Convolution in Frequency Domain MCQ Quiz - Objective Question with Answer for Convolution in Frequency Domain - Download Free PDF

Last updated on Apr 4, 2025

Latest Convolution in Frequency Domain MCQ Objective Questions

Convolution in Frequency Domain Question 1:

A signal is defined as y(t)=(sinπtπt)(sin2πtπt) then value of +Y(ω)dω is _______

  1. 2
  2. 4

Answer (Detailed Solution Below)

Option 1 : 4π

Convolution in Frequency Domain Question 1 Detailed Solution

Concept:

The short-cut for convolution of two rectangular pulses of unequal and equal width can be given as:

F2 S.B 20.8.20 Pallavi D2 (1)

 

F2 S.B 20.8.20 Pallavi D3

Analysis:

Using the property of Fourier transform

Y(t) = x1(t) x2(t)

Taking Fourier transform

Y(ω)=12π[X1(ω)X2(ω)]

x1(t)=(sinπtπt)

X1(ω)=[1|ω|<π0otherwise

x2(t)=(sin2πtπt)

X2(ω)=[1|ω|<2π0otherwise

Y(ω)=12π[X1(ω)X2(ω)]

 

12πF1 Neha.B 25-09-20 Savita D1

+Y(ω)dω=AreaunderY(ω)

(2×12×2π×2π+2π×2×π)2π

 =4π

Top Convolution in Frequency Domain MCQ Objective Questions

Convolution in Frequency Domain Question 2:

A signal is defined as y(t)=(sinπtπt)(sin2πtπt) then value of +Y(ω)dω is _______

  1. 2
  2. 4

Answer (Detailed Solution Below)

Option 1 : 4π

Convolution in Frequency Domain Question 2 Detailed Solution

Concept:

The short-cut for convolution of two rectangular pulses of unequal and equal width can be given as:

F2 S.B 20.8.20 Pallavi D2 (1)

 

F2 S.B 20.8.20 Pallavi D3

Analysis:

Using the property of Fourier transform

Y(t) = x1(t) x2(t)

Taking Fourier transform

Y(ω)=12π[X1(ω)X2(ω)]

x1(t)=(sinπtπt)

X1(ω)=[1|ω|<π0otherwise

x2(t)=(sin2πtπt)

X2(ω)=[1|ω|<2π0otherwise

Y(ω)=12π[X1(ω)X2(ω)]

 

12πF1 Neha.B 25-09-20 Savita D1

+Y(ω)dω=AreaunderY(ω)

(2×12×2π×2π+2π×2×π)2π

 =4π

Convolution in Frequency Domain Question 3:

If h1(t)F.TH1(ω) 

 h2(t)F.TH2(ω) 

 h1(t)h2(t)dt=12πH1(ω)H3(ω)dω 

Then relation which hold true is:

  1. H3(ω) = H2(- ω)
  2. H3(ω) = H2(ω)
  3. Both 1 and 2
  4. None of these

Answer (Detailed Solution Below)

Option 1 : H3(ω) = H2(- ω)

Convolution in Frequency Domain Question 3 Detailed Solution

Concept:

F[h1(t)h2(t)]=12π[H1(ω)H2(ω)]

Where * represents convolution.

Calculation:

h1(t)h2(t)dt=h1(t)h2(t)ejωtdt|ω=0

=12π[H1(ω)H2(ω)]|ω=0

=12πH1(λ)H2(ωλ)dλ|ω=0

=12πH1(λ)H2(λ)dλ=12πH1(ω)H2(ω)dω

∴ H3(ω) = H2(-ω)

Convolution in Frequency Domain Question 4:

If h1(t)F.TH1(ω) 

 h2(t)F.TH2(ω) 

 h1(t)h2(t)dt=12πH1(ω)H3(ω)dω 

Then relation which hold true is:

  1. H3(ω) = H2(- ω)
  2. H3(ω) = H2(ω)
  3. Both 1 and 2
  4. None of these

Answer (Detailed Solution Below)

Option 1 : H3(ω) = H2(- ω)

Convolution in Frequency Domain Question 4 Detailed Solution

Concept:

F[h1(t)h2(t)]=12π[H1(ω)H2(ω)]

Where * represents convolution.

Calculation:

h1(t)h2(t)dt=h1(t)h2(t)ejωtdt|ω=0

=12π[H1(ω)H2(ω)]|ω=0

=12πH1(λ)H2(ωλ)dλ|ω=0

=12πH1(λ)H2(λ)dλ=12πH1(ω)H2(ω)dω

∴ H3(ω) = H2(-ω)

Convolution in Frequency Domain Question 5:

We have x(t)=4sin(t)sin(2t)πt2 then the value of X(jω) at ω=12 is ________

Answer (Detailed Solution Below) 4

Convolution in Frequency Domain Question 5 Detailed Solution

x(t)=4sintsin2tπt2=4π(sintπt)(sin2tπt)

Let x1(t)=sintπt and x2(t)=sin2tπt

Then using the property that

x1(t)x2(t)FT12π[X1(jω)X2(jω)]

We have

EC Signals Subject Test 1 reviewed Images Q21

EC Signals Subject Test 1 reviewed Images Q21.1

EC Signals Subject Test 1 reviewed Images Q21.2

Thus X(jω)|ω=12=4

Convolution in Frequency Domain Question 6:

We have two signals x(t) and y(t) given by

EC Signals Subject Test 1 reviewed Images Q19

EC Signals Subject Test 1 reviewed Images Q19.1

Then, X(jω)G(jω) is

  1. 1jω

  2. 1jω(1ωcotω)

  3. jω

  4. 1jω+πδ(ω)

Answer (Detailed Solution Below)

Option 2 :

1jω(1ωcotω)

Convolution in Frequency Domain Question 6 Detailed Solution

EC Signals Subject Test 1 reviewed Images Q19.2

EC Signals Subject Test 1 reviewed Images Q19.3

Let s(t)=dx(t)dt=g(t)+f(t)

S(jω)=(2sinωω)ejωejω  ________ (1)

Now

x(t)=t[dx(t)dt]dt=ts(t)dtX(jω)=S(jω)jω+πS(0)δ(ω)

from (1) S(0)=2sinωωejωejω|ω=0=0

X(jω)=S(jω)jω=1jω(G(jω)+F(jω))(s(t)=g(t)+f(t)S(jω)=G(jω)+F(jω))

Now G(jω)=2sinωω and F(jω)=ejωejω=2cosω

Now X(jω)G(jω)=1jω(1+F(jω)G(jω))=1jω(1+2cosω2(sinωω))

X(jω)G(jω)=1jω(1ωcotω)

On another approach we see that three options have 1jω term in them, which prompts that x(t) is integral of g(t) (as people usually forget the impulses) and then when you try, you get X(jω)=G(jω)jω+πG(0)δ(ω) (remember you have missed the impulses) X(jω)G(jω)=1jω+π2δ(ω)2sinωω

(G(0)=2 and G(jω)=2sinωω) and then you realise that no option matches. After this you rethink, and if you can discover that

dx(t)dtg(t) but dx(t)dt=g(t)+f(t)

x(t)=t(g(t)+f(t))dtX(jω)=1jω(G(jω)+F(jω))+π(G(0)+F(0)δ(ω)G(jω)+F(jω)=2sinωωejωejω=2sinωω2cosω

And G(j0)+F(j0)=0

X(jω)G(jω)=1jω(1+F(jω)G(jω))

Here you see that a cotω term comes and you can mark option d without solving further.

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