EE-210. Signals and Systems

Homework 7 Solutions∗

Spring 2010

Exercise Due Date

11th May.

Problems

Q1 Let H1 be the causal system described by the difference equation

w[n] =712

w[n− 1]− 112

w[n− 2] + x[n− 1]− 12x[n− 2]

x[n] w[n] y[n]

H1 H2

Figure 1: Q1

(a) Determine the system H2 Fig. 1 so that y[n] = x[n]. Is the inverse system H2 causal?

Solution H2(z) = 1H1(z) . Not causal.

(b) Determine the system H2 Fig. 1 so that y[n] = x[n − 1]. Is the inverse system H2

causal?

Solution H2(z) = z−1

H1(z) . Causal.

(c) Determine the difference equation for system H2 in part (a) and (b)

Solution i. y[n] =12y[n− 1] + w[n + 1]− 7

12w[n] +

112

w[n− 1]

ii. y[n] =12y[n− 1] + w[n]− 7

12w[n− 1] +

112

w[n− 2]

Q2 Let y(k) = sin (ωkT ), determine a so that y satisfies the difference equation

y(k)− ay(k − 1) + y(k − 2) = 0∗LUMS School of Science & Engineering, Lahore, Pakistan.

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Solution if ω = 0 and/or T = 0, a can take any value. Otherwise, consider the characteristicequation.

z2 − az + 1 =⇒ z =a

2±

√a2

4− 1 =

a

2± j

√1− a2

4= e±jωT = cos ωT ± j sinωT

We geta = 2cosωT

Q3 (a) Determine the circular convolution between x[n] = {1̂, 1, 0, 0} and y[n] = {1̂, 1, 1, 1}for N=4. Where α̂ represents the value of signal at n = 0 Verify the result by using4-point DFT and IDFT.

Solution r[n] = {2̂, 2, 2, 2}(b) If you want to calculate the linear convolution of x[n] = {1, 1} and y[n] = {1, 1, 1}

using the fast Fourier transform (FFT). What is required minimum number of datapoints N in the FFT calculation?

Solution N=4

Q4 Determine all possible signals x[n] and corresponding ROC associated with the two-sidedz-transform

X(z) =5z−1

(1− 2z−1)(3− z−1)

Solution Partial fraction expansion gives

X(z) =5z−1

(1− 2z−1)(3− z−1)=

11− 2z−1︸ ︷︷ ︸

X1(z)

+−1

1− 13z−1︸ ︷︷ ︸

X2(z)

X1(z) ⇒

x11 = 2nu(n), ROC11 = |z| > 2,

x12 = −2nu(−n− 1), ROC12 = |z| < 2,

X2(z) ⇒

x21 = −(13)nu(n), ROC21 = |z| > 1

3,

x21 = (13)nu(−n− 1), ROC22 = |z| < 1

3,

x[n] is given by

x[n] = x1i[n] + x2j [n], ROC = ROC1i ∩ROC2j

Combinations with non-empty ROC are:

x[n] = [2n − (1/3)n]u[n], ROC = |z| > 2,

x[n] = −2nu[−n− 1]− (1/3)nu[n], ROC = 1/3 < |z| < 2,

x[n] = [−2n + (1/3)n]u[−n− 1], ROC = |z| < 1/3.

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Q5 (a) The transfer function of a filter is

H(z) =1

z + 4

and is valid of |z| < 4. Is the filter

i. causal? - NOii. stable? - YES

(b) Find the stable impulse response of a system with the transfer function

H(z) =1

(z − 4)(z − 0.1)

Also, calculate the ROC where the expression is valid.

Solution

H(z) =1

(z − 4)(z − 0.1)=

1039

[1

z − 4− 1

z − 0.1]

=1039

[−1/4

1− z/4− z−1

1− 0.1z−1]

= − 10156

∞∑k=0

(z/4)k − 1039

z−1∞∑

k=0

(0.1z−1)k

= − 10156

0∑k=−∞

(4)kz−k − 10039

∞∑k=1

(0.1)kz−k

But H(z)=∑∞

k=−∞ h[k]z−k, therefore

h[k] ={

−(10/156)4k k ≤ 0−(100/39)0.1k k > 0

Region of Convergence: 0.1 < |z| < 4

Q6 Let x[n] and y[n] be two sequences with

x[n] = 0 for n < 0, n ≥ 8y[n] = 0 for n < 0, n ≥ 20

A 20-point DFT is performed on x[n] and y[n]. The two DFT’s are multiplied and aninverse DFT is performed resulting in new sequence r[n].

(a) Which elements of r[n] correspond to a linear convolution of x[n] and y[n]?

Solution The elements r[n], n = 0, . . . , 6 will be incorrect. The elements r[n], n = 7, . . . , 19 willbe correct

(b) How should the procedure be changed so that all elements of r[n] correspond to linearconvolution of x[n] and y[n]?

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Solution The error is caused by the 7 last values of y. This issue can be resolved by increasingthe length of the sequences and the length of the DFTs to 27 by adding zeros.

Q7 A signal is fed to a system that down-sample the input signal by factor D. The input andoutput are related by the equation

y[n] = {. . . , x̂[0], x[D], x[2D], x[3D], x[4D], . . .} = x[nD] n = 0,±1,±2,±3, . . .

(a) Find the DTFT of y[n]

Solution First, define the signal

x̃[k] ={

x[n] n = 0,±D,±2D0 otherwise

which contains only the samples that will be left in the downsampled signal. Then,

Y (f) =∞∑

m=−∞y(m)e−j2πfm

where,

x̃[k] = y(m) if n = Dm

x̃[k] = 0 otherwise

Y (f) =∞∑

m=−∞x̃[n]e−j2πfn/D = X̃(

f

D) (1)

In order to find an expression for X̃(f), define the selection function

s[n] ={

1 n = 0,±D,±2D0 otherwise

and note that s[n] can be written in the form

s[n] =1D

D−1∑k=0

ej2πkn/D

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Since x̃[n] = s[n]x[n], we get

X̃(f) =∞∑

n=−∞x̃[n]e−j2πnf

=∞∑

n=−∞s[n]x[n]e−j2πnf

=1D

∞∑n=−∞

D−1∑n=0

ej2πkn/Dx[n]e−j2πfn

=1D

D−1∑n=0

∞∑n=−∞

x[n]e−j2πn(f−k/D)

=1D

D−1∑n=0

X(f − k

D) (2)

Putting eq. 2 in eq. 1

Y (f) =1D

D−1∑n=0

X(f − k

D)

(b) Find the z-transform of y[n]

Solution

Y (f) =1D

D−1∑n=0

X(z1/De−j2πk/D)

(c) By removing samples in down-sampling, some information is lost. This loss of infor-mation will lead to aliasing problems. Find H(f) so that aliasing can be avoided.

x[n] y[n]

H(f) ↓ D

Figure 2: Q8

Solution

H(f) ={

1 , |f | = 1/2D0 , 1/2D < |f | ≤ 1/2

=⇒

Y (f) =1D

X(f

D) |f | ≤ 1

2

HINT DTFT: X(f) =∑∞

n=−∞ x[n]e−j2πfn

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x[m] y[n]↑ U H(f) ↓ D

Figure 3: Q8

Q8 Consider the system in Fig. 8. If U and D are prime integers, find H(f) so that aliasingcan be avoided. Also find Y (f).

Solution

H(f) ={

U , |f | ≤ min [ 12U , 1

2D ]0 ,min [ 1

2U , 12D ] < |f | ≤ 1

2

=⇒

Y (f) ={

UD X(f U

D ) , |f | ≤ min [ 12 , D2U ]

0 , otherwise

Q9 If the filter h[n] = {1,0,0,5,0,0,3}, find g[n] such that both systems in Fig. 9 produce thesame output.

x[n] y[m]

↑ 2h[n] ↓ 3

x[n] y[m]↑ 2 g[n]↓ 3

Figure 4: Q9

Solution H(z)=1 + 5z−3 + 3z−6. The equivalent diagram is

x[m] y[n]↑ 2 f [n] ↓ 3

for first system to be equivalent to this system,

F (z) = H(z2) = 1 + 5z−6 + 3z−12

similarly the second system is equivalent to this system if,

G(z3) = F (z) ⇒ G(z) = 1 + 5z−2 + 3z−4 ⇒ g[n] = {1, 0, 5, 0, 3}

Note that it is impossible to make the two systems equivalent if, for example, h[n] ={1, 0, 3, 4}

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