Signals And Systems Assignment Help

Signals And Systems Assignment Help

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PROBLEMS P.1 Consider the two feedback systems in Figures P25.1-1 and P25.1-2.

(a) Express the overall system functions in terms of the system functions in the forward and feedback paths.

For the remainder of the problem, assume that K(s) and K(z) are both constant val ues K and that matlabassignmentexperts.com

(b) Plot the pole-zero pattern of Q(s) and Q(z) for K = 0, -1, and 1. (c) Indicate the locus in the s plane of the poles of Q(s) and in the z plane the poles of Q(z) as K increases from K = 0 toward large positive values and as K decreases from K = 0 toward large negative values. (d) Specify the range of values of K for which each of the systems is stable.

P.2 For the systems in Figures P25.2-1 to P25.2-3, derive the transfer functions indicated.

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P.3 Consider the system G(s) with frequency response G(jw) indicated in Figure P25.3-1.

(a) Sketch the approximate overall frequency response of each of the feedback sys tems in Figure P25.3-2.

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(b) Determine which of the two systems in part (a) attempts to equalize the gain G(jw), i.e., to make it more constant over all frequencies, and which system is intended as an approximation to a system that is the inverse of G(jw).

P.4 Consider the basic feedback systems of Figure P25.4. Determine the

closed-loop sys tem impulse response for each of the following specifications of the system functions in the forward and feedback paths.

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P.5 For the system given in Figure P25.5, find Y(s) in terms of X(s) and W(s).

P.6 Consider the causal discrete-time system depicted in Figure P25.6.

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(a) Show that this is not a stable system. (b) (b) Suppose that we allow feedback with one unit of delay. That is, suppose that x[n] = xZ,[n] -Ky[n -1], where xe[n] is an externally applied signal now regarded as the input to the overall closed-loop system. Is it possible to stabilize the system with feedback of this form? If so, find the maximum range of values of K for which the system is stable. (c) Suppose that instead of using feedback with one unit of delay as in part (b), we allow feedback with two units of delay, so that x[n] = x,[n] -Ky[n -21 Specify the full range of values of K (if any such values exist) for which this system is stable.

P.7 In this problem we illustrate how feedback can be used to increase the bandwidth of an amplifier. Consider an amplifier whose gain falls off at high frequencies. Spe cifically, suppose that the system function of this amplifier is matlabassignmentexperts.com

(a) What is the dc gain of the amplifier (i.e., the magnitude of its frequency response to zero frequency)? (b) What is the system time constant? (c) Suppose that we define bandwidth as the frequency at which the magnitude of the amplifier frequency response is 1/ \2 times its magnitude of dc. What is the bandwidth of this amplifier? (d) Suppose that we place this amplifier in a feedback loop as depicted in Figure P25.7. What is the dc gain of the closed-loop system? What are the time constant and bandwidth of the closed-loop system?

(e) Find the value of K that leads to a closed-loop bandwidth that is exactly double the bandwidth of the open-loop amplifier. What are the corresponding closedloop system time constant and dc gain? matlabassignmentexperts.com

SOLUTIONS S.1 (a)

We have and From eq. (S25.1-2),

V(s) = X(s) -Y(s)K(s) Y(s) = V(s)H(s)

Substituting eq. (S25.1-3) into eq. (S25.1-1), we have

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Similarly,

(b)

as shown in Figures S25.1-2 and S25.1-3, respectively.

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For K = -1, as shown in Figures S25.1-4 and S25.1-5, respectively.

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For K = 1, as shown in Figures S25.1-6 and S25.1-7, respectively.

The pole is located at s = 2(1 -K), as shown in Figure S25.1-8.

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Hence, the locus of the pole is the line Re{s} = 0. Similarly, for the locus of the pole is also the line Re{z} = 0, shown in Figure S25.1-9.

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The root location decreases as K moves to infinity and increases as K moves to negative infinity. (d) The system is stable for 2(1 -K) < 0, or K > 1.

The system is stable for -1 < 2(1 -K) < 1, or < K < i.

S.2We use Problem P25.1. (a)

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S.3 (a)

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(b) From the frequency response in part (a), clearly system 1 tends to make the response more constant and system 2 tends to resemble the inverse of G(jw).

S.4 For the system in Figure S25.4-1, we denote the closed-loop system function by

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Therefore,

In this case,

v(t) = te 2tu(t)

v(t) = le -2 tu(t)

(c) The system function G(s) = e-113 corresponds to a delay of 1, i.e., the feedback system of Figure P25.4(a) becomes that shown in Figure S25.4-2.

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We can now recursively obtain the impulse response by inspection. With x(t) = 8(t),

Therefore,

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S.5

y(t) = K2w(t) + KK 2 v(t)

By taking the transform of eq. (S25.5-1), we have Y(s) = K 2 W(s) + KK 2Vs) Also

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Therefore,

and

S.6 (a) The system function of the system given in Figure P25.6 must be determined first. So we write down the difference equation y[n] = x[n] + y[n -1] + 4y[n -2] Taking the z-transform of the equation, we have

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The poles of this system are located at

Since IzI > 1 for at least one pole the system is unstable. (b) With closed-loop feedback, the difference equation is y[n] = x,[n] -Ky[n -1] + y[n -1] + 4y[n -2] Thus,

The poles are now located at

Note that the roots are purely real because the term inside the square root is always positive. For z = 1,

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Thus,

K2 + 2K + 1 = K2 -2K + 17, 4K = 16, or K = 4 We can also calculate z 2: Z2= -4 Similarly, zi = -1, z2 = 4 for K = -2. Observe the root locus in Figure S25.61.

Observe that if one of the poles is inside zI 1, the other is outside. Hence,the system is unstable for all values of K. (c) The difference equation can be written as y[n] = x,[n] + y[n -1] + (4 -K)y[n -2]

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Therefore,

In this case, the poles are located at

For a stable system, we want

If we set 17 -4K > 0, then

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Now suppose 17 -4K < 0. Then

Thus, for K in the range 4 < K < 5, we have a stable system. The root locus is shown in Figure S25.6-2.

S.7 (a) The de gain of the amplifier is IH(O)| = |G|. (b) h(t) = Gae-"'u(t).Therefore, the time constant is 1/a.

Thus wc = ± a. Hence the bandwidth is a. matlabassignmentexperts.com

(d) The closed-loop transfer function is

From part (a), the time constant is

From part (c), the bandwidth is (1 + KG)a. From part (a), the dc gain is

(e) We require (GK + 1)a = 2a. Hence, K = 1/G. So the bandwidth becomes 2a.

The time constant is 1/(2a), and IH(0) = IG/2 1,the de gain.

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