GEORGIA INSTITUTE OF TECHNOLOGY
SCHOOL OF ELECTRICAL AND COMPUTER ENGINEERING
ECE 2026 — Summer 2023
Quiz #1
June 14, 2023
NAME:
GT username:
(FIRST)
(LAST)
(e.g., gtxyz123)
Important Notes:
◦ Do not unstaple the test.
◦ Closed book, except for one two-sided page (8.5”×11”) of hand-written notes.
◦ Calculators are allowed, but no smartphones/readers/watches/tablets/laptops/etc.
◦ JUSTIFY your reasoning CLEARLY to receive partial credit.
◦ Express all angles as a fraction of π. For example, write 0.1π 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. Write your answers in the provided answer boxes.
If more space is needed for scratch work, use the backs of the previous pages.
Problem
Value
1
25
2
25
3
25
4
25
Total
Score Earned
PROB. Su23-Q1.1. Shown below are twelve different diagrams — labeled A through L — that show the
locations of the powers {z0, z1, z2, z3, z4, z5, z6, z7, ... z89, z90} in the complex plane,
for twelve different values of z:
A
B
C
D
E
F
G
H
I
J
K
L
(The horizontal and vertical components represent the real and imaginary parts, respectively.
The axes are not labeled, the different plots are drawn with different scales. Only the shapes matter.)
(a)
(b)
Match each diagram above to the corresponding value of z listed below.
Indicate your answer by writing a letter (from {A ... L}) into each answer box.
(i)
z = 0.99e j0.02π
(ii)
z = 0.99e –j0.02π
(iii)
z = 1.01e j0.02π
(iv)
z = 1.01e –j0.02π
(v)
z = 0.99e j0.025π
(vi)
z = 0.99e –j0.025π
(vii)
z = 1.01e j0.025π
(viii)
z = 1.01e –j0.025π
(ix)
z = 0.99e j0.03π
(x)
z = 0.99e –j0.03π
(xi)
z = 1.01e j0.03π
(xii)
z = 1.01e –j0.03π
Explain your approach.
PROB. Su23-Q1.2.
Define three signals as follows:
• Let x1( t ) = 2cos(20πt + 0.5π) + 2cos(22πt + 0.5π).
• Let x2( t ) = Bsin(22πt)cos(2πFt), where B > 0 and F > 0 are unspecified.
• Let x3( t ) be the signal whose spectrum is shown in the sketch below:
j
–j –j
j
–12 –11
11
0
12
f (Hz)
If the sum of these three signals is a single sinusoid:
x1( t ) + x2( t ) + x3( t ) = Acos(2πf0 t + ϕ),
then it must be that:
B=
> 0,
F=
> 0,
Hz
and where (in standard form)
A=
> 0,
f0 =
> 0,
Hz
and ϕ =
∈(–π, π].
PROB. Su23-Q1.3.
Shown below is a plot of x( t ) =
x( t )
4
k=1
w
W
w
W
480πt
60k
------------- cos --------------- :
k+1
k
T
A
t
0
The scale is not labeled.
(a)
The time between nearest peaks is T =
(b)
The peak value is A =
(c)
The Fourier series for this signal is x( t ) =
.
.
where the fundamental frequency is f0 =
kake
jk2πf0 t
,
Hz.
Identify which of the following FS coefficients are nonzero with an X (Hint: only four are nonzero!):
a0
a1
a2
a3
a4
a5
a6
a7
a8
a9
a10
a11
a12
PROB. Su23-Q1.4. (The two parts of this problem are unrelated.)
(a)
Find numerical values for the constants A, B, C, and D so that the spectrogram of the signal
x( t ) = 2026cos(πAt + Bcos(πCt + D))
looks like this:
2000
FREQUENCY (Hz)
A=
B=
1000
C=
0
0
2
4
6
8
D=
TIME (secs)
(b)
A continuous-time sinusoid x( t ) = cos(2πf0t) whose unspecified frequency f0 is known to be in
the range 700 < f0 < 1700 Hz is sampled with sampling rate fs = 8000 Hz, resulting in a discretetime sequence:
n
fs
x[ n ] = x( ----- ).
Find the unique value for the frequncy f0 in the range f0∈(700, 1700) for which the resulting
discrete-time sequence is periodic with fundamental period N0 = 40,
i.e., so that N0 = 40 is the smallest positive integer N for which x[ n ] = x[n + N ] for all n.
f0 =
Hz
∈(700, 1700) .
GEORGIA INSTITUTE OF TECHNOLOGY
SCHOOL OF ELECTRICAL AND COMPUTER ENGINEERING
ECE 2026 — Summer 2023
Quiz #1
June 14, 2023
NAME:
ANSWER KEY
(FIRST)
GT username:
(LAST)
(e.g., gtxyz123)
Important Notes:
◦ Do not unstaple the test.
◦ Closed book, except for one two-sided page (8.5”×11”) of hand-written notes.
◦ Calculators are allowed, but no smartphones/readers/watches/tablets/laptops/etc.
◦ JUSTIFY your reasoning CLEARLY to receive partial credit.
◦ Express all angles as a fraction of π. For example, write 0.1π 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. Write your answers in the provided answer boxes.
If more space is needed for scratch work, use the backs of the previous pages.
Problem
Value
1
25
2
25
3
25
4
25
Total
Score Earned
PROB. Su23-Q1.1. Shown below are twelve different diagrams — labeled A through L — that show the
locations of the powers {z0, z1, z2, z3, z4, z5, z6, z7, ... z89, z90} in the complex plane,
for twelve different values of z:
0.99e j0.025π
1.01e-j0.025π
j0.02π
j0.025π
1.01e
1.01e
1.01e j0.03π
1.01e-j0.02π
F
E
D
B
C
A
0.99e j0.03π
0.99e-j0.03π
0.99e-j0.02π
H
G
1.01e-j0.03π
0.99e-j0.025π
J
I
K
0.99e j0.02π
L
(The horizontal and vertical components represent the real and imaginary parts, respectively.
The axes are not labeled, the different plots are drawn with different scales. Only the shapes matter.)
(a)
Match each diagram above to the corresponding value of z listed below.
Indicate your answer by writing a letter (from {A ... L}) into each answer box.
(i)
z = 0.99e j0.02π
–j0.02π
(ii)
I
z = 0.99e
(iii)
B
z = 1.01e j0.02π
–j0.02π
(iv)
D
z = 1.01e
(v)
F
z = 0.99e j0.025π
(vi)
J
z = 0.99e –j0.025π
(vii)
E
z = 1.01e j0.025π
(viii)
C
z = 1.01e –j0.025π
(ix)
G
z = 0.99e j0.03π
(x)
H
z = 0.99e –j0.03π
(xi)
A
z = 1.01e j0.03π
(xii)
K
z = 1.01e –j0.03π
Explain your approach.
• starting point is
z0
= 1, real, as highlighted
• subsequent powers determines magnitude:
expand ⇒ | z | = 1.01
contract ⇒ | z | = 0.99
• rotation direction determines sign of phase
B
(b)
L
⇒ phase positive
⇒ phase negative
1• #revolutions ∈{ ----2π 90(0.020π) = 0.9
1----2π 90(0.025π) = 1.125
1----2π 90(0.030π) = 1.35 }
determines
phase magnitude ∈{0.02π, 0.025π, 0.03π}
PROB. Su23-Q1.2.
Define three signals as follows:
• Let x1( t ) = 2cos(20πt + 0.5π) + 2cos(22πt + 0.5π).
• Let x2( t ) = Bsin(22πt)cos(2πFt), where B > 0 and F > 0 are unspecified.
• Let x3( t ) be the signal whose spectrum is shown in the sketch below:
j
–j –j
j
–12 –11
11
0
12
f (Hz)
If the sum of these three signals is a single sinusoid:
x1( t ) + x2( t ) + x3( t ) = Acos(2πf0 t + ϕ),
then it must be that:
B=
4
> 0,
1
F=
> 0,
Hz
and where (in standard form)
A=
4
> 0,
12
f0 =
> 0,
Hz
–0.5π
and ϕ =
∈(–π, π].
CANCEL WHEN B = 4
–j –j
–11 –10
jB/4 jB/4
–12
j
–10
j
–12 –11
SPECTRUM for x1( t )
0
SPECTRUM for x2( t )
WHEN F = 1
0
j
10
j
CANCEL
f (Hz)
11
–jB/4
–jB/4
10
12
–j –j
SPECTRUM for x3( t )
0
2j
11
12
–2j
SPECTRUM for SUM
–12
0
12
PROB. Su23-Q1.3.
Shown below is a plot of x( t ) =
x( t )
4
k=1
w
W
w
W
480πt
60k
------------- cos --------------- :
k+1
k
T
A
t
0
The scale is not labeled.
(a)
The time between nearest peaks is T =
0.05
.
240
The gcd of contributing frequencies at --------- ∈{240, 120, 80, 60}
k
is the fundamental freq ⇒ f0 = 20 Hz
1
= 0.05
⇒ fundamental period T0 = ---f
0
(b)
The peak value is A =
Evaluate at time 0 ⇒ x( 0 )
(c)
.
163
=
4
k=1
w
60k-----------k+1
W
------ + 120
--------- + 180
--------- + 240
--------cos(0) = 60
2
3
4
5
= 30 + 40 + 45 + 48 = 163
The Fourier series for this signal is x( t ) =
where the fundamental frequency is f0 =
kake
jk2πf0 t
,
Hz.
20
Identify which of the following FS coefficients are nonzero with an X (Hint: only four are nonzero!):
Contributing frequencies {240, 120, 80, 60} = {12f0, 6f0, 4f0, 3f0}
a0
a1
a2
a3
a4
X
X
a5
a6
X
a7
a8
a9
a10
a11
a12
X
PROB. Su23-Q1.4. (The two parts of this problem are unrelated.)
(a)
Find numerical values for the constants A, B, C, and D so that the spectrogram of the signal
x( t ) = 2026cos(πAt + Bcos(πCt + D))
looks like this:
2000
FREQUENCY (Hz)
The equation for this
sinusoid is
A=
2000
B=
8000
C=
0.25
D=
–0.5π
1000
fi( t ) = 1000 + 1000sin(0.25πt)
0
0
2
4
6
8
TIME (secs)
Multiply by 2π and integrate
⇒ ϕ( t ) = 2000πt + 8000sin(0.25πt)
= πAt + Bcos(πCt + D)
(b)
A continuous-time sinusoid x( t ) = cos(2πf0t) whose unspecified frequency f0 is known to be in
the range 700 < f0 < 1700 Hz is sampled with sampling rate fs = 8000 Hz, resulting in a discretetime sequence:
n
fs
x[ n ] = x( ----- ).
Find the unique value for the frequncy f0 in the range f0∈(700, 1700) for which the resulting
discrete-time sequence is periodic with fundamental period N0 = 40,
i.e., so that N0 = 40 is the smallest positive integer N for which x[ n ] = x[n + N ] for all n.
2πf n
n
f0 =
0
- ⇒ x[ n ] = cos( --------------Substite t = ----------).
8000
8000
1400
Hz
∈(700, 1700) .
Factor digital frequency as a ratio of integers times 2π:
ω̂ =
2πf -0
----------8000
=
k
---- (2π);
N
when fraction is reduced (k, N share no common factors), the denom N is the period.
⇒ f0 = k 8000
------------ = 200k, where k∈{1, 3, 7, 9, 11 ...} shares no factors with 40.
N
Of these, only k = 7 puts f0 in desired range
⇒
f0 = 200(7) = 1400 Hz.