Chap3StudentSolutions

J roberts 81604 h n 4 j 35 n e j 2214 n 4

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Unformatted text preview: , the impulse response of this system is the same as the impulse response of the system, y [ n ] = x [ n ] minus the impulse response of the system, y [ n ] = x [ n − 1] . The impulse response of the first system is h1 [ n ] = δ [ n ] and the impulse response of the second system is exactly the same except delayed by 1 in discrete time or h1 [ n ] = δ [ n − 1] . The overall impulse resopnse is therefore h [ n ] = h1 [ n ] − h 2 [ n ] = δ [ n ] − δ [ n − 1] , as before. (b) 25 y[ n ] + 6 y[ n − 1] + y[ n − 2] = x[ n ] The homogeneous solution is −3 − j 4 −3 + j 4 y h [ n ] = K1 + K2 25 25 n n and, after discrete-time, n = 0, this is the total solution because the excitation is zero. The first two values of the impulse response are (by direct iteration), y[0] = 1 6 and y[1] = − . 25 625 Solving for the constants, 1 4 + j3 = K1 + K 2 K1 = 25 200 ⇒ 6 4 − j3 −3 − j 4 −3 + j 4 − = K1 K2 = + K2 25 25 625 200 Then the impulse response is 4 + j 3 −3 + j 4 4 − j 3 −3 − j 4 h[ n ] = + 25 200 200 25 n h[ n ] n (4 + j 3)(−3 + j 4) n + (4 − j 3)(−3 − j 4) n = 200(25) n Solutions 3-11 M. J. Roberts - 8/16/04 h[ n ] (4 + j 3)5 n e j 2.214 n + (4 − j 3)5 n e − j 2.214 n = h[ n ] = 200(25) n 4 (e j 2.214 n + e − j 2.214 n ) + j 3(e j 2.214 n − e − j 2.214 n ) = 200(5) n 4 cos(2.214 n ) − 3 sin(2.214 n ) n 100(5) Then, using B A 2 + B 2 cos x − tan −1 A A cos( x ) + B sin( x ) = h[ n ] = cos(2.214 n + 0.644 ) n 20(5) (c) 4 y[ n ] − 5 y[ n − 1] + y[ n − 2] = x[ n ] (d) 2 y[ n ] + 6 y[ n − 2] = x[ n ] − x[ n − 2] The impulse response is the difference of the response, h1[ n ] to a unit impulse at time, n = 0, and the response, h 2 [ n ], to a unit impulse at time, n = 2. 14. Sketch g[ n ]. To the extent possible find analytical solutions. Where possible, compare analytical solutions with the results of using the MATLAB command, conv, to do the convolution. (a) g [ n ] = u [ n ] ∗ u [ n ] = (b) ∞ ∑ m = −∞ u[ m]u[n − m] = g[ n ] = u[ n + 2] ∗ rect 3 [ n ] = ∞ ∑ u[n − m] = m=0 0 ∑ u [ m − n ] = ramp [ n + 1] m = −∞ ∞ ∞ ∑ u[m + 2]rect [n − m] = ∑ rect [n − m] m =−∞ 3 m =−2 ∞ 2 −∞ (c) −2 3 g[ n ] = rect 2 [ n ] ∗ rect 2 [ n ] = ∑ rect 2 [ m] rect 2 [ n − m] = ∑ rect 2 [ n − m] (d) g[ n ] = rect 2 [ n ] ∗ rect 4 [ n ] (e) 3 g [ n ] = 3δ [ n − 4 ] ∗ u [ n ] 4 n Using Aδ [ n − n0 ] ∗ g [ n ] = A g [ n − n0 ] Solutions 3-12 M. J. Roberts - 8/16/04 3 g [ n ] = 3 4 n− 4 u[n − 4 ] n 7 (f) g [ n ] = 2 rect 4 [ n ] ∗ u [ n ] 8 (g) g[ n ] = rect 3 [ n ] ∗ comb14 [ n ] = rect 3 [ n ] ∗ ∞ ∞ m =−∞ m =−∞ ∑ δ[n − 14 m] = ∑ rect [n − 14 m] 3 15. Given the excitations, x[ n ] , and the impulse responses, h[ n ], find closed-form expressions for and plot the system responses, y[ n ] . (a) x[ n ] = e j 2πn 32 h[ n ] = (0.95) u( n ) n , y[ n ] = h[ n ] ∗ x[ n ] = y[ n ] = n ∑e 2πm j 32 ∞ ∑ e Making the change of variable, q = − m j 2π e 32 n y[ n ] = (0.95) ∑ − q =−∞ 0.95 ∞ ∑ rn = n =k j 2π e 32 n = (0.95) ∑ m =−∞ 0.95 m n 2π −j = (0.95) ∑ 0.95e 32 q =− n n ∞ q rk ,...
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