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ECE 216

Final exam

April 25, 20189:30am – 12:00 pm

Circle your lecture section:

LEC101 (Taylor, Mon. 5-6)

LEC102 (Taylor, Mon. 11-12)

LEC103 (Draper, Mon. 11-12)

Guidelines:

• Write your answer in the space provided for each question.

• This exam is closed-note. The last page is an equation sheet. You may remove it.

• You may use a non-programmable calculator.

Problem Score

1 /112 /103 /144 /115 /10

Total /56

1

1. Consider the DT signal x[n] = jej2⇡n + cos(⇡n). Simplify all of your answers.

(a) (2 points) Find the even part of x[n].Solution:

xe

[n] = j cos(2⇡n) + cos(⇡n) = j + (�1)n

(b) (2 points) Find the odd part of x[n].Solution:

xo

[n] = � sin(2⇡n) = 0

(c) (2 points) Is x[n] periodic? If so, find the fundamental period.Solution: Yes. The fundamental period is N = 2.

(d) (3 points) Find the DTFS coe�cients of x[n].Solution: First, observe that x[n] = j + (�1)n. The coe�cients are given by

ak

=1

2

1X

n=0

x[n]e�j

2⇡k2 n

=1

2(j + 1 + (j � 1)e�j⇡k)

=1

2(j + 1 + (j � 1)(�1)k)

=

⇢j, k even1, k odd

.

(e) (2 points) Find the DTFS coe�cients of the complex conjugate of x[n], x[n]⇤.Solution: By inspection, b

0

= �a0

, and b1

= a1

.

2

2. The DTFT of the DT signal x[n] is

X(ej!) =2

1� 0.5e�j!

for ! 2 [0, 2⇡). Simplify all of your answers.

(a) (2 points) Evaluate X(ej!) at ! = �1.Solution: Because X(ej!) has period 2⇡,

X(ej!) =2

1� 0.5e�j!

.

Therefore,

X(ej(�1)) =2

1� 0.5e�j(�1)

=2

1� 0.5ej⇡ 2.0568 + 1.1857i.

(b) (2 points) Find the DT signal x[n].Solution: This is a standard transform pair. x[n] = 2⇥ 0.5nu[n].

(c) (2 points) Find the DTFT of x[n� n0

].Solution: The resulting transform is just X(ej!) multiplied by e�j!n0 ,

2e�j!n0

1� 0.5e�j!

.

(d) (2 points) Find the DTFT of x[n]� x[n� 1].Solution: Due to linearity of DTFT, we can combine the two previous answers toobtain

2(1� e�j!n0)

1� 0.5e�j!

.

(e) (2 points) Find the DTFT ofn+10X

k=n

x[k].

(Hint: Consider the I-DTFT.)Solution: Observe that

n+10X

k=n

x[k] =1

2⇡

Z2⇡

0

X(ej!)n+10X

k=n

ej!kd!

=1

2⇡

Z2⇡

0

X(ej!)ej!n10X

k=0

ej!kd!

=1

2⇡

Z2⇡

0

X(ej!)ej!n1� ej11!

1� ej!d!.

3

From here, we can see that

n+10X

k=n

x[k] ! X(ej!)1� ej11!

1� ej!.

4

3. Consider the DT LTI system

T {x} =n+mX

k=n�m

x[k],

where m is a positive integer. Simplify all of your answers.

(a) (2 points) Find the impulse response, h.Solution:

h[n] =n+mX

k=n�m

�[k] =

⇢1, |n| m0, |n| > m

.

(b) (2 points) Is the system causal? Justify your answer.Solution: No. The output depends on the future of the input. Also, the impulseresponse is nonzero for negative values of n.

(c) (2 points) Is the system bounded-input bounded-output stable? Justify youranswer.Solution: Yes. Observe that the impulse response is absolutely summable:

1X

k=�1

|h[k]| = 2m+ 1 <1.

(d) (3 points) Find the output when the input is x[n] = u[n], the unit step.Solution:

y[n] = (h ⇤ u)[n]

=1X

k=�1

h[k]u[n� k]

=mX

k=�m

u[n� k]

=

8<

:

0 n < �mn+m+ 1, |n| m2m+ 1, n > m

.

(e) (3 points) Find the frequency response, H(ej!).

5

Solution:

H(ej!) =1X

k=�1

h[k]e�j!k

=mX

k=�m

e�j!k

=1� e�j!(m+1)

1� e�j!

+1� ej!(m+1)

1� ej!� ej!0

=ej!/2 � e�j!(m+1/2) � e�j!/2 + ej!(m+1/2)

ej!/2 � e�j!/2

� 1

=�e�j!(m+1/2) + ej!(m+1/2)

ej!/2 � e�j!/2

=sin(!(m+ 1/2))

sin(!/2).

(f) (2 points) Find the output when the input is x[n] = cos(n).Solution:

y[n] = H(ej)ejn

2+H(e�j)

e�jn

2

= H(ej)ejn

2+H(ej)

e�jn

2

=sin(m+ 1/2)

sin(1/2)cos(n),

because H(ej!) is even.

6

4. (Note: All parts of this question (a)-(f) consider the same three signals x, s, and y.)

Let the CT signal x be the cosine defined pointwise as x(t) = cos(2000⇡t).

(a) (1 point) What is the fundamental period T0

of x expressed in terms of secondsand what is the fundamental frequency !

0

expressed in terms of radians per sec-ond? (Show your work/derivation.)

Let s be the CT signal defined pointwise as

s(t) = u (t+ T0

/4)� u (t� T0

/4)

where T0

is the fundamental frequency you found in part (a). Let the third CT signaly be defined pointwise as y(t) = s(t)x(t).

(b) (2 points) In a single plot make clear and well-labeled sketches of both s(t) andof y(t). (Make sure you label clearly which is which.)

(c) (2 points) Compute X(j!) and S(j!). Plot each spectrum (on separate plots)for ! 2 [�28000⇡, 28000⇡], i.e., for the range ±14000 Hz.

(d) (2 points) Compute the spectrum of the CT signal y, i.e., Y (j!).

(e) (2 points) Make a clear and well-labeled sketch of Y (j!) for ! 2 [�28000⇡, 28000⇡].

7

(f) (2 points) Compare your time-domain plots of s and y and your frequency-domainplots of S and Y . Comment on which signal – s or y – has more high-frequencycontent. Your comments should be qualitative (computations are not asked for),should comment on both the time- and frequency-domain plots, and the observa-tions you draw from the time-domain should be consistent with those from thefrequency domain.

8

5. Let the two CT signals x and y have spectrums defined respectively as

X(j!) =

⇢1� |!|

1000⇡

if |!| 1000⇡0 else

,

and

Y (j!) =

⇢1 if |!| 500⇡0 else

.

Let us define the CT signal p as the pointwise product of the time-domain versions ofthe two signals, i.e., p(t) = x(t)y(t).

(a) (2 points) Make a clearly labeled sketch of the spectrum P (j!). (You need notcompute the exact form, but your sketch should clearly reflect the shape of thespectrum.)

(b) (1 points) What is the Nyquist sampling rate (in terms of samples / seconds) for p.

(c) (2 points) Find an expression for the the time-domain signal p(t). (Hint: Thereis an insight here. If you get the insight, this is not hard.)

Define the CT signal z as the convolution of the time-domain versions of the twosignals, i.e., z(t) = (x ⇤ y)(t).

(d) (2 points) Make a clearly labeled sketch of the spectrum Z(j!).

9

(e) (1 points) What is the Nyquist sampling rate (in terms of samples / seconds) forz? (Recall, this is the lowest rate that uniformly-spaced samples can be collectedso that the time domain signal can be exactly recovered from the samples.)

(f) (2 points) Find an expression for the the time-domain signal z(t). (Hint: Thereis an insight here. Observe that the spectrum can be expressed as the sum of twosimpler spectra. If you get the insight this is not too hard.)

10

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11

Equations

Euler’s formula:ejt = cos(t) + j sin(t)

Even and odd parts of the CT signal x(t):

xe

(t) =x(t) + x(�t)

2, x

o

(t) =x(t)� x(�t)

2

Energy and power of the CT signal x(t) in the interval [t1

, t2

]:

E =

Zt2

t1

|x(t)|2dt, P =1

t2

� t1

Zt2

t1

|x(t)|2dt

Units of frequency:

x rad/s =x

2⇡Hz

CT convolution:

(x ⇤ y)(t) =Z 1

�1x(⌧)y(t� ⌧)d⌧.

DT convolution:

(x ⇤ y)[n] =1X

k=�1

x[k]y[n� k].

Geometric series:

n�1X

k=0

↵k =

⇢1�↵

n

1�↵

if ↵ 6= 1n if ↵ = 1

,1X

k=0

↵k =1

1� ↵if |↵| < 1.

CTFS:

x(t) =1X

k=�1

ak

ejk!0t, ak

=1

T0

ZT0

0

x(t)e�jk!0tdt

DTFS:

x[n] =N�1X

k=0

ak

ej2⇡kN n, a

k

=1

N

N�1X

n=0

x[n]e�j

2⇡kN n

Projection of x 2 CN onto y 2 CN :

hx, yihy, yiy, where hx, yi =

N�1X

k=0

x[k]y[k]⇤

CTFT:

x(t) =1

2⇡

Z 1

�1X(j!)ej!td!, X(j!) =

Z 1

�1x(t)e�j!tdt

DTFT:

x[n] =1

2⇡

Z2⇡

0

X(ej!)ej!nd!, X(ej!) =1X

n=�1x[n]e�j!n

12

CTFT pairs:

• x(t) ! X(j!)

• e�atu(t) ! 1

a+j!

, a > 0

• �(t� ⌧) ! e�j!⌧

• ej!0t ! 2⇡�(! � !0

)

• If x ! X(j!) and y ! Y (j!), then

i) (x ⇤ y)(t) ! X(j!)Y (j!)

ii) x(t)y(t) ! (X ⇤ Y )(!) = 1

2⇡

R1�1 X(j�)Y (j(! � �))d�

DTFT pairs:

• x[n] ! X(ej!)

• anu[n] ! 1

1�ae

�j!

• �[n� n0

] ! e�j!n0

Nyquist Sampling Theorem: Suppose x $ X(j!) and X(j!) = 0 for |!| > !max

> 0.Then x[n] is fully determined by its samples, x[n] = x(nT

s

) if !s

= 2⇡/Ts

> 2!max

.

13

Ashley Hung

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