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Tutorial II 1. Evaluate the following discrete-time convolution sums: (i) y[n] = u[n+3] * u[n-3] (ii) 3nu[-n+3] * u[n-2] Tutorial II EKT 230 Signals & Systems Tutorial II EKT 230 Signals & Systems (iii) y[n] = (1/4)n u[n] * u[n+2] for n+2 < 0, n < -2 y[n] = 0 for n+2 >= 0, n >= -2 k 1 y[n] u[k ].u[n k 2] k 4 n2 1 k 0 4 k n 2 1 1 1 4 1 1 4 Tutorial II Tutorial II EKT 230 Signals & Systems n 4 1 1 1 3 4 64 Thus, 4 1 n 1 1 y[n] 3 4 64 , n 2 0, n 2 2. Evaluate the convolution sum for a system with the input signal, x[n] and impulse response, h[n] as in Figure 1. Draw the response y[n]. Use graphical and cross multiplication methods. Cross multiplication method x[n] h[n] 1 2 3 -1 -1 -2 -3 -1 -1 -2 -3 2 2 4 6 1 1 2 3 1 1 2 3 Ans : y[n] = [-1 -3 -3 2 9 5 3] Tutorial II Tutorial II EKT 230 Signals & Systems h[-k] x[k] 1 0 n=0 2 1 1 3 2 3 4 5 h[0-k] 1 -2 2 -1 0 -1 2 1 -3 1 1 -1 y[0] = -1 -3 -2 -1 0 -1 n=1 1 -1 y[1] = -2-1 = -3 h[1-k] 1 -3 2 1 -2 -1 0 1 -1 n=2 y[2] = -3-2+2 = -3 h[2-k] 1 -3 n=3 -1 2 1 -2 -1 0 1 -1 h[3-k] 1 -3 1 -2 2 -1 0 1 -1 n=4 -1 h[4-k] 1 -3 -2 1 y[4] = 6+2+1 =9 2 -1 0 -1 Tutorial II y[3] = -3+4+1 =2 -1 1 -1 Tutorial II EKT 230 Signals & Systems h[k] x[k] 1 0 n=5 2 1 3 2 3 4 6 -3 2 1 -2 -1 0 -1 y[5] = 3+2 =5 1 -1 h[6-k] y[6] = 3 1 -3 2 1 -2 -1 0 1 -1 n=7 -1 y[7] = 0 h[7-k] 1 -3 -2 1 2 -1 0 -1 y[n]=[-1 -3 -3 2 9 5 3] Tutorial II 7 h[5-k] 1 n=6 5 1 -1 Tutorial II EKT 230 Signals & Systems 3. Compute the convolution y[n] = x[n] * h[n] of the following signals. Draw the output signal, y[n]. ¼ for n = 0 ½ for n = 1 ¾ for n = 2 x[n] = 1 0 for n = 3 elsewhere h[n] = 1 1 0 1 1 0 for n = 2 for n = 3 for n = 4 for n = 5 for n = 6 elsewhere y[n] = x[n] * h[n] = x[k ]h[n k ] k = x[0]h[n] + x[1]h[n-1] + x[2]h[n-2] + x[3]h[n-3] = (1/4)h[n] + (1/2)h[n-1] + (3/4)h[n-2] + h[n-3] Thus, y[2] = ¼ y[3] = ¼ + ½ = ¾ y[4] = ½ + ¾ = 5/4 y[5] = ¼ + ¾ + 1 = 2 y[6] = ¼ + ½ + 1 = 7/4 y[7] = ½ + ¾ = 5/4 y[8] = ¾ + 1 = 7/4 y[9] = 1 y[10] = 0 Tutorial II Tutorial II EKT 230 Signals & Systems 2 y[n] 7/4 5/4 7/4 5/4 1 3/4 1/4 Or 0 1 2 3 4 5 6 7 8 9 10 x[n] h[n] ¼ ½ ¾ 1 1 ¼ ½ ¾ 1 1 ¼ ½ ¾ 1 0 0 0 0 0 1 ¼ ½ ¾ 1 1 ¼ ½ ¾ 1 y[n] = [ ¼ ¾ 5/4 2 7/4 5/4 7/4 1] n=2 4. . A linear time invariant system has an impulse response, h(t) and input signal, x(t). Use convolution to find the response, y(t) for following signal: Flip. Tutorial II Tutorial II For t < 1, For 1=< t < 2, EKT 230 Signals & Systems y (t ) 0 t 1 y (t ) (2)e ( ) d 0 2e ( ) t 1 0 2e (t 1) 2 For 2=< t < 3, t 2 y (t ) ( 2) e ( ) d (2)e ( ) d t 2 0 2e t 1 ( ) t 2 1 2e ( ) t 2e ( t 2 ) 2e (1) 2e ( t ) 2e ( 2 ) 2 4e ( t 2 ) 2e ( t 1) 2 For 3=< t < 4, t 2 y (t ) ( ) ( 2) e d t 3 2e ( ) t 2 t 3 2 ( 2 ) e ( ) 2e ( ) 2 t 2 4 e ( t 2 ) 2 e ( t 3 ) 2 Tutorial II d t 2 2 e ( t 2 ) 2 e ( t 3 ) 2 e ( 2 ) 2 e ( t 2 ) Tutorial II EKT 230 Signals & Systems For 4=< t < 5, 2 y (t ) ( 2) e ( ) d t 3 2e ( ) For 5 < t, 2 t 3 2e ( 2 ) 2e ( t 3) y (t ) 0 0 2 2e (t 1) (t 2 ) 2e (t 1) 2 4e y (t ) (t 2 ) 2e (t 3) 2e 2 4e 2e (t 3) 2e 2 0 t 1 1 t 2 2t 3 3t 4 4t 5 t 5 5. A linear time invariant system has an impulse response, h(t) and input signal, x(t). Use convolution to find the response, y(t) for following signals: x(t ) cos(t )u (t 1) u (t 3) h(t ) u (t ) Convolution, y(t) = x(t)*h(t) Tutorial II Tutorial II For t < -1, y(t) = 0. For -1 =< t<0 EKT 230 Signals & Systems t y (t ) (1) cos( )d 1 For -1 =< t<0 Tutorial II sin( ) sin( t ) t 1 sin( t ) sin( ) Tutorial II EKT 230 Signals & Systems 3 y (t ) (1) cos( )d 1 sin( ) 3 1 sin( 3 ) 0 sin( t ) , 1 t 1 y (t ) 0, otherwise 6. Evaluate the following convolution integral: m(t) = y(t) * z(t) Tutorial II sin( ) Tutorial II Tutorial II EKT 230 Signals & Systems