ECE 308 SIGNALS AND SYSTEMS FALL 2017HOMEWORK ASSIGNMENT #8

Solutions

1. (a) H(0) = 0, H(3) = 1, H(6) = 1, and H(13) = 0. Thus,

y(t) = 3 cos(3t)− 5 sin(6t− 30◦)

(b) We have

H(0) = 0

H(2k) =

{1, k = 1, 2, 30, k = 4, 5, 6, . . .

so that

y(t) = cos(2t) +1

2cos(4t) +

1

3cos(6t)

(c) Example 3.2 on p. 103 of the text shows that

x(t) =1

2+

2

π

∞∑k=1k odd

1

ksin

(kπ

2

)cos(kπt).

Since for k = 1, 2, 3, . . . we have 2 ≤ kπ ≤ 7 only for k = 1 and 2, and only k = 1is odd among these,

y(t) =2

πcos(πt).

2. (a) Since

H(ω) =1

jω + 1

|H(1)| =∣∣∣∣ 1

j + 1

∣∣∣∣ =

√2

2

6 H(1) = 61

j + 1= −π

4

we have

y(t) = |H(1)| cos(t+ 6 H(1))

y(t) =

√2

2cos(t− π

4

)(b)

y(t) = |H(1)| cos(t+ 45◦ + 6 H(1))

= |H(1)| cos(t+

π

4+ 6 H(1)

)y(t) =

√2

2cos(t)

3. (a)

H(0) = 1

H(50) =10

j50 + 10= 0.1961e−j1.3734

y(t) = 2 + 2(0.1961) cos(50t+ (π/2)− 1.3734)

y(t) = 2 + 0.3922 cos(50t+ 0.1974)

(b)

X(ω) =3

jω + 4

Y (ω) =30

(jω + 10)(jω + 4)

=5

jω + 4− 5

jω + 10

y(t) = 5e−4tu(t)− 5e−10tu(t)

4. We have

x(t) =∞∑

k=−∞

ckejkπt

with

ck =1

2

∫ 2

1

2(t− 1)e−jkπtdt =

∫ 2

1

(t− 1)e−jkπtdt

=

12, k = 0[(t−1)−jkπ −

1(jkπ)2

]e−jkπt

⌋21, k 6= 0

=

{12, k = 0

1−jkπ + 1

(kπ)2− 1

(kπ)2e−jkπ, k 6= 0

=

{12, k = 0

1−jkπ + 1

(kπ)2− 1

(kπ)2(−1)k, k 6= 0

(a) Using the expression above and the Matlab code posted at the course web site,we get the plot shown in Figure 1.

(b) The line spectra for x(t) are shown in Figure 2 and those for y(t) in Figure 3.

(c) The approximation of y(t) is shown in Figure 4.

Figure 1: Plot for problem 4(a): magnitude and phase of H(ω).

Figure 2: Plot for problem 4(b): line spectra of x(t).

Figure 3: Plot for problem 4(b): line spectra of y(t).

Figure 4: Plot for problem 4(c): approximation of y(t).

5. With

x(t) = 1.5 +∞∑k=1

(1

kπsin(kπt) +

2

kπcos(kπt)

),

we find (referencing Fig. P5.10 from the text) that

y(t) = a0 +∞∑k=1

(ak cos(kπt) + bk sin(kπt))

with

a0 =1

2

ak =

∫ 1

0

cos(kπt)dt =1

kπsin(kπt)

⌋10

= 0

bk =

∫ 1

0

sin(kπt)dt =−1

kπcos(kπt)

⌋10

=1

kπ(1− cos(kπ))

=

{2kπ, k odd

0, k even

Thus,

y(t) =1

2+

∞∑k=1k odd

2

kπsin(kπt)

x(t) = 1.5 +∞∑

k=−∞k 6=0

1

2

(2

kπ− j 1

kπ

)ejkπt

y(t) = 0.5 +∞∑

k=−∞k 6=0

−j2kπ

(1− cos(kπ)) ejkπt

so that

H(0) =0.5

1.5=

1

3

and

H(kπ) =

{ −j2kπ

1kπ−j 1

2kπ

, k odd

0, k even

=

{2−j45, k odd

0, k even

Therefore,

H(0) =1

3

H(π) = H(3π) = H(5π) = · · · = 2− j45

H(2π) = H(4π) = H(6π) = · · · = 0