text:
# HORIZONTALLY HOMOTHETIC HARMONIC MORPHISMS AND STABILITY OF TOTALLY GEODESIC SUBMANIFOLDS

• Yun, Gab-Jin (DEPARTMENT OF MATHEMATICS MYONG JI UNIVERSITY) ;
• Choi, Gun-Don (GARC AND DEPARTMENT OF MATHEMATICS SEOUL NATIONAL UNIVERSITY)
• Published : 2008.03.31

#### Abstract

In this article, we study the relations of horizontally homothetic harmonic morphisms with the stability of totally geodesic submanifolds. Let $\varphi:(M^n,g)\rightarrow(N^m,h)$ be a horizontally homothetic harmonic morphism from a Riemannian manifold into a Riemannian manifold of non-positive sectional curvature and let T be the tensor measuring minimality or totally geodesics of fibers of $\varphi$. We prove that if T is parallel and the horizontal distribution is integrable, then for any totally geodesic submanifold P in N, the inverse set, $\varphi^{-1}$(P), is volume-stable in M. In case that P is a totally geodesic hypersurface the condition on the curvature can be weakened to Ricci curvature.

#### References

1. P. Baird and J. Eells, A Conservation Law for Harmonic Maps, Geometry Symposium, Utrecht 1980 (Utrecht, 1980), pp. 1-25, Lecture Notes in Math., 894, Springer, Berlin-New York, 1981
2. P. Baird and S. Gudmundsson, p-harmonic maps and minimal submanifolds, Math. Ann. 294 (1992), no. 4, 611-624 https://doi.org/10.1007/BF01934344
3. G. D. Choi and Gabjin Yun, Harmonic morphisms and stable minimal submanifolds, Preprint
4. B. Fuglede, Harmonic morphisms between Riemannian manifolds, Ann. Inst. Fourier (Grenoble) 28 (1978), no. 2, vi, 107-144
5. S. Gudmundsson, The Geometry of Harmonic Morphisms, Ph. D Thesis, 1992
6. T. Ishihara, A mapping of Riemannian manifolds which preserves harmonic functions, J. Math. Kyoto Univ. 19 (1979), no. 2, 215-229 https://doi.org/10.1215/kjm/1250522428
7. A. Kasue and T. Washio, Growth of equivariant harmonic maps and harmonic morphisms, Osaka J. Math. 27 (1990), no. 4, 899-928
8. H. B. Lawson, Lectures on Minimal Submanifolds. Vol. I. Second edition. Mathematics Lecture Series, 9. Publish or Perish, Inc., Wilmington, Del., 1980
9. E. Loubeau, On p-harmonic morphisms, Differential Geom. Appl. 12 (2000), no. 3, 219-229 https://doi.org/10.1016/S0926-2245(00)00013-9
10. S. Montaldo, Stability of harmonic morphisms to a surface, Internat. J. Math. 9 (1998), no. 7, 865-875 https://doi.org/10.1142/S0129167X98000361
11. B. O'Neill, The fundamental equations of a submersion, Michigan Math. J. 13 (1966), 459-469 https://doi.org/10.1307/mmj/1028999604

#### Cited by

1. Stability and integrability of horizontally conformal maps and harmonic morphisms 2013, https://doi.org/10.1002/mana.201200183


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DeepAI

# The Integers as a Higher Inductive Type

We consider the problem of defining the integers in Homotopy Type Theory (HoTT). We can define the type of integers as signed natural numbers (i.e., using a coproduct), but its induction principle is very inconvenient to work with, since it leads to an explosion of cases. An alternative is to use set-quotients, but here we need to use set-truncation to avoid non-trivial higher equalities. This results in a recursion principle that only allows us to define function into sets (types satisfying UIP). In this paper we consider higher inductive types using either a small universe or bi-invertible maps. These types represent integers without explicit set-truncation that are equivalent to the usual coproduct representation. This is an interesting example since it shows how some coherence problems can be handled in HoTT. We discuss some open questions triggered by this work. The proofs have been formally verified using cubical Agda.

• 3 publications
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format: plain


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## Results (1-50 of 79 matches)

Next
Label $\alpha$ $A$ $d$ $N$ $\chi$ $\mu$ $\nu$ $w$ prim arith $\mathbb{Q}$ self-dual $\operatorname{Arg}(\epsilon)$ $r$ First zero Origin
2-1400-1400.1021-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.1021 $$0.0 0 -0.339 0 0.210810 Modular form 1400.1.bs.c.1021.1 2-1400-1400.1021-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.1021$$ $0.0$ $0$ $0.160$ $0$ $0.856756$ Modular form 1400.1.bs.a.1021.1
2-1400-1400.1021-c0-0-2 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.1021 $$0.0 0 -0.339 0 1.12233 Modular form 1400.1.bs.c.1021.2 2-1400-1400.1021-c0-0-3 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.1021$$ $0.0$ $0$ $0.160$ $0$ $2.03560$ Modular form 1400.1.bs.b.1021.1
2-1400-1400.1189-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.1189 $$0.0 0 -0.370 0 0.604152 Modular form 1400.1.cc.a.1189.3 2-1400-1400.1189-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.1189$$ $0.0$ $0$ $0.130$ $0$ $0.848698$ Modular form 1400.1.cc.a.1189.1
2-1400-1400.1189-c0-0-2 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.1189 $$0.0 0 -0.370 0 1.23065 Modular form 1400.1.cc.a.1189.4 2-1400-1400.1189-c0-0-3 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.1189$$ $0.0$ $0$ $0.130$ $0$ $1.58018$ Modular form 1400.1.cc.a.1189.2
2-1400-1400.181-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.181 $$0.0 0 0.339 0 1.04958 Modular form 1400.1.bs.c.181.1 2-1400-1400.181-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.181$$ $0.0$ $0$ $-0.160$ $0$ $1.25102$ Modular form 1400.1.bs.a.181.1
2-1400-1400.181-c0-0-2 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.181 $$0.0 0 -0.160 0 1.61389 Modular form 1400.1.bs.b.181.1 2-1400-1400.181-c0-0-3 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.181$$ $0.0$ $0$ $0.339$ $0$ $1.86298$ Modular form 1400.1.bs.c.181.2
2-1400-1400.461-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.461 $$0.0 0 -0.120 0 0.999118 Modular form 1400.1.bs.b.461.1 2-1400-1400.461-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.461$$ $0.0$ $0$ $-0.120$ $0$ $1.43905$ Modular form 1400.1.bs.a.461.1
2-1400-1400.461-c0-0-2 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.461 $$0.0 0 0.380 0 2.22580 Modular form 1400.1.bs.c.461.1 2-1400-1400.461-c0-0-3 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.461$$ $0.0$ $0$ $0.380$ $0$ $2.58009$ Modular form 1400.1.bs.c.461.2
2-1400-1400.629-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.629 $$0.0 0 0.410 0 0.369572 Modular form 1400.1.cc.a.629.1 2-1400-1400.629-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.629$$ $0.0$ $0$ $-0.0899$ $0$ $1.45264$ Modular form 1400.1.cc.a.629.4
2-1400-1400.629-c0-0-2 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.629 $$0.0 0 -0.0899 0 1.68188 Modular form 1400.1.cc.a.629.3 2-1400-1400.629-c0-0-3 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.629$$ $0.0$ $0$ $0.410$ $0$ $2.08441$ Modular form 1400.1.cc.a.629.2
2-1400-1400.69-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.69 $$0.0 0 -0.410 0 0.928049 Modular form 1400.1.cc.a.69.2 2-1400-1400.69-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.69$$ $0.0$ $0$ $0.0899$ $0$ $1.72390$ Modular form 1400.1.cc.a.69.4
2-1400-1400.69-c0-0-2 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.69 $$0.0 0 0.0899 0 1.82697 Modular form 1400.1.cc.a.69.3 2-1400-1400.69-c0-0-3 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.69$$ $0.0$ $0$ $-0.410$ $0$ $2.17091$ Modular form 1400.1.cc.a.69.1
2-1400-1400.741-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.741 $$0.0 0 -0.380 0 0.260034 Modular form 1400.1.bs.c.741.1 2-1400-1400.741-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.741$$ $0.0$ $0$ $0.120$ $0$ $1.36166$ Modular form 1400.1.bs.a.741.1
2-1400-1400.741-c0-0-2 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.741 $$0.0 0 0.120 0 1.36242 Modular form 1400.1.bs.b.741.1 2-1400-1400.741-c0-0-3 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.741$$ $0.0$ $0$ $-0.380$ $0$ $1.38374$ Modular form 1400.1.bs.c.741.2
2-1400-1400.909-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.909 $$0.0 0 -0.130 0 1.13247 Modular form 1400.1.cc.a.909.1 2-1400-1400.909-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.909$$ $0.0$ $0$ $-0.130$ $0$ $1.25997$ Modular form 1400.1.cc.a.909.2
2-1400-1400.909-c0-0-2 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 1400.909 $$0.0 0 0.370 0 1.79634 Modular form 1400.1.cc.a.909.3 2-1400-1400.909-c0-0-3 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 1400.909$$ $0.0$ $0$ $0.370$ $0$ $2.63828$ Modular form 1400.1.cc.a.909.4
2-1400-280.187-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 280.187 $$0.0 0 -0.141 0 0.494814 Modular form 1400.1.cf.a.1307.1 2-1400-280.187-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 280.187$$ $0.0$ $0$ $-0.141$ $0$ $1.49096$ Modular form 1400.1.cf.a.1307.2
2-1400-280.227-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 280.227 $$0.0 0 -0.182 0 0.630451 Modular form 1400.1.cf.a.507.1 2-1400-280.227-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 280.227$$ $0.0$ $0$ $-0.182$ $0$ $1.19156$ Modular form 1400.1.cf.a.507.2
2-1400-280.229-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 280.229 $$0.0 0 -0.0727 0 0.741382 Modular form 1400.1.bl.a.1349.1 2-1400-280.229-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 280.229$$ $0.0$ $0$ $-0.220$ $0$ $1.30472$ Modular form 1400.1.bl.a.1349.2
2-1400-280.243-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 280.243 $$0.0 0 0.182 0 1.63034 Modular form 1400.1.cf.a.243.1 2-1400-280.243-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 280.243$$ $0.0$ $0$ $0.182$ $0$ $2.07627$ Modular form 1400.1.cf.a.243.2
2-1400-280.269-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 280.269 $$0.0 0 0.0727 0 1.31612 Modular form 1400.1.bl.a.549.1 2-1400-280.269-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 280.269$$ $0.0$ $0$ $0.220$ $0$ $2.17577$ Modular form 1400.1.bl.a.549.2
2-1400-280.27-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 280.27 $$0.0 0 0.177 0 1.17540 Modular form 1400.1.y.b.307.2 2-1400-280.27-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 280.27$$ $0.0$ $0$ $-0.104$ $0$ $1.47228$ Modular form 1400.1.y.b.307.4
2-1400-280.27-c0-0-2 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 280.27 $$0.0 0 -0.104 0 1.56004 Modular form 1400.1.y.b.307.3 2-1400-280.27-c0-0-3 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 280.27$$ $0.0$ $0$ $0.177$ $0$ $1.60644$ Modular form 1400.1.y.b.307.1
2-1400-280.27-c0-0-4 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 280.27 $$0.0 0 0.338 0 1.88065 Modular form 1400.1.y.a.307.1 2-1400-280.27-c0-0-5 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 280.27$$ $0.0$ $0$ $0.338$ $0$ $2.36079$ Modular form 1400.1.y.a.307.2
2-1400-280.3-c0-0-0 $0.835$ $0.698$ $2$ $2^{3} \cdot 5^{2} \cdot 7$ 280.3 $$0.0 0 0.141 0 1.25999 Modular form 1400.1.cf.a.843.2 2-1400-280.3-c0-0-1 0.835 0.698 2 2^{3} \cdot 5^{2} \cdot 7 280.3$$ $0.0$ $0$ $0.141$ $0$ $1.79225$ Modular form 1400.1.cf.a.843.1


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## anonymous 4 years ago how do you solve, x/2+x/5=x/6-1/3

1. eyust707

You bring all the terms with x's to one side and the numerical term to the other side. Then you find the least common multiple of the denominators. You multiply each term by the appropriate factor to make the denominators all have the same LCM. Then you add the terms. Finally you multiply each side by the denominator so that x is isolated.

2. anonymous

can you explain further?

3. theEric

Well, you want x to be on only one side of the equation when you're finished, so that's actually a good place to start.

4. theEric

Given$\frac{x}{2}+\frac{x}{5}=\frac{x}{6}-\frac{1}{3}$,

5. anonymous

X=5/8???

6. anonymous

i put all the x values on one side and i made the LCD 60 is that good so far?

7. theEric

Correct, GeekChic_! 60 is a good denominator! 30 also works if you'd like, but 60 will be fine too!

8. anonymous

yea thats what i got but shouldn't it be -5/8?

9. anonymous

yea lol i meant 30 as LCD

10. theEric

So you have$\frac{30x}{60}+\frac{12x}{60}-\frac{10x}{60}=-\frac{1}{3}$

11. anonymous

yup

12. theEric

Oh yeah, definately$-\frac{5}{8}$

13. anonymous

$\frac{ 5x+2x }{ 10 }=\frac{ 3x-6 }{ 18 }$ then cross-multiply $18(5x+2x)=10(5x+2x)$

14. anonymous

x=-5/8

15. anonymous

alright thanks guys!

16. theEric

No problem! I'm glad you could work through the problem! Take care!


text:
# Mesh Method and Node Method Check

1. Feb 18, 2010

### IronBrain

1. The problem statement, all variables and given/known data
Given the following circuit using mesh's method to solve for the power of the dependent source.
Using the same circuit and using the node voltage method solve for the power of the dependent source.

Circuit:

2. Relevant equations
Noting that I have to implement a super mesh for the dependent source to solve the unknown currents. I have arrived to the following equations

$Vx = 4I_1$

$3Vx = I_2 - I_1$

The Mesh equation:

$-4I_1-12I_2+30-8I_1-20 = 0$

Simplifying:

$-12I_1-12I_2+10 = 0$

3. The attempt at a solution
Solving the simulatenous equations by doing various substitutions of

$Vx = 4I_1$

$3Vx = I_2 - I_1$

By solving the mesh equations for $I_1 = \frac{10}{12} - I_2$

I have arrived at the following solutions

$I_1 = \frac{5}{84}$

$I_2 = \frac{65}{84}$

$Vx = \frac{5}{21}$

$3Vx = \frac{5}{7}$

Checking by Kirchoff's voltage law to see if all voltages within the outter loop it does equate to 0.

I need a check to see if this is right and I have no idea how to properly implement the node voltage method because the equations do not seem to come out right. I think my trouble is properly putting the correct polarity/sign convention of all devices

Help! Thanks!

Last edited: Feb 18, 2010
2. Feb 18, 2010

### IronBrain

Sigh bump....

3. Feb 18, 2010

### IronBrain

Dammit! I need this checked by tomorrw morning!

4. Feb 18, 2010

### Staff: Mentor

Watch the attitude please. The PF is a volunteer help site, so you won't always get help in a timely fashion. The right way to check your problem is for you to do it the two ways the problem asks you -- if the answers are the same, then you probably got it right.

Can you be more specific about what part of the KCL equation writing is bothering you?

5. Feb 18, 2010

### IronBrain

Well I am not entirely sure if my sign convention for all devices is correct. And which or how many nodes to pick when implementing the node method and what to use when using KCL for node for the middle branch, would I just put the current down that branch as 3Vx, and I just want to see if my first solutions using Mesh is correct

6. Feb 19, 2010

### Staff: Mentor

On your KVL work, I'm not sure I understand why you wrote an equation around the whole outside loop, instead of writing the 2 equations for the left and right loops. Also, be sure to label the +/- assumed voltage drop direction on R3, so you can keep it consistent in later equations.

For the KCL, I would put ground at the bottom of R4, and write equations for nodes on each side of the V1 source, and the top node at the top end of the dependent current source.

7. Feb 19, 2010

### IronBrain

Thanks
Im in my originating equations I believe when doing super mesh you "pretend" as if the current source in a common branch is there hence the equation for my mesh.

Also thanks for the tips on the node method, ill try that out

8. Feb 19, 2010

### Staff: Mentor

To be honest, I'm not familiar with the super mesh concept. I'd just write the two loop equations for the KVL. But I hardly ever use KVL, so I'm probably not a great help on that. The KCL should be fine, and a good check against your answer for the KVL.

9. Feb 19, 2010

### IronBrain

Thanks for help, so do I label the polarities of the devices based on the direction of each loop I have chosen or based of the direction of the device?

I believe super mesh is used for a dependent source or any device shared between a common branch of currents.

Also would I put a node on top of V2?

10. Feb 19, 2010

### The Electrician

You've got a sign error here:

The Mesh equation:

$-4I_1-12I_2+30-8I_1-20 = 0$

It should be:

$-4I_1-12I_2-30-8I_1+20 = 0$

And you have a sign error here:

$Vx = 4I_1$

$3Vx = I_2 - I_1$

You should have:

$Vx = -4I_1$

$3Vx = I_2 - I_1$

Combine them to get:

$-12I_1 = I_2 - I_1$

Now if you solve your simultaneous equations, you should get:

I1 = 1/12
I2 = -11/12

From which the voltage at the top middle node = 19 volts, assuming the bottom of R4 is the reference, or ground.

For the nodal method, I would move R2 to a position between R1 and the middle node; then the negative terminal of V1 is grounded and the whole thing is easier to conceptualize.

Last edited: Feb 19, 2010
11. Feb 19, 2010

### IronBrain

Thanks!
I always have a problem with getting the right sign convention here, I've been working on this for multiple hours trying to find a comprehendable explainable solution to my mess. Being that the original direction gets confused with the direction I am trying to take. How do you know which convention of the device is correct, I can never seem to get them correct. Also one minor detail left.

To find the power of the dependent source can I simply use the fact that

$P_ds = I^2R$

Which R is the resistance of 6 ohms

12. Feb 19, 2010

### The Electrician

You know the current in R4, so you can calculate the voltage across R4. Then the voltage across the dependent source is (19 - voltage across R4). Then the power delivered by the dependent source is (voltage across the source)*(current through the source).

13. Feb 19, 2010

### IronBrain

I got -1A for the current dependent source, does that seem right?


text:
# Groups between a group and its profinite completion

If $G$ and $H$ are finitely generated residually finite groups such that $G\leq H\leq \hat G$, where $\hat G$ denotes the profinite completion of $G$, does it follow that $$\hat H \cong \hat G$$

Thanks!

-

By the universal property of profinite completions, we have to show that every homomorphism of groups $G \to K$, where $K$ is a profinite group, extends uniquely to a homomorphism of groups $H \to K$.

Well, we have only one choice: $G \to K$ corresponds to a homomorphism of profinite groups $\hat{G} \to K$, and this can be restricted to $H \leq \hat{G}$. This is clearly an extension of $G \to K$.

Now, assume that two homomorphism $f,g : H \to K$ agree on $G$. Then $\hat{f},\hat{g} : \hat{H} \to K$ are homomorphisms of profinite groups which agree on $\hat{G}$. Then they also agree on $H \leq \hat{G}$, i.e. $f=g$.

-
How is $\hat H$ inside $\hat G$? –  Zeppelin Mar 19 '13 at 13:45
You need the inclusion map from $G$ to $\hat H$ be continuous? –  Zeppelin Mar 19 '13 at 13:52
I am sorry I do not understand your proof. –  Zeppelin Mar 19 '13 at 13:57
$G$ is discrete. And I've corrected the proof. Is it OK now? –  Martin Brandenburg Mar 19 '13 at 14:20
G is not discrete, it has the profinite topology. By definition of profinite completion, you need the inclusion map from $G$ to $\hat H$ to be continuous. But I do not think it is the case always. –  Zeppelin Mar 19 '13 at 14:32


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PIFOP markdown interpreter recognizes and formats a subset of LaTeX functions. These functions can be included either inside an inline code span that starts and ends with $, or inside a LATEX code block, like this: ## Markdown + $\LaTeX$ Inline $\LaTeX$ math expression: $\displaystyle \sum_{(i,j) \in E} x_{ij} = 1$ LATEX \textbf{\LaTeX{} code block with math expression:}\\ \displaystyle \sum_{(i,j) \in E} x_{ij} = 1   OUTPUT: LaTeX can be used to create files that are full of complex math expressions. Here is a model formulation for the traveling salesman problem: ## TSP formulation #### Data - $V$: set of vertices $i \in V$ - $n = |V|$: total number of cities in the salesman region - $i = 1$: city where the salesman lives - $E$: set of arcs $(i,j)$, such that $i \in V$ and $j \in V$ - $c_{ij}: (i,j) \in E$: distance between cities $i \in V$ and $j \in V$; $c_{ij}$ may or may not be equal to $c_{ji}$. #### Variables - $x_{ij} \in \{0,1\}$, for all $(i,j) \in E$, such that: - LATEX x_{ij}= \begin{cases} 1 & \text{if the salesman travels directly from city$i$to city$j$} \\ 0 & \text{otherwise} \end{cases}  - $y_{ij} \geq 0$, for all $(i,j) \in E$: number of products the salesman has when going from city $i$ to city $j\$.

#### Model

LATEX
\begin{array}{lrcllr}
\text{minimize }  & z & = & \displaystyle \sum_{(i,j) \in E} c_{ij}x_{ij} & & (1) \\
\text{subject to} & \displaystyle \sum_{(i,j) \in E} x_{ij} & = & 1 & \forall i \in V & (2) \\
& \displaystyle \sum_{(i,j) \in E} x_{ij} & = & 1 & \forall j \in V & (3) \\
& \displaystyle (n-1) x_{ij} & \geq & y_{ij} & \forall (i,j) \in E & (4) \\
& \displaystyle \sum_{(j,1) \in E} y_{j1} + n & = & \displaystyle \sum_{(1,j) \in E} y_{1j} + 1 & & (5) \\
& \displaystyle \sum_{(j,i) \in E} y_{ji} & = & \displaystyle \sum_{(i,j) \in E} y_{ij} + 1 & \forall (i,j) \in E \lor i \neq 1 & (6) \\
& \displaystyle x_{ij} & \in & \{0,1\} & \forall (i,j) \in E & (7) \\
& \displaystyle y_{ij} & \geq & 0 & \forall (i,j) \in E & (8) \\
\end{array}

OUTPUT:
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text:
# Kwon Group, Physics, Kyung Hee University

## Members' Pages

### Former Members

lab:research:mosi

### MoSI nanowires

Mo6S9-xIx nanowires exhibit interesting structural and electronic properties due to the flexible S linkages, which can be either short or long linke in an accordion. We have investigated the structrual and electronic properties of isolated and bundled Mo6S9-xIx nanowires. We find the optimum interwire distance depends on composition and to a smaller degree on the orientation of the wires. Structural order is expected in bundles with x=0 and x=6, since there is no disorder in the decoration of the Mo clusters. In bundles with other stoichiometries we expect structural disorder to occur. We find that the electronic structure of some nanowires can be switched from metallic to semiconducting by changing the interwire separation. The most intriguing changes occur in nanowires with x=4.5 and 6,which change from metal to semiconductor or undergo a magnetic transition upon axially stretching or compressing the nanowires or upon changing the inter-wire separation.


text:
# Rate of flow through a hole

1. May 12, 2006

### mmh37

Hello everyone! I have trouble finding the answer to the following problem - this really bugs me. Can anyone give me a hint?

Here is the problem:

A vessel contains a hole of 0.1mm^2 above the liquid level (contains liquid mercury) at T=0°C. The vessel is contained within a very much larger vessel in which a very high vacuum is maintained. After 30 days it is found that 24mg of mercury have been lost from the inner vessel. The relative atomic mass of mercury is 201.

I have already calculated the average speed of the mercury atoms in the vapor to be

$$169 m/s$$

now the questions:

1) What is the mean rate of flow of mercury atoms through the hole?

2) hence calculate the vapour pressure at T=0°C

note:

1) 2.76 * 10^13 atoms/sec

2) 2.5 * 10^(-2) N/m^2

Last edited: May 12, 2006
2. May 13, 2006

### Tide

You only need to divide the number of atoms of mercury that have been transported by the total time it took for them to be transported to find the average rate of flow. The number you showed is correct.

To find the number of atoms, divide the gram molecular weight of Hg by the number of grams transported and multiply by Avogadro's number.

3. May 13, 2006

### mmh37

Thank you Tide!!

That was indeed easier than expected!
However, I can still not calculate the pressure. I was trying to do it with the relationship p=nkT but that does not work. Would you mint giving me another hint?

4. May 13, 2006

### Staff: Mentor

5. May 14, 2006

### mmh37

thanks for this! Using the flux and the ideal gas law one can now find the pressure - the result it gives is correct!

Last edited: May 15, 2006


text: The root locus of the system for varying K is shown in the figure. 1. The root locus is the part of roots of the characteristic equation traced out in the S plane. View desktop site, Question 5 a) What is root locus analysis of a system? Which one of the following statements is not correct? help_outline. Rule 3 − Identify and draw the real axis root locus branches. Sketch the root locus. 47. Question: Q1) It Is Desired To Sketch The Complete Root Locus For A Single Loop Feedback System With Closed Loop Characteristic Equation: (s) S(s 1 J0.5)(s 1 J0.5) K(s 1 Jl)(s 1 Jl) (s) S? Show how you would apply it here. Tagged with. Or is there any difference between G(s)C(s) and G(z)C(z) in root locus? [4 marks] b) Sketch the root locus for a unity feedback system that has the forward transfer function: G(s) = K(s + 2) (s + 1)(8 +5)(s2 + 4s +5) i. What is the Weightage of Bode and root-locus plots in GATE Exam? Stack Exchange Network. This arises because of the angle criterion (Eq. Solution for For each root locus in Figure 1 to 4; i) Identify whether the system is stable or not based on the poles and zeros location on the s-plane ii)… This set of Control Systems Multiple Choice Questions & Answers (MCQs) focuses on “The Root Locus Concepts”. Electrical Engineering (EE) students definitely take this Test: The Root Locus And Its Concepts exercise for a better result in the exam. Z is given by 0.81+j0.26. Want to see the step-by-step answer? c) Both a and b. d) None of the above . Average marks 1.73. [4 marks] ii. Thank you for reading.. control-theory. [4 marks] b) Sketch the root locus for a unity feedback system that has the forward transfer function: G(s) = K(s + 2) (s + 1)(8 +5)(s2 + 4s +5) i. which will improve your skill. Has bounty. Solution for Which of the following is the root locus plot of the transfer function G(S) = (S+4)/(S^4−9S^2) 2) If poles are added to the system, where will the system tend to shift the root locus? Root Locus Questions And Solutions 2019 Last update: June 2020. asked yesterday. which will improve your skill. Question 5 A) What Is Root Locus Analysis Of A System? Section-B: Routh's Stability (8 Marks) Questions 2: Routh's stability criterion is a method which can be used to find the stability of a linear system. Cancel-1. Section-B: Routh's Stability (8 Marks) Questions 2: Routh's stability criterion is a method which can be used to find the stability of a linear system. So, we can use the magnitude condition for the points, and this satisfies the angle condition. • Full rules in FPE, page 279 (4th edition). Determine the number of branches, asymptotes, their angles and intersection with the real axis, and real-axis sections that belong to the root locus. Want to see this answer and more? Hence, we can identify the nature of the control system. Check out a sample Q&A here. 47. SIMPLE ANSWERS TO USUAL QUESTIONS ABOUT UNUSUAL FORMS OF THE EVANS’ ROOT LOCUS PLOT L.H.A. 10, Enter the code above here : 9 Question 5 a) What is root locus analysis of a system? What is the Weightage of Bode and root-locus plots in GATE Exam? 5) and the symmetry of the root locus. Click here to refresh, © 2015 - 2020 MechanicalTutorial.com This Content is copyrighted By MechanicalTutorial.com |All Rights Reserved. 2 State the open and closed-loop transfer functions [G(s) and T(S)]. Question No. 162. 0 b. e. If the point in part (d) is a closed loop pole, find the gain, K. f. Find the value of gain will make the system marginally stable. 3. The solved questions answers in this Test: The Root Locus And Its Concepts quiz give you a good mix of easy questions and tough questions. If the gain of the system is reduced to a zero value, the roots of the system in the s-plane, A. Solution for Sketch the root locus plot of unity feedback system with an open loop transfer function k G(s) = s(s+5)(s+10) Find the range of k for the system to… Question No. Q20.In second order system, which among the following remains independent of gain (k)? Home >> Category >> Electronic Engineering (MCQ) questions & answers >> Root Locus; 1) For a unity feedback system with G(s) = 10 / s 2, what would be the value of centroid? If the angle of the open loop transfer function at a point is an odd multiple of 180 0 , then that point is on the root locus. How does one prove the piecewise continuity of the roots given that some of the coefficients of the polynomials are a function of a parameter (this also applies to constructing the root locus for the closed loop function)? Provide details and share your research! Provide details and share your research! 6 The root locus is the part of roots of the characteristic equation traced out in the S plane. Which one of the following statements is not correct? share | cite | follow | asked 1 min ago. 2s2 1.25s K(s2 2s 2) Given The Roots Of Dk/ds=0 As S= … May 05, 2019 Add Comment Control Theory, Download Control Theory, Ebook Control Theory, Errol Morris Shocker Baskerville Is The Typeface Of Truth, Free Ebook Control Theory, Free PDF Control Theory, How To Create A Stack Of Books In Adobe Illustrator Vectips, PDF Control Theory, Root Locus Plot Common Questions And Answers Youtube, Self Criticism And Self Enhancement Theory Research And GATE Questions & Answers of Bode and root-locus plots. This is - by definition - a critically damped system. 0 b. GATE students definitely take this The Root Locus Technique - 1 exercise for a better result in the exam. Most frequent. a. Can someone answer this root locus question. Please be sure to answer the question. Root Locus - Control System Multiple Choice Questions and Answers 6. 5 4 A typical transfer function is of the form $$G(s) = Y(s)/U(s)$$. for –∞ < k < 0, has break away/break in points and angle of departure at pole P (w.r.t. [4 marks] iii. Privacy For what range of gain K is the system stable? Therefore, each point on the Root Locus Diagram will represent a set of tuning parameters. Use following: Show transcribed image text. 1+K (s+4)/s (s+2) =0. How does one prove the piecewise continuity of the roots given that some of the coefficients of the polynomials are a function of a parameter (this also applies to constructing the root locus for the closed loop function)? The value of Kat intersection of { = 0.707 line and root locus is fullscreen. The root locus of the plots of the variations of the poles of the closed loop system function with changes in Question: Use Root Locus Design To Find Value Of Beta And Kc. How can I plot it? That means: The root locus splits up into two different parts at a pole quality factor of Qp=0.5. d) Gives infinite poles . | The value of Kat intersection of { = 0.707 line and root locus is . share | improve this answer | follow | answered Dec 20 '18 at 18:33. the Laplacian), as a function of the control variables such as $$Kc$$. Terms Read Free Locus Problems With Answers so you can download it instantly. root locus). No - the root locus is not drawn for the open-loop function. [4 marks], electrical engineering questions and answers. But avoid … Asking for help, clarification, or responding to other answers. Root Locus diagrams are much more difficult to create for PID control. I place my poles, the root locus on the real axis, count the branches (four go to infinity, no zeros), find the centroid then asymptotes, find the breakouts on the real axis, find the intersection of the root locus with the imaginary axis and find the the angle of departure from the two complex poles. But this approach begs the question, why bother to sketch the root locus in the first place? Chu Chu. 2) If poles are added to the system, where will the system tend to shift the root locus? root locus). The system can have, < 47. The root loci of a system has three asymptotes. The following tags: Apply filter. This case applies to your example because we have a 3rd-order system with 3 poles including one pole pair (real or conjugate-complex, depending on the gain block K). © 2003-2021 Chegg Inc. All rights reserved. See Answer. Thank you for reading.. control-theory. The solved questions answers in this The Root Locus Technique - 1 quiz give you a good mix of easy questions and tough questions. 8 Most votes. For each of the following compounds, classify it with the atomic-scale picture that best represents it in solution or as insoluble in aqueous solution. For root loci which of the following are the starting points? Recent activity. Root Locus - Control System Multiple Choice Questions and Answers 46. View Answer For the system shown in the figure below, find the output, c(t), if the input, r(t), is a unit step. Use root locus design to find value of Beta and Kc. What does it show? Explanation: No explanation is available for this question! Root locus plots show the roots of the systems characteristic equation, (i.e. MathJax reference. Apply this method to the system in Question 1, to find the range of K>0 such that the system is asymptotically stable. But avoid … Asking for help, clarification, or responding to other answers. Question No. Ignore everything else because all the real terms equating to zero marks the position on the root locus that it crosses the imaginery axis. Making statements based on opinion; back them up with references or personal experience. Aykut Can Gulen Aykut Can Gulen. Break away point c. Stability point Home; Article; Root Locus Questions And Solutions APK file 2019-2020 [latest version] The best 8 apps for Android root 2019-2020. MathJax reference. 1. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.. Visit Stack Exchange Answer Explanation ANSWER: 0. B. 2. The characteristic equation will contain unknown variables Kc, Ti, and Td. the positive real axis) equal to a) ± √2 and 0 o b) ± √2 and 45 o c) ± √3 and 0 o d) ± √3 and 45 o This is your solution of Root Locus (Control Systems) Notes | EduRev search giving you solved answers for the same. a) Root loci can be used for analyzing stability and transient performance b) Root loci provide insight into system stability and performance c) Shape of the root locus gives idea of type of controller needed to meet design specification G (s)H (s) =K (s+b)/s (s+a) Centre = (-b,0) = (-4,0). Use MathJax to format equations. Determine the number of branches, asymptotes, their angles and intersection with the real axis, and real-axis sections that belong to the root locus. 65) For drawing root locus, the angle of asymptote yields the direction along which _____branches approach to infinity. a. p + z b. p – z c. p / z d. p x z. 1. Average marks 1.73. 3 share | improve this answer | follow | edited Oct 8 '17 at 11:28. answered Oct 8 '17 at 8:06. & To Study Root Locus (Control Systems) Notes | EduRev for this is your one stop solution. Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more ... (i.e. Then drill down to find \$\omega\$. I want to draw the root locus by hand to have a rough idea. Apply this method to the system in Question 1, to find the range of K>0 such that the system is asymptotically stable. 0. votes. How can I plot root locus like these system can you help me? Use Following: This question hasn't been answered yet Ask an expert . polynomials … Bounty ending soon. Can't read the image? Making statements based on opinion; back them up with references or personal experience. Sorted by. with reasonable accuracy. Answer Explanation ANSWER: 0. The root locus for negative values of k, i.e. GATE students definitely take this The Root Locus Technique - 1 exercise for a better result in the exam. The root locus lies at all points on the real axis to the left of an odd number of poles and zeros that lie on the real axis. check_circle Expert Answer. The root locus is the part of roots of the characteristic equation traced out in the S plane. (Note: You can validate the imaginary-axis crossing point using an accurate sketch.) And a happy xmas to you. pole-zeroplot root-locus. State the rule for finding the break-away and break-in points using differential calculus. Verify Your Handwritten Sketch Using MATLAB Or Other Similar Software Tools. Sketch the root locus. Andy aka Andy aka. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. What does it show? If odd number of the open loop poles and zeros exist to the left side of a point on the real axis, then that point is on the root locus branch. • Basic questions: 1. What points are on the root locus? Total 6 Questions have been asked from Root Loci topic of Control Systems subject in previous GATE papers. 7 Hence, you have to … 46. Coincide with zero. The root locus diagram of a unity feedback control system with loop transfer function : G(s) - s(s-2) K is intersecting the line = 0.707 which is shown in figure below. For the root locus plot shown in figure(1) a. In this page you can learn various important control system multiple choice questions answers, mcq on control system, short questions and answers on control system, sloved control system objective questions answers etc. The limitations of root locus method for tuning PID controllers are: Not perform well on a nonlinear system. So i decided to solve question with root locus too for double check and K i found on root locus plot is different than Nyquist plot and i checked with mathlab too.Did i miss something i thought root locus plot and Nyquist plot suppose to give me same answer. 1 Find the angles of departure and arrival from the complex poles and zeros. We can find the value of K for the points on the root locus branches by using magnitude condition. b) Exhibit variations. b. votes . Our book servers hosts in multiple countries, allowing you to get the most less latency time to download any of our books like this one. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. a) Construct the root locus for the system represented by the block diagram in Figure 2. a. At which of the following root loci will end ? [4 marks] ii. a. Solution for For each root locus in Figure 1 to 4; i) Identify whether the system is stable or not based on the poles and zeros location on the s-plane ii)… Solution for Figure 2 shows the root locus of a discrete-time control system with the following characteristic equation: (z + 0.9672) A=1+ D(z)G(z) =1+0.00484K… The root locus diagram of a unity feedback control system with loop transfer function : G(s) - s(s-2) K is intersecting the line = 0.707 which is shown in figure below. ANSWER: (c) Real but not equal. a) Open loop poles. You don't need to calculate the break-in and break-out points. polynomials … b) Closed loop poles. ANSWER: (b) p – z. The points on the root locus branches satisfy the angle condition. Loses significance at high frequencies or high degrees of damping. Home >> Category >> Electronic Engineering (MCQ) questions & answers >> Root Locus; 1) For a unity feedback system with G(s) = 10 / s 2, what would be the value of centroid? Making statements based on opinion; back them up with references or personal experience. c) Gives zero feedback . The main objective of drawing root locus plot is : a) To obtain a clear picture about the open loop poles and zeroes of the system. Find the imaginary-axis crossing point and the gain, K, at the imaginary-axis crossing point. Solution for Sketch the root locus plot of unity feedback system with an open loop transfer function k G(s) = s(s+5)(s+10) Find the range of k for the system to… GATE Questions & Answers of Bode and root-locus plots. No accepted answer. Image Transcriptionclose. 2 c. 5 d. 10. Use MathJax to format equations. Use MathJax to format equations. Question. Answer: c. Explanation: The root locus is the locus of the change of the system parameters of the characteristic equation traced out in the s-plane. MathJax reference. Newest. It is the main purpose of the root locus of a system with feedback to show if resp. Contact us: info@mechanicaltutorial.com. I can't figure out how to find the root locus centroid for the poles of this simple equation in a positive feedback system. What Does It Show? The constant damping ratio line, for ζ = 0.5, intersects the root locus at point A. Provide details and share your research! So, the angle condition is used to know whether the point exist on root locus branch or not. Please be sure to answer the question. Expert Answer . 50. But avoid … Asking for help, clarification, or responding to other answers. Do check out the sample questions of Root Locus (Control Systems) Notes | EduRev for , the answers and examples explain the meaning of chapter in the best manner. A control system has G ( s ) H ( s ) = K / [ s ( s + 4 ) ( s2 + 4s + 20 ) ( 0 < K )< ∞ ) What is the number of breakaway points in the root locus diagram ? Root Locus - Control System Multiple Choice Questions and Answers 46. A con-sequence of choosing the parameter for the integral control, K i, is that it a ects the location of the poles and zeros of the cruise control system on the s-plane. Question: A) Sketch The Root Locus Of The Pure Gain Compensated System In Figure Q2 When The Gain K Varies From 0 To 0. Solution for Sketch the root locus plot of unity feedback system with an open loop transfer function k s(s+5)(s+ 10) G(s) = Find the range of k for the system… Average marks 1.83. A have G(s)C(s)=K/(s*(s+1)) and I want a root locus in G(z)C(z). hence, Root locus, can help us in the design by selecting the operation point (that yield the required characteristics) on the root locus. Answer: c. Explanation: s (s+2)+K (s+4) =0. The solved questions answers in this The Root Locus Technique - 1 quiz give you a good mix of easy questions and tough questions. Hans Muller. 49. 1. Stack Exchange Network. Which one of the following is correct ? By examining these graphs it is possible to determine the stability of different values of the control variable. Root Locus Analysis • General root locus is hard to determine by hand and requires Matlab tools such as rlocus(num,den) to obtain full result, but we can get some important insights by developing a short set of plotting rules. a. Centroid b. What is the Weightage of Root Loci in GATE Exam? So i decided to solve question with root locus too for double check and K i found on root locus plot is different than Nyquist plot and i checked with mathlab too.Did i miss something i thought root locus plot and Nyquist plot suppose to give me same answer. GATE Questions & Answers of Root Loci. Solution for In root locus plot, the loci at poles of G(s)H(s) O starts with k=0 and terminate with k=00 O starts with k=o and terminate with k= -∞ O starts… The distance from the origin to … You do not need to calculate the intersection points with the jo-axis. Monteiro∗ luizm@mackenzie.br, luizm@usp.br J.J. Da Cruz† jaime@lac.usp.br ∗Escola de Engenharia da Universidade Presbiteriana Mackenzie e Escola Polite´cnica da Universidade de Sa˜o Paulo †Escola Polite´cnica da Universidade de Sa˜o Paulo ABSTRACT After a ﬁrst contact with Evans’ root … Explanation: No explanation is available for this question! Q19.Root locus specifies the movement of closed loop poles especially when the gain of system _____ a) Remains constant. Total 11 Questions have been asked from Bode and root-locus plots topic of Control Systems subject in previous GATE papers. In this technique, we will use an open loop transfer function to know the stability of the closed loop control system. Determine all the asymptotes, points of intersection, etc. In the root locus diagram, we can observe the path of the closed loop poles. Visit Stack Exchange. Root Locus Signals and Systems: 3C1 Control Systems Handout 3 Dr. David Corrigan Electronic and Electrical Engineering corrigad@tcd.ie • Recall, the example of the PI controller car cruise control system. 1answer 65 views Root Locus of the system. d. Determine if the point -8 + j0 is a closed loop pole. 2 c. 5 d. 10. a. Use MathJax to format equations. As the output of the system is changed. Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more ... (i.e. Please be sure to answer the question. Find the asymptotes. Z Is Given By 0.81+j0.26. Also draw the root locus with reasonable accuracy. [4 marks] iv. Show Clearly All The Rules And Calculations Used To Sketch The Root Locus Diagram And To Find The Important Points. QUESTION: 17. 1 2 2 bronze badges. locus Questions and Answers - TopperLearning locus problems with answers is available in our digital library an online access to it is set as public Page 1/3. Which one of the following statements is not correct? 66) Which point on root locus specifies the meeting or collision of two poles? MathJax reference. 48. My watched tags. Provide details and share your research! In this page you can learn various important control system multiple choice questions answers, mcq on control system, short questions and answers on control system, sloved control system objective questions answers etc. But avoid … Asking for help, clarification, or responding to other answers. Total 11 Questions have been asked from Bode and root-locus plots topic of Control Systems subject in previous GATE papers. c. Find the departure and arrival angles. Can someone answer this root locus question. Please be sure to answer the question. Root Locus Concepts Objective Questions. for which constant gain values the CLOSED-LOOP function will have poles in the "forbidden" region (right half of the s-plane). Root Locus Concepts Objective Questions. Making statements based on opinion; back them up with references or personal experience. By MechanicalTutorial.com |All Rights Reserved the  forbidden '' region ( right half of the following statements not! Into two different parts at a pole quality factor of Qp=0.5 this Content is by... The poles of this simple equation in a positive feedback system personal experience your one stop solution for values! The jo-axis explanation is available for this question point -8 + j0 is a loop! = 0.707 line and root locus plots show the roots of the system the... Are: not perform well on a nonlinear system file 2019-2020 [ latest ]! Root-Locus plots Questions and answers 6 65 ) for drawing root locus Design to find of. Solutions 2019 Last update: June 2020 root locus like these system can you help me Control.., clarification, or responding to other answers exercise for a better result in the S plane how to the! Loop transfer function to know the stability of different values of the equation... But root locus questions and answers equal verify your Handwritten Sketch using MATLAB or other Similar Software Tools total 6 Questions have asked! Privacy & Terms | View desktop site, question 5 a ) Construct the root at... J0 is a closed loop poles ( Kc\ ) d. p x z show All. And draw the real axis root locus ca n't figure out how to find value of intersection. And closed-loop transfer functions [ G ( S ) and the symmetry of the locus... A and b. d ) None of the root locus branches satisfy the angle of departure and arrival from complex... So, we can find the imaginary-axis crossing point using an accurate Sketch. at. Plots in GATE Exam z b. p – z c. p / z p! Exist on root locus branch or not the asymptotes, points of intersection, etc ( MCQs ) focuses “! Technique - 1 exercise for a better result in the Exam 66 ) which point root! Sketch the root locus ( Control Systems subject in previous GATE papers hence, we can use magnitude! Questions and answers 46 a set of Control Systems subject in previous GATE papers if poles are added to system... June 2020 locus Concepts ” of Control Systems subject in previous GATE papers points with the.. Using magnitude condition for the poles of this simple equation in a positive system. ) real but not equal | answered Dec 20 '18 at 18:33 Rights Reserved observe. Be sure to answer the question how can I PLOT root locus plots show the roots the. Best 8 apps for Android root 2019-2020 asymptote yields the direction along which _____branches approach to infinity this answer follow. Is Used to Sketch the root locus in the S plane branch or not for help, clarification, responding. By the block Diagram in figure 2 in the s-plane, a Rights Reserved  forbidden '' region ( half! The figure to answer the question, why bother to Sketch the root locus Diagram, we can use magnitude! Are added to the system is reduced to a zero value, the angle criterion ( Eq be... The value of Beta and Kc angle of departure and arrival from the complex poles and.... An accurate Sketch. can download it instantly high degrees of damping an... Concepts ”: use root locus Technique - 1 quiz give you a good mix easy... S ( s+2 ) +K ( s+4 ) /s ( s+2 ) =0 unknown Kc. Or other Similar Software Tools the open-loop function at 8:06 find the root at. [ 4 marks ], electrical engineering Questions and Solutions APK file 2019-2020 latest. For finding the break-away and break-in points using differential calculus Similar Software Tools follow | edited Oct '17... An open loop transfer function to know the stability of the characteristic equation out... Following remains independent of gain ( K ) ; Article ; root locus Technique - 1 quiz give a. Where will the system for varying K is shown in the  forbidden region. ) if poles are added to the system in the S plane - by definition - a damped! Quality factor of Qp=0.5 question: use root locus click here to refresh, © 2015 2020. From Bode and root-locus plots and Solutions 2019 Last update: June 2020 c ) Both a and b. )... To show if resp in previous GATE papers Questions & answers of Bode and root-locus plots topic Control! By examining these graphs it is the Weightage of root locus Diagram will a! So, we will use an open loop transfer function to know the stability of values... Angle condition is Used to Sketch the root locus Design to find the imaginary-axis crossing point using accurate. Break away/break root locus questions and answers points and angle of departure at pole p ( w.r.t limitations of root Diagram... Two poles View desktop site, question 5 a ) Construct the root locus for negative values the... Loci topic of Control Systems subject root locus questions and answers previous GATE papers can find the value of Kat intersection of =... Free locus Problems with answers so you can download it instantly point -8 + is. 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text:
# Bundesliga: Is the East-West segregation still present?

This week the Economist had a very interesting article on the Bundesliga, where the newspaper touches upon the underepresentation of teams from the ex-East Germany in the soccer league.  While the article was certainly eye-opening (I had never thought of even examining this question!) one thing that it was missing (in my opinion) was a baseline for comparing the number of teams from the ex-East region in the league.  And this is what I ought to do in this post.

I collected data from Wikipedia for the teams participating in the Bundesliga from the fall of the Berlin wall until the current season.  In the following figure I present the fraction $\rho$ of teams from the ex-East Germany and the corresponding 95% confidence intervals of these percentages.

I have further considered 3 baselines for comparisson; (a) the fraction of the area covered from ex-East Germany region, (b) the fraction of the population in the ex-East Germany region, and (c) the fraction of the cities in the ex-East Germany region.  All these baselines are intuitive in the sense that you expect that the number of teams a region has might be proportional to the number of cities it includes (compared to the total region that the league serves).

As we can see considering the area baseline for all the 25 years after the reunification of the country, the East region has been under-represented significantly (at the 5% confidence level).  In other words, if we perform the following statistical test:

we can reject the null hypothesis every season. When considering the other two baselines, there are 6 seasons, where the baseline expected falls within the 95% confidence intervals of the actual representation of the East region, and hence, we cannot reject the null hypothesis that the actual fraction is equal to the baseline. Even these cases though can be false negatives due to low statistical power cause by the small sample size (i.e., 20 teams in the 1991-92 season and 18 in the rest of the seasons).  Also as the Economist article mentions as well, there has not been a single team from the East region sicne 2009!

I did the same excercise for the basketball Bundesliga using data from the 2008-2009 season and onwards (during the 1990s and the beginning of 2000s the teams in the league were even less than the current 18 and hence, any statistics obtained are tricky).

The situation in the basketball Bundesliga appears to be better (again though the area baseline is never reached).  However, I will leave the conclusions with regards to the geo-sports segregation in Germany today to you; I just wanted to give out more data and a baseline that was missing from the Economist’s article! Any comments are always welcome.


text:
## Constant speed in the X axis of a body

Questions about the LÖVE API, installing LÖVE and other support related questions go here.
Forum rules
Z_ReC
Prole
Posts: 7
Joined: Sat Nov 14, 2015 2:48 am

### Constant speed in the X axis of a body

Hello everyone!

As the title says, i need a object having a constant speed in the X axis!

I leave an attachment example.
Constant speed cube would have to be 480, but no.

I hope you help me!
Attachments
cube.love

Doctory
Party member
Posts: 441
Joined: Fri Dec 27, 2013 4:53 pm

### Re: Constant speed in the X axis of a body

in love.physics, bodies have velocity which is decreased when they hit another body, this is why the velocity isnt constant
"This is what videogames should be like; they should have meaning and triple machine guns"
- Jan Willem Nijman, 2013
github

ivan
Party member
Posts: 1549
Joined: Fri Mar 07, 2008 1:39 pm
Contact:

### Re: Constant speed in the X axis of a body

Yep, Doctory is right.
You can overwrite the linear velocity manually each step, but then the collisions would look weird.

Beelz
Party member
Posts: 234
Joined: Thu Sep 24, 2015 1:05 pm
Location: New York, USA
Contact:

### Re: Constant speed in the X axis of a body

Not getting into collisions... This will move you along at a constant rate:

Code: Select all

love.update(dt)
object.x = object.x + distancePerSecond * dt
object.y = 500
end

Code: Select all

if self:hasBeer() then self:drink()
else self:getBeer() end

GitHub -- Website

Z_ReC
Prole
Posts: 7
Joined: Sat Nov 14, 2015 2:48 am

### Re: Constant speed in the X axis of a body

Beelz wrote:Not getting into collisions... This will move you along at a constant rate:

Code: Select all

love.update(dt)
object.x = object.x + distancePerSecond * dt
object.y = 500
end

try your code, this is the result.

it can be improved? Thanks!

ivan
Party member
Posts: 1549
Joined: Fri Mar 07, 2008 1:39 pm
Contact:

### Re: Constant speed in the X axis of a body

If you don't want your body to respond to collisions but move at a constant rate you can make it a "kinematic" body and set its velocity.
Kinematic bodies push other objects but do not respond to collisions themselves.
it can be improved? Thanks!

Code: Select all

love.update(dt)
local lvx, lvy = object:getLinearVelocity()
object:setLinearVelocity(10, lvy)
end

Z_ReC
Prole
Posts: 7
Joined: Sat Nov 14, 2015 2:48 am

### Re: Constant speed in the X axis of a body

ivan wrote:If you don't want your body to respond to collisions but move at a constant rate you can make it a "kinematic" body and set its velocity.
Kinematic bodies push other objects but do not respond to collisions themselves.
it can be improved? Thanks!

Code: Select all

love.update(dt)
local lvx, lvy = object:getLinearVelocity()
object:setLinearVelocity(10, lvy)
end
Thanks, but I want to respect collisions!

With "kinematic" this happens:

In the left top corner see that the speed is constant!
but does not respect the collision!

I need collision respect, at the same time, keep constant speed.

-----------------------------------
This respects the collision, but no speed!

also thanks.

Doctory
Party member
Posts: 441
Joined: Fri Dec 27, 2013 4:53 pm

### Re: Constant speed in the X axis of a body

switch from the love.physics, it is impossible to get a constant velocity in love.physics without screwing something up
"This is what videogames should be like; they should have meaning and triple machine guns"
- Jan Willem Nijman, 2013
github

s-ol
Party member
Posts: 1077
Joined: Mon Sep 15, 2014 7:41 pm
Location: Cologne, Germany
Contact:

### Re: Constant speed in the X axis of a body

Doctory wrote:switch from the love.physics, it is impossible to get a constant velocity in love.physics without screwing something up
Not really, he just needs to do a few things every frame:
• get current velocity vector (:getLinearVelocity())
• normalize and scale it by the target velocity
Now you can either just set it directly (setLinearVelocity()), which might look weird, or instead apply a force:
• subtract the actual velocity from the target velocity vector
• apply a force proportional to that difference

s-ol.nu /blog  -  p.s-ol.be /st8.lua  -  g.s-ol.be /gtglg /curcur

Code: Select all

print( type(love) )
if false then
baby:hurt(me)
end

ivan
Party member
Posts: 1549
Joined: Fri Mar 07, 2008 1:39 pm
Contact:

### Re: Constant speed in the X axis of a body

Z_ReC wrote:This respects the collision, but no speed!
1.you set the linear velocity
2.you update the simulation
3.the linear velocity changes due to collisions with the floor

So you're going in the update 'step' with the desired velocity
and collisions happen which slow down the body.
There is no simple way to work around this
if you want to have 'realistic'-looking collisions.

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text:
Conditionals and Operators

$r /; c$ : Relation $r$ holds under the condition $c$ .
$a=b$ : The expression $a$ is mathematically identical to $b$ .
$a \ne b$ : The expression a is mathematically different from $b$ .
$x > y$ : The quantity $x$ is greater than quantity $y$ .
$x \ge y$ : The quantity $x$ is greater than or equal to the quantity $y$ .
$x < y$ : The quantity $x$ is less than quantity $y$ .
$x \le y$ : The quantity $x$ is less than or equal to quantity $y$ .
$P := Q$ : Statement $P$ defines statement $Q$ .
$a \wedge b$ : a and b.
$a \vee b$ : a or b.
$\forall a$ : for all $a$ .
$\exists$ : [there] exists.
$\iff$ : If and only if.
Sets & Domains
$\{ a_1, a_2, \ldots, a_n \}$ : A finite set with some elements $a_1, a_2, \ldots, a_n$ .
$\{ a_1, a_2, \ldots, a_n \ldots \}$ : An infinite set with elements $a_1, a_2, \ldots$
$\mathrm{\{ listElement /; domainSpecification\}}$ : A sequence of elements listElement with some domainSpecifications in the set. For example, $\{ x : x=\frac{p}{q} /; p \in \mathbb{Z}, q \in \mathbb{N^+}\}$ $a \in A$ : $a$ is an element of the set A.
$a \notin A$ : a is not an element of the set A.
$x \in (a,b)$ : The number x lies within the specified interval $(a,b)$ .
$x \notin (a,b)$ : The number x does not belong to the specified interval $(a,b)$ . Standard Set Notations
$\mathbb{N}$ : the set of natural numbers $\{0, 1, 2, \ldots \}$
$\mathbb{N}^+$ : The set of positive natural numbers: $\{1, 2, 3, \ldots \}$
$\mathbb{Z}$ : The set of integers $\{ 0, \pm 1, \pm 2, \ldots\}$
$\mathbb{Q}$ : The set of rational numbers
$\mathbb{R}$ : The set of real numbers
$\mathbb{C}$ : The set of complex numbers
$\mathbb{P}$ : The set of prime numbers.
$\{ \}$ : The empty set.
$\{ A \otimes B \}$ : The ordered set of sets $A$ and $B$ .
$n!$ : Factorial of n: $n!=1\cdot 2 \cdot 3 \ldots (n-1) n /; n \in \mathbb{N}$


text:
# Power Spectrum Estimation of three sinusoids in white noise

Let's assume we have a random process consisting of three sinusoids in white noise: $$x[n] = 3 \cdot \sin(ω_1 \cdot n + ϕ_1) + 5 \cdot \cos(ω_2 \cdot n + ϕ_2) + 2 \cdot \sin(ω_3 \cdot n + ϕ_3) + v[n]$$ where $$ϕ_1$$, $$ϕ_2$$, $$ϕ_3$$ are uncorrelated uniformly distributed random variables in $$[0,2\pi]$$

I want to estimate the power spectral density of $$x[n]$$. For that purpose I apply two different methods:

1. Periodiagram of $$x[n]$$
2. Welch method (Averaged Modified Periodogram) using Hamming window

I generated the following code and ran in Matlab:

N = 1024;
w1 = 0.3 * pi;
w2 = 0.6 * pi;
w3 = 0.7 * pi;
phi1 = unifrnd(0, 2*pi);
phi2 = unifrnd(0, 2*pi);
phi3 = unifrnd(0, 2*pi);
u = randn(1, N);
n = 0:N-1;
x = 3 * sin(w1*n + phi1) + 5*cos(w2*n + phi2) + 2*sin(w3*n + phi3) + u;

fig = figure;
plot(n,x);
title('signal x[n]');
xlabel('samples');
ylabel ('Magnitude');

fig=figure;
[p,f]=periodogram(x);
plot(f,p);

fig=figure
[pw,f]= pwelch(x, N/8, N/16);
plot(f,pw);

Which is the difference in these two methods about the resolution and the variance of the power spectrum ?

## 1 Answer

The standard Periodogram uses all the data and computes its power spectrum spectrum estimatation at once; therefore it provides more spectral resolution but less estimation reliability (larger estimation variance).

The Welch's modification to the standard Periodogram is the concept of dividing the signal into shorter blocks and averaging the computed per block Periodograms, and therefore it has less spectral resolution (due to shorter blocks) but provides smaller estimation variance, due to averaging.

• Thanks for the response! I would like to ask something more. Whis is the correct way to produce the standard periodogram and the Welch's modification to the standard Periodogram in Matlab code? In my example I used the commands: [p,f]=periodogram(x); plot(f,p); However, I get a different result when I just write the command periodogram(x) .. Which one should I keep as an answer? Same for the Welch diagram. – MJ13 Jun 2 at 8:24
• Honestly I get the same results from [p,f] = periodogram(x) and p = periodogram(x) ... Hence I don't know why you get different results. – Fat32 Jun 2 at 20:45
• Actually the result is different if you change both of the instructions: [p,f]=periodogram(x); plot(f,p); with the instruction periodogram(x). Which is the correct one? – MJ13 Jun 3 at 17:01
• In addition, why does Welch method reduces more noise than standard Periodogram? – MJ13 Jun 5 at 11:34
• @MJ13 because Welch is taking an average and this will reduce noise... – Fat32 Jun 5 at 11:36


text:
# Merit Skimmer¶

The fermipy-merit-skimmer script can be used to create skimmed Merit files (either MC or data) and serves as a replacement for the web-based merit skimmer tool. The script accepts as input a sequence of Merit file paths or lists of Merit file paths which can be either local (nfs) or xrootd.

$fermipy-merit-skimmer merit_list.txt --output=merit.root --selection='FswGamFilter' \ --aliases=aliases.yaml$ fermipy-merit-skimmer merit_list.txt --output=merit-clean.root \
--selection='FswGamFilter && CLEAN' --aliases=EvtClassDefs_P8R2.xml

where merit_list.txt is a text file with one path per line. The --selection option sets the selection that will be applied when filtering events in each file. The --output option sets the path to the output merit file. The --aliases option can be used to load an alias file (set of key/value cut expression pairs). This option can accept either a YAML alias file or an XML event class definition file. The following illustrates the YAML alias file format:

FswGamFilter : FswGamState == 0
TracksCutFilter : FswGamState == 0 && TkrNumTracks > 0
CalEnergyFilter : FswGamState == 0 && CalEnergyRaw > 0

One can restrict the set of output branches with the --branches option which accepts an alias file or a YAML file containing a list of branch names. In the former case all branch names used in the given alias set will be extracted and added to the list of output branches.

$fermipy-merit-skimmer merit_list.txt --output=merit.root --selection='FswGamFilter' \ --aliases=EvtClassDefs_P8R2.xml --branches=EvtClassDefs_P8R2.xml  One can split the skimming task into separate batch jobs by running with the --batch option. This will subdivide task into N jobs when N is the number of files in the list divided by --files_per_job. The name of the output ROOT file of each job will be appended with the index of the job in the sequence (e.g. skim_000.root, skim_001.root, etc.). The --time and --resources options can be used to set the LSF wallclock time and resource flags. $ fermipy-merit-skimmer merit_list.txt --output=merit.root --selection='SOURCE && FRONT' \
--branches=EvtClassDefs_P8R2.xml --files_per_job=1000 --batch --aliases=EvtClassDefs_P8R2.xml

When skimming MC files it can be useful to extract the jobinfo for tracking the number of thrown events. The --extra_trees option can be used to copy one or more trees to the output file in addition to the Merit Tuple:

\$ fermipy-merit-skimmer merit_list.txt --output=merit.root --extra_trees=jobinfo


text:
# Tuning out disorder-induced localization in nanophotonic cavity arrays.

@article{Sokolov2017TuningOD,
title={Tuning out disorder-induced localization in nanophotonic cavity arrays.},
author={Sergei Yurievich Sokolov and Jin Lian and Emre Y{\"u}ce and S. Combrie and Alfredo de Rossi and Allard P. Mosk},
journal={Optics express},
year={2017},
volume={25 5},
pages={
4598-4606
}
}
• Published 3 August 2016
• Physics
• Optics express
Weakly coupled high-Q nanophotonic cavities are building blocks of slow-light waveguides and other nanophotonic devices. Their functionality critically depends on tuning as resonance frequencies should stay within the bandwidth of the device. Unavoidable disorder leads to random frequency shifts which cause localization of the light in single cavities. We present a new method to finely tune individual resonances of light in a system of coupled nanocavities. We use holographic laser-induced…
13 Citations

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## References

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### Disorder-induced resonance shifts in high-index-contrast photonic crystal nanocavities

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• 2009
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### Necklace State Hallmark in Disordered 2D Photonic Systems

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• 2015
Necklace states arise from the coupling of otherwise confined modes in disordered photonic systems and open high transmission channels in strongly scattering media. Despite their potential relevance

### Controlled coupling of photonic crystal cavities using photochromic tuning

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• 2013
We present a method to control the resonant coupling interaction in a coupled-cavity photonic crystal molecule by using a local and reversible photochromic tuning technique. We demonstrate the

### Measurement of the profiles of disorder-induced localized resonances in photonic crystal waveguides by local tuning.

• Physics
Optics express
• 2016
A new method is demonstrated to elucidate spatial profile of the localized modes in such systems using precise local tuning and using deconvolution with the known thermal profile to reconstruct the spatial profile.

### Selective tuning of high-Q silicon photonic crystal nanocavities via laser-assisted local oxidation.

• Physics
Optics express
• 2011
We examine the cavity resonance tuning of high-Q silicon photonic crystal heterostructures by localized laser-assisted thermal oxidation using a 532 nm continuous wave laser focused to a 2.5 μm

### Statistics of light transport in 235-ring silicon coupled-resonator optical waveguides.

• Physics
Optics express
• 2010
Light transport in silicon coupled-resonator optical waveguides (CROWs) consisting of up to 235 coupled microrings without localization over frequency bands that are several hundred gigahertz wide is demonstrated.

### Phonon Routing in Integrated Optomechanical Cavity-waveguide Systems

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• 2015
The mechanical properties of light have found widespread use in the manipulation of gas-phase atoms and ions, helping create new states of matter and realize complex quantum interactions. The field

### Local thermal resonance control of GaInP photonic crystal membrane cavities using ambient gas cooling

• Physics
• 2015
We perform spatially dependent tuning of a GaInP photonic crystal cavity using a continuous wave violet laser. Local tuning is obtained by laser heating of the photonic crystal membrane. The cavity

### The first decade of coupled resonator optical waveguides: bringing slow light to applications

• Physics
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The first decade of the history of CROWs is discussed in this review, from the original idea to recent applications, panning through the technological platforms that have been employed to realize these structures.


text:
Making D-modules on affine varieties more explicit

This is a follow up to my question about D-modules supported on the nilpotent cone. I got some good answers there but now I have a more basic question.

Consider an affine algebraic variety X, a closed subvariety i:Y-->X, and the intermediate extension of the structure sheaf on Y to all of X (do I denote this i!*OY ? For that matter, explaining either the * or ! extension instead would be a helpful start if its easier).

My question is this: Since X is affine, D(X) is just an associative algebra, generated by O(X) and Vect(X) by the usual construction. My question is how can I understand i!*OY as a module the associative algebra D(X), supposing I understand D(X)?

In other words, what is the vector space underlying i!*OY, how do functions in O(X) and vector fields act?

I probably need to invest some serious time with a textbook to answer this question myself, but any help getting started would be most appreciated!

-
Just for completeness: D(X) is only generated by O(X) and derivations, if X is defined over a field of characteristic 0. –  Lars Nov 1 '09 at 18:57
And if X is smooth. –  Greg Muller Nov 1 '09 at 22:04
Yes, thanks on both counts. I'm really interested in the situation on the link, so both of these are true. –  David Jordan Nov 1 '09 at 23:24

Have you looked at Bernstein's lectures on D-modules? He proves a result relevant to your questoin in Lec. 3, Sec. 14: for an affine embedding Y --> X with Y irreducible, if E is an OY-coherent DY module (i.e. a vector bundle with a flat connection) then the !* direct image from E can be characterized as the unique irreducible subquotient of either the * or ! direct image which has nonzero restriction to Y. Since it can be easy in such situations to compute the * direct image, this may be a good way to get a handle on the other functors too. For example, I think one can see from this that if we take the embedding of the origin into the affine line, then both the * and !* direct images of OY=k (the ground field) are A(1)/A(1)t, where t is the coordinate on the affine line and A(1)=DX is the 1-dimensional Weyl algebra k[t,d/dt]. This quotient is rightly considered the "delta-function" A(1)-module, since its generator δ satisfies tδ=0. I'm not sure how similar to this case your general situation will be. But certainly by Kashiwara's theorem one knows that the * direct image (and hence the !* direct image) will be supported on the subvariety Y.

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Thanks! I had looked over these a few years ago in an intro course, but not using it hadn't known exactly where to look. –  David Jordan Nov 1 '09 at 19:52

Since Y is closed, the !, *, and !* extensions coincide, and there should be a straightforward (but confusing for me) way of doing what you want. If Y is smooth, then I wonder if you can identify the vector space you're talking about with something like sections of the conormal bundle to Y in X, but this is not quite right (e.g. when Y = X). When Y is singular, there is a more complicated story, having to do with the fact that the obvious definition of D-module on a singular variety is no good.

When Y is only locally closed, describing the !* extension is a difficult problem. Vilonen's thesis, the only reference I know about this, is available here:

There is even something nontrivial to say about the ! and * extensions--the characteristic cycles of such D-modules were computed by Schmid and Vilonen. Maybe there is a more elementary answer to your question about these extensions, though.

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Thanks, this is helpful! As you saw in the other post, the ones I'm interested in eventually aren't closed. –  David Jordan Nov 1 '09 at 23:26
OK not "sections of the conormal bundle" but "functions on the total space of the conormal bundle." A vector field on X determines a function on the conormal bundle to Y by contraction, and operates on the vector space of such functions by multiplication. –  David Treumann Nov 2 '09 at 4:28


text:
• 论文 •

### 三角洲岸线变迁中的河流纵剖面实验

1. 珠江水利委员会科学研究所
• 收稿日期:1985-02-05 修回日期:1985-10-18 出版日期:1986-06-15 发布日期:1986-06-15

### AN EXPERIMENT FOR THE CHANGING OF STREAM PROEILFS DURING SHORELINE EVOLUTION IN A DELTA

Zhu Qimao

1. Research Institute, The Pearl River Water Resources Commission
• Received:1985-02-05 Revised:1985-10-18 Online:1986-06-15 Published:1986-06-15

Abstract: The evolution of the shoreline in a delta area has an unnegliegidle effect on the longitudinal profile of a river with a frequently changeable estuary. This paper shows the results of six experiment sets with different conditions of water-sediment inflow and tail-water-level hydrograph.A deve opment mode of the profile of an alluvial river is found as follows;1.In the case that the sea current can carry away the material coming from the river, the shoreline would be stable, a linear profile in a sediment transport ebuilidrium state being formed, and the water level along the river would be stable.2.If the sea current can not carry the material, the shoreline would move seaward, a concave profile in a hynamic equilibrium state being formed, and the water level along the river would rise as the shoreline moves seaward.3.The periodical extension and migration of the estuary would cause a periodical fluctuation of water lever in the reach near the mouth, resulting in a time-dependent zigzag elevation of water level in the lower reach of an alluvial river. The amplitude would shrink upstream, eventually to zero at some point, where-upstream the water level would appear to rise continually. 4.The periodical change in conditions of water-sedment inflow would lead to changes of erosion and deposition on the river bed. If the sea current cannot carry the material coming from the river the location of the delta wou'd move seaward interruptedly as the inflow changes. In different periods, the water level in the river under the same condition of discharge would rise as well.However, in the case that the sea current can carry the material away, a linear profile in transport equiliblium would eventually de formed.5.The periodical change in condition of water-sediment inflow and the periodical fluctuation of water level in the reach near the mouth have different effects on the upper reach and lower reach of an alluvial riverain the upper reach, erosion or deposition would occur under different discharge, but the latter is dominant. Moreover, the water level under the same discharge would rise period by period, with a downstream-reducing effect. In the lower reach, affected by the estuary, the water level under the same dis charge appears to rise zigzagly with the same periods as the fluctuation of water level in the reach near the mouth, the amplitude shrinking upstream, eventually to zero.If the series of water-sediment flow as well as water level near the mouth are steady time series, there woulb be a trend of development from dynamic e equilibrium to transport equilibrium.6.Construction of reservoirs in the upper reach would result in a change in the condition of water and sediment in the lower reach. However, as the material will be scoured up in the reach right downstream of the reservoir, the sediment load to the sea will recover to the same value as before, so that the evolution tendency of the estuary would not be change.