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

Last updated on Apr 17, 2025

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

Latest Fourier Transform MCQ Objective Questions

Top Fourier Transform MCQ Objective Questions

Fourier Transform Question 1:

The Fourier transform of ex22 is

  1. 12eω22
  2. eω22
  3. π2
  4. π

Answer (Detailed Solution Below)

Option 2 : eω22

Fourier Transform Question 1 Detailed Solution

Concept:

d[f(x)]dxFouriertransformjωF(ω)

xf(x)Fouriertransformjd[F(ω)]dω

Where F(ω) is Fourier transform of f(x)

Calculation:

We are given f(x)=ex22

d[f(x)]dx=xex22       ---(1)

Take Fourier transform of above equation (1)

jωF(ω)=jd[F(ω)]dω

1F(ω)d[F(ω)]=ωdω        ---(2)

By integrating the above equation, we get

ln[F(ω)]=ω22

F(ω)=eω22

Fourier Transform Question 2:

The Fourier sin transform of xm-1 is defined by

  1. 0xm1sinsx
  2. xm1sinsx
  3. 0xm1sinx
  4. xm1sinx

Answer (Detailed Solution Below)

Option 2 : xm1sinsx

Fourier Transform Question 2 Detailed Solution

Fourier sine transform of the function f(x) is defined as

F(s)=0f(x)sinsxdx

=xm1sinsxdx

Fourier Transform Question 3:

If \(f(x)= \rm \left\{\begin{matrix}\rm x\,\,\,\,\,\, if -\pi / 2 < x < \pi / 2 \\ \rm \pi-x \,\, if \pi / 2 Then a0 is equal to

  1. -2π
  2. 0
  3. 1

Answer (Detailed Solution Below)

Option 3 : 0

Fourier Transform Question 3 Detailed Solution

Concept:

A Fourier series is an expansion of a periodic function f(x) in terms of an infinite sum of sines and cosines.

Calculation:

a0 = 1ππ/2π/2f(x)dx+1ππ/23π/2f(x)dx

a0 = 1ππ/2π/2xdx+1ππ/23π/2(πx)dx

= 0 + 0 = 0

The correct answer is option (3).

Fourier Transform Question 4:

A continuous-time signal x(t) has the Fourier transform as shown in the figure. Value of signal x(t) at t = π4 sec is _______.

F1 Madhuri Engineering 26.09.2022 D13

Answer (Detailed Solution Below) 6.48 - 6.50

Fourier Transform Question 4 Detailed Solution

Calculation:

X(ω) can be represented in terms of the triangular signal as 

X(ω) = 8π tri (ω4)

Fourier transform of a triangular function is given by:

A tri(tτ)Aτ Sa2(ωτ2)

Applying duality property:

Aτ Sa2(tτ2)2πA tri(ωτ)

X(ω) = 8π tri (ω4)= 2π A tri (ω4)

∴ A = 4, τ = 4

Inverse signal x(t) is given as:

x(t)=Aτ Sa2(tτ2)

x(t)=16(sin 2t2t)2

At t = π4

x(t) = 6.49

Fourier Transform Question 5:

The Fourier transform of a unit step function is given by

  1. F(jw) = 1 / w
  2. F(jw) = π δ(ω)+1jω
  3. F(jw) = 1 / jw
  4. None of these

Answer (Detailed Solution Below)

Option 2 : F(jw) = π δ(ω)+1jω

Fourier Transform Question 5 Detailed Solution

The unit step signal is defined as:

u(t)={1for t00for t<0

|u(t)| dt=

Since the unit step signal is not absolutely integrable, we cannot find the Fourier transform using the standard formula. 

Hence, we will derive the Fourier transform of the unit step signal starting from the Fourier transform of the signum function. 

The signum function can be defined as follows:

sgn(t)={1for t>00for t=01for t<0

|sgn(t)| dt=

F9 Neha B 5-10-2020 Swati D12

∴ We can see that the signum function is also not absolutely integrable. 

The signum function can be made absolutely integrable by multiplying with the exponential function.

sgn(t)=lima0 [eatu(t)eatu(t)]

Taking Fourier transform we get:

F[sgn(t)]=lima0 [1a+jω1ajω]

F[sgn(t)]=lima0 [2jωa2+ω2]

F[sgn(t)]=2jω

Now, the unit step signal can be represented in terms of signum function as follows:

u(t)=1+sgn(t)2

Taking Fourier transform we get:

F[u(t)]=F[12]+12F[sgn(t)]

We know, the Fourier transform of a DC signal 'A' is given as:

AF.T2πA δ(ω)

∴ F[u(t)]=2π×12×δ(ω)+12×2jω

 F[u(t)]=π δ(ω)+1jω

Fourier Transform Question 6:

The Fourier Cosine transform of the function F(x)={1,|x|<a0,|x|>a

  1. 2πsinass
  2. 1πcosass
  3. 12πcosass
  4. 1πsinass

Answer (Detailed Solution Below)

Option 1 : 2πsinass

Fourier Transform Question 6 Detailed Solution

Concept:

Fourier Sine transform:

Fs[f(x)]=2π0f(t)sinstdt

Fourier Cosine transform:

FC[f(x)]=2π0f(t)cosstdt

Calculation:

Given function is F(x)={1,|x|<a0,|x|>a

The Fourier Cosine transform is,

Fs[f(x)]=2π0f(t)cosstdt

=2πaacosstdt

=2π[sinsts]aa

=2πsinass

Fourier Transform Question 7:

A string of length (ℓ) is stretched tightly between two fixed ends. Find the displacement y(x, t) if it is released from its initial position given by y = y0 sin3 (πx/ℓ) 

  1. sinπxLcosπxLcosπctL+sin3πxLcos3πctL
  2. 3y04sinπxLcosπctLy04sin3πxLcos3πctL
  3. sinπxLcosπctL3y04sin3πxLcos3πctL
  4. None of these

Answer (Detailed Solution Below)

Option 2 : 3y04sinπxLcosπctLy04sin3πxLcos3πctL

Fourier Transform Question 7 Detailed Solution

Concept:

Equation describing transverse vibrations of a string is

2yt2=c22yx2

The boundary conditions are y(0, t) = y(ℓ, t) = 0

Initial condition, y(x, 0) = y0 sin3 (πx/ℓ)

yt=0

The vibration is periodic and the solution is given by,

y(x1 t) = (c1 cos px + c2 sin px) (c3 cos cpt + c4 sin pt)

Find the constants by applying the conditions

Calculation:

y(0, t) = y(L, t) = 0

c1 (c3 cos cpt + c4 sin cpt) = 0

c1    = 0

y(x, t) = c2 sin px (c3 cos cpt + c4 sin cpt)

y(1, t) = 0

c2 sin pL (c3 cos cpt + c4 sin cpt) = 0

PL = nπ

P = nπ/ℓ

∴ y(x,t)=c2sinnπxL[c3cosnπct+c4sinnπct]

yt=0 at t = 0

⇒ (c2sinnπx)nπc×c4=0

c4 = 0

y(x,t)=c2c3sinnπxcosnπct

bnsinnπxLcosnπct

The solution may be expressed as

y=n=1bnsinnπxLcosnπct

Also,

y(x,0)=y0sin3πxL

∴ n=1bnsinnπxL=y0sin3πxL

⇒ y0[3sinπxLsin3πxL4]=b1sinπxL+b2sin2πxL+b3sin3πxL+

Now,

Comparing both sides.

b1=3y04,b2=0,b3=y04,b4=b5=b6==0

∴ y(x,t)=3y04sinπxLcosπctLy04sin3πxLcos3πctL

Fourier Transform Question 8:

The amplitude spectrum |X(jω)| of a real signal x(t) is ______

  1. an even function 
  2. an odd function
  3. an even and odd function both
  4. neither even nor odd function

Answer (Detailed Solution Below)

Option 1 : an even function 

Fourier Transform Question 8 Detailed Solution

Fourier spectra:

The Fourier transform X(ω) of x(t) is, in general, complex, and it can be expressed as

X(ω) = |X (ω)| ejϕ(ω)

Fourier transform X(ω) of a nonperiodic signal x(t) is the frequency-domain specification of x(t) and is referred to as the spectrum(or Fourier spectrum) of x(t).

The quantity |X(ω)| is called the magnitude spectrum of x(t), and ϕ(ω) is called the phase spectrum of x(t).

If x(t) is a real signal, then we get

X(-ω) = x(t)ejωtdt

Then it follows that

X(-ω) = X*(ω)

and |X(-ω)| = |X(ω)|, ϕ(-ω) = - ϕ(ω)

Hence, the amplitude spectrum of |X(ω)| is an even function and continuous, the phase spectrum ϕ (ω) is an odd function.

Note:

For periodic signals, the amplitude spectrum is an even function of ω, and the phase spectrum is an odd function of ω.

Fourier Transform Question 9:

Using the Fourier expansion of f(x)={0πx0sinx0xπ 

Find the sum of series,

11.313.5+15.717.9+ (Correct up to 3 decimal).

Answer (Detailed Solution Below) 0 - 0.5

Fourier Transform Question 9 Detailed Solution

Concept:

Fourier series of f(x)

f(x)=a02+n=1ancosnπxL+n=1nnsinnπxl

a0=1lLLf(x)dx,an=1lllf(x)cosnπxLdx,bn=1lLLf(x)sinnπxLdx

Calculation:

a0=1πππf(x)dx=1π0πsinxdx=2π

an=1π0πsinxcosnxdx=12π0π[sin(n+1)xsin(n1)x]dx

an=12π[cos(n+1)xn+1+cos(n1)xn1]0π

= 0, n is odd and 2π(n21), n is even

Now,

bn=1π0πsinxsinnxdx

bn=12π0π[cos(n1)xcos(n+1)x]dx

bn=12π[sin(n1)xn1sin(n+1)xn+1]0π=0(n1)

When n = 1

b1=12

f(x)=1π2π[cos2x221+cos4x421+]+12sinx

Putx=π2

1=1π2π(11.3+13.515.7+)+12

11.313.5+15.7=14(π2)=0.285

Fourier Transform Question 10:

If G(jω) is the Fourier transform of g(t), then the Fourier transform of g(t+ln2) will be ________.

  1. 2jωG(jω)
  2. 2jωG(jω)
  3. 2ejωG(jω)
  4. 2ejωG(jω)

Answer (Detailed Solution Below)

Option 2 : 2jωG(jω)

Fourier Transform Question 10 Detailed Solution

Concept:

Shifting property of Fourier transform:

If F(jω) is the complex Fourier transform of f(t), then

F(f(tt0))=ejωtoF(jω)

Calculation:

F(g(t+ln2))=ejω(ln2)G(jω)

=ejωln2G(jω)

Since a lnb = ln ba, the above expression can be written as:

=eln2jωG(jω)

This can be written as:

=2jωG(jω)

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