EE102B Spring 2018 GoldsmithSignal Processing and Linear Systems II

Problem Set #7 – SolutionsDue: Friday June 1st, 2018 at 5 PM.

1. Laplace Transform Convergence (10 pts)Determine whether each of the following statements is true or false. If a statement is true, construct aconvincing argument for it. If it is false, give a counterexample.

(a) The Laplace transform of t2u(t) does not converge anywhere on the s-plane.Solution:False

(b) The Laplace transform of et2u(t) does not converge anywhere on the s-plane.

Solution:True

(c) The Laplace transform of e jω0t does not converge anywhere on the s-plane.Solution:True

(d) The Laplace transform of e jω0tu(t) does not converge anywhere on the s-plane.Solution:False

(e) The Laplace transform of |t| does not converge anywhere on the s-plane.Solution:True

2. Bode Plots (10 pts)Sketch the Bode plots for the following frequency responses:

(a)s−10

10(s+1)Solution:Find plot on next page

(b)100(s+10)(10s+1)(s+100)(s2 + s+1)Solution:Find plot on next page

3. Cascade and parallel realizations (20 pts)This problem shows how a higher-order LTI system can be constructed from lower-order subsystems,using the equivalences of interconnected systems presented on page 183 of the Laplace Transformchapter in course reader. Consider an LTI system with transfer function of the form

H(s) =s−β1

(s− γ1)(s− γ2), ℜ(s)> min{ℜ(γ1),ℜ(γ2)},

where γ1 6= γ2 6= β1.

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(a) Find rational transfer functions H1(s) and H2(s) of order 1 (i.e., their numerator and denominatorare polynomial in s of degree up to 1), such that the following two diagrams are equivalent:

x(t) H1(s) H2(s) y(t)

x(t) H(s) y(t)

Solution:The two figures are equivalent provided

H(s) = H1(s)H2(s).

This is satisfied (for example) by

H1(s) =s−β1

s− γ1, ℜ(s)> ℜ(γ1)

andH2(s) =

1s− γ2

, ℜ(s)> ℜ(γ1).

(b) Find rational transfer functions H1(s) and H2(s) of order 1 (i.e., their denominator is a degree 1polynomial in s), such that the following two diagrams are equivalent:

x(t)

H1(s)

H2(s)

+ y(t)

x(t) H(s) y(t)

Hint: Use the partial-fraction expansion representation of H(s).Solution:In order to obtain a parallel realization we do partial-fraction expansion of H(s):

H(s) =A1

s− γ1+

A2

s− γ2,

where A1 =γ1−β1γ1−γ2

, and A2 =γ2−β1γ2−γ2

. We now may choose:

Ha(s) =γ1−β1

γ1− γ2

1s− γ1

, ℜ(s)> ℜ(γ1)

Hb(s) =γ2−β1

γ2− γ1

1s− γ2

, ℜ(s)> ℜ(γ2)

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4. Stabilizing unstable system by feedback

This problem makes reference to the Feedback Systems lecture notes, pages 197-198. H(s) consistsof a proportional-plus-derivative transfer function.

x(t) + G(s) = 1/s y(t)

H(s) = K

(a) Assume K = 0. Find the response of the system to a step input x(t) = u(t). Is te system stable?Explain.

(b) Find a real K such that the system is stable.

(c) Find the response to a step input x(t) = u(t) with the K as in (b).

(d) Extra credit: using the final value theorem, find

limt→∞

y(t)

when x(t) = u(t) as in (c). Make sure your answer agrees with the limit evaluated directly fromthe step response in (c).

Solution:

(a) Note: There was a typo in this question and we forgot to add a - sign at the summation in theblock diagram. Hence we have decided to accept the solutions that assume both - and + sign,the solution for + sign is shown in redWe have

Y (s) = X(s)H(s) =1s2 ,R(s)> 0.

The inverse Laplace transform o the above leads y(t) = u(t)t. The system is not stable since theROC o its transer unction does not contain the jω axis. Indeed, u(t)t is an unbounded responseto a bounded input u(t).

(b) The close-loop transfer function is

Y (s)X(s)

=G(s)

1+G(s)H(s)=

1s+K

We observe that the poles of this transfer function are at s = −K. We require that the poles areto the left of the jω axis, or K > 0.The close-loop transfer function is

Y (s)X(s)

=G(s)

1−G(s)H(s)=

1s−K

We observe that the poles of this transfer function are at s = K. We require that the poles are tothe left of the jω axis, or K < 0.

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(c) With the system as in (b) we have

Y (s) = X(s)1s=

1s(s+K)

=1K

(1s− 1

s+K

),

which leads to

y(t) =u(t)K

(1− e−kt

).

With the system as in (b) we have

Y (s) = X(s)1s=

1s(s−K)

=1K

(1

s−K− 1

s

),

which leads to

y(t) =u(t)K

(ekt −1)

).

(d) Since x(t) = 0 or t < 0 and since the (c) implies that the limit exists, the finite value theoremimplies

Y (s) = lims→0

sX(s)1s= lim

s→0

ss(s+K)

=1K.

Y (s) = lims→0

sX(s)1s= lim

s→0

ss(s−K)

=− 1K.

5. Computing z-transforms

For the following signals, determine bilateral z-transforms and regions of convergence. Note that ifthe region of convergence is null, the z-transform does not exist.

(a) x[n] = δ [n]−3u[n−5].

(b) x[n] = β n

n! u[n].

(c)(4

3

)nu[n]+

(12

)nu[−n].

(d) x[n] = anu[n−1].

(e) x[n] =−anu[−n].

Solution:

(a) 1− 3z−1

1− z−1 , ROC: {z : |z|> 1}.

(b)

X(s) =n=∞

∑n=−∞

x[n]z−n =n=∞

∑n=−∞

β n

n!u[n]z−n =

n=∞

∑n=0

β n

n!z−n =

n=∞

∑n=0

(β z−1)n

n!= eβ/z,

ROC; entire z plane except z = 0

(c) X(z) =1

1−4/(3z)+

z−1

z−1−2, ROC: {z : |z|−1 < 3/4 and |z|−1 > 2}, i.e., ROC is null.

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(d)az−1

1−az−1 ,ROC:{z : |az−1|< 1} .

(e)az−1

1−az−1 ,ROC:{z : |az−1|> 1} .

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MATLAB Assignment

General Instructions

(30 pts)Answer all questions asked. Your submission should include all m-file listings and plots requested. All

plots should have a title and x- and y-axes properly labeled.

In the following problems, you will examine the pole locations for second-order systems of the form

H(s) =ω2

n

s2 +2ηωns+ω2n

(1)The values of the damping ratio η and undamped natural frequency ωn specify the locations of the poles,

and consequently the behavior of this system. In this exercise, you will see how the locations of the polesaffect the frequency response.

First, you will examine the pole locations and frequency responses for four different choices of η whileωn remains fixed at 1.

a) Define H1(s) through H4(s) to be the system functions that result from fixing ωn = 1 in Eq. (1)above while η is 0, 1

4 ,1, and 2, respectively. Define a1 through a4 to be the coefficient vectors forthe denominators of H1(s) through H4(s). Find and plot the locations of the poles for each of thesesystems. Note that the command roots([a,b,c]) returns the values of the roots (in this casepoles) of an equation of the form ax2 + bx+ c. To plot the poles, display the imaginary part on they-axis and the real part on the x-axis.

b) Define omega = [-5:0.1:5] to be the frequencies at which you will compute the frequencyresponses of the four systems. Use the Matlab command freqs to compute and plot |H( jω)| for eachof the four systems you defined in Part (a). How are the frequency responses for η < 1 qualitativelydifferent from those for η ≥ 1? Can you explain how the pole locations for the systems cause thisdifference?

Next, you will trace the locations of the poles as you vary η and ωn and see how varying these parametersaffects the frequency response of the system.

c) First, you will vary η in the range 0≤ η ≤ 10 while holding ωn = 1. Defineetarange=[0 logspace(-1,1,99)] to get 100 logarithmically spaced points in 0≤ η ≤ 10.Define a_eta to be a 3 x 100 matrix where each column is the denominator coefficients for H(s)when η has the value in the corresponding column of etarange. Notice that only the middlecoefficient is changing with each new value of η . Define etapoles to be a 2 x 100 matrix whereeach column is the roots of the corresponding column of a_eta. On a single figure, plot the realversus imaginary parts for each row of etapoles and describe the loci they trace. On your plot,indicate the following points: η = 0, 1

4 ,1, and 2 (you can use something like the scatter commandto show these individual points). In order to get a square aspect ratio with equal length axes in yourplot, you can type:

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>> axis(’equal’)>> axis([-4 0 -2 2])

Describe qualitatively how you expect the frequency response to change as η goes from 0 to 1 andthen from 1 to 10.

d) In this problem, you will hold η = 14 and examine the effect of increasing ωn from 0 to 10. Define

omegarange=[0 logspace(-1,1,99)] to get 100 logarithmically spaced points in the regionof interest. Define a_omega and omegapoles analogously to the way you defined a_eta andetapoles in Part (c). On a single figure, plot the real versus imaginary parts for each columnof omegapoles. How would you expect changing ωn to change the frequency response H( jω)?Use freqs to evaluate the frequency response when ωn = 2 and η = 1

4 and plot the magnitude ofthis frequency response. Compare this with the plot you made in Part (b) for ωn = 1 and η = 1

4 .How are they different? Does this match what you expected from your plot of the loci traced byomegapoles?

e) Extra Credit: There is no reason that η needs to be positive. Repeat Part (c) for η between -10 and0. When η is negative, can the system described by H(s) be both causal and stable? Also, plot thefrequency response magnitude for the system when η =−1

4 and ωn = 1 using freqs. Is the systemwith the frequency response computed by freqs causal? Also explain any similarities or differencesbetween this plot and the frequency response magnitude plotted in Part (b) for η = 1

4 .

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Table of ContentsPart a ................................................................................................................................ 1Part b ................................................................................................................................ 2Part c ................................................................................................................................ 5Part d ................................................................................................................................ 6Part e ................................................................................................................................ 9

Part a% H(s) = (omega_n)^2 / (s^2 + 2*eta*omega*s + omega^2)% H1(s) --> eta = 0% H2(s) --> eta = 1/4% H3(s) --> eta = 1% H4(s) --> eta = 2

a1 = [1 0 1];a2 = [1 1/2 1];a3 = [1 2 1];a4 = [1 4 1];

% Find the poles of these systemspoles1 = roots(a1);poles2 = roots(a2);poles3 = roots(a3);poles4 = roots(a4);

figurehold onscatter(real(poles1), imag(poles1), 'b')scatter(real(poles2), imag(poles2), 'g')scatter(real(poles3), imag(poles3), 'k')scatter(real(poles4), imag(poles4), 'r')legend('H_1', 'H_2', 'H_3', 'H_4')title('Locations of Poles')ylabel('Imaginary')xlabel('Real')

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Part bomega = -5:0.1:5;b = 1;

h1 = freqs(b,a1,omega);figureplot(omega, abs(h1),'b')title('Frequency Response for eta = 0')

h2 = freqs(b,a2,omega);figureplot(omega, abs(h2),'g')title('Frequency Response for eta = 1/4')

h3 = freqs(b,a3,omega);figureplot(omega, abs(h3),'k')title('Frequency Response for eta = 1')

h4 = freqs(b,a4,omega);figureplot(omega, abs(h4),'r')title('Frequency Response for eta = 2')

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% Answers to questions% For eta > 0, the frequency response is symmetric about omega = 0. For eta% < 0, the frequency response is symmetric but not at 0. For eta > 0, the% poles are real while for eta < 0 they have imaginary parts.

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Part cetarange = [0 logspace(-1,1,99)];

% Coefficients of denominatora_eta = zeros(100,3);a_eta(:,1) = 1;a_eta(:,2) = 2*etarange;a_eta(:,3) = 1;

% Poles for different eta valuesetapoles = zeros(100,2);for i = 1:100 etapoles(i,:) = roots(a_eta(i,:));end

figurehold onscatter(real(etapoles(:,1)), imag(etapoles(:,1)), 'b')scatter(real(etapoles(:,2)), imag(etapoles(:,2)), 'g')% Mark particular values of eta% eta = 0 is sample 1% eta = 1/4 is sample 21% eta = 1 is sample 51% eta = 2 is sample 66

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etas = [1,21,51,66];for eta = 1:length(etas) scatter(real(etapoles(etas(eta),1)), imag(etapoles(etas(eta),1)), 'r') scatter(real(etapoles(etas(eta),2)), imag(etapoles(etas(eta),2)), 'r')endxlabel('Real Part')ylabel('Imaginary Part')title('Poles for Changing Eta')axis('equal')axis([-4 0 -2 2])

% The values asked for are marked in red.% As eta goes from 0 to 1, the frequency response will be closer to% centered at 0. As eta goes from 1 to 10, the frequency response will be% sharper and centered at 0.

Part domegarange=[0 logspace(-1,1,99)];eta = 1/4;

% Coefficients of denominatora_omega = zeros(100,3);

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a_omega(:,1) = 1;a_omega(:,2) = 2*eta*omegarange;a_omega(:,3) = omegarange.^2;

% Poles for different omega valuesomegapoles = zeros(100,2);for i = 1:100 omegapoles(i,:) = roots(a_omega(i,:));end

figurehold onscatter(real(omegapoles(:,1)), imag(omegapoles(:,1)), 'b')scatter(real(omegapoles(:,2)), imag(omegapoles(:,2)), 'g')xlabel('Real Part')ylabel('Imaginary Part')title('Poles for Changing Omega')axis('equal')axis([-4 0 -2 2])

% The frequency response will be more spread out (centered at wider points)% when omega is larger.

omega = -5:0.1:5;b = 4;a = [1 1 4];

h = freqs(b,a,omega);figureplot(omega, abs(h),'b')title('Frequency Response for eta = 1/4, omega = 2')

% Yes, this matches what we would expect

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Part enetarange = [0 logspace(-1,1,99)]*-1;

% Coefficients of denominatora_neta = zeros(100,3);a_neta(:,1) = 1;a_neta(:,2) = 2*netarange;a_neta(:,3) = 1;

% Poles for different eta valuesnetapoles = zeros(100,2);for i = 1:100 netapoles(i,:) = roots(a_neta(i,:));end

figurehold onscatter(real(netapoles(:,1)), imag(netapoles(:,1)), 'b')scatter(real(netapoles(:,2)), imag(netapoles(:,2)), 'g')xlabel('Real Part')ylabel('Imaginary Part')title('Poles for Changing Eta')axis('equal')

% All the poles are in the right hand side of the figure so it cannot be% causal and stable.

omega = -5:0.1:5;b = 1;a = [1 -1/2 1];

h = freqs(b,a,omega);figureplot(omega, abs(h),'b')title('Frequency Response for eta = -1/4, omega = 1')

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Published with MATLAB® R2016a