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Hint: Write x(n) as (1 + (-1)") u(n) and use linearity.
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Consider a system with input x(n) and output y(n) that satisfies the difference equation y(n) = ny(n - I) If x(n) = 6(n), determine y(n) for all n.
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+ x(n)
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A linear shift-invariant system is described by the LCCDE
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Find the step response of the system (i.e., the response to the input x(n) = u(n)).
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A system is characterized by the difference equation
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If the input is x(n) = 2u(n) - 3nu(n), find the response of the system assuming initial conditions of y(-1) = 2 and y(-2) = 1.
Consider the system described by the difference equation y(n) - y(n
- 1) + 0.25y(n - 2) = x(n) - 0.25x(n - 1)
(a) Find the unit sample response of the system. (b) Find the response of the system to x(n) = (0.25)"u(n).
For a savings account that pays interest at the rate of I percent per month, if deposits are made on the first of each month at the rate of $50 per month, how much money will there be in the account at the end of 1 year
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Answers to Supplementary Problems
(a) If x(n) = x(n
+ N), by shift-invariance, y(n) = y(n + N). Therefore, y(n) is periodic with period N.
(b) No. (c) Yes.
CHAP. 11
SIGNALS AND SYSTEMS
1.45 1.46
Even. (a) is true.
(a) Linear, shift-varying, stable, noncausal, noninvertible. (6) Linear, shift-varying, unstable, noncausal, invertible. ( c ) Linear, shift-invariant, stable. noncausal, invertible. (d) Nonlinear, shift-invariant, stable, causal, invertible. ( 0 ) Nonlinear, shift-invariant, stable, noncausal, noninvertible. (a) la1
1. (6) Any finite a . ( c ) la[ < 1.
1.50 1.51
No. Consider the system y(n) = x(n)cos(nr/2).
1.53 1.54
(a) Yes. (6) No. ( c ) No. The sequence values. beginning at index n = -I, are y(n) = {2,0, -4, 1,6,0, I , -1, -2,6,3).
(a) s(n) =
k=-cc
h(k). With h(n) = u(n) - u(n
- 6)
the step response is
(6) h(n) = s(n) - s(n - 1). If s(n) = (-0.5)nu(n), then h(n) = S(n) + 3(-0.5)"u(n - 1).
(a) y(n) = S(n - 2)
+ 2S(n - 3) - 2S(n
- 5) - S(n
- 6).
( b ) y(n) = S(n
+ 2) - 26(n) + S(n - 2).
bn+l - an+l y(n) = - u(n).
h(n) = S(n) - S(n - I).
y(n) = [(4n
SIGNALS AND SYSTEMS
[CHAP. 1
+ 3)(-i)"
3]u(n).
y(-2) = 17 - j n s * max(y(n)]= y(n) = 4 n ) . Y @ )= u(n - 4 ) y(n) = [4(-1)" y(n) = n! u(n). ~ ( n= )
y , which occurs at index n = 8.
+ f (n - 2)(n - 1) u(n - 2).
- !(-f)"]u(n).
[i + (;)(3)"
- 2(2")]u(n).
~ ( n= [lj .[4" 11 - 4n )
+ 12(2)"]u(n).
h(n) = [ i n
+ ~ ] ( i ) ~ u ( (nb)). y(n) = (n + I ) ( f ) n u ( n ) .
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2
Fourier Analysis
2.1 INTRODUCTION
The Fourier representation of signals plays an extremely important role in both continuous-time and discrete-time signal processing. It provides a method for mapping signals into another "domain" in which to manipulate them. What makes the Fourier representation particularly useful is the property that the convolution operation is mapped to multiplication. In addition, the Fourier transform provides a different way to interpret signals and systems. In this chapter we will develop the discrete-time Fourier transform (i.e., a Fourier transform for discrete-time signals). We will show how complex exponentials are eigenfunctions of linear shift-invariant (LSI) systems and how this property leads to the notion of a frequency response representation of LSI systems. Finally, we will explore how the discrete-time Fourier transform may be used to solve linear constant-coefficient difference equations and perform convolutions.
2.2 FREQUENCY RESPONSE
Eigenfunctions of linear shift-invariant systems are sequences that, when input to the system, pass through with only a change in (complex) amplitude. That is to say, if the input is x(n), the output is y(n) = kx(n), where A, the eigenvalue, generally depends on the input x(n).
Signals of the form ~ ( n = ejnw )
-00 n<ca
where w is a constant, are eigenfunctions of LSI systems. This may be shown from the convolution sum:
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