Signals and systems Assignment 01 -...

Signals and systems – Assignment 01 - Solutions

Instructor: Prof. K. C. Veluvolu

Question 1

Q1. Graph these functions

a. g(t) = −2.5 ramp (t−1

2)

b. g(t) = 5 rect (2t)

c. g(t) = −1.5 rect (t−1

2) + 2.5 rect (

2t+3

2)

Solutions (a)

(b)

-10 -8 -6 -4 -2 0 2 4 6 8 10-12

-10

-8

-6

-4

-2

0



(c)

Question 2

Graph the signal given below with shifted and/or scaled versions.

where g(t) is periodic with fundamental period, 4

a. t → (t + 8); g(t) → −2g(t−1

2)

-10 -8 -6 -4 -2 0 2 4 6 8 10-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5



b. t → (2t + 1); g(t) → −2g(2t+1

3)

Solutions (a)

(b)



Question 3

Describe the following signal mathematically as functions of ramp and step function.

Solution

From the figure, if the first line is extended, we have a ramp function

𝑥1 =15

4𝑅𝑎𝑚𝑝(𝑡) ,

𝑥(𝑡) =15

4𝑅𝑎𝑚𝑝(𝑡) − 15 ∑ 𝑈(𝑡 = 4𝑘)∞

𝑘=1

Question 4

Graph the time derivatives of these functions. a. g(t) = 10 cos(20𝜋𝑡) 𝑠𝑔𝑛(𝑡) b. g(t) = |sin (2𝜋𝑡)|

Solutions

(a)

𝑔(𝑡) = 10 cos(20𝜋𝑡) 𝑠𝑔𝑛(𝑡)

Then

𝑔′(𝑡) = {200𝜋𝑠𝑖𝑛(20𝜋𝑡); 𝑡 < 0

0; 𝑡 = 0−200𝜋 sin(20𝜋𝑡) ; 𝑡 > 0

Thus,



(b) 𝑔(𝑡) = |sin (2𝜋𝑡)|

𝑔′(𝑡) = 2𝜋|cos (2𝜋𝑡)|

Then,

-0.25 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25-10

-5

0

5

10

-0.25 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25-1000

-500

0

500

1000



Question 5

Find even and odd parts

a. g(t) = 12 +sin(4𝜋𝑡)

4𝜋𝑡

b. g(t) = (8 + 7𝑡)cos (32𝜋𝑡 − 3𝜋

2)

c. g(t) = 8𝑡 sin (12𝜋𝑡 +𝜋

8)

Solutions (a)

𝑔(𝑡) = 12 +sin (4𝜋𝑡)

4𝜋𝑡

Obviously, the function is an even function, hence

𝑔𝑜𝑑𝑑(𝑡) = 0

And

𝑔𝑒𝑣𝑒𝑛(𝑡) = 12 +sin (4𝜋𝑡)

4𝜋𝑡

(b)

𝑔(𝑡) = (8 + 7𝑡)cos (32𝜋𝑡 −3𝜋

2)

𝑔(𝑡) = 8 cos (32𝜋𝑡 −3𝜋

2) + 7𝑡𝑐𝑜𝑠(32𝜋 −

3𝜋

2)

Obviously,

𝑔𝑒𝑣𝑒𝑛(𝑡) = 8 cos (32𝜋𝑡 −3𝜋

2)

𝑔𝑜𝑑𝑑(𝑡) = 7𝑡 cos (32𝜋𝑡 −3𝜋

2)

(c)

𝑔(𝑡) = 8𝑡sin (12𝜋𝑡 +𝜋

8)

𝑔(𝑡) = 8𝑡𝑠𝑖𝑛(12𝜋𝑡)𝑐𝑜𝑠(𝜋

8) + 8𝑡𝑐𝑜𝑠(12𝜋𝑡)𝑠𝑖𝑛(

𝜋

8)

𝑔(−𝑡) = 8𝑡𝑠𝑖𝑛(12𝜋𝑡)𝑐𝑜𝑠 (𝜋

8) − 8𝑡𝑐𝑜𝑠(12𝜋𝑡)𝑠𝑖𝑛(

𝜋

8)

Thus



𝑔𝑒𝑣𝑒𝑛(𝑡) = 8𝑡𝑠𝑖𝑛(12𝜋𝑡)𝑐𝑜𝑠(𝜋

8)

𝑔𝑜𝑑𝑑(𝑡) = 8𝑡𝑐𝑜𝑠(12𝜋𝑡)𝑠𝑖𝑛(𝜋

8)

Question 6

Find whether the signal is periodic and, if it is, find the period: a. g(t) = 28 sin(400𝜋𝑡) + 12 cos(500𝜋𝑡)

b. g[n] = 𝑒2𝜋𝑗𝑛

3 − 𝑒2𝜋𝑗𝑛

4

(a) g(t) = 28 sin(400𝜋𝑡) + 12 cos(500𝜋𝑡)

g(t) is periodic and the period of the function is the lcm of the period of the two components

i.e. lcm of 1

200 𝑎𝑛𝑑

1

250, Thus the period is

1

50 seconds.

(b)

𝑔[𝑛] = 𝑒2𝜋𝑗𝑛

3 − 𝑒2𝜋𝑗𝑛

4 Since the function is periodic, the period of the function becomes the LCM of the periods of the two components i.e. the lcm of 3 𝑎𝑛𝑑 4; thus, the period is 12 seconds.

Question 7 Find signal energy and average signal power for both these signals: (a)

g(t) = δ(t − 2)

(b)

g(t) = 2 sin (200𝜋𝑡 −3𝜋

2)

Solutions

(a) 𝐸 = ∫ |𝑥(𝑡)|2𝑑𝑡∞

−∞

But

𝛿(𝑡) = lim𝐴→0

1

𝐴𝑅𝑒𝑐𝑡(

𝑡

𝐴)

∫ |𝛿(𝑡 − 2)|2𝑑𝑡 = ∫ lim𝐴→0

|1

𝐴𝑟𝑒𝑐𝑡 (

𝑡−2

𝐴) |2𝑑𝑡 = ∞

∞

−∞

∞

−∞

Since the function is aperiodic, its power is undefined. (c)

The period of the function is 1

100𝑠𝑒𝑐



Then, the energy of the signal is infinite. Also

𝑃𝑜𝑤𝑒𝑟 =1

𝑇∫ |𝑔(𝑡)|2𝑑𝑡

𝑇

2

−𝑇

2

𝑃𝑜𝑤𝑒𝑟 = 100 ∫ |(2 sin (200𝜋𝑡 −3𝜋

2) |2𝑑𝑡

1

50

−1

50

𝑃𝑜𝑤𝑒𝑟 = 2

Question 8 Graph these functions:

a. g[𝑛] = −4δ [2𝑛

3]

b. g[𝑛] = 8δ4[𝑛] c. g[𝑛] = δ2[𝑛 + 1] − δ2[𝑛]

d. g[𝑛] = ∑ cos (2𝜋𝑚

12) 𝑢[𝑚]𝑛

𝑚=0

e. g[𝑛] = 5δ[2𝑛] + 3δ[4(𝑛 − 2)] Solutions (a)

(b)



(c)

(d)



(e)



Question 9 (a)

∑ 𝛿3[𝑛] = 1 + 1 + 1 + 1 + 1 + 1 + 110

𝑛=−10 = 7 (b)

∑ 𝛿3[2𝑛] = 1 + 1 + 1 + 1 + 110

𝑛=−20 = 5



Question 10 Find the signal energy of each of these signals:

a. x[𝑛] = (−1/3)𝑛𝑢[𝑛 − 2] b. x[𝑛] = (−1)𝑛

Solutions

(a)

𝐸 = ∑ |(−1

3)𝑛𝑈[𝑛 − 2]|2∞

𝑛=−∞

𝐸 = ∑ |(−1

3)𝑛|2∞

𝑛=2

𝐸 = ∑ (1

9)𝑛 =

1

81+

1

729+

1

6561+ ⋯ =

1

72∞𝑛=2

(b)

𝐸 = ∑ |(−1)𝑛|2∞𝑛=−∞ = ∞

Question 11

Show that a system with excitation x[n] and response y[n] described by y[n] = nx[n]

is linear, static and time-variant.

Solutions

Let 𝑦1[𝑛] = 𝑛𝑥1[𝑛] and 𝑦2[𝑛] = 𝑛𝑥2[𝑛]

Then 𝑦3[𝑛] = 𝑎𝑛𝑥1[𝑛] +b 𝑛𝑥2[𝑛]

If 𝑦4[𝑛] = 𝑓[𝑎𝑥1[𝑛] + 𝑏𝑥2[𝑛]] = 𝑎𝑛𝑥1[𝑛] +b 𝑛𝑥2[𝑛]

Thus

𝑦3[𝑛] = 𝑦4[𝑛] and so the system is linear.

Also, to check for its time invariance

𝑦[𝑛] = 𝑛𝑥[𝑛]



Then

𝑦[𝑛, 𝑘] = 𝑛𝑥[𝑛 − 𝑘]

And 𝑦[𝑛 − 𝑘] = [𝑛 − 𝑘]𝑥[𝑛 − 𝑘]

Since 𝑦[𝑛, 𝑘] ≠ 𝑦[𝑛 − 𝑘], the system is time variant.

Question 12

Show that the system described by

y(t) = {10, 𝑥(𝑡) > 2

5𝑥(𝑡), −2 < 𝑥(𝑡) ≤ 2−10, 𝑥(𝑡) ≤ −2

is non-linear, static, stable, non-invertible and time-invariant.

Solution

If 𝑥(𝑡) = 1.5, 𝑦(𝑡) = 7.5. If 𝑥(𝑡) = 3, 𝑦(𝑡) = 10 Therefore the system is not homogeneous and not

linear.

The system is static because the value of y depends only on the value of x at the same time.

The system is stable because the response can never lie outside the range -10 to 10.

The system is non-invertible because, if the response is 10 the excitation can be any value from 2 to

infinity.

The system is time invariant because if the excitation is shifted in time, the response, which occurs at

the same time as the excitation because it is static, shifts the same amount in time.



Question 13

A system is described by y[n] = ∑ 𝑥[𝑚]𝑛+1𝑚=−∞ . Classify this system as to time invariance, BIBO stability

and invertibility.

Solutions

Let, 𝑥1[n] = g[n] then 𝑦1[n] = ∑ 𝑔[𝑚]𝑛+1𝑚=−∞

Let, 𝑥2[n] = g[n − 𝑛0] then 𝑦2[n] = ∑ 𝑔[n − 𝑛0]𝑛+1𝑚=−∞

Letting m = q − 𝑛0, the first equation can be rewritten as

𝑦1[n − 𝑛0] = ∑ 𝑔[𝑚]n−𝑛0+1𝑚=−∞ = ∑ 𝑔[q − 𝑛0]n+1

𝑞=−∞ = 𝑦2[n]

So, it is Time invariant

Stability:

If the excitation is a constant, the response increases without bound. So, Unstable.

Invertibility:

Inverting the functional relationship, y[n] = ∑ 𝑥[𝑚]𝑛+1𝑚=−∞ . Taking the first backward difference of both

sides of the original system equation, we will have that

y[n] − y[n − 1] = ∑ 𝑥[𝑚]

𝑛+1

𝑚=−∞

− ∑ 𝑥[𝑚]

𝑛

𝑚=−∞

x[n + 1] = y[n] − y[n − 1] ⇒ x[n] = y[n − 1] − y[n − 2]

Thus, the excitation is determined by the response. So, the system is Invertible.

Signals and systems Assignment 01 -...

Documents

Transcript of Signals and systems Assignment 01 -...