ECE2111 review questions for topics 5–8
The following questions cover concepts and skills introduced in topics 5–8. They are indica- tive of the style of problems on the second mid-semester test for ECE2111. (The test will have fewer problems though!) Any of the problems on tutorial sheets 6–9 are also useful for review, as are problems at the end of chapters 3, 4, and 5 of the recommended text A. V. Oppenheim and A. S. Willsky, ‘Signals and Systems’.
1. Suppose the input x and the output y of a continuous-time LTI system are related by the differential equation
y(t) = −dy(t) + 0.5x(t) − dx(t) + 0.5dx2(t) for all t dt dt dt2
subject to the condition of initial rest.
(a) Find the frequency response H of the system.
(b) If the input to the system is x(t) = 2 + cos(t), find an expression for y(t) for all t.
(a) The frequency response of the system is
0.5 − (jω) + 0.5(jω)2
H(ω) = 1 + jω for all ω.
(b) If the input were x1(t) = 2 = 2ej0t then the output would be y1(t) = 2H(0) = 1 for all t.
If the input were x2(t) = cos(t), then the output would be y2(t) = |H(1)| cos(t + ∠H(1)).
From this we can see that |H(1)| = 1/√2 and that ∠H(1) = −3π/4. So y2(t) =
√1 cos(t − 3π/4). Since the system is linear the overall output is 2
y(t)=y1(t)+y2(t)=1+√1 cos(t−3π/4). 2
H(1)= −j . 1+j
2. Suppose the input x and the output y of a discrete-time LTI system are related by the difference equation
y[n]=ay[n−1]+x[n−1]−0.5x[n−2] foralln subject to the condition of initial rest.
(a) Assuming that |a| < 1, find the frequency response H of the system. Express your answer in terms of a. (b) Use the table of discere-time Fourier transforms, together with the properties of the DTFT, to find the impulse response of the system. Express your answer in terms of a. (c) If the input to the system is x[n] = jn for all n then the output is y[n] = jn+3 for all n. Find the value of a. (a) The frequency response of the system is e−jω − 0.5e−2jω H(ω) = 1 − ae−jω for all ω. (b) The impulse response is the inverse DTFT of H. We first rewrite H as H(ω) = e−jω where W (ω) = 1 − 0.5e−2jω 1 = e−jωW(ω) − 0.5e−2jωW(ω) 1 − ae−jω 1 − ae−jω 1 . From the tables, we know that the inverse DTFT of W 1−ae−j ω w[n] = anu[n] for all n. By the time-shifting property of the DTFT, together with linearity of the DTFT, we can see that h[n] = w[n − 1] − 0.5w[n − 2] = an−1u[n − 1] − 0.5an−2u[n − 2]. (c) If the input to the system is x[n] = jn = ejπn/2 then the output is y[n] = H(π/2)jn = jn+3. So we know that H(π/2) = j3 = −j. This means that e−jπ/2 − 0.5e−jπ −j + 0.5 H(π/2)= 1−ae−jπ/2 = 1+aj =−j. Multiplying through by the denominator we obtain and so that a = 0.5. −j + 0.5 = −j + a (a) Find the fundamental frequency ω0 of x. x(t) = find X−3, X−2, X−1, X0, X1, X2, X3. Xkejω0kt k=−∞ x(t) = cos(2t) + sin(t) − 3 cos(3t) ∞ for all t. (c) Let the frequency response of a continuous-time LTI system be 0 if ω ≤ −1.5 1.5+ω if−1.5≤ω≤0 H(ω)= 1.5−ω if0≤ω≤1.5 . 0 if ω ≥ 1.5 Sketch H(ω) vs ω for −2 ≤ ω ≤ 2. (d) If x is the input to an LTI system with frequency response H, find the output y of the system. (a) The fundamental frequency is ω0 = 1 rad/sec. (b) We rewrite the sinusoids in terms of complex exponentials to obtain x(t) = (−3/2)e−j3t + (1/2)e−j2t + (j/2)e−jt + (−j/2)ejt + (1/2)ej2t + (−3/2)ej3t. Comparing coefficients with the Fourier series decomposition we see that X−3 =−3/2, X−2 =1/2, X−1 =j/2, X0 =0, (c) A sketch of H(ω) vs ω is shown below X1 =−j/2, H(ω) −1.5 0 (d) If x is the input then the output is y(t) = X−3H(−3)e−j3t + X−2H(−2)e−j2t + X−1H(−1)e−jt+ X1H(1)ejt + X2H(2)ej2t + X3H(3)ej3t since all of the other Fourier coefficients of x are zero. Note that H(−3) = H(−2) = H(2) = H(3) = 0 and that H(−1) = H(1) = 0.5. Hence y(t) = (j/4)e−jt − (j/4)ejt = 0.5 sin(t) for all t. 4. A discrete-time LTI system has impulse response h1[n] = u[n] − u[n − 3]. (a) Find the frequency response H1 of the system. (b) A second discrete-time LTI system has impulse response h2[n] = δ[n] + δ[n − 2]. Find the frequency response H2 of the system. (c) The system with impulse response h1 and the system with impulse response h2 are connected in series to form a single system with frequency response H. Find an expression for H(ω) for all ω. (d) Find h1 ∗ h2 without using the convolution formula. (e) Find a difference equation relating the input x and the output y of the system with frequency response H. (a) The frequency response is H1(ω) = n=−∞ h1[n]e−jωn = 1 + e−jω + e−j2ω. H2(ω) = h2[n]e−jωn = 1 + e−j2ω. H(ω) = H1(ω)H2(ω) = (1 + e−jω + e−j2ω)(1 + e−j2ω) =1+e−jω +2e−j2ω +e−j3ω +e−j4ω. (d) We know that h1 ∗h2 is the inverse DTFT of H and so 1 ifn=0,1,3,4 (h1∗h2)[n]=2 ifn=2 0 otherwise. (e) A difference equaiton relating the input and output is y[n]=x[n]+x[n−1]+2x[n−2]+x[n−3]+x[n−4] foralln. 5. A continuous-time LTI system has frequency response that satisfies a for |ω| ≤ 5π/4 H(ω) = b for |ω| > 7π/4 and its value for 5π/4 < |ω| < 7π/4 is left unspecified. The input to the system is x(t) = Xkejω0kt k=−∞ where ω0 = π and Xk = (0.5)|k| for all k. (a) Find the fundamental period of x. (b) Is x a real signal? Explain why or why not. (c) If the output of the system is y(t) = x(t) − (1 + cos(πt)) for all t find the values of a and b. (a) The fundamental period of x is 2π/π = 2. (b) x is a real signal. This is because its Fourier coefficients satisfy X∗ = (0.5)|−k| = (0.5)|k| = X for all k. −k k (c) First, we know that the output and the input and the frequency response are related by y(t) = H(kπ)Xkejkπt k=−∞ =b Xkejkπt+bXkejkπt+a(X−1e−jπt+X0+X1ejπt) BecauseX0 =1andX−1 =X1 =0.5wecanseethat 1 + cos(πt) = X0 + X−1e−jπt + X1ejπt so we can write y(t) = x(t) − (1 + cos(πt)) = Xkejkπt − (X0 + X−1e−jπt + X1ejπt) = Xkejkπt+Xkejkπt. k=−∞ k=2 Equating our two descriptions of y we see that a = 0 and b = 1.