EE 123 University of California, BerkeleyAnant Sahai February 15, 2007

Digital Signal Processing

Midterm 1 SolutionInstructions

• Total time allowed for the exam is 80 minutes

• Some useful formulas:

– Discrete Time Fourier Transform (DTFT)

X(ejω) =∞∑

n=−∞

x[n]e−jωn

– Inverse Fourier Transform

x[n] =1

2π

∫ 2π

−2πX(ejω)ejωndω

– Z Transform

X(z) =∞∑

n=−∞

x[n]z−n

1. (20 points) Match each of the pole-zero plots to the corresponding magnitude and phaseplots given below. Write your answer in the space provided below each magnitude and phaseplot.

−1/4

1/4

−1/43/4

(a) (b)

21/2

(c) (d)

Doublepole

1/4

1/4

−1/4

(2 − 2j)

(2 + 2j)

(e) (f)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−400

−300

−200

−100

0

Normalized Frequency (×π rad/sample)

Ph

ase

(d

eg

ree

s)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−5

0

5

10

15

Normalized Frequency (×π rad/sample)

Ma

gn

itud

e (

dB

)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−200

−100

0

100

200

Normalized Frequency (×π rad/sample)

Ph

ase

(d

eg

ree

s)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−15

−10

−5

0

5

10

Normalized Frequency (×π rad/sample)

Ma

gn

itud

e (

dB

)

(1)-(b) (2)-(d)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−800

−600

−400

−200

0

Normalized Frequency (×π rad/sample)

Ph

ase

(d

eg

ree

s)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−1

−0.5

0

0.5

1

Normalized Frequency (×π rad/sample)

Ma

gn

itud

e (

dB

)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−800

−600

−400

−200

0

Normalized Frequency (×π rad/sample)

Ph

ase

(d

eg

ree

s)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 217

17.5

18

18.5

19

19.5

Normalized Frequency (×π rad/sample)

Ma

gn

itud

e (

dB

)

(3)-(e) (4)-(f)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−800

−600

−400

−200

0

Normalized Frequency (×π rad/sample)

Ph

ase

(d

eg

ree

s)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−5

0

5

Normalized Frequency (×π rad/sample)

Ma

gn

itud

e (

dB

)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−400

−300

−200

−100

0

Normalized Frequency (×π rad/sample)

Ph

ase

(d

eg

ree

s)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−1

−0.5

0

0.5

1

Normalized Frequency (×π rad/sample)

Ma

gn

itud

e (

dB

)

(5)-(a) (6)-(c)

2. (70 points) Consider a causal linear time-invariant system with impulse response h[n]. Thez-transform of h[n] is

H(z) =1 − z−1

1 − 0.5z−1

Consider the cascade configuration given in the figure below. Here we assume that α is a realnumber.

h[n]

LTI, Causal

α−n

αn

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

a. (10 pts) Draw the pole-zero plot of H(z). Also, sketch the region of convergence. Isthis system stable?

1/2 1

ROC: z > 12. Since the unit circle lies in the region of convergence, the system is stable.

b. (10 pts) Assume that α = 13 and x[n] =

(

14

)nu[n]. Compute the output y[n] for the

above choice of input x[n].

From the system diagram, we have v[n] = α−nx[n], w[n] = v[n] ∗ h[n] and y[n] = w[n]αn. Thisimplies

y[n] = (v[n] ∗ h[n])αn

=

∞∑

k=−∞

v[n − k]h[k]

αn

=

∞∑

k=−∞

x[n − k]α−(n−k)h[k]

αn

=

∞∑

k=−∞

x[n − k]αkh[k]

= x[n] ∗ (αnh[n])

Taking the z-transform, we get

Y (z) = X(z)H(

α−1z)

=1

1 − 14z−1

·1 − αz−1

1 − 0.5αz−1

=1

1 − 14z−1

·1 − 1

3z−1

1 − 16z−1

=−1

1 − 14z−1

+2

1 − 16z−1

(1)

ROC of Y (z) is the intersection of the ROC of X(z) and H(z), which is |z| > 14 . Now, taking

the inverse z-transform of (1) we have

y[n] = −

(

1

4

)n

u[n] + 2

(

1

6

)n

u[n]

c. (10 pts) Is the overall system enclosed by the dashed box linear? Explain your reasoning.

We have already shown that the output y[n] = x[n] ∗ (αnh[n]). Since the input and output arerelated via a convolution relationship, the system is a linear and time-invariant. Moreover, theimpulse response of the overall system enclosed by the dashed box is g[n] = αnh[n].

d. (10 pts) Is the overall system enclosed by the dashed box time-invariant? Explain yourreasoning.

See solution for part (c).

e. (10 pts) Find the response g[n] of the overall system within the enclosed box to theinput x[n] = δ[n].

Again from part (c), we have g[n] = αnh[n].

f. (10 pts) Plot the pole-zero plot of G(z), the z-transform of g[n]. Also, sketch the regionof convergence.

We know that g[n] = αnh[n]. This implies

G(z) = H(

α−1z)

=1 − 1

3z−1

1 − 16z−1

, ROC : |z| >1

6

The pole-zero plot of G(z) is shown in the figure below.

1/3

1/6

g. (10 pts) What happens to the pole-zero plot in part (f) if α = e−jω0 , for some realnumber ω0.

If α = e−jω0 , then the poles and zeros of G(z) are obtained by rotating the correspondingpoles and zeros of H(z) by ω0 degrees in the clockwise direction. The region of convergence is|z| > 0.5.

−jw0e

−jw0e1/2

3. (40 points) A real-valued signal x(t) is known to be band-limited, i.e., X(jω) = 0, for|ω| > W . x(t) is first mixed by multiplying by cos(ω0t). The resulting signal v(t) is sampleduniformly (sampling time T ) using the architecture given in the figure below. The periodicimpulse train used for sampling is

s(t) =∞∑

n=−∞

δ(t − nT )

Conversion fromimpulse train

to discrete−timesequence

cos(w t)0

v (t)sv[n] = v (nT)s

s(t)

v(t)

C/D converter

x(t)

Assume that x(t) = sin WtWt

and ω0 = 2W .

a. (10 pts) Compute and draw X(jω), the Fourier transform of x(t).

X(jω) =

πW

, −W ≤ ω ≤ W

0 otherwise

π/W

X(j )ω

0ω

W−W

b. (10 pts) Compute and draw V (jω), the Fourier transform of v(t).

v(t) = x(t) cos(jω0)

⇒ V (jω) =1

2[X(j(ω + ω0)) + X(j(ω − ω0))]

ωV(j )

π/2W

0ω

2WW 3W−W−3W −2W

c. (10 pts) Compute and draw Vs(jω), the Fourier transform of vs(t).

vs(t) = v(t)s(t)

⇒ Vs(jω) =1

2π[V (jω) ∗ S(jω)]

=1

T

∞∑

k=−∞

V (j(ω − Ωk))

where Ω = 2πT

. The Fourier transform Vs(jω) is shown in Fig. 1.(d) What is the minimum sampling rate Ω = 2π

Trequired to reconstruct the original

signal x(t) from the discrete time signal vs[n].The maximum frequency component of v(t) is 3W . Hence, from the Nyquist sampling theoremit is obvious that we can reconstruct x(t) for sampling rates Ω > 6W . Again from Nyquistsampling theorem, it is clear that we cannot recover x(t) at a sampling rate Ω < 2W . Now, theonly question is whether we can recover x(t) for sampling rates 2W ≤ Ω ≤ 6W . From Fig. 1it is clear that if Ω = 2W , the adjacent copies of X(jω) don’t overlap. In this case we canreconstruct x(t) by passing vs[n] through an ideal low-pass filter.For Ω > 2W , it is possible to reconstruct x(t) from the samples vs[n], inspite of the overlap ofthe adjacent copies of X(jω). The reconstruction process in this case is more than just low-passfiltering.

V (j )s ω

.. . .. .π/2W

2W

ω−Ω Ω0−2Ω 2Ω

2W

Figure 1: The Fourier transform Vs(jω). The figures corresponds to the case when the samplingfrequency Ω = 3.5W . The different colors represent the copies of V (jω) repeated periodicallydue to sampling.