Time Differentiation MCQ Quiz - Objective Question with Answer for Time Differentiation - Download Free PDF

Last updated on May 16, 2025

Latest Time Differentiation MCQ Objective Questions

Time Differentiation Question 1:

Fourier transform of tett2/2 is

  1. (1ip)eipe(p21)/2
  2. (1ip)eipe(p21)/2
  3. (1+ip)eipe(p21)/2
  4. (1+ip)eipe(p21)/2

Answer (Detailed Solution Below)

Option 4 : (1+ip)eipe(p21)/2

Time Differentiation Question 1 Detailed Solution

Concept:

Fourier Transform:

  • The Fourier transform of a function converts it from time domain to frequency domain.
  • If f(t) is a time-domain function, its Fourier Transform is defined as:
  • F(p)=f(t)eiptdt
  • For Gaussian functions like (et2/2) , their Fourier transform is also Gaussian.
  • The Fourier transform of  et2/2  is  2πep2/2.
  • Multiplying by t in time domain corresponds to taking derivative with respect to p in frequency domain:
  • F[tf(t)]=iddpF(p)

 

Calculation:

Given,

Let f(t) = t e−t²⁄2

Let F(p) = Fourier transform of e−t²⁄2 = e−p²⁄2

⇒ Fourier transform of t f(t) = i × d/dp (e−p²⁄2)

⇒ = i × (−p e−p²⁄2) = −i p e−p²⁄2

⇒ Now t e−t²⁄2 = f(t),

so full FT is i × d/dp (F(p))

⇒ Add F(p) itself:

Final result = (1 + i p) e−(p² − 1)/2

∴ The correct Fourier transform is: (1ip)eipe(p21)/2  

Top Time Differentiation MCQ Objective Questions

Time Differentiation Question 2:

The Fourier transform of x(t) is X(jω). Then, the Fourier transform of d2x(t1)dt2

  1. ω2X(jω)ejω
  2. ω2X(jω)ejω
  3. ω2X(jω)ejω
  4. ω2X(jω)ejω

Answer (Detailed Solution Below)

Option 2 : ω2X(jω)ejω

Time Differentiation Question 2 Detailed Solution

Let x1(t)=x(t1)

The x1(t)FTX1(jω)=X(jω)ejω

Now,

Let x2(t)=d2x1(t)dt2=d2x(t1)dt2

Taking Fourier transform

X2(jω)=(jω)2X1(jω)X2(jω)=ω2X(jω)ejω

Thus,

d2x(t1)dt2FTω2X(jω)ejω

Time Differentiation Question 3:

Fourier transform of tett2/2 is

  1. (1ip)eipe(p21)/2
  2. (1ip)eipe(p21)/2
  3. (1+ip)eipe(p21)/2
  4. (1+ip)eipe(p21)/2

Answer (Detailed Solution Below)

Option 4 : (1+ip)eipe(p21)/2

Time Differentiation Question 3 Detailed Solution

Concept:

Fourier Transform:

  • The Fourier transform of a function converts it from time domain to frequency domain.
  • If f(t) is a time-domain function, its Fourier Transform is defined as:
  • F(p)=f(t)eiptdt
  • For Gaussian functions like (et2/2) , their Fourier transform is also Gaussian.
  • The Fourier transform of  et2/2  is  2πep2/2.
  • Multiplying by t in time domain corresponds to taking derivative with respect to p in frequency domain:
  • F[tf(t)]=iddpF(p)

 

Calculation:

Given,

Let f(t) = t e−t²⁄2

Let F(p) = Fourier transform of e−t²⁄2 = e−p²⁄2

⇒ Fourier transform of t f(t) = i × d/dp (e−p²⁄2)

⇒ = i × (−p e−p²⁄2) = −i p e−p²⁄2

⇒ Now t e−t²⁄2 = f(t),

so full FT is i × d/dp (F(p))

⇒ Add F(p) itself:

Final result = (1 + i p) e−(p² − 1)/2

∴ The correct Fourier transform is: (1ip)eipe(p21)/2  

Time Differentiation Question 4:

If x(t) is real valued signal whose Fourier transform is defined as X(ω)=jω(3+jω)2.Then value of x(t) at t = 1/3 sec is ______

Answer (Detailed Solution Below) 0

Time Differentiation Question 4 Detailed Solution

e3tu(t)13+jωte3tu(t)1(3+jω)2(differentiation property in frequency domain)

ddt[te3tu(t)]jω(3+jω)2( differentiation property in time domain)

x(t)=ddt[te3tu(t)]

=tddt[e3tu(t)]+e3tu(t)dtdt

=t[e3tδ(t)3e3tu(t)]+e3tu(t)te3tδ(t)3te3tu(t)+e3tu(t)

at  t = 1/3, δ(t) = 0

3(13)e3(13)4(t)+e3(13)4(t)=0

Time Differentiation Question 5:

Find the inverse Fourier transform of x(ω)=jω(4+jω)2

  1. e-4t u(t) – 4t e-4t u(t)
  2. e-4t u(t) + 4t e-4t u(t)
  3. e4t u(t) – 4t e4t u(t)
  4. e4t u(t) + 4t e4t u(t)

Answer (Detailed Solution Below)

Option 1 : e-4t u(t) – 4t e-4t u(t)

Time Differentiation Question 5 Detailed Solution

We know that,

eatu(t)1a+jω

e4tu(t)1(4+jω)

te4tu(t)1(4+jω)2

ddt[te4tu(t)]jω(4+jω)2

x(t)ddt[te4tu(t)]

tddt[e4tu(t)]+e4tu(t)

=t[e4tδ(t)+(4)e4tu(t)]+e4tu(t)

From the property of delta function

F (t) δ (t – t0) = f (t0) δ (t – t0)

⇒ t e-4t δ (t) = 0 e-4(0) δ (t) = 0

⇒ x (t) = -4te-4t u(t) + e-4t u(t)

Time Differentiation Question 6:

The Fourier transfer of the signal x(t) given below be x(ω) then the value of |x(ω)|ω=1rad is ________.

03.08.2018.013

Answer (Detailed Solution Below) 0.2 - 0.3

Time Differentiation Question 6 Detailed Solution

Differentiating given signal w.r.t time

03.08.2018.014

Again differentiating w.r.t time we get

03.08.2018.016

The Fourier transform of above signal is

d2x(t)dt2=δ(t2)+2δ(t3)δ(t4)+δ(t+2)+δ(t+4)2δ(t+3) 

(jω)2x(ω)=ej2ω+2e3jωej4ω+ej2ω+ej4ω+ej4ω2ej3ω

X(ω)=2jω2[sin(2ω)2sin(3ω)+sin(4ω)] 

|X(1)|=|21[sin(2)2sin(3)+sin(4)]| 

≃ 0.26

Time Differentiation Question 7:

The Fourier transform of x(t) is X(jω). Then, the Fourier transform of d2x(t2)dt2

  1. ω2X(jω)e2jω
  2. ω2X(jω)e2jω
  3. ω2X(jω)e2jω
  4. ω2X(jω)e2jω

Answer (Detailed Solution Below)

Option 3 : ω2X(jω)e2jω

Time Differentiation Question 7 Detailed Solution

Let x1(t)=x(t2)

The x1(t)FTX1(jω)=X(jω)ej2ω

Now,

Let x2(t)=d2x1(t)dt2=d2x(t2)dt2

Taking Fourier transform

X2(jω)=(jω)2X1(jω)X2(jω)=ω2X(jω)e2jω

Thus,

d2x(t2)dt2FTω2X(jω)e2jω

Time Differentiation Question 8:

The input x(t)and output y(t) of causal LTI system are related by:-

d2y(t)dt2+6dy(t)dt+8y(t)=2dx(t)dt+6x(t)

Then, the impulse response of the system is

  1. e2tu(t)e4tu(t)

  2. e2tu(t)+e4tu(t)

  3. [e4te2t]u(t)

  4. e2tu(t)+e4tu(t)

Answer (Detailed Solution Below)

Option 2 :

e2tu(t)+e4tu(t)

Time Differentiation Question 8 Detailed Solution

We have d2y(t)dt2+6dy(t)dt+8y(t)=2dx(t)dt+6x(t)

Taking Fourier transform we have,

(jω)2Y(jω)+6(jω)Y(jω)+8Y(jω)=2X(jω)(jω)+6X(jω)H(jω)=Y(jω)X(jω)=2(jω)+6(jω)2+6(jω)+8H(jω)=2(jω)+6(jω+2)(jω+4)=1(jω+2)+1(jω+4)x(t)=e2tu(t)+e4tu(t).

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