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Introduction to Digital Signal Processing
Test 1
Fall 2006
時間： 2006 年 11 月 7 日，140 分鐘 (2:30 – 4:50 PM)
答案卷上請標示清楚題號與頁碼，勿忘記寫上姓名學號。
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1. (5%) Let x[n] = [n+1]+2[n]+ 3[ n1]+ 2[ n2]+ [ n3], and y[n] =
2[n+1]+[n]5[ n1] [ n2]+ 3[ n3]. Find their convolution x[n]y[n].
2. To prove the property that time-domain convolution implies z-domain
multiplication, let us consider x[n] and y[n] (  n  ) that are two
discrete-time signals. Their z-transforms are as follows:
X(z) = … + z3x[3] + z2x[2] + zx[1] + x[0] + z1x[1] + z2x[2] + z3x[3] + …
Y(z) = … + z3y[3] + z2y[2] + zy[1] + y[0] + z1y[1] + z2y[2] + z3y[3] + …
The multiplication of X(z) and Y(z) results in a new polynomial W(z) = X(z)Y(z).
Let w[n] be the coefficients of this polynomial.
W(z) =… + z3w[3] + z2w[2] + zw[1] + w[0] + z1w[1] + z2w[2] + z3w[3] +…
(a) (5%) Write down w[1], w[0], w[1], and w[2] by multiplying X(z) and Y(z).
(b) (5%) Show that, in general, w[n] is the convolution of x[n] and y[n].
3. Suppose that three LTI systems are connected in cascade; i.e., the output of S1 is
the input of S2, and the output of S2 is the input of S3. The three systems are
specified as follows:
S1: y1[n] = x1[n]  x1[n1],
S2: y2[n] = x2[n]  2x2[n1] + x2[n2],
S3: y3[n] = x3[n1] + x3[n2],
where the output of Si is yi[n] and its input is xi[n].
(a) (10%) Consider the equivalent system that is a single operation from the input
x[n] (into S1) to the output y[n] (the output of S3). Thus, x[n] is x1[n] and y[n]
is y3[n]. Write down the system function (in z-transform) of the equivalent
system.
(b) (5%) What is the region of convergence (ROC) of the system function (except
(c)
(d)
(e)
(f)
(g)
possible z = 0 and z =  )?
(5%) Write down the impulse response of this system.
(5%) Is this system FIR or IIR? Explain your answer.
(5%) Write down the frequency response of this system.
(5%) Is this a causal system? Explain your answer.
(5%) Is this system stable? Explain your answer.
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4. An LTI system is described by the difference equation
y[n] = 0.8y[n1]  0.8x[n] + x[n1]
(a) (5%) Determine the system function H(z) for this system.
(b) (5%) Plot the poles, zeros, and ROC of H(z) on the z-plane.
(c) (5%) Is the system stable?
(d) (5%) If the input is x[n] = cos[(/4)n], calculate the DTFT Y(ejw) of the output.
(hint: decompose cos[(/4)n] into eigenfunctions of an LTI system).
(e) (5%) If the input is x[n] = 0.5nu[n], calculate the output y[n] by using the inverse
z-transform.
5. Consider the DFT and IDFT that are defined as the following transform pair:
1 N 1
forward DFT:
Y [k ] 
x[n]WNkn ,
k  0,1..., N  1

N n 0
inverse DFT (IDFT):
x[n] 
1
N 1
 Y [k ]W
N
k 0
 kn
N
,
n  0,1..., N  1
where WN = ej2/N.
(a) (5%) Let x be the N-dimensional vector consisting of the time domain sequence, x
= [x[0], x[1], …, x[N1]]T and Y be the vector consisting of the frequency
components, Y = [Y[0], Y[1], …, Y[N1]]T (where T is the matrix transpose). We
can represent the forward transform as Y = Wx, where W is an NN matrix.
Similarly, we can also represent the inverse transform as x = AY, where A is an
NN matrix. Question: Please write down the NN matrices W and A.
(b) (3%) Show that A = W*T, where W* is the matrix obtained by replacing every
element in W by its complex conjugate.
(c) (7%) The matrix WH  W*T is usually called the conjugate-transpose matrix of W.
A matrix W satisfying WHW = I is called a unitary matrix, where I is the identity
matrix,
1 0   0
0 1
0

I  




 

0 0   1
Question: Prove that W is a unitary matrix by showing that the multiplication of
A and W is identity, i.e., I = AW = WHW (or equivalently, A=W1= WH).
(hint: You may need to use the sum of a geometric sequence, 1 + r + r2 + … + rN1
= (1 rN)/(1 r).)
(continue)
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(d) (5%) Show that it also holds that WWH = I, and use this property to prove the
Parseval theorem ||Y||2 = ||x||2, i.e., the lengths of the vectors Y and x are the same.
(hint: You may need to use some properties about conjugate-transpose matrix:
Let v be a vector with complex values, then ||v||2 = vHv.
Given two squared matrices B and C, (BC)H = CHBH and (BH) H = B).