ECE 421 Introduction to Signal Processing

Midterm 2 – Spring 2013

March 27, 2013

Question 1Consider the signal

x(n) = {1, 3,−5↑, 0,−1}

with Fourier transform X(ω). Compute the following quantities:(Hint: you need not explicitly compute X(ω).)(a) X(0)Solution: X(0) =

∑n x(n)e0 =

∑n x(n) = 1 + 3− 5− 1 = −2.

(b)∫ π−πX(ω)dω

Solution: Recall that x(n) = 12π

∫ +π

−π X(ω)ejωndω. Therefore,

x(0) = −5 =1

2π

∫ +π

−πX(ω)ejωndω =

1

2π

∫ +π

−πX(ω)dω,

which implies that∫ +π

−π X(ω)dω = −10π.

(c) X(π)Solution: X(0) =

∑n x(n)e−jπn =

∑n x(n)(−1)n = 1(−1)−2 +

3(−1)−1 − 5(−1)0 − 1(−1)2 = 1− 3− 5− 1 = −8.

(d)∫ π−π |X(ω)|2dω

1

Solution: This part uses the Parsevel theorem,

1

2π

∫ π

−π|X(ω)|2dω =

∑n

|x(n)|2.

Therefore,∫ π

−π|X(ω)|2dω = 2π

[∑n

|x(n)|2]

= 2π[12 + 32 + 52 + 12

]= 72π.

Question 2(a) Compute and sketch the magnitude response of the followingsystem:

y(n) =1

3(x(n− 1) + x(n− 2) + x(n− 3)).

Solution: In the z-domain,

Y (z) =1

3

[X(z)z−1 +X(z)z−2 +X(z)z−3

],

which yields H(z) = Y (z)/X(z) = 13 [z−1 + z−2 + z−3]. To com-

pute the Fourier response, we substitute z = ejω, which yields

H(ω) =1

3[e−jω + e−2jω + e−3jω].

To sketch the magnitude response, it may help to realize that|e−2jω| = 1. Therefore,

|H(ω)| =∣∣∣∣13e−2jω[ejω + 1 + e−jω]

∣∣∣∣ =1

3|1 + 2 cos(ω)| .

To sketch the magnitude response, we can work out some valuesat interesting frequencies: (i) at ω = 0, |H(0)| = 1

3(1 + 2) = 1;(ii) at ω = ±π, |H(0)| = |13(1 − 2)| = 1

3 ; (iii) |H(ω)| = 0 when

2

cos(ω) = −12 , which implies ω = ±2

3π.

(b) Consider the following inputs

(1) x(n) = cos(π

2n),

(2) x(n) = 3 sin(3π

4n)u(n),

where u(n) is the step function. Will either input have an outputthat is always 0? Why? (Hint: no need to calculate the output.)Solution: The output will be zero if and only if the Fourier re-sponse that corresponds to the input contains non-zeros only atfrequencies where |H(ω)| = 0. This means that the input musthave zero Fourier response everywhere except for ω = ±2

3π. It iseasy to see that both inputs above do not satisfy these require-ments; the output will sometimes be non-zero.

Question 3Given a continuous-time signal xa(t) with Xa(F ) = 0 for |F | >B, determine the minimum sampling rate Fs for the followingsignals. Please explain your answer.(a) x4a(t)Solution: Taking a signal to the power of 4 in the time domainmeans taking a convolution between 4 Fourier responses, and soF(x4a(t)) is limited to the range |F | ≤ 4B, which implies thatFs > 8B.

(b) xa(−t)xa(t)Solution: Owing to the time reversal property, F(xa(−t)) =X(−ω), which is also band-limited between −B and B. A prod-uct in the time domain is a convolution in the frequency domainbetween two signals that are band-limited between −B and B.Therefore, the output signal is band-limited between −2B and

3

2B, and the sampling rate must be Fs > 4B.

Question 4In this question, you will design a high pass filter with one zeroand one pole. The zero will be at ω = 0, and it will lie on the unitcircle. The pole will be at ω = π, and it will be inside the unitcircle (we want the filter to be BIBO stable). This filter shouldhave unit gain at ω = π, i.e., H(π) = 1), and have a magnituderesponse of

√0.1 at ω = π

2 , i.e., H(π2 ) =√

0.1. Please derive thesystem function and justify your design carefully.Solution: First, the zero is at ω = 0, which implies the positivepart of the real axis. Because the zero is on the unit circle, thezero appears at z = +1. Second, the pole at ω = π is on thenegative part of the real axis, let us say that the location of thepole is −p, where p ∈ (0, 1). Therefore, the filter will have theform,

H(ω) = G1− e−jω

1 + pe−jω,

where G is some gain parameter.We have two constraints: (i) H(π) = 1 and (ii) H(π2 ) =

√0.1.

Let us investigate these constraints,

H(π) = G1− e−jπ

1 + pe−jπ= G

1− (−1)

1 + p(−1)=

2G

1− p.

Therefore, G = 1−p2 . The second constraint yields

H(π

2

)= G

1− e−j π21 + pe−j

π2

= G1− (j)

1 + p(j).

Taking the magnitude,∣∣∣H (π2

)∣∣∣ = G

√12 + (−1)2√12 + (p)2

=1− p

2

√2/(1 + p2) =

√0.1.

4

Squaring this value yields (1−p)2·24·(1+p2) = 0.1. This expression can be

simplified,

1− 2p+ p2 = 0.2(1 + p2)

0.8− 2p+ 0.8p2 = 0

p2 − 2.5p+ 1 = 0

(p− 2)(p− 0.5) = 0.

The correct solution (inside the unit circle) is p = 0.5.Let us verify that this solution is correct. The gain parametersatisfies G = 1−p

2 = 0.25. Therefore, the overall transfer functionis

H(ω) = 0.251− e−jω

1 + 0.5e−jω.

At ω = π, the transfer function is

H(π) = 0.251 + 1

1− 0.5= 1.

At ω = π2 ,

|H(π

2

)| = 0.25| 1− j

1− 0.5j| = 0.25

√1 + 1

1 + 0.25=

√8 · 0.252

5=√

0.1.

5