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# 数字信号处理代写 | EE 453 EXAM #1 Fundamentals of Digital Signal Processing

### 数字信号处理代写 | EE 453 EXAM #1 Fundamentals of Digital Signal Processing

1.
Consider the sampling and reconstruction system shown above. This system achieves lter-
ing of a continuous-time signal by rst sampling, then performing a ltering operation in the
discrete-time domain, before converting the ltered discrete-time signal back to continuous-time.
Assume that the sampling is ideal and that the reconstruction lter is also ideal. Suppose that
the discrete-time system is represented by the non-recursive di erence equation
y[n] = x[n] x[n 3]
(a) (10 points) Find the lter’s system function H(z)
Solution:
Y (z) = X(z)(1 z 3
)
H(z) = Y (z)
X(z)
= 1 z 3
H(z) = 1 z 3

(b) (5 points) Identify and sketch the poles and zeros of H(z) in the complex plane
Solution: There are zeros at the cubed roots of unity zk = f0;
2
3 ;
4
3 g

(c) (10 points) Find an expression for the magnitude response jH(ej!)j from ! = [ ; ].
Leave your answer in terms of sine or cosine.
Solution: First we must calculate the frequency response
H(ej!) = 1 e j3!
= ej3!=2
2j
ej3!=2
e j3!=2
2j
= ej3!=2
2j sin(3!=2)
jH(ej!)j = j2 sin(3!=2)j
(d) (10 points) Find the system output y(t) if the input is x(t) = cos(21250t + 
3 ) and the
sample rate Fs = 10000. You DO NOT need to simplify any sine or cosine terms (So for
instance 2 cos(

6 ) does not need to be simpli ed to 1:732)
Solution: If the continuous frequency
= 21250t, and the sampling rate Fs =
10; 000, then the discrete frequency ! = 21250
10000 = 
4 . jH(ej!)j!==4 = 2 sin(
3
8 ), while
\H(ej!)j!==4 = 3
8 + 
2 = 7
8 . The nal answer is then
y(t) = 2 sin(
3
8
) cos(2  1250t + 
3
+ 7
8
)