Properties of Fourier Transform MCQ Quiz in मराठी - Objective Question with Answer for Properties of Fourier Transform - मोफत PDF डाउनलोड करा

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पाईये Properties of Fourier Transform उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Properties of Fourier Transform एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Properties of Fourier Transform MCQ Objective Questions

Top Properties of Fourier Transform MCQ Objective Questions

Properties of Fourier Transform Question 1:

The Fourier transform of the signal shown below is jksinω such that the value of k is ________.

F1 Shubham Madhu 24.09.20 D7

Answer (Detailed Solution Below) 2

Properties of Fourier Transform Question 1 Detailed Solution

Concept:

δ(t)FT1

δ(tt0)FTejωt0

Analysis:

The given signal can be written as:

y(t) = δ(t + 1) + (-δ(t – 1))

y(t) = δ(t + 1) – δ(t – 1)

Taking the Fourier transform, we get:

y(jω) = e – e-jω

= cos ω + j sin ω – (cos ω – j sin ω)

= 2j sin ω

∴ k = 2        

Properties of Fourier Transform Question 2:

Consider the discrete time signal x(n)=2[k=0{δ(n1k)δ(n6k)}] having discrete time Fourier transform X(e). Which of the following is / are correct?

  1. 1πππ|X(ejω)|2dω=40
  2. 1πππ|X(ejω)|2dω=20
  3. ππ|ddωX(ejω)|2dω=440π
  4. ππ|ddωX(ejω)|2dω=220π

Answer (Detailed Solution Below)

Option :

Properties of Fourier Transform Question 2 Detailed Solution

Concept:

The energy of a discrete time signal using Parsevel’s theorem is given by:

n=|x(n)|2=12πππ|X(ejω)|2dω

Calculation:

Given discrete time signal is:

x(n)=2[k=0δ(n1k)δ(n6k)]

u(n)=k=0δ(nk)

u(n1)=k=0δ(n1k)

u(n6)=k=0δ(n6k)

∴ x(n) = 2[u(n - 1) – u(n - 6)]

= 2, 1 ≤ n ≤ 5

F1 Shubham 7.12.20 Pallavi D18

1πππ|X(ejω)|2dω=2n=|x(n)|2 

1πππ|X(ejω)|2dω=2[22+22+22+22+22] 

1πππ|X(ejω)|2dω=40 

Also,

n=|nx(n)|2=12πππ|ddωX(ejω)|2dω 

ππ|ddωX(ejω)|2dω=2πn=|nx(n)|2 

|ddωX(ejω)|2dω=2π[12×22+22×22+32×22+42×22+52×22]

ππ|ddωX(ejω)|2dω=440π 

Hence, the correct options are (1) and (3).

Properties of Fourier Transform Question 3:

Let z(t) be the output of the first system and the input to the second system in the cascade. Find the output y(t).

quesOptionImage905

  1. y(t) = x(t) ⊗ b2(t)
  2. y(t) = z(t) ⊗ b2(t)
  3. y(t) = z(t) ⊗ x(t)
  4. y(t) = z(t) + b2(t)

Answer (Detailed Solution Below)

Option 2 : y(t) = z(t) ⊗ b2(t)

Properties of Fourier Transform Question 3 Detailed Solution

Analysis:

When two blocks are cascaded their impulse responses are to be convolved.

quesImage3891

Here two blocks are cascaded, then equivalent impulse response of the whole system will be:

b1(t) ⊗ b2(t)

In the above figure:

z(t) = x(t) ⊗ b1(t)

y(t) = z(t) ⊗ b2(t)

The expression for y(t) in terms of x(t) will be:

y(t) =x(t) ⊗ (b1(t)⊗b2(t))

Important Points

Properties of convolution:

Associativity: f* (f* f3) = (f* f2) * f3

Commutativity: f* f2 = f* f

Distribitivity: f* (f2 + f3) = f* f2 + f* f3

Multilinearity: a(f* f2) = (af1) * f2 = f1(af2

Properties of Fourier Transform Question 4:

The Fourier transform of the signal shown below is jksinω such that the value of k is_______.

F1 Shubham Madhu 24.09.20 D7

Answer (Detailed Solution Below) 2

Properties of Fourier Transform Question 4 Detailed Solution

Concept:

δ(t)FT1

δ(tt0)FTejωt0

Analysis:

The given signal can be written as:

y(t) = δ(t + 1) + (-δ(t – 1))

y(t) = δ(t + 1) – δ(t – 1)

Taking the Fourier transform, we get:

y(jω) = e – e-jω

= cos ω + j sin ω – (cos ω – j sin ω)

= 2j sin ω

∴ k = 2        

Properties of Fourier Transform Question 5:

The differential equation description for the system with the given impulse response is:

H(jω)=1+jω(jω+2)(jω+3)

  1. y(t)+y˙(t)=x ¨(t)+5x˙(t)+6x(t)
  2. y ¨(t)+3y˙(t)+2y(t)=x(t)+x˙(t)
  3. y˙(t)+5y(t)=x ¨(t)+y˙(t)+x(t)
  4. y ¨(t)+5y˙(t)+6y(t)=x(t)+x˙(t)

Answer (Detailed Solution Below)

Option 4 : y ¨(t)+5y˙(t)+6y(t)=x(t)+x˙(t)

Properties of Fourier Transform Question 5 Detailed Solution

Concept:

The transfer function of a system is defined as:

H(jω)=Y(jω)X(jω)

Y(jω) = H(jω) × X(jω)

Taking the inverse Fourier transform of the above, we get the equation of the system in the time domain.

Also,

Ifx(t)F.T.X(jω)

thendnx(t)dt(jω)n.X(jω)

Calculation:

GivenH(jω)=1+jω(2+jω)(3+jω)

This can be written as:

Y(jω)X(jω)=1+jω(2+jω)(3+jω)

Y(jω)X(jω)=1+jω6+2(jω)+3(jω)+(jω)2

Y(jω)X(jω)=1+jω(jω)2+5jω+6

⇒ (jω)2 Y(jω) + 5(jω) Y(jω) + 6Y(jω) = X(jω) + (jω) × (jω)

⇒ Taking the inverse Fourier Transform of the above, we get:

d2y(t)dt2+5dy(t)dt+6y(t)=x(t)+dx(t)dt

y ¨(t)+5y˙(t)+6y(t)=x(t)+x˙(t)

So, Option (4) is correct.

Properties of Fourier Transform Question 6:

The output of a continuous-time system y(t) is related to its input x(t) as y(t) = x(t) + 12 x(t − 1). If the Fourier transforms of x(t) and y(t) are X(ω) and Y(ω) respectively, and |X(0)|2 = 4, the value of |Y(0)|2 is ______

Answer (Detailed Solution Below) 9

Properties of Fourier Transform Question 6 Detailed Solution

y(t)=x(t)+12x(t1)

By applying Fourier transform,

Y(ω)=X(ω)+12ejωX(ω)

Y(ω)=(1+12ejω)X(ω)

Y(ω)X(ω)=1+12ejω

H(ω)=1+12ejω

H(0)=1+12=32

Y(ω) = X(ω).H(ω)

⇒ |Y(ω)|2 = |X(ω)|2.|H(ω)|2

⇒ |Y(0)|2 = |X(0)|2.|H(0)|2

|Y(0)|2=4.(32)2=4×94=9

Properties of Fourier Transform Question 7:

The energy of the signal x(t)=sin(4πt)4πt  is______

Answer (Detailed Solution Below) 0.25

Properties of Fourier Transform Question 7 Detailed Solution

We know,

x(t)=sin4πt4πt

Integrating mod square of x(t) to find the energy will be a tedious process. 

 

Fourier transform of  sin4πt4πt

Gate EC 2016 paper 2 Images-Q10

From Parseval’s theorem

|x(t)|2dt=12π|X(ω)|2dω

we have,

12π4π4π(14)2dω=12π4π4π116dω⇒=12π.116ω|4π4π=12π.116.8π=14

Thus,

signal's energy |x(t)|2dt=14

Properties of Fourier Transform Question 8:

Consider the signal x(t) shown below whose Fourier transform is X(jω).

F2 S.B 29.6.20 Pallavi D2

What is the value of X(jω)2sinωωej2ωdω? 

Answer (Detailed Solution Below) 21.97 - 22.00

Properties of Fourier Transform Question 8 Detailed Solution

Concept:

The Fourier transform of a rectangular pulse is a sinc pulse, i.e.

rect(t)F.T2sinωω 

F2 S.B 29.6.20 Pallavi D3

F2 S.B 29.6.20 Pallavi D4

Considering another signal y(t) = y1 (t + 2), its Fourier transform will be:

y1(t+2)2sinωωe2jω 

Y(jω)=2sinωωe2jω    ---(1)

We can write:

x(t)y(t)F.TX(jω)Y(jω) 

From the definition Fourier transform, we have:

x(t)y(t)=12πX(jω)Y(jω)ejωtdω 

For t = 0, the above expression can be written as:

X(jω)Y(jω)dω=2π[x(t)y(t)]t=0 

Using Equation (1):

X(jω)2sinωωe2jωdω=2π(x(t)y(t))t=0 

∴ We need to evaluate the value of x(t) * y(t) at t = 0, to get the required integral value.

The convolution in the time domain is defined as:

x(t)y(t)=x(τ)y(tτ)dτ 

x(t)y(t)|t=0=x(τ)y(τ)dτ 

x(τ) and y(-τ) are represented as shown:

F2 S.B 29.6.20 Pallavi D5

The required integral is nothing but the shaded portion, i.e.

x(t)y(t)|t=0=(1×1)+12×1×1+(1×2) 

= 3.5

X(jω)2sinωωej2ωdω=2π×3.5 

= 7π

= 21.991

Properties of Fourier Transform Question 9:

The Fourier transform of the signal ept2 is X(ω), then value of 1ωX(ω)dX(ω)dω is ______ p-1.

Answer (Detailed Solution Below) -0.5

Properties of Fourier Transform Question 9 Detailed Solution

Concept:

The normalized Gaussian pulse is given as:

f(t)=Keat2

And its Fourier transform is given as:

F.T.{f(t)}=πaeπ2f2a

=πae(π2ω24π2a)

Analysis:

ept2πpeω2/4p=X(ω)

ddωX(ω)=πpddωeω24p

=πpeω24p[2ω4p]

Substituting the values of ω and X(ω) in the given equation, the equation evaluates to

=12p

Properties of Fourier Transform Question 10:

The given mathematical representation belongs to:

y(t) = x(t - T)

  1. time multiplication
  2. time shifting
  3. time scaling
  4. time reversal

Answer (Detailed Solution Below)

Option 2 : time shifting

Properties of Fourier Transform Question 10 Detailed Solution

Time-shifting property: When a signal is shifted in time domain it is said to be delayed or advanced based on whether the signal is shifted to the right or left.

For example:

F9 Neha B 5-10-2020 Swati D7

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