Signals and systems HW1 solution

1.

Probs. 1.6 (b)

The signal is not periodic.

A periodic continuous-time signal 𝑥(𝑡) has the property that there is a

positive value of 𝑻 for which

𝑥(𝑡) = 𝑥(𝑡 + 𝑇)

for all values of 𝑡.

For this problem,

𝑥2(𝑡) = 𝑥2(𝑡 + 𝑇)

⟹ 𝑒(−1+𝑗)𝑡 = 𝑒(−1+𝑗)(𝑡+𝑇)

⟹ 𝑒(−1+𝑗)𝑡 = 𝑒(−1+𝑗)𝑡+(−1+𝑗)𝑇

⟹ 1 = 𝑒(−1+𝑗)𝑇

⟹ 0 = (−1 + 𝑗)𝑇

⟹ 𝑇 = 0

Since 𝑻 is not a positive value, we conclude that the signal 𝑥2(𝑡) is not

periodic.

Probs. 1.6 (d)

The signal is periodic.

𝑥4[𝑛] = 3𝑒𝑗3𝜋(𝑛+12

) 5⁄= 3𝑒𝑗

3𝜋10 ∙ 𝑒𝑗

3𝜋5

𝑛

The general discrete-time complex exponential signal can be expressed in the

form

𝑥[𝑛] = 𝐶𝛼𝑛

where

𝐶 = |𝐶|𝑒𝑗𝜃

and

α = |𝛼|𝑒𝑗𝜔0

Then

𝑥4[𝑛] = 3𝑒𝑗3𝜋10 ∙ 𝑒𝑗

3𝜋5

𝑛 = 𝐶𝛼𝑛

where

𝐶 = 3𝑒𝑗3𝜋10 = |𝐶|𝑒𝑗𝜃

and

α = 𝑒𝑗3𝜋5 = |𝛼|𝑒𝑗𝜔0 ⟹ 𝜔0 =

3𝜋

5

Since |α| = 1, the fundamental period is given by

𝑁 = 𝑚 (2𝜋

𝜔0) = 𝑚 (

2𝜋

3𝜋 5⁄) = 𝑚 (

10

3)

By choosing 𝑚 = 3, we obtain the fundamental period to be 10.

Probs. 1.25 (b)

The signal is periodic.

𝑥(𝑡) = 𝑒𝑗(𝜋𝑡−1) = 𝑒−𝑗 ∙ 𝑒𝑗𝜋𝑡

The general continuous-time complex exponential signal can be expressed in

the form

𝑥(𝑡) = 𝐶𝑒𝛼𝑡

where

𝐶 = |𝐶|𝑒𝑗𝜃

and

α = 𝑟 + 𝑗𝜔0

Then,

𝑥(𝑡) = 𝑒−𝑗 ∙ 𝑒𝑗𝜋𝑡 = 𝐶𝑒𝛼𝑡

where

𝐶 = 𝑒−𝑗 = |𝐶|𝑒𝑗𝜃

and

α = 𝑗𝜋 = 𝑟 + 𝑗𝜔0 ⟹ 𝜔0 = 𝜋

Since α is purely imaginary, the fundamental period is given by

𝑇 =2𝜋

𝜔0=

2𝜋

𝜋= 2

We obtain the fundamental period to be 2.

Probs. 1.25 (c)

The signal is periodic.

𝑥(𝑡) = [cos (2𝑡 −𝜋

3)]

2

=1 + cos (4𝑡 −

2𝜋3 )

2=

1

2+

1

2cos (4𝑡 −

2𝜋

3)

The general continuous-time sinusoidal signal can be expressed in the form

𝑥(𝑡) = 𝐴 cos(𝜔0𝑡 + 𝜙)

Then, the frequency for 1

2cos (4𝑡 −

2𝜋

3) is 𝜔0 = 4.

The fundamental period is given by

𝑇 =2𝜋

𝜔0=

2𝜋

4=

𝜋

2

We obtain the fundamental period to be 𝝅

𝟐.

Probs. 1.25 (f)

The signal is not periodic.

𝑥(𝑡) = ∑ 𝑒−(2𝑡−𝑛)𝑢(2𝑡 − 𝑛)

∞

𝑛=−∞

𝑥(𝑡) = 𝑥(𝑡 + 𝑇)

⇒ ∑ 𝑒−(2𝑡−𝑛)𝑢(2𝑡 − 𝑛)

∞

𝑛=−∞

= ∑ 𝑒−(2(𝑡+𝑇)−𝑛)𝑢(2(𝑡 + 𝑇) − 𝑛)

∞

𝑛=−∞

⇒ ∑ 𝑒−(2𝑡−𝑛)𝑢(2𝑡 − 𝑛)

∞

𝑛=−∞

= 𝑒−2𝑇 ∑ 𝑒−(2𝑡−𝑛)𝑢(2(𝑡 + 𝑇) − 𝑛)

∞

𝑛=−∞

⇒ ∑ 𝑒−(2𝑡−𝑛)[𝑢(2𝑡 − 𝑛) − 𝑒−2𝑇𝑢(2(𝑡 + 𝑇) − 𝑛)]

∞

𝑛=−∞

= 0

⇒ 𝑢(2𝑡 − 𝑛) − 𝑒−2𝑇𝑢(2(𝑡 + 𝑇) − 𝑛) = 0

⇒ 𝑇 = 0

Since 𝑻 is not a positive value, we conclude that the signal 𝑥(𝑡) is not

periodic.

Probs. 1.26 (c)

The signal is periodic.

𝑥[𝑛] = cos (𝜋

8𝑛2)

𝑥[𝑛] = 𝑥[𝑛 + 𝑁]

⟹ cos (𝜋

8𝑛2) = cos (

𝜋

8(𝑛 + 𝑁)2)

⟹ cos (𝜋

8𝑛2) = cos (

𝜋

8𝑛2 +

𝜋

4𝑛𝑁 +

𝜋

8𝑁2)

⟹𝜋

4𝑛𝑁 +

𝜋

8𝑁2 = 2𝜋𝑘, 𝑘 = 0, ±1, ±2, …

⟹𝜋

4𝑛𝑁 = 2𝜋𝑘1, 𝑘1 = 0, ±1, ±2, … 𝑎𝑛𝑑

𝜋

8𝑁2 = 2𝜋𝑘2, 𝑘2 = 0, ±1, ±2, …

For 𝜋

4𝑛𝑁 = 2𝜋𝑘1, 𝑘1 = 0, ±1, ±2, …

𝑛𝑁

8= 𝑘1 ⟹ 𝑁 = 8

For 𝜋

8𝑁2 = 2𝜋𝑘2, 𝑘2 = 0, ±1, ±2, …

𝑁2

16= 𝑘2 ⟹ 𝑁 = 4

Thus, the fundamental period is the least common multiple of 8 and 4

which is

lcm(8,4) = 8

The answer is 8.

Probs. 1.26 (d)

The signal is periodic.

𝑥[𝑛] = cos (𝜋

2𝑛) cos (

𝜋

4𝑛)

= (1

2𝑒𝑗

𝜋2

𝑛 +1

2𝑒−𝑗

𝜋2

𝑛) (1

2𝑒𝑗

𝜋4

𝑛 +1

2𝑒−𝑗

𝜋4

𝑛)

=1

4𝑒𝑗

3𝜋4

𝑛 +1

4𝑒𝑗

𝜋4

𝑛 +1

4𝑒−𝑗

𝜋4

𝑛 +1

4𝑒−𝑗

3𝜋4

𝑛

For 1

4𝑒𝑗

3𝜋

4𝑛

, 𝑁 = 𝑚2𝜋

(3𝜋 4⁄ )= 𝑚

8

3= 8

For 1

4𝑒𝑗

𝜋

4𝑛

, 𝑁 = 𝑚2𝜋

(𝜋 4⁄ )= 8

For 1

4𝑒−𝑗

3𝜋

4𝑛 =

1

4𝑒𝑗(−

3𝜋

4+2𝜋)𝑛 =

1

4𝑒𝑗

5𝜋

4𝑛

, 𝑁 = 𝑚2𝜋

(5𝜋 4⁄ )= 𝑚

8

5= 8

For 1

4𝑒−𝑗

𝜋

4𝑛 =

1

4𝑒𝑗(−

𝜋

4+2𝜋)𝑛 =

1

4𝑒𝑗

7𝜋

4𝑛

, 𝑁 = 𝑚2𝜋

(7𝜋 4⁄ )= 𝑚

8

7= 8

The answer is 8.

Probs. 1.26 (e)

The signal is periodic.

𝑥[𝑛] = 2 cos (𝜋

4𝑛) + sin (

𝜋

8𝑛) − 2 cos (

𝜋

2𝑛 +

𝜋

6)

2 cos (𝜋

4𝑛) has a fundamental period of 𝑁 = 𝑚

2𝜋

(𝜋 4⁄ )= 8.

sin (𝜋

8𝑛) has a fundamental period of 𝑁 = 𝑚

2𝜋

(𝜋 8⁄ )= 16.

−2 cos (𝜋

2𝑛 +

𝜋

6) has a fundamental period of 𝑁 = 𝑚

2𝜋

(𝜋 2⁄ )= 4.

Thus, the fundamental period is the least common multiple of 8, 16 and 4

which is

lcm(8,16,4) = 16

The answer is 16.

2.

Probs. 1.21 (c)

Probs. 1.21 (d)

Probs. 1.21 (e)

Probs. 1.21 (f)

Probs. 1.22 (d)

Probs. 1.22 (e)

The answer is the same as the original signal.

𝑥[𝑛]𝑢[3 − 𝑛] = 𝑥[𝑛]

Probs. 1.22 (f)

Probs. 1.22 (g)

3.

Probs. 1.23 (a)

Probs. 1.23 (b)

Probs. 1.24 (b)

Even

Odd

Probs. 1.24 (c)

Even

Odd

4.

Probs. 1.27 (b)

y(𝑡) = [cos(3𝑡)]𝑥(𝑡)

(1)

The system is memoryless, as the value of y(𝑡) at any particular time

depends only on the value of 𝑥(𝑡) at that time, not 𝑥(𝑡 + 𝜏) or 𝑥(𝑡 − 𝜏).

(2)

y(𝑡 − 𝑡0) = [cos(3(𝑡 − 𝑡0))]𝑥(𝑡 − 𝑡0)

Let 𝑥2(𝑡) = 𝑥(𝑡 − 𝑡0)

𝑥2(𝑡) → 𝑦2(𝑡) = [cos(3𝑡)]𝑥2(𝑡) = [cos(3𝑡)]𝑥(𝑡 − 𝑡0)

≠ [cos(3(𝑡 − 𝑡0))]𝑥(𝑡 − 𝑡0) = y(𝑡 − 𝑡0)

The system is not time invariant, as 𝑥(𝑡 − 𝑡0) ↛ y(𝑡 − 𝑡0).

(3)

Assume

𝑥1(𝑡) → 𝑦1(𝑡) = [cos(3𝑡)]𝑥1(𝑡)

𝑥2(𝑡) → 𝑦2(𝑡) = [cos(3𝑡)]𝑥2(𝑡)

Let

𝑥3(𝑡) = α𝑥1(𝑡) + 𝛽𝑥2(𝑡)

Then

𝑥3(𝑡) → 𝑦3(𝑡) = [cos(3𝑡)]𝑥3(𝑡) = [cos(3𝑡)][α𝑥1(𝑡) + 𝛽𝑥2(𝑡)]

= α[cos(3𝑡)]𝑥1(𝑡) + 𝛽[cos(3𝑡)]𝑥2(𝑡) = α𝑦1(𝑡) + 𝛽𝑦2(𝑡)

The system is linear, as α𝑥1(𝑡) + 𝛽𝑥2(𝑡) → α𝑦1(𝑡) + 𝛽𝑦2(𝑡).

(4)

The system is causal, as the value of y(𝑡) at any particular time depends only

on the value of 𝑥(𝑡) at present and past time, not 𝑥(𝑡 + 𝜏) for 𝜏 > 0.

(5)

Let

|𝑥(𝑡)| < 𝛽

Since

|cos(3𝑡)| < 1

Then

|cos(3𝑡)||𝑥(𝑡)| < 𝛽 ⇒ |𝑦(𝑡)| < 𝛽

The system is stable.

Probs. 1.27 (e)

y(𝑡) = {0, 𝑥(𝑡) < 0𝑥(𝑡) + 𝑥(𝑡 − 2), 𝑥(𝑡) ≥ 0

(1)

The system is not memoryless, as the value of y(𝑡) depends on the value of

𝑥(𝑡) at 𝑥(𝑡 − 2).

(2)

y(𝑡 − 𝑡0) = {0, 𝑥(𝑡 − 𝑡0) < 0

𝑥(𝑡 − 𝑡0) + 𝑥(𝑡 − 𝑡0 − 2), 𝑥(𝑡 − 𝑡0) ≥ 0

Let 𝑥2(𝑡) = 𝑥(𝑡 − 𝑡0)

𝑥2(𝑡) → 𝑦2(𝑡) = {0, 𝑥2(𝑡) < 0

𝑥2(𝑡) + 𝑥2(𝑡 − 2), 𝑥2(𝑡) ≥ 0

= {0, 𝑥(𝑡 − 𝑡0) < 0

𝑥(𝑡 − 𝑡0) + 𝑥(𝑡 − 𝑡0 − 2), 𝑥(𝑡 − 𝑡0) ≥ 0= y(𝑡 − 𝑡0)

The system is time invariant, as 𝑥(𝑡 − 𝑡0) → y(𝑡 − 𝑡0).

(3)

Assume

𝑥1(𝑡) → 𝑦1(𝑡) = {0, 𝑥1(𝑡) < 0

𝑥1(𝑡) + 𝑥1(𝑡 − 2), 𝑥1(𝑡) ≥ 0

𝑥2(𝑡) → 𝑦2(𝑡) = {0, 𝑥2(𝑡) < 0

𝑥2(𝑡) + 𝑥2(𝑡 − 2), 𝑥2(𝑡) ≥ 0

Let

𝑥3(𝑡) = α𝑥1(𝑡) + 𝛽𝑥2(𝑡)

Then

𝑥3(𝑡) → 𝑦3(𝑡) = {0, 𝑥3(𝑡) < 0

𝑥3(𝑡) + 𝑥3(𝑡 − 2), 𝑥3(𝑡) ≥ 0

= {0, α𝑥1(𝑡) + 𝛽𝑥2(𝑡) < 0

α𝑥1(𝑡) + 𝛽𝑥2(𝑡) + α𝑥1(𝑡 − 2) + 𝛽𝑥2(𝑡 − 2), α𝑥1(𝑡) + 𝛽𝑥2(𝑡) ≥ 0

= α {0, α𝑥1(𝑡) + 𝛽𝑥2(𝑡) < 0

𝑥1(𝑡) + 𝑥1(𝑡 − 2), α𝑥1(𝑡) + 𝛽𝑥2(𝑡) ≥ 0

+ 𝛽 {0, α𝑥1(𝑡) + 𝛽𝑥2(𝑡) < 0

𝑥2(𝑡) + 𝑥2(𝑡 − 2), α𝑥1(𝑡) + 𝛽𝑥2(𝑡) ≥ 0= α𝑦1(𝑡) + 𝛽𝑦2(𝑡)

The system is linear, as α𝑥1(𝑡) + 𝛽𝑥2(𝑡) → α𝑦1(𝑡) + 𝛽𝑦2(𝑡).

(4)

The system is causal, as the value of y(𝑡) at any particular time depends only

on the value of 𝑥(𝑡) at present and past time, not 𝑥(𝑡 + 𝜏) for 𝜏 > 0.

(5)

Let

|𝑥(𝑡)| < ∞ 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡

Since

|𝑥(𝑡 − 2)| < ∞

Then

|y(𝑡)| = {0, |𝑥(𝑡)| < 0|𝑥(𝑡)| + |𝑥(𝑡 − 2)|, |𝑥(𝑡)| ≥ 0

< ∞ ⇒ |𝑦(𝑡)| < ∞

The system is stable.

Probs. 1.28 (b)

y[𝑛] = 𝑥[𝑛 − 2] − 2𝑥[𝑛 − 8]

(1)

The system is not memoryless, as the value of y[𝑛] depends on the value of

𝑥[𝑛] at 𝑥[𝑛 − 2] and 𝑥[𝑛 − 8].

(2)

y[𝑛 − 𝑛0] = 𝑥[𝑛 − 𝑛0 − 2] − 2𝑥[𝑛 − 𝑛0 − 8]

Let 𝑥2[𝑛] = 𝑥[𝑛 − 𝑛0]

𝑥2[𝑛] → 𝑦2[𝑛] = 𝑥2[𝑛 − 2] − 2𝑥2[𝑛 − 8]

= 𝑥[𝑛 − 2 − 𝑛0] − 2𝑥[𝑛 − 8 − 𝑛0]

= 𝑥[𝑛 − 𝑛0 − 2] − 2𝑥[𝑛 − 𝑛0 − 8] = y[𝑛 − 𝑛0]

The system is time invariant, as 𝑥[𝑛 − 𝑛0] → y[𝑛 − 𝑛0].

(3)

Assume

𝑥1[𝑛] → 𝑦1[𝑛] = 𝑥1[𝑛 − 2] − 2𝑥1[𝑛 − 8]

𝑥2[𝑛] → 𝑦2[𝑛] = 𝑥2[𝑛 − 2] − 2𝑥2[𝑛 − 8]

Let

𝑥3[𝑛] = α𝑥1[𝑛] + 𝛽𝑥2[𝑛]

Then

𝑥3[𝑛] → 𝑦3[𝑛] = 𝑥3[𝑛 − 2] − 2𝑥3[𝑛 − 8]

= α𝑥1[𝑛 − 2] + 𝛽𝑥2[𝑛 − 2] − 2α𝑥1[𝑛 − 8] − 2𝛽𝑥2[𝑛 − 8]

= α(𝑥1[𝑛 − 2] − 2𝑥1[𝑛 − 8]) + 𝛽(𝑥2[𝑛 − 2] − 2𝑥2[𝑛 − 8])

= α𝑦1[𝑛] + 𝛽𝑦2[𝑛]

The system is linear, as α𝑥1[𝑛] + 𝛽𝑥2[𝑛] → α𝑦1[𝑛] + 𝛽𝑦2[𝑛].

(4)

The system is causal, as the value of y[𝑛] at any particular time depends only

on the value of 𝑥[𝑛] at present and past time, not 𝑥[𝑛 + 𝑛0] for 𝑛0 > 0.

(5)

Let

|𝑥[𝑛]| < ∞ 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡

Since

|𝑥[𝑛 − 2]| < ∞ and |𝑥[𝑛 − 8]| < ∞

Then

|𝑥[𝑛 − 2]| − 2|𝑥[𝑛 − 8]| < ∞ ⇒ |𝑦(𝑡)| < ∞

The system is stable.

Probs. 1.28 (e)

y[𝑛] = {𝑥[𝑛], 𝑛 ≥ 10, 𝑛 = 0

𝑥[𝑛 + 1], 𝑛 ≤ −1

(1)

The system is not memoryless, as the value of y[𝑛] depends on the value of

𝑥[𝑛] at 𝑥[𝑛 + 1].

(2)

y[𝑛 − 𝑛0] = {

𝑥[𝑛 − 𝑛0], 𝑛 − 𝑛0 ≥ 10, 𝑛 − 𝑛0 = 0

𝑥[𝑛 − 𝑛0 + 1], 𝑛 − 𝑛0 ≤ −1

Let 𝑥2[𝑛] = 𝑥[𝑛 − 𝑛0]

𝑥2[𝑛] → 𝑦2[𝑛] = {𝑥2[𝑛], 𝑛 ≥ 10, 𝑛 = 0

𝑥2[𝑛 + 1], 𝑛 ≤ −1= {

𝑥[𝑛 − 𝑛0], 𝑛 ≥ 10, 𝑛 = 0

𝑥[𝑛 + 1 − 𝑛0], 𝑛 ≤ −1

≠ {

𝑥[𝑛 − 𝑛0], 𝑛 − 𝑛0 ≥ 10, 𝑛 − 𝑛0 = 0

𝑥[𝑛 − 𝑛0 + 1], 𝑛 − 𝑛0 ≤ −1= y[𝑛 − 𝑛0]

The system is not time invariant, as 𝑥[𝑛 − 𝑛0] ↛ y[𝑛 − 𝑛0].

(3)

Assume

𝑥1[𝑛] → 𝑦1[𝑛] = {𝑥1[𝑛], 𝑛 ≥ 10, 𝑛 = 0

𝑥1[𝑛 + 1], 𝑛 ≤ −1

𝑥2[𝑛] → 𝑦2[𝑛] = {𝑥2[𝑛], 𝑛 ≥ 10, 𝑛 = 0

𝑥2[𝑛 + 1], 𝑛 ≤ −1

Let

𝑥3[𝑛] = α𝑥1[𝑛] + 𝛽𝑥2[𝑛]

Then

𝑥3[𝑛] → 𝑦3[𝑛] = {𝑥3[𝑛], 𝑛 ≥ 10, 𝑛 = 0

𝑥3[𝑛 + 1], 𝑛 ≤ −1

= {α𝑥1[𝑛] + 𝛽𝑥2[𝑛], 𝑛 ≥ 1

0, 𝑛 = 0α𝑥1[𝑛 + 1] + 𝛽𝑥2[𝑛 + 1], 𝑛 ≤ −1

= α ({𝑥1[𝑛], 𝑛 ≥ 10, 𝑛 = 0

𝑥1[𝑛 + 1], 𝑛 ≤ −1) + 𝛽 ({

𝑥2[𝑛], 𝑛 ≥ 10, 𝑛 = 0

𝑥2[𝑛 + 1], 𝑛 ≤ −1)

= α𝑦1[𝑛] + 𝛽𝑦2[𝑛]

The system is linear, as α𝑥1[𝑛] + 𝛽𝑥2[𝑛] → α𝑦1[𝑛] + 𝛽𝑦2[𝑛].

(4)

The system is not causal, as the value of y[𝑛] depends on the future value

𝑥[𝑛 + 1].

(5)

Let

|𝑥[𝑛]| < ∞ 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡

Since

|𝑥[𝑛 + 1]| < ∞

Then

{|𝑥[𝑛]|, 𝑛 ≥ 10, 𝑛 = 0

|𝑥[𝑛 + 1]|, 𝑛 ≤ −1< ∞ ⇒ |𝑦(𝑡)| < ∞

The system is stable.

5.

Probs. 1.31

Probs. 1.42