Post on 30-May-2018
ECE 308 SIGNALS AND SYSTEMS FALL 2017Answers to selected problems on prior years’ examinations
Answers to problems on Midterm Examination #1, Spring 2009
1. x(t) = r(t+ 1)− r(t− 1)− u(t− 1)− 2 r(t− 2) + 2 r(t− 3) + u(t+ 3)
2. x[n] = δ[n+ 1] + 2 δ[n]− δ[n− 1] + δ[n− 3]
3.
Memoryless Linear Causal Time-invariant
System 1 YES NO YES NO
System 2 NO YES NO YES
4.
y[n] =
2, n = 04, n = 16, n = 2, 3, 4, 54, n = 62, n = 70, otherwise
5. (a)
y(t) =1
2
(1− e−2t
)u(t)
(b) One way to express the answer is
y(t) =1
2
(1− e−2(t+1)
)u(t+ 1)− 1
2
(1− e−2(t−3)
)u(t− 3)
and another is
y(t) =
0, t < −1
12
(1− e−2e−2t) , −1 ≤ t < 3
12
(e6 − e−2) e−2t, t ≥ 3
Answers to problems on Midterm Examination #2, Spring 2009
1. (a) X(ω) =∫∞−∞ x(t)e−jω tdt
(b) x(t) = 12π
∫∞−∞X(ω)ejω tdω
2. x(t) = 12
+∑∞
k=16π k
sin(π6k)
cos(k π
6t)
3. V (Ω) = 6ω
sin(
32ω)e−j(9/2)ω
4. (a)
(b) x(t) = 3 cos(6t+ π
4
)5. (a) V (ω) = 2
(jω+3)(jω+2)
(b) v(t) = [−2e−3t + 2e−2t]u(t)
Answers to problems on Midterm Examination #3, Spring 2009
1.
W (Ω) =(1− e−jΩN)2e−jΩ
(1− e−jΩ)2
2.
3.
y(t) = 3 +
√2
2cos (3t)−
√3
2sin(√
3t− π
6
)4. |H(ω)| is given by
-
6
−9 −6 −3 3 6 9
1
0
|H(ω)|
ω
5. y[n] = 23(−1)n
Answers to problems on the Final Examination, Spring 2009
1. x(t) = 2u(t+ 1)− r(t+ 1) + 2 r(t− 1)− r(t− 3)− u(t− 3)− r(t− 4) + r(t− 5)
2.
3. (a)
Xk =
0, k = 0, 2, 4, 62, k = 1, 3, 5, 7
(b)
x[n] =1
2cos(π
4n)
+1
2cos
(3π
4n
)4. y(t) = 3 cos(2t)
5. y(t) = (e−t − e−2t)u(t) + (e−(t−2) − e−2(t−2))u(t− 2)
6.
y(t) =
((t− 3
2
)e−t +
7
2e−3t
)u(t)
Answers to problems on Midterm Examination #1, Fall 2009
1. (a) x[n] = 2u[n] + 4u[n− 1]− 3u[n− 5]− 3u[n− 6]
(b) x[n] = 2 δ[n] + 6 (δ[n− 1] + δ[n− 2] + δ[n− 3] + δ[n− 4]) + 3 δ[n− 5]
2. (a) IS MEMORYLESS
(b) HAS MEMORY
(c) LINEAR
(d) LINEAR
(e) NONCAUSAL
(f) CAUSAL
(g) TIME-INVARIANT
(h) TIME-VARYING
3.
y[n] =
1, n = 13, n = 26, n = 34, n = 40, n = 5−6, n = 6−5, n = 7−3, n = 80, otherwise
4.
-
6y(t)
t @
@@@
−2 2 4 6 8
1
2
5. (a) y(t) = e−t (1− e−t)u(t)
(b) y(t) = e−(t−2)(1− e−(t−2)
)u(t− 2)
Answers to problems on Midterm Examination #2, Fall 2009
1. Sum / Sinusoids / Periodic / Finite
2.
x(t) =2
π+∞∑k=1
4
π(1− 4k2)cos(k2t)
3.
x(t) =3
πcos
(5t+
3π
4
)4.
x(t) = 3 e−2t u(t)− 3 e2t u(−t)
5. α = 1/2, ∆ = 1
Answers to problems on Midterm Examination #3, Fall 2009
1.
v[n] =
(1
2
)nu[n] +
(1
2
)−nu[−n]
or
v[n] =
(1
2
)nu[n] + 2nu[−n]
2. (a) X0 = 4, X1 = 2, X2 = 0, X3 = 2
(b) x[n] = 1 + cos(π2n)
3. (a) X(Ω) = 2 + e−jΩn + e−j3Ωn
(b)v[n] = x[n] + x[n− 4] + x[n− 8] + · · ·+ x[n− 76]
or
v[n] =19∑k=0
x[n− 4k]
(c)
V (Ω) = X(Ω)e−j38Ω sin(40Ω)
sin(2Ω)
4.y(t) = 8− 4 cos
(3t− π
2
)+ 4 cos(2000t− π)
5. Set H(ω) to be
-
6
−2 −1 1 2
1
0
H(ω)
ω
Answers to problems on the Final Examination, Fall 2009
1.
n y[n]
−4 0
−3 0
−2 0
−1 0
0 0
n y[n]
1 1
2 2
3 1
4 0
5 -1
n y[n]
6 -2
7 -1
8 0
9 0
10 0
2. (a) y1(t) 6= y2(t)
(b) Y1(ω) = 12e−j(ω+1)X(ω + 1) + 1
2e−j(ω−1)X(ω − 1)
Y2(ω) = 12e−jω [X(ω + 1) +X(ω − 1)]
(c) Both systems are linear.
(d) Both systems are time-varying.
3. y[n] = 23
cos(π n)
4. y(t) =[−e−5(t−1) + e−3(t−1)
]u(t− 1)
5. y(t) = 1 + e−t sin(2t), t ≥ 0
Answers to problems on the Midterm Examination #1, Spring 2011
1. x(t) = −r(t+ 1) + r(t− 1) + r(t− 2)− r(t− 4) + u(t− 4)− u(t− 6)
2.
y[n] =
0, n ≤ 02, n = 14, n = 23, n = 32, n = 42, n = 50, n = 6−2, n = 7−1, n = 80, n =≥ 9
3. (a)
y(t) =1
2e−
12tu(t)
(b)
y(t) =1
2e−
12
(t−2)u(t− 2)
(c)
y(t) =(
1− e−12t)u(t)
(d)
y(t) = 3(
1− e−12
(t−2))u(t− 2)
4.y(t) =
(1− e−6(t−3)
)u(t− 3)−
(1− e−6(t−6)
)u(t− 6)
5.
y[n] =
(1
3
)nu[n]
Answers to problems on the Midterm Examination #2, Spring 2011
1.
x1(t) =1
2+∞∑k=1
(−2
(kπ)2
)(1− (−1)k
)cos(kπt)
x2(t) =∞∑k=1
2
kπ
(1− (−1)k
)sin(kπt)
2. (a)
sgn(t) = u(t)− u(−t)
SGN(ω) =
(1
jω+ πδ(ω)
)−(
1
−jω+ πδ(−ω)
)=
2
jω
(b)1
t←→ jπsgn(−ω)
3.
Y (ω) = −jAπδ(ω − ω0) + jAπδ(ω + ω0)
y(t) = A sin(ω0t)
4.
X(Ω) =−1
2e−jΩ(
1− 12e−jΩ
)2
5. X(4π) = 2
6. Even
Answers to problems on the Final Examination, Spring 2011
1. (a) y(t) = (e−t − e−2t)u(t)
(b) y(t) = (e−(t−2) − e−2(t−2))u(t− 2)
(c) y(t) = (e−t − e−2t)u(t)− e−4(e−(t−2) − e−2(t−2))u(t− 2)
2. y[n] = 12 +√
2 cos(π2n− π
4
)3. (a) y(t) = a0 +
√2
2
(a1 cos
(4t− π
4
)+ b1 sin
(4t− π
4
))(b) y(t) = 3
2+ 3
√2
πcos(4t− π
4
)4.
Y (ω) =3ej4
(ω − 2)2 + 9+
3e−j4
(ω + 2)2 + 9
5. y(t) = 1− 2e−2t + 2e−4t for t ≥ 0
Answers to problems on the Midterm Examination #1, Fall 2011
1. (a) TRUE
(b) TRUE
(c) TRUE
(d) TRUE
(e) TRUE
(f) TRUE
2.
-
6
• • •
• •
•
• •
• •
• • •
-6 -4 -2 2 4 6
2
4
-2
-4
n
x[n]
3. x(t) = r(t+ 3)− 2u(t+ 2)− r(t+ 1) + u(t+ 1) + r(t− 2)− r(t− 3)− 2u(t− 4)
4. y[n] =(2−
(12
)n)u[n]
5. (a) y(t) = (1− e−t)u(t)− (1− e−(t−3))u(t− 3)
(b) y(t) = 3(1− e−(t−2))u(t− 2)
(c) y(t) = (1− e−t)u(t)
Answers to problems on the Midterm Examination #2, Fall 2011
1.
x(t) =∞∑k=1
2
kπ(1− cos(kπ∆)) sin(kπt)
2. (a) X(ω) = 2/(1 + ω2)
(b) y(t) = 2/(1 + t2)
3.
-
6
@@@@@@@@
-2 -1 1 2
−π2
π2
ω
Y (ω)
4. x(t) = 1π
cos(ω0t+ θ)
5. (a)
y(t+ 2) = x(t+ 2) cos(2π(t+ 2))
= x(t) cos(2πt+ 4π)
= x(t) cos(2πt)
= y(t)
(b) dk = 12
(ck−2 + ck+2)
Answers to problems on the Midterm Examination #3, Fall 2011
1. (a)
Y (Ω) =12e−j2Ω(
1− 12e−jΩ
)2
(b) y[n] =(
12
)ncos(π2n)u[n]
2. (a) X5 = 12
(b) For k = 0, 1, . . . , 4 and for k = 6, 7, . . . , 11, Xk = 0
3. (a)
H(Ω) =1
1− 14e−jΩ
(b)
y[n] =
(5
(1
4
)n− 4
(1
5
)n)u[n]
4.
y(t) =27
4+
5
4cos(
2t− π
4
)5. y[n] = 1
2cos(π2n)
Answers to problems on the Final Examination, Fall 2011
1. (a) H(Ω) = 1 + cos(Ω)
(b) y[n] = 2
2. Y (ω) = 1jω+3
3. y(t) = 1 + 4π
cos(πt)
4. (a) ω0 = 1
(b) y(t) = −6 cos(t)
5. (a) y(t) = (1 + e−2t − 2e−3t)u(t)
(b) y(t) = u(t)
Answers to problems on the Midterm Examination #1, Fall 2012
1. (a) TRUE
(b) TRUE
(c) TRUE
(d) TRUE
(e) FALSE
(f) TRUE
2. (a)
-
6
• •
•
•
•
•
•
•
•
• •
-4 -2 2 4 6
2
4
n
x[n]
(b) The plot for part (b) is the same as for part (a).
3.
x(t) = 2u(t+ 3)− 2r(t+ 3) + 4r(t+ 2)− 2u(t+ 1)− 2r(t+ 1)
+ u(t− 1)− 2u(t− 2) + 3r(t− 2)− 3r(t− 3)− 2u(t− 4)
4. y[n] = 503
((1.06)n+1 − 1)u[n]
5. (a) y(t) = (e−t − e−2t)u(t)
(b) y(t) = (e−t − e−2t)u(t)− e−4(e−(t−2) − e−2(t−2))u(t− 2)
Answers to problems on the Midterm Examination #2, Fall 2012
1. (a)
x(t) =∞∑
k=−∞k 6=0
− j
kπ
(1− cos
(kπ
3
))ejkπt
(b)
y(t) =∞∑k=1
2
kπ
(cos
(kπ
3
)− 1
)sin(k2πt)
2. (a)e−1/2e−j2ω
jω + (1/4)
(b)e−1/2
2π(jt− (1/4))
3. (a) In this problem you derive that x(t) = 1.
(b) x(t) = 3 cos(6t+ (π/4))
4. (a) v(t) = −jtx(t)
(b) y(t) = −jtej2tx(t)
5. (a) We have
Z(ω) = XL(ω) +XR(ω) +XL(ω− ω1) +XL(ω + ω1)−XR(ω− ω1)−XR(ω + ω1)
where ω1 = 76000π.
(b)
-
6
@@@@
@@
@@
@@@@
−30, 000π 30, 000π
3
1
ω
Z(ω)
(c) y1(t)
Answers to problems on the Midterm Examination #3, Fall 2012
1. (a)
Y (Ω) = X(Ω) +1
2X(−Ω)ejΩ
(b)
Y (Ω) =3/4
(5/4)− cos(Ω)
Y (Ω) is even and purely real.
2. (a) x[n] is periodic with period N = 4.
(b) 4c2
(c) X2 = 4c2 and X3 = 4c3
3.
2
(1
2
)nu[n]
4.
yA(t) = −5
2cos(
2t+π
4
)yB(t) = −16
9+
5
6cos(3t− π)
5.
y(t) = 2 cos(6t) +30
35cos(10t) +
10
21cos(14t)
Answers to problems on the Final Examination, Fall 2012
1. (a) y(t) = y0(t)− y0(t− 1) = u(t)− u(t− 1)− u(t− 4) + u(t− 5)
(b) x2(t) = x0(t+1)−2x0(t−1) so that y(t) = u(t+1)−2u(t−1)−u(t−3)+2u(t−5)
2. y[n] = 4A cos(π2n)
if x[n] = x1[n], and y[n] = 0 if x[n] = x2[n], so that we differentiatebetween the two possibilities based on whether or not y[n] = 0 or a multiple of cos
(π2n).
3.
y(t) = 2 +6√
3
πcos(πt)
4.y(t) = 4
(e−2t − e−4t
)u(t)− 4e−12
(e−2(t−3) − e−4(t−3)
)u(t− 3)
5.
y(t) =
(1− 5
2e−3t +
3
2e−5t
)u(t)
Answers to problems on Midterm Examination #1, Spring 2013
1.
x(t) = −r(t+ 1) + 2r(t)− 2r(t− 1) + r(t− 2) + 2u(t− 2)
− r(t− 3) + r(t− 4) + u(t− 4)− u(t− 6)
2.
-
6x[n]
n• • • •
• •
•• •
• •
•• • • • • • •
−6 −4 −2 2 4 6 8−1
1
2
3. (a) Causal, time-invariant
(b) Noncausal, time-invariant
(c) Causal, time-invariant
(d) Causal, time-varying
4.
y[n] =
9, n = 13, n = 27, n = 33, n = 42, n = 50, otherwise
5. (a)y(t) = 6
(e−2t − e−3t
)u(t)
(b)y(t) = 2
(e−2(t−1) − e−3(t−1)
)u(t− 1)− 2
(e−2(t−2) − e−3(t−2)
)u(t− 2)
6. (a)
y(t) =
t, 0 ≤ t < 24− t, 2 ≤ t < 40, t < 0 or t ≥ 4
(b)x(t) = p(t− 1)− p(t− 4)− p(t− 7)
Answers to problems on Midterm Examination #2, Spring 2013
1.
x(t) = 6 +∞∑
k=−∞k 6=0
8
kπsin
(3kπ
4
)ej
kπ4t
2.
xe(t) =3
2+∞∑`=1
4
(2`− 1)2π2cos ((2`− 1)πt)
xo(t) =∞∑k=1
(−1)k2
kπsin(kπt)
3. (a)
Y (ω) =12ω2
(9 + ω2)2
(b) TRUE
4. The sketches for (a) and (b) are as follows.
(The curved portion of the plot in (b) has formula 2 + 2 cos(πω/4).)
5. The sketch of Y (ω) is as follows.
-
6
@@@@AAAA
-2 -1 1 2 3 4
4π2
ω
Y (ω)
Answers to problems on Midterm Examination #3, Spring 2013
1.
Y (Ω) =1/8(
1 + 14e−jΩ
)2
2. (a) X0, X1, X2, X3 = 4, j2, 0,−j2(b) c0 = 1, c1 = j/2, c2 = 0, c3 = −j/2(c) x[n] = 1− sin(πn/2)
3.
y(t) = 8− 4 cos(
0.5t− π
2
)+ 4 sin
(√3
2t− π
3
)4.
y(t) = 4 cos(5t) + cos(10t) +4
9cos(15t)
5.
y[n] =3
4(−1)n
Answers to problems on the Final Examination, Spring 2013
1. (a)
y(t) =1
2
(1− e−2t
)u(t)
(b)
y(t) =1
2
(1− e−2(t−2)
)u(t− 2)− 1
2
(1− e−2(t−4)
)u(t− 4)
2. y(t) = 3 cos(2t)
3. X(ω) is purely real and with phase equal to 0 for all ω. The magnitude |X(ω)| is thesame as X(ω), shown in the plot below.
4. The output that results when p[n] is the input is shown below.
-
6
• • • • •••• •
••• • •
y[n]
n
4
2
−2 2 4 6 8
Therefore, the smallest n0 with no overlap of pulse responses is 6.
5. (a) X(0) = 4, X(π/2) = 2, X(π) = 0, X(3π/2) = 2
(b) X0 = 4, X1 = 2, X2 = 0, X3 = 2
Answers to problems on Midterm Examination #1, Fall 2014
1.x(t) = r(t+ 1)− r(t)− 2u(t− 2) + r(t− 4)− r(t− 5)
2.
-
6x(t)
t
AAAA
@@−5 −3 −1 1 3 5 7
−2
2
4
3. (a) y(t) = 2r(t)− 4r(t− 2) + 2r(t− 4)
(b)
-
6y(t)
t@
@@@
1 3 5
2
4
4.
y(t) =
4π
(1 + sin
(π2t)), −1 < t < 3
0, otherwise
5. (a)
H(ω) =2
2 + jω
(b)
y(t) = 3√
2 + cos
(t− 3π
4
)6. y(t) = 2u(t− 2)− 2u(t− 4)
Answers to problems on Midterm Examination #2, Fall 2014
1. (a)
y(t) =
(3
2e−3t − 3e−4t +
3
2e−5t
)u(t)
(b)
y(t) =
(3
2e−3(t−2) − 3e−4(t−2) +
3
2e−5(t−2)
)u(t− 2)
2.y(t) = 1− 2e−2t + 2e−4t, t ≥ 0
3.
y(t) =
(3t− 3
4+
3
4e−4t
)u(t)
4. A sketch of the block diagram is as follows.
5. (a) No. There is a pole at s = 0.
(b) y(t) = 2tu(t)
Answers to problems on Midterm Examination #3, Fall 2014
1.
x(t) =∞∑n=1
2
nπ
(1− cos
(nπ3
))sin(nπt)
2. (a)
X(ω) =4
4 + ω2
(b)
y(t) =4
4 + t2
3.
Y (ω) =
4(ω + 3) (1− (ω + 3)2) , |ω + 3| ≤ 1−4(ω − 3) (1− (ω − 3)2) , |ω − 3| ≤ 10, otherwise
4.
y(t) =27
4+
5
4cos(
2t− π
4
)
Answers to problems on the Final Examination, Fall 2014
1.
y(t) = (1− e2t)u(t) + (1− e2(t−1))u(t− 1)− (1− e2(t−2))u(t− 2)− (1− e2(t−3))u(t− 3)
2.
y(t) =
(7
2− 2e−t − 3
2e−2t
)u(t)
3. (a)
H(ω) =
∫ ∞−∞
h(t)e−jωtdt =
∫ 10
0
e−jωtdt =1
−jω[e−j10ω − 1
]=
1
jω(1− e−j10ω) =
1
jω(ej5ω − e−j5ω)e−j5ω
=j2 sin(5ω)
jωe−j5ω =
2 sin(5ω)
ωe−j5ω
(b)
y(t) =10
ln(2)
4.
y[n] =3
2cos
(2π
5n
)
Answers to problems on Midterm Examination #1, Fall 2015
1.x(t) = −r(t+ 1) + r(t) + 2u(t− 2)− r(t− 4) + r(t− 5)
2.
-
6x(t)
t
−5 −3 −1 1 3 5 7
1
2
3. (a) Linear, time-invariant, causal, has memory
(b) Linear, time-invarient, non-causal, has memory
4.
-
6y[n]
n• •
•••• • •
•••• •
−2 2 4
6 8
−1
1
2
5.
y(t) =
0, t < 012t2, 0 ≤ t < 1
−12t2 + 2t− 1, 1 ≤ t < 2
1, 2 ≤ t < 4
5− t, 4 ≤ t < 5
0, 5 ≤ t
6.
y(t) =
0, t < 212
(1− e−2(t−2)
), 2 ≤ t < 6
12(1− e−8), 6 ≤ t
7.y(t) = 2
[e−(t−1)u(t− 1)− e−(t−3)u(t− 3)
]
Answers to problems on Midterm Examination #2, Fall 2015
1. (a) T = 2
(b)
x(t) =1
2+∞∑k=1
2
kπsin
(kπ
2
)cos(kπt)
(c) Even symmetry
2. (a)
X(ω) =
(12
+ jω)e−( 1
2+jω)
374
+ jω − ω2
(b)
y(t) =
(12− jt
)e−( 1
2−jt)
374− jt− t2
3.x(t) = 8 cos
(4t+
π
6
)4. (a) v(t) = 2x(t− 2)
(b) y(t) = x(t) ∗ (e−tu(t))
5.
X(Ω) =3/4
(5/4)− cos(Ω)
Answers to problems on Midterm Examination #1, Fall 2016
1. (a) TRUE
(b) FALSE
(c) TRUE
(d) TRUE
(e) TRUE
(f) TRUE
2. (a)
-
6x[n]
n• •
•
•
•
•
•
•• • •
−4 −2 1 3 5
3
1
2
(b)
-
6x[n]
n• •
•
• • •
•• •
• • •−4 −2 2 4 6
3
1
2
3. x(t) = 2u(t+ 3)− r(t+ 1) + 4r(t− 2)− 3r(t− 3)− 2u(t− 4)
4. (a)
y[n] =
(1−
(1
2
)n)u[n− 1]
(b)
y[n] =
(1−
(1
2
)n)u[n− 1]−
(1−
(1
2
)n−2)u[n− 3]
5. (a)y(t) =
(e−2t − e−3t
)u(t)
(b)y(t) =
(e−2t − e−3t
)u(t)
Answers to problems on the Final Examination, Fall 2015
1.
y(t) = 2 +9
4cos(
4t− π
2
)2.
y(t) =
√3
2cos(π
2n− π
6
)3.
y(t) =8
3+
8√
3
πcos(π
2t)
4.y(t) = 6(e−t − e−2t)u(t)− 6(e−(t−3) − e−2(t−3))u(t− 3)
5.y(t) =
(1− 2e−2t + e−4t
)u(t)
Answers to problems on Midterm Examination #2, Fall 2016
1. (a)
X(ω) =
∫ ∞−∞
x(t)e−jωtdt
(b)
x(t) =1
2π
∫ ∞−∞
X(ω)ejωtdω
(c) ck is given by
ck =1
T
∫ T
0
x(t)e−jkω0tdt
where ω0 = 2π/T
(d) x(t) is given by
x(t) =∞∑
k=−∞
ckejkω0t
where ω0 = 2π/T .
2.
x(t) =4
3+∞∑k=1
8
kπsin
(kπ
3
)cos
(kπ
3t
)3. (a) v(t) = t x(t)
(b)
V (ω) = jω cos
(ω2
)− 2 sin
(ω2
)ω2
4. x(t) = 1− 3 sin(5t)
5. (a)
Y (ω) =6
(jω + 4)(jω + 1)
(b)y(t) = 2(e−t − e−4t)u(t)
Answers to problems on Midterm Examination #3, Fall 2016
1.
Y (Ω) =e−j4Ω(
1− 13e−jΩ
)2
2.
x[n] =3
4cos(π
2n)
3.
y(t) = 6 + cos
(1
2t− 7π
12
)− 6 sin
(√3
2t− π
2
)4.
y(t) = 2 cos(πt)
5.
y[n] =√
2 cos
(π
2n− 5π
16
)
Answers to problems on the Final Examination, Fall 2016
1. (a)
y(t) =1
3(1− e−3t)u(t)
(b)y(t) = (1− e−3(t−2))u(t− 2)
(c)(1− e−3(t−2))u(t− 2)− (1− e−3(t−6))u(t− 6)
2.y(t) = 3
3.
y(t) =1
πsin(πt)
4. (a) X(0) = 6, X(π/2) = 0, X(π) = −2, X(3π/2) = 0
(b) X0 = 6, X1 = 0, X2 = −2, X3 = 0