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Unformatted text preview: ∑ x[m] . m =−∞ Taking the first backward difference of both sides of the original system equation,
y[ n ] − y[ n − 1] = n +1 n +1−1 m =−∞ m =−∞ ∑ x[m] − ∑ x[m] x[ n + 1] = y[ n ] − y[ n − 1]
The excitation is uniquely determined by the response.
Invertible.
40. A DT system is described by
n y[ n ] − 8 y[ n − 1] = x[ n ] .
Classify this system as to time invariance, BIBO stability and invertibility.
Stability:
The homogeneous equation is
or n y[ n ] = 8 y[ n − 1]
y[ n ] = 8
y[ n − 1] .
n Thus, as n increases without bound, y[ n ] must be decreasing because it is
previous value and 8
times its
n 8
approaches zero. Rearranging the original equation,
n
y[ n ] = x[ n ] 8
+ y[ n − 1] .
n
n x[ n ]
8
must be bounded and y[ n − 1]
n
n
must be getting smaller because it is a decreasing fraction of its previous value. Therefore
for a bounded excitation, the response is bounded.
Stable. For any bounded excitation, x[ n ] , as n gets larger, Solutions 329 M. J. Roberts  8/16/04 41. A DT system is described by y[ n ] = x[ n ] . Classify this system as to linearity, BIBO stability, memory and invertibility.
Invertibility:
Inverting the functional relationship,
Invertible. x[ n ] = y 2 [ n ] . 42. Graph the magnitude and phase of the complexsinusoidal response of the system
described by
1
y[ n ] + y[ n − 1] = e − jΩn
2
as a function of Ω.
This is the steadystate solution so all we need is the particular solution of the
difference equation. The equation can be written as
n
1
y[ n ] + y[ n − 1] = (e − jΩ ) = α n
2 where α = e − jΩ The particular solution has the form, y[ n ] = Kα n . 43. Find the impulse response, h[ n ], of the system in Figure E43. x[n] y[n] 2
0.9 D Figure E43 DT system block diagram
or y[ n ] = 2 x[ n ] + 0.9 y[ n − 1]
y[ n ] − 0.9 y[ n − 1] = 2 x[ n ] The homogeneous solution (for n ≥ 0) is of the form,
y[ n ] = K hα n
therefore the characteristic equation is
K hα n − 0.9K hα n −1 = 0 . Solutions 330 M. J. Roberts  8/16/04 and the eigenvalue is α = 0.9 and, therefore, y[ n ] = K h (0.9) n We can find an initial condition to evaluate the constant, K h , by directly solving the
difference equation for n = 0.
y[0] = 2 x[0] + 0.9 y[−1] = 2 .
Therefore 2 = K h (0.9) ⇒ K h = 2 .
0 Therefore the total solution is y[ n ] = 2(0.9) n which is the impulse response.
44. Find the impulse responses of these systems.
(a) 3 y[ n ] + 4 y[ n − 1] + y[ n − 2] = x[ n ] + x[ n − 1] (b) 5
y[ n ] + 6 y[ n − 1] + 10 y[ n − 2] = x[ n ]
2 45. Plot g[ n ]. Use the MATLAB conv function if needed.
(a) 2πn g[ n ] = rect1[ n ] ∗ sin 9
Write out the convolution sum. Then use
sin ( x + y ) = sin ( x ) cos ( y ) + cos ( x ) sin ( y )
write out the entire summation and simplify what you get. You should ultimately get 2π n g [ n ] = 2.5321sin 9 (c) 2πn g[ n ] = rect 2 [ n ] ∗ sin 9
g[n] = 0 (d) g[ n ] = rect 3 [ n ] ∗ rect 3 [ n ] ∗ comb14 [ n ] (b) First convolve the two rectangles. Then convolve the result with the comb, thereby
periodically repeating it.
(e) Similar to (d) but with a different result. Solutions 331 M. J. Roberts  8/16/04 2π n 7 g [ n ] = 2 cos ∗
u[n] 7 8 n (f) Write the convolution sum. Express the cos in exponential form and combine with
other terms. Then use
∞
1
∑ rn = 1 − r , r < 1
n=0
to put the result in closed form, and simplify using
cos ( x + y ) = cos ( x ) cos ( y ) − sin ( x ) sin ( y )
Finally use B
A cos ( x ) +...
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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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