ECE 216 Final examexams.skule.ca/exams/ECE216H1_20181_631555460258Final...ECE 216 Final exam April...
Transcript of ECE 216 Final examexams.skule.ca/exams/ECE216H1_20181_631555460258Final...ECE 216 Final exam April...
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Last name: First name: ID number:
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
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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
.
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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!.
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From here, we can see that
n+10X
k=n
x[k] ! X(ej!)1� ej11!
1� ej!.
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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!).
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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.
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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⇡].
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(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.
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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!).
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(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.)
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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
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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
.
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