ece_351_hwset1_sol_fall_20_v23

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ECE 351 D. van Alphen Fall 19 1/19 Homework Set 1 Solutions 1 – 59 1. answer: c. discrete 2. System response H(  ) or H(f) or H(s): ratio of output over input, in the frequency domain; H(  ) gives gain for sinusoids of frequency  . 3. MATLAB Code: % program ece_351_hw1_prob3 t_e = 0: .01 : 2; x_e = 3*sin(8*pi*t_e)+ 3; n_f = 0 : .05 : 2 + 3; x_f = 3*sin(8*pi*n_f) + 3; subplot(2,1,1), plot(t_e, x_e), xlabel( 'x' ), grid subplot(2,1,2), stem(n_f, x_f), xlabel( 'n' ) 4. Input: a sine, at frequency 30  rad/sec., a cosine at frequency 5  rad/sec., and a dc term of 5; System: Ideal LPF, with passband (-10  rad/sec., 10  rad/sec.), gain of 5 over passband; will only pass the cosine and the dc term, not the sine Output: y(t) = 5*(input in passband) = 20 cos(5  t) + 25 5. a/(1-r) (Be sure to know this formula. It will re-occur many times in this course.) 6.  (t) 7. See Lathi, p. 29 0 0.5 1 1.5 2 0 2 4 6 x 0 1 2 3 4 5 0 2 4 6 n Frequency : 8  rad/sec = 4 Hz. Period T 0 = 1/4 sec.

ECE 351 D. van Alphen Fall 19 2/19 8a. 4 y(t) b. 3 y(t – 2) c. 4 y(t) + 3 y(t – 2) 9a/b. T 0 = 2 sec., f 0 = 1/T 0 = .5 Hz.,  0 = 2  f 0 =  rad/sec. 9c. a 0 = .5(4) + .5(0) = 2 10. True or false – give counterexamples (or explanation) for statements that are false: a. If a signal is continuous in time, it is analog. False Counterexample: the binary (hence digital) waveform b. If a signal is discrete in time, it is digital. False Counterexample: the un-quantized waveform with sample level values {  , 3.2222…, e, …} c. y[n] = 3 x[n] + 2 x[n – 1] is time-invariant. True (constant coefficients) d. false – sampling makes the signal discrete, not digital e. false – it’s a decaying exponential f. false – they assume an infinite number of amplitude signals g. false – they are zero for negative values of time. 11. y[n] = 3 x[n] is linear, because it displays proportionality and superposition: Proportionality: Say x[n]  y[n] = 3 x[n]; Then a x[n]  3 (ax[n]) = a (3 x[n]) = a y[n] Superposition: Say x 1 [n]  y 1 [n] = 3 x 1 [n]; And x 2 [n]  y 2 [n] = 3 x 2 [n]; Then x 1 [n] + x 2 [n]  3 (x 1 [n] + x 2 [n]) = 3 x 1 [n] + 3 x 2 [n] = y 1 [n] + y 2 [n] 12. x(t) = sinc(2t)  (  /2) rect(  /4) by Table entry 18, p. 702, with W = 2 0 … t  X(  ) -2 2  /2

ECE 351 D. van Alphen Fall 19 3/19 13. F -1 { rect(  T/(2  )) } = (W/  ) sinc(Wt), where W =  /T  x(t) = (  /T) sinc(  t/T) (Table entry 18, p. 702) 14. After reading about the pinouts at: http://pinouts.ru/Slots/USB_pinout.shtml A connection: to the computer; B connection: to the peripheral Current < .5 A, Voltage < 5 V (dc);  Power = I V < 2.5 W. In a hub, upstream ports connect to the host (usually a computer); downstream ports are for plugging in your peripheral devices (e.g., printer, mp3, ram stick, …) 15a. x[n] = .9 n 15b. y[n] = x[-n] 15c. y[n] = x[n-2] MATLAB Code: % Program ece_351_hw1 % This code solves Problem 15 n = -4:4; x_a = .9 .^ n; x_b = .9 .^ (-n); x_c = .9 .^ (n-2); subplot(3,1,1), stem(n,x_a), xlabel( 'n' ),ylabel( 'x_a' ), axis([-4.5 4.5 0 2]) subplot(3,1,2),stem(n, x_b), xlabel( 'n' ),ylabel( 'x_b' ), axis([-4.5 4.5 0 2]) subplot(3,1,3), stem(n, x_c), xlabel( 'n' ),ylabel( 'x_c' ), axis([-4.5 4.5 0 2]) Plots generated by the above code: MATLAB Function Code to be used for 15d and 15e: Note: this code calls the function unit_step as given in lecture 2 (p. 27). % Program ece_351_hw1_more % This codes solves problems 15d and 15e >> n = -5:10; -4 -2 0 2 4 0 1 2 n x a -4 -2 0 2 4 0 1 2 n x b -4 -2 0 2 4 0 1 2 n x c

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ECE 351 D. van Alphen Fall 19 4/19 >> x = 5*(unit_step(n) - unit_step(n-3)); >> y = (2*n) .* (unit_step(n-1) - unit_step(n-4)); >> subplot(2,1,1), stem(n, x) >> subplot(2,1,2), stem(n, y) Plots generated by the above code: 16. a. False: consider x[n] = e -n u[n]: energy E = ... 1 4 2 2 0           e e e n n (geo. series, with ratio r = e -2  sum = a/(1-r) = 1/(1-e -2 ) = 1.156, finite b. True c. True d. False – it’s neither even nor odd; e. True 17a.  [n] +  [n-1] +  [n-2] 17b. 3  [n+2] + 3  [n+1] + 3  [n] 17c. –  [n] –  [n-1] 18a. u[n-1] – u[n-3] 18b. 4 { u[n] – u[n – 3] } 18c. 2 u[n] + u[n – 1] – 3 u[n – 2] 19a. E = sum of squares = 16 19 b. E = 16, same as part a (shifting doesn’t change the energy; changing the sign doesn’t change the energy) 19 c. E = 9 (16)= 144, since tripling the amplitude increases the “squares” by a factor of 9. 20. Problem 8.1-1 -5 0 5 10 0 5 n x d -5 0 5 10 0 5 n y

ECE 351 D. van Alphen Fall 19 5/19 a. x 1 (t)  X 1 (  ) with  m = 2  x 10 5  f m = 10 5  f N = f s = 2 x 10 5 = 200 KHz b. x 2 (t)  X 2 (  ) with  m = 3  x 10 5  f m = 1.5 x 10 5  f N = f s = 3 x 10 5 = 300 KHz c. x 1 2 (t)  X 1 (  )*X 1 (  ) with  m = 4  x 10 5  f m = 2 x10 5  f N = f s = 4 x 10 5 = 400 KHz d. x 2 3 (t)  X 2 (  )*X 2 (  )*X 2 (  ) with  m = 9  x 10 5  f m =4.5 x 10 5  f N = f s = 9 x 10 5 = 900 KHz e. x 1 (t) x 2 (t)  X 1 (  )*X 2 (  ) with  m = 5  x 10 5  f m = 2.5 x 10 5  f N = f s = 5 x 10 5 = 500 KHz 21. Problem 8.1-2abc (using FT table, p. 702) a. sinc(100  t)  (1/100) rect(  /200  ), with  m = 100  f m    sinc 2 (100  t) has f m = 100, f N = f s = 200 Hz, T s = 1/200 sec. = 5 ms b. (same as a) c. sinc(100  t) + 3 sinc 2 (60  t); sinc(100pt) has bandwidth f m = 50, while 3 sinc 2 (60  t) has bandwidth f m = 60 Hz; over-all f m for the sum is thus 60 Hz.  f N =f s = 120 Hz, T s = 1/120  8.33 ms 22. Problem 8.1-7a (using FT table, p. 702) x(t) = 5 sinc 2 (5  t) + cos(20  t)   (  /(20  )) +  {  (  – 20  ) +  (  + 20  )},  m = 20  , f m = 10 Hz; Undersampled at 10 Hz (vs. Nyquist rate of > 20 Hz)  repeat spectral replicas every 10 Hz (or every 20  rad/sec) Note: LPF would recover a dc term that was not part of the original signal, due to the delta at the origin 23a. Ts = .2 sec. 23b. x s (t) =  (t) + 3  (t - .2) + 5  (t – .4) + 2  (t – .6) + 3  (t - .8) x r (t) = sinc(  t / T s ) * x s (t) = sinc(5  t) * xs(t) = sinc(5  t) * {  (t) + 3  (t - .2) + 5  (t – .4) + 2  (t – .6) + 3  (t - .8)} = sinc(5  t) + 3 sinc(5   t - .2)) + 5 sinc(5   t - .4)) + 2 sinc(5   t - .6) + 3 sinc(5   t - .8)) -30  -20  -10  10      … … Spectrum prior to sampling shown in black

ECE 351 D. van Alphen Fall 19 6/19 23c. MATLAB Code: % program ece_351_hw1 % % This code solves problem 23c in Homework Set 1. It calls on the function % mysinc which was given in lecture 2, p. 16. % t = -.2 : .01 : 1.2; xr1 = mysinc(5*pi*t) + 3 * mysinc(5*pi*(t - .2)); xr = xr1 + 5 * mysinc(5*pi*(t - .4)) + 2 * mysinc(5*pi*(t - .6) + 3 * … mysinc(5*pi*(t - .8)); n = [0 .2 .4 .6 .8]; xs = [1 3 5 2 3]; plot(t, xr, '-k' , n, xs, 'or' ) xlabel( 't, sec.' ), title( 'Recovered Signal and Samples' ) 24. Problem 8.3-1abc a. Nyquist rate = 2f m = 30 K samples/sec. b. L = 65,536 levels  l = log 2 (L) = 16 bits required c. sec 000 , 48 16 sec 000 , 30 bits sample bits samples   25. Problem 8.3-2 a. Nyquist rate = 2f m = 9 M samples/sec.  f s = 1.2 (9M) = 10.8 M samples/sec. b. log 2 (1024) = 10 levels c. sec 10 8 . 10 10 sec 10 8 . 10 7 6 bits x sample bits samples x   26. Problem 3.1-1 a. E = 1 2 + 2 2 + 3 2 + 2 2 + 1 2 = 19 b. E = 19, same as answer a (shifting does not change the energy) c. E = 2 (3 2 + 6 2 + 9 2 ) = 252 d. E = 2(4 2 + 2 2 ) = 40 27. Problem 3.1-2 (using periodicity, find energy in one cycle, and divide by N) a. P = E/N = (1 2 + 2 2 + 3 2 + 2 2 + 1 2 )/6 = 19/6 b. P = E/N = { 2 (3 2 + 2 2 + 1 2 ) }/12 = 28/12 = 7/3 c. sequence: 1, a, a 2 , ... , a N 0 -1 ; E = 1 + a 2 + a 4 + ... a 2(N 0 -1) 2 2 1 1 0 a a N    ; 0 2 N 2 N ) a 1 ( a 1 P 0   

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ECE 351 D. van Alphen Fall 19 7/19 28. Problem 3.2-1 a. E x b. E x c. E x d. K 2 E x 29. Problem 3.2-2 (power only) a. P x b. P x c. P x d. c 2 P x e. P x 30. Problem 3.2-3abc a. horizontal flip or x[n] b. shift x[n] left by 6 c. shift x[n] right by 6 31. Problem 3.3-1 a. (1) n P = 1 b. (-1) n P = 1 c. u[n] P = (1/2) 0 + (1/2) 1 = ½ d. (-1) n u[n] P = (1/2) 0 + (1/2) 1 = ½ e. cos[ (  /3)n +  /6]  =  /3, F = 1/6, N 0 = 6 (period) 32. Problem 3.3-3abc (for n values from 0 to 10) a. u[n-2] – u[n-6] (on at n = 2, off at n = 6) b. n { u[n] – u[n – 7] } : on at n = 0, off at n = 7 0 0.2 0.4 0.6 0.8 1 0 2 4 6 8 10 n x[n] n cos((pi/3)n + pi/6) cos 2 () 0 0.866025404 0.75 1 6.12574E-17 0.00 2 -0.866025404 0.75 3 -0.866025404 0.75 4 -1.83772E-16 0.00 5 0.866025404 0.75 Energy = sum: 3.00 Excel: P = E/N = 1/2

ECE 351 D. van Alphen Fall 19 8/19 c. (n – 2) { u[n – 2] – u[n – 6] } : on at n = 2, off at n = 6 33. Problem 3.3-5ac a. e -.5n = (e -.5 ) n = (.6065) n c. e -j  n = cos(  n) – j sin(  n) = cos(  n) = (-1) n 34. Problem 3.3-6ac a. e -(1+j  )n = (e -1 ) n (e -j  ) n = (1/e) n (cos(  ) – j sin(  )) n = (-1/e) n c. e (1+j  )n = (-e) n 35a. periodic, since  = 2.4  is a rational multiple of  To find period : F =  /2  = 1.2; period N = m/ F (m: smallest integer such that N is integer) 1/ F = non-integer, 2/ F = non-integer, ... , 5/ F = non-integer, 6/ F = 5 = N, the period 35b. not periodic, since  = 2 is not a rational multiple of  35c. not periodic, since  = ½ is not a rational multiple of  35d. periodic, since  = 3/4  is a rational multiple of  To find period : F =  /2  = 3/8; period N = m/ F (m: smallest integer such that N is integer) 1/ F = non-integer, 2/ F = non-integer, 3/ F = 8 = N, the period 35e. not periodic, since  =  2 is not a rational multiple of  36. MATLAB Code for parts a, b, and c: >> n = -8: 8; xa = [zeros(1, 6) -1 0 2 zeros(1, 8)]; 0 2 4 6 8 10 0 2 4 6 8 10 n x[n] 0 2 4 6 0 2 4 6 8 10 n x[n]

ECE 351 D. van Alphen Fall 19 9/19 >> xb = [zeros(1, 9) -1 0 2 zeros(1, 5)]; xc = fliplr(xa); >> subplot(3, 1, 1), stem(n, xa); >> subplot(3, 1, 2), stem(n, xb); >> subplot(3, 1, 3), stem(n, xc); MATLAB code for parts d and e, using the function (unit_step) defined in Lecture 2, and continuing the code from above: >> n = -8:8; >> xd = 4*(unit_step(n) - unit_step(n-4)); >> xe = exp(-.3*n) .* (unit_step(n) - unit_step(n-5)); >> figure >> subplot(2,1,1), stem(n, xd) >> subplot(2,1,2), stem(n, xe) 37a. y[n] = (n + 3) x[n] is causal. True – since the output y[n] = n x[n] + 3 x[n] only depends on the current input, not future input. 37b. y[n – 2] = x[n] is causal. False, since y[n] = x[n+2] depends on future input. -8 -6 -4 -2 0 2 4 6 8 -2 0 2 n x, part a -8 -6 -4 -2 0 2 4 6 8 -2 0 2 n x, part b -8 -6 -4 -2 0 2 4 6 8 -2 0 2 n x, part c -8 -6 -4 -2 0 2 4 6 8 0 1 2 3 4 n x, part d -8 -6 -4 -2 0 2 4 6 8 0 0.2 0.4 0.6 0.8 1 n x, part e

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ECE 351 D. van Alphen Fall 19 10/19 37c. A discrete-time signal with finite power is always a power signal and not an energy signal. False; consider the signal u[n] – u[n-4], with finite (average) power 0, and energy 4, which is an energy signal. 37d. If a discrete-time signal has infinite energy, then it is a power signal. False. Consider 2n u[n], which has infinite energy and infinite power; it is neither an energy signal or a power signal. 37e. u[n] – u[n-1] =  [n-1]; False; it equals  [n]. 38. Problem 3.4-9 c, e Note: y[n] = = .5 x[n] + .5 x[-n] , the average of a signal “x” and the horizontally flipped version of “x”. c. Linear – yes; see lecture 1 or 2, showing that a system that flips the input is linear e. Causal – no; for example, y[-1] = .5 x[-1] + .5 x[1] depends on future input 39. Problem 3.5-1 40. Problem 3.5-2 y[n] - .6 y[n – 1] - .16 y[n-2] = 0, y[-1] = -25, y[-2] = 0 n = 0: y[n] = .6 y[n – 1] + .16 y[n-2] = .6 (-25) + .16 (0) = -15 n = 1: y[n] = .6 y[n – 1] + .16 y[n-2] = .6 (-15) + .16 (-25) = -9 -4 = -13                    k k k n k x k n k x ] [ ] [ ] [ ] [ 5 .  

ECE 351 D. van Alphen Fall 19 11/19 n = 2: y[n] = .6 y[n – 1] + .16 y[n-2] = .6 (-13) + .16 (-15) = -10.2 41. Problem 3.6-1 42. Problem 3.6-2 y[n + 2] + 2 y[n+1] + y[n] = 0, y[-1] = 1, y[-2] = 1 Char. eq: E 2 + 2E + 1 = 0   2 + 2  + 1 = 0  (  +1)(  + 1) = 0,   = -1 (double root) Char. modes: (-1) n , n(-1) n Solution: y[n] = c 1 (-1) n + c 2 n (-1) n ; find c 1 and c 2 with IC’s n = -1: y[-1] = -c 1 + c 2 = 1; n = -2; y[-2] = c 1 - 2c 2 = 1; Thus, y[n] = -3 (-1) n - 2 n (-1) n 43. Problem 3.6-3: y[n+2] – 2 y[n+1] + 2 y[n] = 0; y[-1] = 1, y[-2] = 0 (E 2 – 2E + 2) y[n] = 0;  2 – 2  + 2 = 0  g = 1  j 4 / 2  j e   ) n 4 cos( ) 2 ( c ] n [ y n     ; Use IC’s n = -1: y[-1] = 1 = ) sin 2 1 cos 2 1 ( 2 c ) 4 cos( 2 c        

ECE 351 D. van Alphen Fall 19 12/19 n = -2: y[-2] = 0 =       sin 2 c ) 2 cos( 2 c Solving the 2 equations for c cos  , c sin    c cos  = 2, c sin   c = 2,  = 0 ) n 4 cos( ) 2 ( 2 ] n [ y n   44. Problem 3.6-5 45. Problem 3.7-1 a. (E + 2) y[n] = x[n]; A 0 = b 0 /a 0 = 1/2  h[n] = .5  [n] + y 0 [n] u[n] Char. Eq. to find y 0 [n]:  + 2 = 0      h[n] = .5  [n] + c(-2) n (*) Find IC iteratively, from given diff. eq: h[n+1] = -2h[n] +  [n] Let n = -1: h[0] = -2h[-1] +  [-1] = 0 Return to (*), and let n = 0 to use IC: n = 0: h[0] = .5  [0] + c(-2) 0 = 0  .5 + c = 0  c = -.5  h[n] = .5  [n] - .5(-2) n u[n] b. y[n] + 2 y[n-1] = x[n]; adv. op. form: y[n+1] + 2 y[n] = x[n+1] (E + 2) y[n] = E x[n] A 0 = b 0 /a 0 = 0  h[n] = y 0 [n] u[n] Char. Eq. to find y 0 [n]:  + 2 = 0      h[n] = c(-2) n (**) Find IC iteratively, from given diff. eq: h[n+1] = -2h[n] +  [n+1] Let n = -1: h[0] = -2h[-1] +  [0] = 1 Return to (**), and let n = 0 to use IC:

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ECE 351 D. van Alphen Fall 19 13/19 n = 0: h[0] = c(-2) 0 = 1  c = 1  h[n] = (-2) n u[n] 46. x[n] * y[n] if x[n] = { 1, 1, 1} and y[n] = {1, 1, 1, 1, 1}. 47a. {  [n] + 3  [n – 2] } * sin(3  n/10) = sin(3  n/10) + 3 sin(3  (n-2)/10) 47b. {  [n] + 3  [n – 3] } *  [n + 1] =  [n + 1] + 3  [n - 2] 47c. 2  [n] * { cos(3  n/10) + sin(3  n/10) } = 2 cos(3  n/10) + 2 sin(3  n/10) 48. x[n] = {2, 2, 2} = 2 {  [n] +  [n-1] +  [n-2] } h[n] = {1, 2}  [n] * h[n] = 1 2  [n-1] * h[n] = 1 2  [n-2] * h[n] = 1 2 {1 3 3 2 } Twice Total: {2 6 6 4 } 49. Let x[n] = u[n] – u[n – 2]. Let h[n] = n {u[n – 2] – u[n – 4]} Find y[n] = x[n] * h[n]. x[n] = u[n] – u[n-2] =  [n] +  [n-1] h[n] = n { u[n – 2] – u[n – 4] } = n { 0 0 1 1} = { 0 0 2 3 }  [n] * h[n] = 0 0 2 3  [n-1] * h[n] = 0 0 2 3 {1, 1, 1} * {1, 1, 1, 1, 1} = {  [n| +  [n-1] +  [n-2] } * {1, 1, 1, 1, 1} {1, 1, 1, 1, 1} original seq. {0, 1, 1, 1, 1, 1} shifted one {0, 0, 1, 1, 1, 1, 1} shifted two ans = sum: {1, 2, 3, 3, 3, 2, 1}

ECE 351 D. van Alphen Fall 19 14/19 Total = { 0 0 2 5 3 } 50. x[n] = .3 n u[n], h[n] = .5 n u[n] y[n] = x[n] * h[n] (Using Talbe 3.1, entry 4, which was derived in class) 51. Consider the system with input/output relationship as shown: 51a. Find the impulse response, h[n]. Let x =  , y = h Ans: h[n] = 3  [n] + 2  [n – 3] 51b. Find the output if the input is x[n] = .2 n y[n] = x[n] * h[n] = (.2) n * {3  [n] + 2  [n – 3] } = 3 (.2) n + 2 (.2) n-3 (keep going to combine like terms) = 3 (.2) n + 2 (.2) n (.2) -3 = 3 (.2) n + 2 (.2) n (125) = 3 (.2) n + 250 (.2) n = 253 (.2) n 51c. Is the system causal? Yes; Output y depends only on present and past input 51d. … linear? Yes; proportionality & superposition hold Proportionality: Input ax[n] yields output y[n] = 3 ax[n] + 2 ax[n-3] = a y[n]; Superposition: Input x 1 [n] + x 2 [n] yields y[n] = 3{x 1 [n] + x 2 [n]} + 2{x 1 [n-3] + x 2 [n-3]} = {3x 1 [n] + 2 x 1 [n-3]} + {3 x 2 [n] + 2 x 2 [n-3]} = y 1 [n] + y 2 [n] 51 e. … time-invariant? Yes; constant coefficients 52. Lathi 3.8-1 h[n] = (-2) n u[n – 1]; x[n] = e -n u[n + 1] Recall shift property: If x 1 [n] * x 2 [n] = y[n], then x 1 [n-m] * x 2 [n-p] = y[n-m-p]; Now let p = -m; x 1 [n-m] * x 2 [n+m] = y[n-m+m] = y[n] h[n] x[n] y[n] = 3 x[n] + 2 x[n – 3] ] n [ u ) 3 . 5 (. 5 ] n [ u 5 . 3 . 5 . 3 . 1 n 1 n 1 n 1 n         

ECE 351 D. van Alphen Fall 19 15/19  Advancing h by one and delaying x by one will have no effect! x[n] * h[n] = (-2) n+1 u[n] * (e) -(n-1) u[n] = (-2) (-2) n u[n] * e (e -1 ) n u[n] = (-2e) { (-2) n u[n] * (e -1 ) n u[n] } = (-2e) = (2e 2 ) 53. Lathi 3.8-3 h[n] = [ (2) n-2 + 3(-5) n+2 ] u[n – 1]; x[n] = 3 n+2 u[n + 1] Using shift property, advance h by 1, delay x by 1: x[n] * h[n] = 3 n+1 u[n] * [ (2) n-1 + 3(-5) n+3 ] u[n] = 3  3 n u[n] * [ .5 (2) n + 3(-5) 3 (-5) n u[n] ] = 3/2 [ 3 n u[n] * (2) n u[n] ] - 1125 [3 n u[n] * (-5) n u[n] ] = 3/2 - 1125 = 1.5 {3 n+1 – 2 n+1 } u[n] - (140.635) [3 n+1 – (-5) n+1 ] u[n] x[n] * h[n] = 1.5 {3 n+1 – 2 n+1 } u[n] - (1125/8) [3 n+1 – (-5) n+1 ] u[n] = { -417.375  3 n – 3  2 n – 703.125 (-5) n } u[n] 54. Lathi 3.8-10 y[n + 1] + 2 y[n] = x[n + 1]; y[-1] = 10, x[n] = e -n u[n] Getting the zero-input response: (E + 2) = 0   + 2 = 0   = -2 y 0 [n] = c (-2) n ; y 0 [-1] = 10 = c (-2) -1 = -c/2  c = -20 Zero-input response: y 0 [n] = -20 (-2) n u[n] Zero-state response: e e n n / 1 2 ) / 1 ( ) 2 ( 1 1       1 2 ) / 1 ( ) 2 ( 1 1      e e n n u[n] u[n] ] n [ u ) ) 3679 (. ) 2 (( 296 . 2 1 n 1 n      Table 3.1, Entry 4 ] [ 2 3 2 3 1 1 n u n n     ] [ ) 5 ( 3 ) 5 ( 3 1 1 n u n n      

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ECE 351 D. van Alphen Fall 19 16/19 Finding h[n]: y[n + 1] = - 2 y[n] + x[n + 1] h[n + 1] = -2 h[n] +  [n + 1] ; solve iteratively n = -1: h[0] = 0 + 1; n = 0: h[1] = -2h[0] = -2; n = 1; h[2] = - 2 h[1] = 4; … h[n] = (-2) n u[n] y[n] = x[n] * h[n] = (e -1 ) n u[n] * (-2) n u[n] = ] [ ) 2 3679 (. ) 2 ( ) 3679 (. 1 1 n u n n      Total Response: y[n] = -20 (-2) n u[n] + .1554 (.3679) n u[n] + .8446 (-2) n u[n] = -19.1554 (-2) n u[n] + .1554 (.3679) n u[n] 55. Lathi 3.8-11: Find the zero-state response: h[n] = .5 n u[n], x[n] = 2 n u[n]; y[n] = x[n] * h[n] = 2 n u[n] * .5 n u[n] = x[n] = 2 n-3 u[n] = (1/8) 2 n u[n]; y[n] = Find the zero-state response: h[n] = .5 n u[n] c. x[n] = 2 n u[n-2] = 4 2 n-2 u[n-2] (so we can use the shift property) y[n] = answer a, delayed by 2, and scaled by 4 = ] n [ u ) 3679 . 2 ( ) 2 ( ) 3679 (. 1 n 1 n     ] n [ u } ) 2 ( ) 3679 {(. 4223 . 1 n 1 n      y[n] = (zero-state response) ] n [ u } ) 2 ( 8446 . ) 3679 (. 1554 {. n n    ] n [ u ) 5 . 2 ( 3 2 ] n [ u 2 / 1 2 ) 2 / 1 ( 2 1 n 1 n 1 n 1 n         ] n [ u ) 5 . 2 ( 12 1 ] n [ u ) 5 . 2 ( 3 2 8 1 1 n 1 n 1 n 1 n         (using the answer from part a, and linearity) ] 2 n [ u ) 5 . 2 ( 3 8 1 n 1 n    

ECE 351 D. van Alphen Fall 19 17/19 56. Let x[n] =  [n] to find y[n] = h[n]; note that h 2 simplifies to 4(  [n] +  [n-1]) by rewriting the window in terms of delta functions. Output of h 1 is h 1 , since the input is a delta; Output of h 2 is h 1 *h 2 , since the input is h 1 : h 1 *h 2 = {1, 2, 1} * 4(  [n] +  [n-1]) = {4, 8, 4} + {0, 4, 8, 4} = {4, 12, 12, 4} Output of h 3 is h 1 *h 3 , since the input is h 1 : {1, 2, 1} * (2  [n] - 3  [n-3]) = {2, 4, 2} – {0, 0, 0, 3, 6, 3} = {2, 4, 2, -3, -6, -3} y[n] = output of summer = {6, 16, 14, 1, -6, -3} 57. A certain mobile communications channel is modeled as having impulse response: h[n] =  [n] + .5  [n – 1] + .2  [n – 3] Find the output, y[n], if the input is x[n] = u[n] – u[n – 4] = {1, 1, 1, 1} y[n] = x[n] * h[n] =x[n] * {  [n] + .5  [n – 1] + .2  [n – 3]} = x[n] + .5 x[n-1] + .2 x[n-3] { 1 1 1 1 } (original x[n]) .5 .5 .5 .5 (½ of original, shifted by 1) .2 .2 .2 .2 (.2 times original, shifted by 3) 58. Find the characteristic modes of the system with difference equations: a. y[n] – (5/6) y[n – 1] + (1/6) y[n – 2] = x[n] y[n+2] – (5/6) y[n+1] + (1/6) y[n ] = 0 (advanced form, with right-side = 0)  2 – (5/6)  + (1/6) = 0; (  -1/3)(  -1/2) = 0;  = ½, 1/3 Characteristic modes: (1/2) n , (1/3) n b. y[n + 2] + (2/3) y[n + 1] + (1/9) y[n] = x[n + 1] Characteristic equation:  2 + (2/3)  + (1/9) = 0; (  +1/3)(  + 1/3) = 0; Repeated roots:  = 1/3, 1/3 Characteristic modes: (1/3) n , n(1/3) n Sum: {1 1.5 1.5 1.7 .7 .2 .2}

ECE 351 D. van Alphen Fall 19 18/19 59. Convolve x[n] with y[n]. (Call the output z[n]) a. x[n] = {1, 2, 3}, y[n] =  [n] + 2  [n – 3] = {1, 0, 0, 2} 1 2 3 2 4 6 Ans: z[n] = {1, 2, 3, 2, 4, 6} b. x[n] = {1, 2, 3}, y[n] = u[n] – u[n – 4] = {1 1 1 1} {  [n] + 2  [n-1] + 3  [n-2]} * { 1 1 1 1} Solution: 1 1 1 1 (original) 2 2 2 2 (twice orig., shifted by 1) 3 3 3 3 (3 times orig., shifted by 2) Ans: z[n] ={1, 3, 6, 6, 5, 3} c. x[n] = {1, 2, 3}, y[n] = {1, 2, 3} {  [n] + 2  [n-1] + 3  [n-2]} * { 1 2 3} Solution: 1 2 3 (original) 2 4 6 (twice orig., shifted by 1) 3 6 9 (3 times orig., shifted by 2) Ans: z[n] = {1, 4, 10, 12, 9} d. x[n] = {1, 1, 2, 1}, y[n] = {2, 1, 2} Ans: z[n] = {2, 3, 7, 6, 5, 2 } e. x[n] = 2n u[n], y[n] = 3n u[n] z[n] = x[n] * y[n] = 6 {n u[n] * n u[n]} = 6 (1/6) n (n-1) (n+1) u[n], by table entry 7, p. 291 = n (n-1) (n+1) u[n] 1 2 3 2 4 6 1 3 6 6 5 3 1 4 10 12 9

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ECE 351 D. van Alphen Fall 19 19/19 f. x[n] = y[n] = u[n] - u[n – 4] = {1, 1, 1, 1} Solution: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 {1 2 3 4 3 2 1} Ans: z[n] =