Problem Set Solution 4: Convolution

4 Convolution Solutions to Recommended Problems S4.1 The given input in Figure S4.1-1 can be expressed as linear combinations of xi[n], x 2[n], X3 [n]. x,[ n] 0 2 Figure S4.1-1 = 2x 1 [n] - 2x 2 [n] + x3[n] (b) Using superposition, y 4[n] = 2yi[n] - 2y 2[n] + y3 [n], shown in Figure S4.1-2. (a) x 4 [n] -1 0 1 Figure S4.1-2 (c) The system is not time-invariant because an input xi[n] + xi[n - 1] does not produce an output yi[n] + yi[n - 1]. The input x,[n] + xi[n - 11 is xi[n] + xi[n - 1] = x2[n] (shown in Figure S4.1-3), which we are told produces y 2[n]. Since y 2[n] # yi[n] + yi[n - 1], this system is not time-invariant. x 1 [n] +x 1 [n-1] =x2[n] 0 1 n Figure S4.1-3 S4-1 Signals and Systems S4-2 S4.2 The required convolutions are most easily done graphically by reflecting x[n] about the origin and shifting the reflected signal. (a) By reflecting x[n] about the origin, shifting, multiplying, and adding, we see that y[n] = x[n] * h[n] is as shown in Figure S4.2-1. (b) By reflecting x[n] about the origin, shifting, multiplying, and adding, we see that y[n] = x[n] * h[n] is as shown in Figure S4.2-2. y[n] 3 2 0 1 2 3 4 5 6 Figure S4.2-2 Notice that y[n] is a shifted and scaled version of h[n]. S4.3 (a) It is easiest to perform this convolution graphically. The result is shown in Fig­ ure S4.3-1. Convolution / Solutions S4-3 y(t) = x(t) * h(t) 4-­ t | 4 8 Figure S4.3-1 (b) The convolution can be evaluated by using the convolution formula. The limits can be verified by graphically visualizing the convolution. y(t) e-'-Ou(r - = t+ T - r + 1)dr t> dr, 0, t < 0, -0, = 1)u(t - 1 e (- Let r' r)dr (t - 7x(r)h = 1. Then -e e- d r y( ) 0 0 t > 0, t < 0 , (c) The convolution can be evaluated graphically or by using the convolution formula. y(t) x(r)6(t - = , - 2) dr = x(t - 2) So y(t) is a shifted version of x(t). y(t) It- / \, I 1 3 Figure S4.3-2 t 5 Signals and Systems S4-4 S4.4 (a) Since y[n] = E=-oox[m]h[n - m], b[m 6 - no]h[n - y[n] = m= m] = h[n - no] -oO We note that this is merely a shifted version of h[n]. y [n] = h1[12­ I ae|41 (n 8 n 1) no (n1+ 1) Figure S4.4-1 (b) y[n] = E =_c(!)'u[m]u[n ­ m] For n > 0: 1 y[n] = y[n] = Forn < 0: 2 - + = 2( 1 , (i)" y[n] = 0 Here the identity N-i _ N T am Mr=O 1­ a has been used. y[n] 2--­ 140 0 1 2 Figure S4.4-2 (c) Reversing the role of the system and the input has no effect on the output because y[n] = E m=-oo x[m]h[n ­ m] = L h[m]x[n ­ m] The output and sketch are identical to those in part (b). Convolution / Solutions S4-5 S4.5 (a) (i) Using the formula for convolution, we have y 1 (t) r) dr x(r)h(t - = r(-)-2 u(t = r) dr - t e -( = 2e 2 - t > 0, dr, e 2(1 10 = t > 0, t< 0 y(t) = 0, y 1 (t) - 2 - t 0 Figure S4.5-1 (ii) Using the formula for convolution, we have =4(1 - e-t/2), y 2 (t) 3 2e-(t-r)/2 dr, y2(t) = = { 2e-(-­ /2 3 t>- 0, t : 0, t >_3, d-, 3 S4e (t-r)/2 = 4e- t/ 2 (e'/ y 2(t) = 0, 0 0 2 - t 0 3 4 4(e -(t-3)/2 t 1), y 2 (t) 0 1 2 5 6 Figure S4.5-2 _ 3, e-t/2 Signals and Systems S4-6 (b) Since x 2 (t) = 2[xl(t) - xl(t - 3)] and the system is linear and time-invariant, y 2(t) = 2[yi(t) - y1(t - 3)]. For 0 s t s 3: For 3 y 2(t) = 2yi(t) = 4(1 - e-'/2) y 2(t) = 2y,(t) - 2yi(t - 3) t = 4(1 - 3 = 4e- t/ e / Fort< 0: 4(1 e-1/2 2 2 _ - e- t -3)2) 1 y 2(t) = 0 We see that this result is identical to the result obtained in part (a)(ii). Solutions to Optional Problems S4.6 (a) x(T) 1 T 0 1 Figure S4.6-1 h(-1 -r) 2, -2 T --1 0 --1 ' Figure S4.6-2 Convolution / Solutions S4-7 h(0-r) Figure S4.6-3 h(1 -- ) Figure S4.6-4 Signals and Systems S4-8 Using these curves, we see that since y(t) = x(t) * h(t), y(t) is as shown in Figure S4.6-6. y(t) W 1 2 --1 0 1 --- 4 Figure S4.6-6 (b) f0 Consider y(t) = x(t) * h(t) = x(t - r)h(r)dr. h(r) 2 1 0 1 2 Figure S4.6-7 For 0 < t < 1, only one impulse contributes. x(t -r) Figure S4.6-8 For 1 < t < 2, two impulses contribute. x(t ­ ) Figure S4.6-9 t Convolution / Solutions S4-9 For 2 < t < 3, two impulses contribute. x(t- r) Figure S4.6-10 For 3 < t < 4, one impulse contributes. t) x(t ­ Figure S4.6-11 For t < 0 or t > 4, there is no contribution, so y(t) is as shown in Figure S4.6-12. y(t) 3 -­ 2 I-­ U' 1 2 3 4 Figure S4.6-12 S4.7 y[n] = x[n] * h[n] =1 x[n - m]h[m] nO =anmm, M=0 n > 0, Signals and Systems S4-10 = a" y[n] = a" a n+1 n+1 _ (la) n > 0, # a - L n < 0 y[n] = 0, S4.8 (a) x(t) = E_= - kT) is a series of impulses spaced T apart. x(t) -2T -T 0 T t 2T Figure S4.8-1 - (b) Using the result x(t) *(t to) = x(to), we have y(t) 3 -2 2 -1 0 2 1 2 1 3 2 2 Figure S4.8-2 So y(t) = x(t) * h(t) is as shown in Figure S4.8-3. y(t) 2 -2 -3 -1 22 -j 0 1 2 Figure S4.8-3 1 3 2 2 .. t Convolution / Solutions S4-11 S4.9 (a) False. Counterexample: Let g[n] b[n]. Then = x[n] * {h[n]g[n]} = x[n] {x[n]* h[n]}g[n] h[0], b[n] = [x[n] *h[n]] n=0 and x[n] may in general differ from b[n]. (b) True. y(2t) = fx(2t Let r' = T/ 2 - r)h(r)dr . Then y(2t) = f x(2t - 2r')h(2r')2dr' = 2x(2t)* h(2t) (c) True. y(t) = x(t) * h(t) y(- t) = x(-t) * h(-t) -f x(t - r)h(r) dr =f - + r)h(-r)dr = f x(-t [-x(t - r)][ -h(r)] dT since x(-) and h(-) are odd fu ictions y(t) Hence y( t) = y(-t), and y(t) is even. (d) False. Le t = b(t - x(t) 1) h(t) = b(t + 1) Ev{ y (t)} = 6(t) y(t) = b(t), Then x(t) * Ev{h(t)} =b(t 1) * i[b(t + 1) + b(t - - = i[b(t) + b(t - Ev{x(t)} * h(t) =1[6t = 1[b(t) + b(t + 2)] But since 1[6(t - 2) + b(t + 2)] # 0, Ev{y(t)} # x(t) *Ev{h(t)} + Ev{x(t)} *h(t) S4.10 (a) 9(t) = Jro J 2 (r)22(t TO2 1 - r) dr, O D(t + T0 ) = = TO 1 (T)- 2 (t 1 (r) 2 + To - r) d-r (t - r)dr = 1)] 2)], 1) + b(t + 1)] * b(t + 1) (t) Signals and Systems S4-12 (b) a+TO 9a(t) 2 1 (i)2 T = a = kTo + b, 2 (t -) dr, - where 0 b - To, (k+1)T0 +b 9a(t) = 2 1 (r)A - 2 (t dr, i) kT0+b TO+b Pa(t) = 2 1 (i)± fb FTo = T1()b t - T)TO 21 t(-r = (c) For 0 s t - 9(t ) T7 r) dr + s r) di - 1&)2(t ) 2- di ) )12( t ­ T ) dr t) =q I = ft e- di 0e + + Ti±e-'di I e/2+t1/2+1 + (-e-T 11/2+t) 0 For +b r) d1 - =(-e-' t (t) 0 i - b = i' TO ()A Tb = 2 r) di, - 2 (t = 1 - e~' + e-(*±1/2) - e-1 = 1 - e-1 + (e 1/2 - 1)e- t < 1: e-' di = e- ( ft = t(t) - 1/2) - e­ )e 1/2 (d) Performing the periodic convolution graphically, we obtain the solution as shown in Figure S4.10-1. X1[n] *x2[n] 2 19 0 3 1 -16 Figure S4.10-1 4 5 (one period) Convolution / Solutions S4-13 (e) S4.11 (a) Since y(t) = x(t) *h (t) and x(t) = g(t) *y(t), then g(t) *h(t) = 6(t). But g(t )*h (t) 1=0 wk=0 = T gkhlo(t - (1+ k)T) 1=0 k=O Let n = 1 + k. Then hboTr - kT grob t - = = n - k and Ygkhn g(t)* h (t) = t- n n=0 \k=0 F, n n = 0, n k=0 Therefore, go = 1/h 0 , gi = -- hi/ho, 1 (-hl 2 (b) We are given that h0 = 1, hi = I, hi g1 h2) ho ho o = = 1, = -i) 0. So g2=+(1)2, g2 = -() O lT d, Signals and Systems S4-14 Therefore, g(t) = ( (-) (tkT) k=O (c) (i) Each impulse is delayed by T and scaled by a, so h(t) (t = -kT ) k=O (ii) If 0 < a < 1, a bounded input produces a bounded output because y(t) = x (t)* h (t), | y(t)| < Zak k=0 < a k=O 6(r ­ kT)x(t - r) di f -w - kT)Ix(t _(T - T)I dr -w Let M = maxlx(t)|. Then 1 ak= M I y(t) < M , al < 1 k=O If a > 1, a bounded input will no longer produce a bounded output. For example, consider x(t) = u(t). Then t 00 yt) = Ta Since f 6(,r - kT) dr f k=O -w 6(r - kD di = u (t-kT ), y(t) = ( aku(t-kT) k=O Consider, for example, t equal to (or slightly greater than) NT: N y(NT) = Z ak k=O If a > 1, this grows without bound as N (or t) increases. (iii) Now we want the inverse system. Recognize that we have actually solved this in part (b) of this problem. gi = 1, g2 = -a gi = 0, i # 0, 1 So the system appears as in Figure S4.11. y(t) x(t) _0_ Delay T Figure S4.11 Convolution / Solutions S4-15 (d) If x[n] = 6[n], then y[n] = h[n]. If x[n] = go[n] + iS[n-N], then y[n] = -h[n] + -h[n], y[n] = h[n] S4.12 (a) b[n] = #[n] - -4[n - 1], x[n] = ( x[k][n - k] = ( k=-- x[n] = E x[k]Q([n - k] - -p[n - k - 1]), k=--w ix[k (x[k] - - 1])4[n- k] k= -w So ak = x[k] - lx[k - 1]. (b) If r[n] is the response to #[n], we can use superposition to note that if x[n] = ( akp[n - k], k=­ then y[n] = Z akr[n - k] k= -w and, from part (a), y[n] = (x [k] - ( fx[k - 1])r[n - k] k= ­ (c) y[n] = i/[n] * x[n] * r[n] when [n] = b[n] - in- 1] -[- 1] and, from above, 3[n] 4[n] - So /[n] = #[n] - -#[n - 11 - 1(#[n - 11 0[n] = *[n] - *[n - 1] + {1[n - 2] (d) 4[n] #[n - b[n] = 1] - -$[n - 2]), r[n], r[n - 1], 4[n] - -1[n- 1] - r[n] - fr[n -1] So h[n] = r[n] - tr[n -1], where h[n] is the impulse response. Also, from part (c) we know that y[n] = Q[n] *x[n] *r[n] and if x[n] = #[n] produces r[n], it is apparent that #[n] * 4[n] = 6[n].