jpg”, scale $=0.7, \mathrm{dpi}=$ “retina”). In this case, scale $=$ $0.7$ will generate a figure whose dimensions are $70 \%$ of what you can currently see in RStudio. Alternatively, you can manually specify the width and height of the figure by using the width and height arguments. To generate a plot with the exact same size as Fig. 2.3, use ggsave(file $=$ “figures/plot.jpg”, width $=4$, height $=2.5, \mathrm{dpi}=1000)^{23}$ If you realize the font size is too small in the figure, you can either change the dimensions in ggsave() (e.g., $3.5 \times 2$ instead of $4 \times 2.5$ will make the font look larger), or you can specify the

text size within ggplot()_an option we will explore later in the book (in chapter 5). In later chapters, code blocks that generate plots will have a ggsave() line, so you can easily save the plot.

As mentioned earlier, you can run the ggsave() line after running the lines that generate the actual plot (you may have already noticed that by pressing Cmd + Enter, RStudio will take you to the next line of code, so you can press Cmd + Enter again). Alternatively, you can select all the lines that generate the plot plus the line containing ggsave() and run all of them together. Either way, you will now have a file named plot.jpg in the figures directory (folder) of your R Project. ${ }^{24}$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。


text:
# Squeezed gaussian expectation

1. Mar 7, 2008

### noospace

I'm trying to evaluate the expectation of position and momentum of

$\exp\left(\xi (\hat{a}^2 - \hat{a}^\dag^2)/2\right) e^{-|\alpha|^2} \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} |n\rangle}$

where $\hat{a},\hat{a}^\dag$ are the anihilation/creation operators respectively.

Recall $\hat{x} = \sqrt{\hbar/(2m\omega)}(\hat{a}^\dag + \hat{a})$.

The calculation of $\langle \hat{x} \rangle$ and $\langle \hat{p} \rangle$ are easy if one uses the Hadamard lemma

$e^X Y e^{-X} = Y+ [X,Y] + (1/2)[X,[X,Y]] + \cdots$.

I'm running into touble in the evaluation of $\langle\hat{x}^2\rangle,\langle\hat{p}^2\rangle$. In particular in the calculation of

$\exp\left(\xi (\hat{a}^\dag^2-\hat{a}^2)/2\right)(\hat{a}^\dag + \hat{a})^2\exp\left(-\xi (\hat{a}^\dag^2-\hat{a}^2)/2\right)$.

I was able to show that

$\exp\left(\xi (\hat{a}^\dag^2-\hat{a}^2)/2\right)(\hat{a}^\dag + \hat{a})^2\exp\left(-\xi (\hat{a}^\dag^2-\hat{a}^2)/2\right) = (\hat{a}^\dag + \hat{a})^2 + \xi([\hat{a}^2,\hat{a}^\dag^2]+[\hat{a}^\dag\hat{a},\hat{a}^\dag^2-\hat{a}^2]) + \frac{\xi}{2!}[(\hat{a}^\dag+\hat{a})^2,[\hat{a}^2,\hat{a}^\dag^2]+[\hat{a}^\dag\hat{a},\hat{a}^\dag^2-\hat{a}^2]]+\cdots$

but I can't figure out how to simplify this.

Last edited: Mar 7, 2008
2. Mar 8, 2008

### soarce

You may try to expand $(a^\dagger +a)^2=1+a^\dagger a + (a^\dagger)^2 + a^2$, then $a^\dagger a$ is the photon number operator...

Few years ago I did similar calculations and I was using the book of J.S.Peng and G.X. Li "introduction to modern quantum optics" where they give many results of some expectation values (however, you must check carefully their formulas against typing errors)
Also W.H. Louisell "Quantum statistical properties of radiation" may contain similar calculations.

but you may try to have a look directly at "the bible of quantum optics": Mandel & Wolf "Optical coherence and quantum optics"

3. Mar 8, 2008

### olgranpappy

oh?
it is also easy, if one knows how to do it.


text:
# Convolutional Coding

Convolutional codes differ from block codes by means of method of operation. A convolutional encoder operates over serial data, whereas block codes operates over a block of input data. Also different is the utilization of memory elements in the convolutional encoders. In the case of block codes, there is no memory element involved in the generation of encoded data.

Convolutional codes are specified as (n,k,L) , where n is the number of output bits from the encoder, k is the number of input bits to the encoder and L is the constraint length of the encoder. Disparate expressions for constraint length are often found in different text books but the fundamental idea is same The constraint length is used to calculate the number of memory stages or flipflops used in the encoder. As long as we know L and the underlying formula we can calculate the number of memory stages (m). So it does not really matter which expression for L is used. The constraint length is expressed as

$$L = k(m+1)$$

In some text books it is expressed as L=k x (m) and in some other books it is even expressed as L=m+1.We will use the first expression throughout our discussion.

We will take up a simple convolutional code (2,1,3) where n=2, k=1 and L=3 ( the expression L=k(m+1) is used).Lets construct the encoder from the above information. The encoder is constructed with 1 input bit, 2 output bits and 2 memory elements. Note that the L=k(m+1) expression leads to 2 memory elements. Here these two memory elements are used to store the past 2 input bits. If the expression L=k*m is used and for a (2,1,3) encoder (L=3), the number of memory elements will be 3, where these 3 memory elements are used to store past 3 input bits . So the expression for constraint length has to be carefully interpreted, otherwise any misinterpretation will lead to a different encoder structure altogether.

Now we know the number of bits going into the encoder , number of bits coming out from it and the number of memory elements. Till now the encoder is like a black box to us in the sense that we don’t know how the memory elements are utilized to generate the output bits from the input. To fully understand the encoder structure we need something called “generator polynomials” that tell us how the memory elements are linked to achieve encoding. The generator polynomials for a specific convolutional encoder set (n,k,L) are usually found through simulation. The set (n,k,L) along with n generator polynomials completely describes a convolutional encoder.

For our (2,1,3) encoder , we use two generator polynomials, one for each output.

$$g_0=\left [1 0 1 \right ]$$

$$g_1=\left [1 1 1\right ]$$

Lets put everything to make the encoder structure.

### Encoder structure,State Diagram and Trellis :

The convolutional encoder can also be represented using a finite state machine. The entire behavior of a convolutional encoder is depicted by a state diagram. The number of states in a state diagram depends on the number of memory elements in the encoder. If number of memory elements is m then the number of states in the state diagram will be 2m.For the (2,1,3) convolutional encoder the number of states will be 4 , i.e., the last two memory elements are used to store the past inputs I -1 and I -2. Transition between the states depends on the present input I 0.The solid line in the state diagram indicates the transitions due to the input I0 =0 and dotted lines are used to represent the transitions due to the input I0= 1. The output bits generated during each state transition is written in dark red color (along the transistion lines).

State diagrams can be easily constructed with the help of a state table as shown below.

## State Table - (2,1,3) Convolutional Encoder

Input
Current State
Next State
Output
0
00
00
00
0
01
00
11
0
10
01
01
0
11
01
10
1
00
10
11
1
01
10
00
1
10
11
10
1
11
11
01

A trellis diagram is constructed from the given state diagram. Trellis diagram helps us to understand the concept of Viterbi Decoding algorithm that is used to decode the convolutional encoded data.

A single stage Trellis is illustrated for understanding. The four possible states of the encoder are depicted as four horizontal rows. Present states are considered to be on the left side and the possible next states on the right side of the trellis. The connections between the present states and the next states exactly follow the state diagram representation. Again, the output bits generated during each state transition are indicated in dark red color. A multiple stage Trellis is constructed by replicating the single stage and juxtaposing each of the single stage side by side as shown in the figure.

The concept of Viterbi Decoding will be discussed in the next article

### Recommended Books

• Abdulbagi Elsanousi

Hello sir I’m looking for Matlab code for BER of an convolutionally coded OFDM system using different modulation schemes over an AWGN channel.
could you help me

• Abdulbagi Elsanousi

hello sir this code for BER of convolution coded ofdm system using 16-QAM … i have tried to play on modulation section to change modulation from 16-qam to 8-qam but there was problem in interleaving part the length of data will change .. could u help me

this is code
% OFDM Code
% Author: Ihsan Ullah,
% Ms-55 Electrical,
% College of EME,
% NUST Pakistan
% No.of Carriers: 64
% coding used: Convolutional coding
% Single frame size: 96 bits
% Total no. of Frames: 100
% Modulation: 16-QAM
% No. of Pilots: 4
% Cylic Extension: 25%(16)

close all
clear all
clc

%%
% Generating and coding data
t_data=randint(9600,1)’;
x=1;
si=1; %for BER rows
%%
for d=1:100;
data=t_data(x:x+95);
x=x+96;
k=3;
n=6;
s1=size(data,2); % Size of input matrix
j=s1/k;

%%
% Convolutionally encoding data
constlen=7;
codegen = [171 133]; % Polynomial
trellis = poly2trellis(constlen, codegen);
codedata = convenc(data, trellis);

%%
%Interleaving coded data

s2=size(codedata,2);
j=s2/4;
matrix=reshape(codedata,j,4);

intlvddata = matintrlv(matrix’,2,2)’; % Interleave.
intlvddata=intlvddata’;

%%
% Binary to decimal conversion

dec=bi2de(intlvddata’,’left-msb’);

%%
%16-QAM Modulation

M=16;
y = qammod(dec,M);
% scatterplot(y);

%%
% Pilot insertion

lendata=length(y);
pilt=3+3j;
nofpits=4;

k=1;

for i=(1:13:52)

pilt_data1(i)=pilt;

for j=(i+1:i+12);
pilt_data1(j)=y(k);
k=k+1;
end
end

pilt_data1=pilt_data1′; % size of pilt_data =52
pilt_data(1:52)=pilt_data1(1:52); % upsizing to 64
pilt_data(13:64)=pilt_data1(1:52); % upsizing to 64

for i=1:52

pilt_data(i+6)=pilt_data1(i);

end

%%
% IFFT

ifft_sig=ifft(pilt_data’,64);

%%

cext_data=zeros(80,1);
cext_data(1:16)=ifft_sig(49:64);
for i=1:64

cext_data(i+16)=ifft_sig(i);

end

%%
% Channel

% SNR

o=1;
for snr=0:2:50

ofdm_sig=awgn(cext_data,snr,’measured’); % Adding white Gaussian Noise
% figure;
% index=1:80;
% plot(index,cext_data,’b’,index,ofdm_sig,’r’); %plot both signals
% legend(‘Original Signal to be Transmitted’,’Signal with AWGN’);

%%
%%
%Removing Cyclic Extension

for i=1:64

rxed_sig(i)=ofdm_sig(i+16);

end

%%
% FFT

ff_sig=fft(rxed_sig,64);

%%
% Pilot Synch%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

for i=1:52

synched_sig1(i)=ff_sig(i+6);

end

k=1;

for i=(1:13:52)

for j=(i+1:i+12);
synched_sig(k)=synched_sig1(j);
k=k+1;
end
end

% scatterplot(synched_sig)

%%
% Demodulation
dem_data= qamdemod(synched_sig,16);

%%
% Decimal to binary conversion

bin=de2bi(dem_data’,’left-msb’);
bin=bin’;

%%
% De-Interleaving

deintlvddata = matdeintrlv(bin,2,2); % De-Interleave
deintlvddata=deintlvddata’;
deintlvddata=deintlvddata(:)’;

%%
%Decoding data
n=6;
k=3;
decodedata =vitdec(deintlvddata,trellis,5,’trunc’,’hard’); % decoding datausing veterbi decoder
rxed_data=decodedata;

%%
% Calculating BER
rxed_data=rxed_data(:)’;
errors=0;

c=xor(data,rxed_data);
errors=nnz(c);

% for i=1:length(data)
%
%
% if rxed_data(i)~=data(i);
% errors=errors+1;
%
% end
% end

BER(si,o)=errors/length(data);
o=o+1;

end % SNR loop ends here
si=si+1;
end % main data loop

%%
% Time averaging for optimum results

for col=1:25; %%%change if SNR loop Changed
ber(1,col)=0;
for row=1:100;

ber(1,col)=ber(1,col)+BER(row,col);
end
end
ber=ber./100;

%%
figure
i=0:2:48;
semilogy(i,ber);
title(‘BER vs SNR’);
ylabel(‘BER’);
xlabel(‘SNR (dB)’);
grid on

• Abdulbagi Elsanousi

sir could you send me ur email .. i need to contact u

• Kumud Altmayer

Mr. Mathuranathan,

Can you please add an appendix in you book with sample Convolutional codes. I bought the book as well. Thank you!

• Thanks for the suggestion. Sure I will add it.

• Anna

sir, i bought the book but i do not find code for convolutional encoding and decoding.. could you please send the code to my mail id.. It is so urgent.. It will be very much helpful if you send as soon as possible.

• Hi Anna,

The code for convolutional decoder is a bit complex and is not available in the ebook. Sorry if the book missed your expectation.

However, if you have a communication toolbox, you can try this

Encoding:

t = poly2trellis([4 3],[4 5 17;7 4 2]); % Define trellis.
code = convenc(ones(100,1),t); % Encode a string of ones.

Decoding (Hard decision):
tb = 2; % Traceback length for decoding
decoded = vitdec(code,t,tb,’trunc’,’hard’); % Decode.

• Anna

Sir,if you could please say the error in the code given below it will be helpful.

close all
clear all
clc

nbitpersym = 52; % number of bits per OFDM symbol (same as the number of subcarriers for BPSK)
nsym = 10^4; % number of symbols
len_fft = 64; % fft size
sub_car = 52; % number of data subcarriers
EbNo = 0:2:12;
Td =3.2*10e-6;%%data symbol duration
Tcp =.8*10e-6;%%cyclic prefix duration= .25Td

EsNo= EbNo + 10*log10(sub_car /len_fft)+ 10*log10(Td/(Td+Tcp)); % symbol to noise ratio

snr=EsNo – 10*log10(sub_car/len_fft); % snr as to be used by awgn fn.

M = modem.pskmod(2); % modulation object

% Generating data

t_data=randint(nbitpersym*nsym,1);
t = poly2trellis([4 3],[4 5 17;7 4 2]); % Define trellis.
enc = convenc(t_data,t); % Encode a string of ones.

% modulating data

mod_data = modulate(M,enc);

% serial to parallel conversion

par_data = reshape(mod_data,nbitpersym,nsym).’;

% pilot insertion

pilot_ins_data=[zeros(nsym,6) par_data(:,[1:nbitpersym/2]) zeros(nsym,1) par_data(:,[nbitpersym/2+1:nbitpersym]) zeros(nsym,5)] ;

% fourier transform time doamain data and normalizing the data

IFFT_data = (len_fft/sqrt(sub_car))*ifft(fftshift(pilot_ins_data.’)).’;

% parallel to serial coversion

% passing thru channel

no_of_error=[];
ratio=[];
for ii=1:length(snr)

chan_awgn = sqrt(80/52)*awgn(ser_data,snr(ii)); % awgn addition

ser_to_para = reshape(chan_awgn,80,nsym).’; % serial to parallel coversion

cyclic_pre_rem = ser_to_para(:,[17:80]); %cyclic prefix removal

FFT_recdata =(sqrt(52)/64)*fftshift(fft(cyclic_pre_rem.’)).’; % freq domain transform

rem_pilot = FFT_recdata (:,[6+[1:nbitpersym/2] 7+[nbitpersym/2+1:nbitpersym] ]); %pilot removal

ser_data_1 = reshape(rem_pilot.’,nbitpersym*nsym,1); % serial coversion

z=modem.pskdemod(2); %demodulation object

demod_Data = demodulate(z,ser_data_1); %demodulating the data
tb = 2; % Traceback length for decoding
decoded = vitdec(code,ser_data,tb,’trunc’,’hard’); % Decode

[no_of_error(ii),ratio(ii)]=biterr(t_data,decoded) ; % error rate calculation

end
ratio
figure(1)
semilogy(EbNo,ratio,’-g’);
theoryBer = (1/2)*erfc(sqrt(10.^(EbNo/10)));
semilogy (EbNo,theoryBer,’-k’);
axis([0 12 10^-5 1])
xlabel(‘EbNo’);
ylabel(‘BER’)
title(‘Bit error probability curve for BPSK using OFDM’);
legend(‘simulated awgn’,’theoretical awgn’)

hello sir i am doing a project on MIMO-OFDM system. Is that book ” Simulation of Digital Communication Systems Using Matlab [eBook] – Second Edition ” contains about ‘convolutional coding and Interleaving of MIMO-OFDM’ so that I will buy that

• Hi,
MIMO-OFDM is not discussed in the ebook. Only basic OFDM and its simulation is available.


text:
# [NTG-context] m-letter backslash codes in xml database

Rob Ermers r.ermers at hccnet.nl
Sat Jan 15 02:31:32 CET 2005

h h extern wrote:
> if you want to move things around in an encoding neutral way:
>
> \defineXMLsingular
>   [c]
>   [n=unknowncharacter]
>   {\executeifdefined {\XMLop{id}} \unknowncharacter}
>
> <c n='eacute'/>

Thanks, Hans, for the example. I do agree, but I'm sorry I don't quite
understand how to apply this code. Could you be more specific?

Do I still have to use the unicode number, or is this a method to use
\"e again in the database?

Should I code each diacritic in this way, such as ouml (ö) in my
example:
J. Schöttelndreher
)?

My solution by the way does not work with, e.g., a Turkish scedil
(U0015F) in my database, while Tex excells in all kinds of diacritics.

Kind regards,

Robert


text:
## Stream: general

### Topic: bat

#### Julian Berman (Dec 21 2020 at 17:58):

If anyone else likes/uses bat (https://github.com/sharkdp/bat/ -- a cat replacement), it was pretty easy to add Lean syntax support to it. The file is here: https://gist.github.com/Julian/062e5ae6a6a5edfea23f8c89a7ff5d41 which was just converted from the vscode-lean file. Stick it in \$(bat --config-dir)/syntaxes, run bat cache --build and yeah should work then. Trying to see if a PR to bat would succeed but will see if it meets their inclusion criteria.

#### Kevin Buzzard (Dec 21 2020 at 19:05):

What does it _do_? A clueless mathematician.

#### Julian Berman (Dec 21 2020 at 19:09):

cat is a program whose function is essentially "print the contents of a file" (technically I'm lying, its real purpose is "conCATenate" a bunch of files together, but everyone uses it essentially for the former). bat is cat but with modernities like "highlight the file you outputted". Basically you use either one when you want to view stuff and aren't sitting inside an editor already (or when say, you are piping outputs from some commands to each other, which seems a bit less likely in this case).

#### Julian Berman (Jan 02 2021 at 15:25):

This got merged, yay: https://github.com/sharkdp/bat/pull/1446

Last updated: May 17 2021 at 22:15 UTC


text:
A TSP to HamCycle Reduction

I'm referring to the decision version of both $TSP$ and $HamCycle$. The first is, given a graph $G=(V,E)$, a weight function $w:E\rightarrow \mathbb R^+$ and an integer $k$, is there a simple cycle with weight $\leq k$ which passes, once, through every vertex? (The weight of the cycle is the sum of the weights of the edges). The second problem is finding a Hamiltoninan Cycle (a cycle that passes through every vertex only once).

I know both are $NPC$. I am interested in finding reduction between both these problems. The first reduction $HamCycle \leq TSP$ is quite easy: for every edge in the original graph you give a weight $0$ and for every pair of vertices who did not have an edge you give a weight of $1$. Then you ask whether the agent can travel with sum of path $\leq 0$. The correctness of this follows pretty quickly.

However, my question if for the reduction the other way around: $TSP\leq HamCycle$. I could not come up with an idea for such. I am aware that since both are $NPC$ there exists a reduction through Cook-Levin Theorem and $SAT$, but I am wondering if there is a direct reduction $TSP\leq HamCycle$?

TSP is run on a fully connected graph with $n$ nodes and $n^2$ edges while HamCycle is run on a graph with $n$ nodes and $m < n^2$ edges (otherwise it would be trivial). A solution to TSP must pass all nodes and contain no cycles and therefore must contain exactly $n$ edges. We could run HamCycle each on all possible paths of length $n$, starting from the one with the lowest edge-sum and continuing until we reach one that contains a Hamiltonian Cycle, but as there are ${{n^2}\choose{n}} = O(n^{2n})$ such possible paths, this would mean running HamCycle an exponential number of times. I can not see how you can avoid this.
• Well, such a reduction must exist; simply compose the reductions for $TSP \le SAT$ and $SAT \le HamCycle$. When you compose them, this "never leaves the graph world". I think the problem is that the notion of "direct reduction" is not well-defined. – D.W. May 25 '18 at 15:58


text:
• Browse all
Measurements of normalized differential cross sections for $t\bar{t}$ production in pp collisions at $\sqrt{s}=7$  TeV using the ATLAS detector

The collaboration
Phys.Rev. D90 (2014) 072004, 2014

Abstract (data abstract)
CERN-LHC. Measurements of normalized differential cross-sections for top-quark pair production as a function of the top-quark transverse momentum, and of the mass, transverse momentum, and rapidity of the ttbar system, in proton--proton collisions at a center-of-mass energy of sqrt(s) = 7 TeV. The dataset corresponds to an integrated luminosity of 4.6 fb$^{-1}$, recorded in 2011 with the ATLAS detector at the CERN Large Hadron Collider. Events are selected in the lepton+jets channel, requiring exactly one lepton and at least four jets with at least one of the jets tagged as originating from a b-quark. The measured spectra are corrected for detector efficiency and resolution effects and are compared to several Monte Carlo simulations and theory calculations. The results are in fair agreement with the predictions in a wide kinematic range. Nevertheless, data distributions are softer than predicted for higher values of the mass of the $t\bar{t}$ system and of the top-quark transverse momentum. The measurements can also discriminate among different sets of parton distribution functions. UPDATE (15 JUN 2015): increased the number of significative digits and added full covariance matrices. The reported values of the normalized differential cross-sections and corresponding covariance matrices are given with a large number of significative digits since the constraints set by the normalization condition need to be accurately fulfilled for a correct evaluation of the chi2. In particular, since the normalization condition lowers by one the rank of the covariance matrix, the chi2 should be calculated removing one measurement and the corresponding row and column of the covariance matrix (see Section XI). The normalization constraints guarantee that the chi2 does not depend on the removed measurement.

• #### Table 1

Data from T 5

10.17182/hepdata.67128.v1/t1

Normalized differential cross-sections for the hadronically decaying top-quark PT. The cross-section in each bin is given as the integral of...

• #### Table 2

Data from T 5

10.17182/hepdata.67128.v1/t2

Normalized differential cross-sections for the mass of the ttbar system. The cross-section in each bin is given as the integral...

• #### Table 3

Data from T 5

10.17182/hepdata.67128.v1/t3

Normalized differential cross-sections for the PT of the ttbar system. The cross-section in each bin is given as the integral...

• #### Table 4

Data from T 5

10.17182/hepdata.67128.v1/t4

Normalized differential cross-sections for the absolute rapidity (|y|) of the ttbar system. The cross-section in each bin is given as...

• #### Table 5

Data from T 6

10.17182/hepdata.67128.v1/t5

Bin-wise full covariance matrices for the normalized differential cross-sections of the hadronically decaying top-quark PT. The elements of the covariance...

• #### Table 6

Data from T 6

10.17182/hepdata.67128.v1/t6

Bin-wise full covariance matrices for the normalized differential cross-sections of the mass of the ttbar system. The elements of the...

• #### Table 7

Data from T 6

10.17182/hepdata.67128.v1/t7

Bin-wise full covariance matrices for the normalized differential cross-sections of the PT of the ttbar system. The elements of the...

• #### Table 8

Data from T 6

10.17182/hepdata.67128.v1/t8

Bin-wise full covariance matrices for the normalized differential cross-sections of the absolute rapidity (|y|) of the ttbar system.