ECE 308 SIGNALS AND SYSTEMS FALL 2017gaw/ece308/ECE308_old_exam_answers.pdfECE 308 SIGNALS AND...

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


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)


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)


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


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


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)


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)


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)


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)


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)


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)


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)


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)


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)


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)


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


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


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)


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)


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

)


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

)


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)

]


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)


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)


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)


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)


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

)


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

ECE 308 SIGNALS AND SYSTEMS FALL 2017gaw/ece308/ECE308_old_exam_answers.pdfECE 308 SIGNALS AND...

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