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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 discretetime, 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 311 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 312 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 closedform
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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This note was uploaded on 06/19/2013 for the course ENSC 380 taught by Professor Atousa during the Spring '09 term at Simon Fraser.
 Spring '09
 ATOUSA

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