GEORGIA INSTITUTE OF TECHNOLOGY

SCHOOL OF ELECTRICAL AND COMPUTER ENGINEERING

ECE 2026 — Summer 2016

Final Exam
August 1, 2016

NAME:
ANSWER KEY GT username:
(FIRST) (LAST) (e.g., gtxyz123)

Tue
Circle your recitation section (otherwise you lose 3 points!):
10 – 11:45am L01 (Stuber)
12 – 1:45pm L02 (Stuber)
2 – 3:45pm L03 (Zhang)
4 – 5:45pm L04 (Zhang)
Important Notes:
• DO NOT unstaple the test.
• One two-sided page (8.5”×11”) of hand-written notes permitted.
• JUSTIFY your reasoning CLEARLY to receive partial credit.
• You must write your answer in the space provided on the exam paper itself.
Only these answers will be graded. Circle your answers, or write them in the boxes provided.
If more space is needed for scratch work, use the backs of the previous pages.

Problem Value Score Earned

1 10

2 10

3 10

4 10

5 10

6 10

7 10

8 10

9 10

10 10

No/Wrong Rec –3

Total
PROB. SU16-Final-1.

Suppose that a sinusoidal signal x( t ) = 0.707cos(100t + 0.3) is added to a delayed version of

itself, resulting in the zero signal:

x( t ) + x( t – t ) = 0.
This equation does not uniquely specify the value of the delay t0 ; there are many possible values it
might take. Name any three. Make them all positive (t0 > 0).

t0 = 0.01 sec, or t0 = 0.03 sec, or t0 =

0.05 sec.

Phasor addition: 1 + e–j100t0 = 0

⇒ 100t0 is an odd multiple of 
⇒ 100t0 = (2m + 1)
⇒ t0 = (2m + 1)/100 ∈{0.01, 0.03, 0.05, 0.07, ... }
PROB. SU16-Final-2.

Consider an LTI system with input x[ n ] and output y[ n ] defined by the difference equation:
y[ n ] = x[ n ] + x[n – 3] + x[n – 6],
where  is a real constant to be determined. (It might take a different value in each part below.)

(a) If a constant input x[ n ] = 7 results in a constant output y[ n ] = 3.5,

then it must be that = –1.5 .

DC gain 1 +  + 1 = 0.5

(b) If a sinusoidal input x[ n ] = cos(n/3) results in a zero output y[ n ] = 0,

then it must be that = 2 .

H(e j/3) = 1 + e –j3/3+ e –6j/3 = 1 –  + 1 = 0

(c) Let s[ n ] denote the step response (the output in response to unit step u[ n ]).
If s[ 5 ] = 2.6, then it must be that  = 1.6 .
y[ 5 ] = x[ n ] + x[n – 3] + x[n – 6],
⇒ s[ 5 ] = u[ 5 ] + u[5 – 3] + u[5 – 6]
= 1 + = 2.6
(d) If the magnitude response is as shown below,
|H(e jˆ)|
 then it must be that  = 0 .

   ̂

DC gain 1 +  + 1 = 2
(e) If the pole-zero plot is as shown below, then the value of  is (circle one):
Im{ z }
 = –2.01
ˆ  = –1.99
|H(e j)| = | 1 + e –3j̂ + e –6j̂ | =  + 2cos( 3 ̂ )
 = –1.01
 = –0.99
(6)
 = –0.01
Shift down the sinusoid by  = –1.99  = 0.01
Re{ z }
to get the two zeros near DC:
 = 0.99
 = 1.01
)
le
irc


tc

= 1.99
̂
ni
(u

 = 2.01
PROB. SU16-Final-3. Consider the six lines of MATLAB code shown below:

RAMBLIN = 4 ; x( t ) = Acos(( t ) + ), where

( t ) = 2Wcos( t )
2 ´( t ) = – Wsin( t )
1
⇒ fi( t ) = -----
-
WRECK =
1600 ;
If no aliasing, maximum
instantaneous frequency would be:
fsamp = 2400 ;
W = WRECK = 1200 + 400
= 1600 Hz
tt = 0:(1/fsamp):RAMBLIN;
xx = real((20 + 26*j)*exp(j*2*pi*WRECK*cos(tt)));
spectrogram(xx, ... ); % see footnote1

Find numerical values for the three unspecified parameters RAMBLIN, WRECK, and fsamp
so that running the above code produces the following spectrogram:

⇒ 1200

fsamp = 2400
400 Hz

1000

800
FREQUENCY (Hz)

600

400

200

0
0 1 2 3 4
⇒
RA
TIME (seconds) MB
LI
N
=
4

1. To avoid confusion, the remaining arguments of spectrogram are not shown. They are not relevant.
If you are curious, however, the complete command is spectrogram(xx,128,120,512,fsamp,'yaxis').
PROB. SU16-Final-4. Shown on the left are the pole-zero plots for ten LTI systems. Shown on the right are
the corresponding magnitude responses, labeled A through K, but in a scrambled order. Match the pole-
zero plot to its corresponding magnitude response by writing a letter (A through K) in each answer box.

|H(e jˆ)|
Im{ z }
C D
̂
B

̂
C
E F
̂

̂
E

̂
A B F

̂
G

̂
G K
H
̂

̂
H J Re{ z }
K

   ̂
PROB. SU16-Final-5.

Shown below is the magnitude response of an FIR filter:

|H(e jˆ )|
10
8
6

   ̂

If the difference equation for this FIR system is:

y[ n ] = x[ n ] + b1x[ n – 1 ] + b2x[ n – 2 ] + b3x[ n – 3 ] + b4x[ n – 4 ],
with b = 1, then the remaining FIR filter coefficients are:

b1 = 0 , b2 = 8 , b3 = 0 , b4 = 1 .

From the plot: |H(e j̂ )| = 8 + 2cos( 2̂ ) = 8 + e 2j̂ + e –2j̂

which would arise from: h[–2] = 1, h[ 0 ] = 8, h[ 2 ] = 1

⇒ delay by 2 to get causal system: h[ 0 ] = 1, h[ 2 ] = 8, h[ 4 ] = 1

(delay does not change magnitude response)

PROB. SU16-Final-6. Consider the discrete-time signal x[ n ] shown below (it is zero for all time n < 0):

1
x[ n ]
0
0 20 40 60 80 n

As shown below, this signal is fed as an input to six different FIR systems (labeled A through F),
producing the six different outputs yA[ n ] through yF[ n ] shown in the stem plots below:
1

yA[ n ]
A
0
0 20 40 60 80 n

1
yB[ n ]
B
0
0 20 40 60 80 n
1

yC[ n ]
C 0
0 20 40 60 80 n
x[ n ] 2

1
yD[ n ]
0
n
D 20 40 60 80

1
yE[ n ]
0
20 40 60 80
n
E -1

1
yF[ n ]

F 0
0 20 40 60 80 n

Match each system above to its description below by writing a letter (A through F) in each answer box:

D first difference filter B 5-point running average filter

E y[ n ] = x[ n ] – 2cos(0.2)x[n – 1] + x[n – 2] A 10-point running average filter

F 2-point running average filter C 20-point running average filter

PROB. SU16-Final-7. The signal x( t ) = cos(240t) + cos(250t)
is periodic with fundamental frequency f = 5 Hz.

Suppose we pass this signal through the cascade of an ideal C-to-D converter and an ideal D-to-C
converter, as shown below, where both have the same (unspecified) sampling rate parameter fs:

IDEAL IDEAL y( t )
x( t ) = cos(240t) + cos(250t) C-to-D D-to-C
CONVERTER CONVERTER

fs

(a) Specify a value for fs in the range 240 < fs < 250 Hz
so that y( t ) is periodic with fundamental frequency f = 120 Hz: fs = 245 samples/sec.

(b) Specify a value for fs in the range 240 < fs < 250 Hz
so that y( t ) is periodic with fundamental frequency f = 4 Hz: fs = 249 samples/sec.

(c) Specify a value for fs in the range 240 < fs < 250 Hz
so that y( t ) is periodic with fundamental frequency f = 3 Hz: fs = 248 samples/sec.

(d) Specify a value for fs in the range 240 < fs < 250 Hz
so that y( t ) is periodic with fundamental frequency f = 2 Hz: fs = 247 samples/sec.

(e) Specify a value for fs in the range 240 < fs < 250 Hz
so that y( t ) is periodic with fundamental frequency f = 1 Hz: fs = 246 samples/sec.
PROB. SU16-Final-8.

Shown below is the serial cascade of three LTI systems:

x[ n ] LTI LTI
w[ n ] LTI y[ n ]
SYSTEM #1 SYSTEM #2 SYSTEM #3

• The first system is defined by an impulse response whose Z transform is H1( z ) = 2 – 0.4z –1.
• The second system is defined by the frequency response H2 (e j̂ ) = 5 – 1.5e – j̂ .
• The third system (with input w[ n ] and output y[ n ]) is defined by the difference equation:
y[ n ] = a1y[n – 1] + a2y[n – 2] + bw[ n ] + b1w[ n – 1].
If the third system inverts the previous two, so that its output y[ n ] is the same as the original input x[ n ]
(satisfying y[ n ] = x[ n ]), then its coefficients must be:

a1 = 0.5 , a2 = –0.06 , b = 0.1 , b1 = 0 .

First two systems: H12( z ) = (2 – 0.4z –1)(5 – 1.5z –1)

= 10 – 5z–1 + 0.6z–2
= 10(1 – 0.5z–1 + 0.06z–2)

1 0.1 Yz
- –2 = -------------
Third system: H3( z ) = ----------------- = -----------------------------------------------------
H 12  z  1 – 0.5z + 0.06z –1 Wz

⇒ (1 – 0.5z–1 + 0.06z–2)Y( z ) = 0.1W( z )

⇒ y[ n ] – 0.5y[n – 1] + 0.06y[n – 2] = 0.1w[ n ]

⇒ y[ n ] = 0.5y[n – 1] – 0.06y[n – 2] + 0.1w[ n ]

a1 a2 b0
PROB. SU16-Final-9.

Consider the following system with input x( t ) and output y( t ):

x( t ) IDEAL x[ n ] y[ n ] IDEAL y( t )
LTI
C-to-D D-to-C
CONVERTER SYSTEM CONVERTER

fs = 1000 samples/s

The sampling rate for both the C-to-D and D-to-C is fs = 1000 samples/s. The LTI system is defined
by the difference equation:

y[ n ] = 0.5x[ n ] + 4x[n – 1] + 0.5x[n – 2].

If the input is x( t ) = 2cos(400t) + 4cos(500t), then the output is:

y( t ) = A1cos(2f1 t + 1) + A2cos(2f2 t + 2), where:

A1 = 8.618 , f1 = 200 Hz, 1 = –0.4 rads,

A2 = 16 , f2 = 250 Hz, 2 = –0.5 rads.

Sampling rate high enough to avoid aliasing

⇒ output frequencies same as input frequencies, 200 and 250 Hz

Digital frequencies after sampling:

̂1 = 400/1000 = 0.4
̂2 = 500/1000 = 0.5

Evaluating frequency response H(e j̂ ) = e–j̂(4 + cos( ̂ )) at these frequencies yields:

H(e j̂1 ) = e–j(4 + cos(0.4)) = 4.39e–j ⇒ A1e j1 = 2( . ) = 8.618e–j

H(e j̂2 ) = e–j(4 + cos(0.5)) = 4e–j0.5 ⇒ A2e j2 = 4( . ) = 16e–j0.5

PROB. SU16-Final-10.

Shown below are eight different outcomes that result from executing the MATLAB code:
stem(abs(fft(ones(1,L),N)));
Match each plot with the corresponding values for the variables N and L by writing a letter
(A through H) in each answer box.

A
L = 2, N = 100 G
0
0 50 100 150 200 L = 2, N = 200 H
B L = 4, N = 100 C
0 L = 4, N = 200 A
0 50 100 150 200

L = 8, N = 100 D
C

0
L = 8, N = 200 B
0 20 40 60 80 100

L = 16, N = 200 E
D
L = 32, N = 200 F
0
0 20 40 60 80 100

0
0 50 100 150 200

0
0 50 100 150 200

0
0 20 40 60 80 100

0
0 50 100 150 200