GEORGIA INSTITUTE OF TECHNOLOGY

SCHOOL OF ELECTRICAL AND COMPUTER ENGINEERING

ECE 2026 — Summer 2018

Final Exam
July 27, 2018

NAME:

Important Notes:
• Do not unstaple the test.
• Closed book and notes, except for three double-sided pages (8.5” £11”) of hand-written notes.
• Calculators are allowed, but no smartphones/WiFI/etc.
• JUSTIFY your reasoning CLEARLY to receive partial credit.
• Express all angles as a fraction of p. For example, write 0.1p as opposed to 18° or 0.3142 radians.
• 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 Points Score

1 10

2 15

3 15

4 15

5 15

6 10

7 20

TOTAL: 100
PROB. Su18-Fin.1.

Let x1( t ) = 2cos(100p(t – 0.002)).

Let x2( t ) be the signal whose spectrum is shown below:

1–j 1+j

–100p 0 100p w [rad/s]

Let x3( t ) be the sinusoidal signal sketched below:

1.5

.. .
1
. ..

0.5

–0.02 –0.01 0.01 0.02 t

–0.5

–1

–1.5

The sum of these three signals can be written as x1( t ) + x2( t ) + x3( t ) = Acos(100pt + j),

where A= and j= .
PROB. Su18-Fin.2.

Suppose the output of a linear and time-invariant discrete-time filter in response to a finite-length
input of x[ n ] = f... 0 0 2 2 2 2 0 0 ... g is y[ n ] = f... 0 0 1 2 3 3 2 1 0 0 ... g, as shown below:

y[ n ]
x[ n ] LTI y[ n ] 3 3
x[ n ] SYSTEM
2 2
2

1 1
1

n n
0 5 0 5

The filter’s impulse response h[ n ] has values:

h[ 0 ] = , h[ 1 ] = , h[ 2 ] = , h[ 3 ] = , h[ 4 ] =
PROB. Su18-Fin.3.

Consider the following system in which a continuous-time input x( t ) is sampled, producing x[ n ], then
filtered, producing y[ n ], and finally converted back to continuous time, yielding an overall output y( t ):

x( t ) IDEAL x[ n ] LTI SYSTEM y[ n ] IDEAL y( t )

C-to-D D-to-C
CONVERTER H( z ) CONVERTER

SAMPLE RATE fs

The system function of the discrete-time LTI system is H( z ) = 1 – z– 1 + z– 2.

(a) Which of the plots (A... H) on the next page shows the magnitude response of this filter?

(b) A constant input x( t ) = 1 (for all t) will result in the constant output y( t ) = (for all t).

(c) Specify a sample rate fs = samples/s

so that a C-to-D input of x( t ) = cos(10pt) + cos(14pt)

results in a zero output, y( t ) = 0 (for all t).
 E

–p 0 p w
^ –p 0 p w
^

–p 0 p w
^ –p 0 p w
^

–p 0 p w
^ –p 0 p w
^

–p 0 p w
^ –p 0 p w
^
PROB. Su18-Fin.4.

Consider an FIR filter whose difference equation is:

y[ n ] = (1 + b)x[ n] + (1 – b)x[n – 1],

where b is an unspecified real constant.

If the output of this system in response to the sinusoidal input x[ n ] = cos(0.7pn)

is y[ n ] = cos(0.7pn + q),

then it must be that b = and q= .

PROB. Su18-Fin.5. Consider an LTI system described by the recursive difference equation:

y[ n ] = x[ n ] – 0.2y[n – 1].
(a) The system is [ IIR ][ FIR ] (circle one).

(b) The system transfer function is H( z ) = .

(c) Sketch below the pole-zero plot for H( z ), carefully labeling the locations of all zeros and poles:
Imfz g

Refz g

)
LE
C
IR
C
IT
N
(U

(d) If the output of this system in response to the input

x[ n ] = 10d[ n ] + Bd[ n – 1] is y[ n ] = 10d[ n ]

(or in other words, as lists of numbers, the input .... 0 10 B 0 ... results in the output ... 0 10 0 ... ),

then it must be that the constant B is

B= .
PROB. Su18-Fin.6.

Shown below are eleven 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 K) in each answer box.

B L = 2, N = 21

L = 2, N = 81
C

L = 3, N = 21

D
L = 3, N = 81

E
L = 4, N = 21

F L = 4, N = 81

L = 5, N = 21
G

L = 5, N = 81

H
L = 8, N = 81

I
L = 21, N = 21

J
L = 21, N = 81

K
PROB. Su18-Fin.7. (The different parts are unrelated.)

(a) There are an infinite number of values for the delay parameter t0 below such that:

cos(100pt) + cos(100p(t – t0)) + cos(100p(t – 2t0)) = 0, for all time t.

Name any two, both positive:

t0 = , or t0 = .

P1 sin ( 0.9pk ) sin ( 0.75p ( n – k ) )

(b) Simplify --------------------------- . --------------------------------------------- = .
k= –1
pk p(n – k)
a function of n
(c) The signal x( t ) = cos(88pt)cos(8pt)) + cos(240pt)
is periodic with fundamental frequency f0 = Hz.

(d) If fX[ 0 ], ... X[ 511]g are the DFT coefficients that result from taking a 512-point DFT
of the length-12 signal x[ n ] = [1 1 1 2 2 2 3 3 3 4 4 4] for n2f0, ... 11g,
then find numerical values for the following two DFT coefficients:

X[ 0 ] = ,

X[ 256] = .
 Table of DTFT Pairs

Time-Domain: xŒn Frequency-Domain: X.e j !O /

ıŒn 1

ıŒn n0  e j !n
O 0

sin. 21 L!/
O j !.L
O 1/=2
uŒn uŒn L e
sin. 21 !/
O
(
sin.!O b n/ 1 j!j
O  !O b
u.!O C !O b / u.!O !O b / D
n 0 !O b < j!j
O 
1
an uŒn .jaj < 1/ j !O
1 ae

Table of DTFT Properties

Property Name Time-Domain: xŒn Frequency-Domain: X.e j !O /

Periodic in !O X.e j.!C2/

O / D X.e j !O /

Linearity ax1 Œn C bx2 Œn aX1 .e j !O / C bX2 .e j !O /

Conjugate Symmetry xŒn is real X.e j !O / D X  .e j !O /

Conjugation x  Œn X  .e j !O /

Time-Reversal xŒ n X.e j !O /

j !d
O
Delay (d =integer) xŒn d e X.e j !O /

Frequency Shift xŒne j !O 0 n X.e j.!O !O 0 / /

1 j.!O !O 0 / / C 1 X.e j.!C

O !O 0 / /
Modulation xŒn cos.!O 0 n/ 2 X.e 2

Convolution xŒn  hŒn X.e j !O /H.e j !O /

1 Z 
X 1
Parseval’s Theorem jxŒnj2 jX.e j !O /j2 d !O
nD 1
2 

Date: 28-Apr-2013
 Table of Pairs for N -point DFT

Time-Domain: xŒn; n D 0; 1; 2; : : : ; N 1 Frequency-Domain: XŒk; k D 0; 1; 2; : : : ; N 1

If xŒn is finite length, i.e., ˇ

X Œk D X.e j !O /ˇ
ˇ
xŒn ¤ 0 only when n 2 Œ0; N 1 (frequency sampling the DTFT)
and the DTFT of xŒn is X.e j !O /
ˇ
!D2k=N
O

ıŒn 1

ıŒn n0  e j.2k=N /n0

e j.2 n=N /k0 N ıŒk k0 , when k0 2 Œ0; N 1

sin. 12 L.2k=N // j.2k=N /.L 1/=2

uŒn uŒn L, when L  N e
sin. 21 .2k=N //

sin. 21 L.2 n=N //

e j.2 n=N /.L 1/=2
N.uŒk uŒk L/, when L  N
sin. 12 .2 n=N //

Table of z-Transform Pairs

Signal Name Time-Domain: xŒn z-Domain: X.z/

Impulse ıŒn 1

n0
Shifted impulse ıŒn n0  z

1
Right-sided exponential an uŒn 1
; jaj < 1
1 az
1 r cos.!O 0 /z 1
Decaying cosine r n cos.!O 0 n/uŒn
1 2r cos.!O 0 /z 1 C r 2 z 2

1
cos.'/ r cos.!O 0 '/z
Decaying sinusoid Ar n cos.!O 0 n C '/uŒn A
1 2r cos.!O 0 /z 1 C r 2 z 2

Table of z-Transform Properties

Property Name Time-Domain xŒn z-Domain X.z/

Linearity ax1 Œn C bx2 Œn aX1 .z/ C bX2 .z/

d
Delay (d =integer) xŒn d z X.z/

Convolution xŒn  hŒn X.z/H.z/

Date: 28-April-2013
 GEORGIA INSTITUTE OF TECHNOLOGY
SCHOOL OF ELECTRICAL AND COMPUTER ENGINEERING

ECE 2026 — Summer 2018

Final Exam
July 27, 2018

ANSWER KEY
NAME:

Important Notes:
• Do not unstaple the test.
• Closed book and notes, except for three double-sided pages (8.5” £11”) of hand-written notes.
• Calculators are allowed, but no smartphones/WiFI/etc.
• JUSTIFY your reasoning CLEARLY to receive partial credit.
• Express all angles as a fraction of p. For example, write 0.1p as opposed to 18° or 0.3142 radians.
• 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 Points Score

1 10

2 15

3 15

4 15

5 15

6 10

7 20

TOTAL: 100
PROB. Su18-Fin.1.

Let x1( t ) = 2cos(100p(t – 0.002)). ) X1 = 2e–j0.2p

Let x2( t ) be the signal whose spectrum is shown below:

1–j 1+j

) X2 = 2 2e j0.25p

–100p 0 100p w [rad/s]

Let x3( t ) be the sinusoidal signal sketched below:

1.5 1.5
) x3( t ) = 1.5cos(100p(t – t0))

.. .
1
) X3 = 1.5e –j100p t0
. ..

= 1.5e–j0.25p
0.5

–0.02 –0.01 0.01 0.02 t

= 0.0025

–0.5
t0

–1

–1.5

The sum of these three signals can be written as x1( t ) + x2( t ) + x3( t ) = Acos(100pt + j),

where A= 4.68 and j= –0.016p .

(= –0.05 rads = –2.89°)

Phasor addition rule:

Aejj = X1 + X2 + X3

= 2e–j0.2p + 2 2 e j0.25p + 1.5e–j0.25p = 4.68e–j0.016p

PROB. Su18-Fin.2.

Suppose the output of a linear and time-invariant discrete-time filter in response to a finite-length
input of x[ n ] = f... 0 0 2 2 2 2 0 0 ... g is y[ n ] = f... 0 0 1 2 3 3 2 1 0 0 ... g, as shown below:

y[ n ]
x[ n ] LTI y[ n ] 3 3
x[ n ] SYSTEM
2 2
2

1 1
1

n n
0 5 0 5

The filter’s impulse response h[ n ] has values:

h[ 0 ] = 0 , h[ 1 ] =
0.5 , h[ 2 ] =
0.5 , h[ 3 ] =
0.5 , h[ 4 ] = 0

y[ n ] = Skh[k]x[n – k]
2
But y[ 0 ] = 0 ) 0 = h[ 0 ] x[ 0 ] = 2h[ 0 ] ) h[ 0 ] = 0
0 2 2
y[ 1 ] = 1 ) 1 = h[ 0 ] x[ 1 ] + h[ 1 ] x[ 0 ] = 2h[ 1 ] ) h[ 1 ] = 0.5

0 0.5 2 2
y[ 2 ] = 2 ) 2 = h[ 0 ] x[2] + h[ 1 ] x[ 1 ] + h[ 2 ] x[ 0 ] = 1 + 2h[ 2 ]
) h[ 2 ] = 0.5
0.5 2 0.5 2 2
y[ 3 ] = 3 = h[ 1 ] x[ 2 ] + h[ 2 ] x[ 1 ] + h[ 3 ] x[ 0 ] = 2 + 2h[ 3 ]
) h[ 3 ] = 0.5
0.5 2 0.5 2 0.5 2 2
y[ 4 ] = 3 = h[ 1 ] x[ 3 ] + h[ 2 ] x[ 2 ] + h[ 3 ] x[ 1 ] + h[ 4 ] x[ 0 ]
) h[ 4 ] = 0
PROB. Su18-Fin.3.

Consider the following system in which a continuous-time input x( t ) is sampled, producing x[ n ], then
filtered, producing y[ n ], and finally converted back to continuous time, yielding an overall output y( t ):

x( t ) IDEAL x[ n ] LTI SYSTEM y[ n ] IDEAL y( t )

C-to-D D-to-C
CONVERTER H( z ) CONVERTER

SAMPLE RATE fs

The system function of the discrete-time LTI system is H( z ) = 1 – z– 1 + z– 2.

(a) Which of the plots (A... H) on the next page shows the magnitude response of this filter?
C
Recognize this as a nulling filter, where –2cos( w
^ ) = –1
) look for nulls at w
^ = ±p/3
(b) A constant input x( t ) = 1 (for all t) will result in the constant output y( t ) = 1 (for all t).

The DC gain is b0 + b1 + b2 = 1–1+1 = 1

(c) Specify a sample rate fs = 6 samples/s

so that a C-to-D input of x( t ) = cos(10pt) + cos(14pt)

results in a zero output, y( t ) = 0 (for all t).

There are an infinite number of solutions, the largest being fs = 6 Hz:

^ 1 = 10p 5p –p
) w --------- = ------ = ------ + 2p
fs 3 3
both reduce to w
^ = ±p/3 ) both nulled
^ 2 = 14p
) w ------ = p
--------- = 7p --- + 2p
fs 3 3
 E

–p 0 p w
^ –p 0 p w
^

–p 0 p w
^ –p 0 p w
^

NULL at p/3
G

–p 0 p w
^ –p 0 p w
^

–p 0 p w
^ –p 0 p w
^
PROB. Su18-Fin.4.

Consider an FIR filter whose difference equation is:

y[ n ] = (1 + b)x[ n] + (1 – b)x[n – 1],

where b is an unspecified real constant.

If the output of this system in response to the sinusoidal input x[ n ] = cos(0.7pn)

is y[ n ] = cos(0.7pn + q),

then it must be that b = +0.2351 and q= –0.21p .

–0.2351 (alternative answer) –0.49p

The frequency response must have magnitude 1 at 0.7p

) 1 = jH(ej0.7p)j2 = j(1 + b) + (1 – b)e –j0.7pj2

= (1 + b)2 + (1 – b)2 + 2(1 + b)(1 – b)cos(0.7p)
= 2 + 2b2 + 2(1 – b2)cos(0.7p)

1 – 2 cos ( 0.7p -) = 0.0553

) b2 = –----------------------------------------- ) b = ±0.2351
2 – 2 cos ( 0.7p )

With b = +0.2351, the frequency response simplifies to

H(ej0.7p) = (1 + b) + (1 – b)e –j0.7p

= 1.2351 + 0.7649e –j0.7p = e –j0.2124p

Alternative answer: With b = –0.2351, the frequency response simplifies to

H(ej0.7p) = (1 + b) + (1 – b)e –j0.7p

= 0.7649 – 1.2351e –j0.7p = e –j0.4876p
PROB. Su18-Fin.5. Consider an LTI system described by the recursive difference equation:

y[ n ] = x[ n ] – 0.2y[n – 1].
(a) The system is [ IIR ][ FIR ] (circle one).

1
(b) The system transfer function is H( z ) = -------------------------- .
1 + 0.2z – 1

(c) Sketch below the pole-zero plot for H( z ), carefully labeling the locations of all zeros and poles:
Imfz g

z -:
H ( z ) = -----------------
z + 0.2

Refz g

)
LE
C
IR
C
IT
N
(U

(d) If the output of this system in response to the input

x[ n ] = 10d[ n ] + Bd[ n – 1] is y[ n ] = 10d[ n ]

(or in other words, as lists of numbers, the input .... 0 10 B 0 ... results in the output ... 0 10 0 ... ),

then it must be that the constant B is

B= 2 .

æ 1 + ----- B –1 ö
ç -z ÷
1
Y( z ) = X( z )H( z ) = (10 + Bz–1) -------------------------- = 10 ç -------------------------
10
-÷
1 + 0.2z – 1 ç
è 1 + 0.2z
– 1÷
ø

B
But Y( z ) = 10 ) ------ = 0.2 ) B=2
10
PROB. Su18-Fin.6.

Shown below are eleven 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 K) in each answer box.

B L = 2, N = 21 B

L = 2, N = 81 E
C

L = 3, N = 21 A
D
L = 3, N = 81 K

E
L = 4, N = 21 F

F L = 4, N = 81 H

L = 5, N = 21 G
G

L = 5, N = 81 C
H
L = 8, N = 81 J
I
L = 21, N = 21 D

J
L = 21, N = 81 I

K
PROB. Su18-Fin.7. (The different parts are unrelated.)

(a) There are an infinite number of values for the delay parameter t0 below such that:

cos(100pt) + cos(100p(t – t0)) + cos(100p(t – 2t0)) = 0, for all time t.

Name any two, both positive:

1-
-------- 1
t0 = , or t0 = ------ .
150 75

Phasor addition rule, where q = 100pt0:

In order for these
1 phasors to add to
0 = 1 + e –jq + e –j2q zero, they must
e –jq form a “peace
sign”, with equally
e –2jq spaced angles

) add to zero when either

2p 1 + 3m 1 4 7 10
q = ------ + m2p ) t0 = ------------------ 2f --------- , --------- , --------- , --------- , ...g
3 150 150 150 150 150
4p 2 + 3k 2 5 8 11
or q = ------ + k2p ) t0 = --------------- 2f --------- , --------- , --------- , --------- , ...g
3 150 150 150 150 150

P1 sin ( 0.9pk ) sin ( 0.75p ( n – k ) ) sin ( 0.75pn )

(b) Simplify --------------------------- . --------------------------------------------- = ------------------------------- .
pk p(n – k) pn
k= –1

a function of n

Recognize as the convolution of

sin ( 0.9pn )-
h[ n ] = --------------------------- with sin ( 0.75pn -) :
x[ n ] = ------------------------------
pn pn

y[ n ] = x[ n ] * h[ n ] ) Y( e jw^ ) = X( e jw^ )H( e jw^ ) = X( e jw^ )

(narrow)(wide)
(multiplying a narrow rectangle by a wide rectangle yields a narrow rectangle) ) y[ n ] = x[ n ].
 (c) The signal x( t ) = cos(88pt)cos(8pt)) + cos(240pt)
is periodic with fundamental frequency f0 = 8 Hz.

Multiplying the 44-Hz sinusoid by the 4-Hz sinusoid leads to the sum of two
sinusoids, one with frequency 40 Hz and the other with frequency 48 Hz, so
overall x( t ) has contributations at three frequencies:

40 Hz = (8)(5)
48 Hz = (8)(6)
120 Hz = (8)(15) ) greatest common factor is 8

(d) If fX[ 0 ], ... X[ 511]g are the DFT coefficients that result from taking a 512-point DFT
of the length-12 signal x[ n ] = [1 1 1 2 2 2 3 3 3 4 4 4] for n2f0, ... 11g,
then find numerical values for the following two DFT coefficients:

X[ 0 ] = 30 ,

P –jk2pn/N
X[ k ] = nx[n]e
X[ 256] = –2 .

P
)X[ 0 ] = nx[n] = 1 + 1 + 1 + 2 + 2 + 2 + 3 + 3 + 3 + 4 + 4 + 4 = 30
P –jpn
)X[256] = nx[n]e = 1 – 1 + 1 – 2 + 2 – 2 + 3 – 3 + 3 – 4 + 4 – 4 = –2