Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Solutions of Chapter 5 Part 2/2 Problem 5.3-18 (a) Find the zero-state response of an LTID system with transfer function H[z] = z (z + 0.2)(z − 0.8) and the input x[n] = (e)(n+1) u[n]. (b) Write the difference equation relating the output y[n] to input x[n]. Solution: (a) x[n] = (e)(n+1) u[n] = e(en u[n]) ⇔ X[z] = e Y [z] = H[z]X[z] = Therefore z z−e ez2 (z + 0.2)(z − 0.8)(z − e) Y [z] ez 1.32 0.186 1.13 = = − − z (z + 0.2)(z − 0.8)(z − e) z − e z + 0.2 z − 0.8 z z z Y [z] = 1.32 − 0.186 − 1.13 z−e z + 0.2 z − 0.8 y[n] = [1.32(e)n − 0.186(−0.2)n − 1.13(0.8)n ] u[n] (b) From H[z] = Y [z] z = X[z] (z + 0.2)(z − 0.8) we have (z2 − 0.6z − 0.16)Y [z] = zX[z] Hence, the corresponding difference equation of the system is y[n + 2] − 0.6y[n + 1] − 0.16y[n] = x[n + 1] or y[n] − 0.6y[n − 1] − 0.16y[n − 2] = x[n − 1] Problem 5.3-23 Find h[n], the unit impulse response of the systems described by the following equations: (a) y[n] + 3y[n − 1] + 2y[n − 2] = x[n] + 3x[n − 1] + 3x[n − 2] (b) y[n + 2] + 2y[n + 1] + y[n] = 2x[n + 2] − x[n + 1] (c) y[n] − y[n − 1] + 0.5y[n − 2] = x[n] + 2x[n − 1] Page 1 of 4 Solution: (a) Write the system in advanced form y[n + 2] + 3y[n + 1] + 2y[n] = x[n + 2] + 3x[n + 1] + 3x[n] or (E 2 + 3E + 2)y[n] = (E 2 + 3E + 3)x[n] Thus, H[z] = z2 + 3z + 3 Y [z] z2 + 3z + 3 = 2 = X[z] z + 3z + 2 (z + 1)(z + 2) H[z] z2 + 3z + 3 3/2 1 1/2 = = − + z z(z + 1)(z + 2) z z+1 z+2 3 z 1 z H[z] = − + 2 z+1 2 z+2 [ ] 3 1 n n h[n] = δ [n] − (−1) + (−2) u[n] 2 2 (b) z2Y [z] + 2zY [z] +Y [z] = 2z2 X[z] − zX[z] H[z] = Therefore 2z2 − z z(2z − 1) Y [z] = 2 = X[z] z + 2z + 1 (z + 1)2 H[z] 2z − 1 2 3 = = − 2 z (z + 1) z + 1 (z + 1)2 z z H[z] = 2 −3 z+1 (z + 1)2 h[n] = [2(−1)n − 3n(−1)n ] u[n] = (2 − 3n)(−1)n u[n] (c) Performing z-transform of the system yields 1 0.5 2 Y [z] − Y [z] + 2 Y [z] = X[z] + X[z] z z z Thus, H[z] = 1 + 2z Y [z] z(z + 2) = 2 = 1 0.5 X[z] 1 − z + z2 z − z + 0.5 H[z] (z + 2) = 2 z z − z + 0.5 Using the pair 12c in Table 1, with 1 A = 1, B = 2, a = −0.5, |γ |2 = 0.5, |γ | = √ 2 √ −2.5 π r = 5.099, β = cos−1 (0.5 5) = , θ = tan−1 ( ) = −1.373 4 0.5 we have π 1 h[n] = 5.099( √ )n cos( n − 1.373) u[n] 4 2 Page 2 of 4 Problem 3.4-3 (used in Problem 5.5-3) A moving average is used to detect a trend of a rapidly fluctuating variable such as the stock market average. A variable may fluctuate (up and down) daily, masking its longterm (secular) trend. We can discern the long-term trend by smoothing or averaging the past N values of the variable. For the stock market average, we may consider a 5-day moving average y[n] to be the mean of the past 5 days’ market closing values x[n], x[n − 1], · · · , x[n − 4]. (a) Write the difference equation relating y[n] to the input x[n]. Solution: (a) y[n] = 15 (x[n] + x[n − 1] + x[n − 2] + x[n − 3] + x[n − 4]) Problem 5.5-3 Find the frequency response for the moving average system in Prob.3.4-3. The input-output equation of this system is given by 1 4 y[n] = ∑ x[n − k] 5 k=0 Solution: The advance operator form of the equation is 1 E 4 y[n] = (E 4 + E 3 + E 2 + E + 1)x[n] 5 Thus, the transfer function of the system is [ ] Y [z] 1 z4 + z3 + z2 + z + 1 H[z] = = X[z] 5 z4 The frequency response is jΩ H[e ] = = = [ ] 1 e j4Ω + e j3Ω + e j2Ω + e jΩ + 1 5 e j4Ω 1 − j2Ω j2Ω e [e + e jΩ + 1 + e− jΩ + e− j2Ω ] 5 1 − j2Ω e [1 + 2 cos Ω + 2 cos 2Ω] 5 Page 3 of 4 Problem 5.5-5 For an LTID system specified by the equation y[n + 1] − 0.5y[n] = x[n + 1] + 0.8x[n] (a) Find the amplitude and the phase response. (b) Find the system response y[n] for the input x[n] = cos(0.5n − π3 ). Solution: For the given system, we have the transfer function H[z] = z + 0.8 z − 0.5 (a) The frequency response is H[e jΩ ] = e jΩ + 0.8 (cos Ω + 0.8) + j sin Ω = e jΩ − 0.5 (cos Ω − 0.5) + j sin Ω Amplitude response: From |H[e jΩ ]|2 = |H[e jΩ ]|H ∗ [e jΩ ] = |H[e jΩ ]|H[e− jΩ ] = we have ( |H[e ]| = jΩ (e jΩ + 0.8)(e− jΩ + 0.8) 1.64 + 1.6 cos Ω = (e jΩ − 0.5)(e− jΩ − 0.5) 1.25 − cos Ω 1.64 + 1.6 cos Ω 1.25 − cos Ω ) 12 The phase response is sin Ω sin Ω ) − tan−1 ( ) cos Ω + 0.8 cos Ω − 0.5 ∠H[e jΩ ] = tan−1 ( (b) For the given input, Ω = 0.5. The amplitude response is |H[e j0.5 ]|2 = Thus, 1.64 + 1.6 cos(0.5) = 8.174 1.25 − cos(0.5) 1 H[e j0.5 ] = (8.174) 2 = 2.86 The phase response is ∠H[e j0.5 ] = 0.2784 − 0.9037 = −0.6253 rad Therefore, the system response is y[n] = 2.86 cos(0.5n − π − 0.6253) = 2.86 cos(0.5n − 1.6725) 3 Page 4 of 4