Signals & Systems Multiple Choice Questions on “Fourier Transforms”.
1. Which of the following is the Analysis equation of Fourier Transform?
A. (F(ω) = int_{-∞}^∞ f(t)e^{jωt} ,dt)
B. (F(ω) = int_0^∞ f(t)e^{-jωt} ,dt)
C. (F(ω) = int_0^∞ f(t)e^{jωt} ,dt)
D. (F(ω) = int_{-∞}^∞ f(t)e^{-jωt} ,dt)
Answer: D
Clarification: For converting time domain to frequency domain, we use analysis equation. The Analysis equation of Fourier Transform is (F(ω) = int_{-∞}^∞ f(t)e^{-jωt} ,dt).
2. Choose the correct synthesis equation.
A. (f(t) = frac{1}{2π} int_{-∞}^∞ F(ω) e^{-jωt} ,dω)
B. (f(t) = frac{1}{2π} int_{-∞}^∞ F(ω) e^{jωt} ,dω)
C. (f(t) = frac{1}{2π} int_0^∞ F(ω) e^{-jωt} ,dω)
D. (f(t) = frac{1}{2π} int_0^∞ F(ω) e^{jωt} ,dω)
Answer: B
Clarification: Synthesis equation converts from frequency domain to time domain. The synthesis equation of fourier transform is (f(t) = frac{1}{2π} int_{-∞}^∞ F(ω) e^{jωt} ,dω).
3. Find the fourier transform of an exponential signal f(t) = e-at u(t), a>0.
A. (frac{1}{a+jω})
B. (frac{1}{a-jω})
C. (frac{1}{-a+jω})
D. (frac{1}{-a-jω})
Answer: A
Clarification: Given f(t)= e-at u(t)
We know that (
u(t)=begin{cases}
0 &text{(t0)} \
end{cases})
Fourier transform,
(F(ω) = int_{-∞}^∞ f(t)e^{-jωt} ,dt = int_{-∞}^∞ e^{-at} u(t)e^{-jωt} ,dt = int_0^∞ e^{-(a+jω)t} ,dt)
F(ω) = (frac{1}{a+jω}), a>0.
4. Find the fourier transform of the function f(t) = e-a|t|, a>0.
A. (frac{2a}{a^2-ω^2})
B. (frac{2a}{a^2+ω^2})
C. (frac{2a}{ω^2-a^2})
D. (frac{a}{a^2+ω^2})
Answer: B
Clarification: The given two-sided exponential function f(t) = e-a|t|, a>0 can be expressed as
(
f(t)=begin{cases}
e^{-at} &text{(t≥0)} \
e^{at} &text{(t≤0)} \
end{cases})
The Fourier transform is
(F(ω) = int_{-∞}^∞ f(t)e^{-jωt} ,dt = int_{-∞}^0 f(t)e^{-jωt} ,dt + int_0^∞ f(t)e^{-jωt} ,dt)
(F(ω) = frac{1}{a+jω} + frac{1}{a-jω} = frac{2a}{a^2+ω^2}).
5. Gate function is defined as ______________
A. (
G(t)=begin{cases}
1 &text{(|t|
end{cases} )
B. (
G(t)=begin{cases}
1 &text{(|t|>frac{τ}{2})} \
0 &text{elsewhere} \
end{cases})
C. (
G(t)=begin{cases}
1 &text{(|t|≤frac{τ}{2})} \
0 &text{elsewhere} \
end{cases})
D. (
G(t)=begin{cases}
1 &text{(|t|≥frac{τ}{2})} \
0 &text{elsewhere} \
end{cases})
Answer: A
Clarification: A gate function is a rectangular function defined as
( G(t) = rect(frac{t}{τ}) = begin{cases}
1 &text{(|t|
end{cases} )
Where τ is pulse width.
6. Find the fourier transform of the gate function.
A. (frac{1}{ω} sin(frac{ωτ}{2}))
B. (frac{1}{ω} cos(frac{ωτ}{2}))
C. (frac{2}{ω} sin(frac{ωτ}{2}))
D. (frac{2}{ω} cos(frac{ωτ}{2}))
Answer: C
Clarification: Gate function is defined as
(
G(t)=begin{cases}
1 &text{(|t|
end{cases} )
The fourier transform is (F(ω) = int_{-∞}^∞ f(t)e^{-jωt} ,dt = int_{-τ/2}^{τ/2} e^{-jωt} ,dt = frac{2}{ω} sin(frac{ωτ}{2})).
7. Choose the wrong option.
A. G(t) = rect((frac{t}{τ}))
B. G(t) = u(t + (frac{τ}{2})) – u(t-(frac{τ}{2}))
C. G(ω) = τ sa((frac{wτ}{2}))
D. G(f) = τ sinc(f)
Answer: D
Clarification: Fourier transform of gate function, G(ω) = (frac{2}{ω} sin(frac{wτ}{2}))
Multiplying and dividing by τ we get
(G(ω) = τ frac{sin(frac{wτ}{2})}{frac{wτ}{2}} = τ frac{sin(frac{2πfτ}{2})}{frac{2πfτ}{2}}= τ frac{sin(πτf)}{πτf} = τ sinc(τf)).
8. Bandwidth of the gate function is __________
A. τ Hz
B. (frac{1}{τ}) Hz
C. 2τ Hz
D. (frac{2}{τ}) Hz
Answer: B
Clarification: The practical bandwidth of the gate function corresponds to the first zero crossing in the spectrum. Therefore, the bandwidth of the pulse or gate function is (frac{2π}{τ}) or (frac{1}{τ}) Hz.
9. Which of the following is not a fourier transform pair?
A. (u(t) leftrightarrow πδ(ω) + frac{1}{jω})
B. (sgn(t) leftrightarrow frac{2}{jω})
C. (A leftrightarrow 2πδ(frac{ω}{2}))
D. (G(t)leftrightarrow sa(frac{ωτ}{2}))
Answer: D
Clarification: (G(t)leftrightarrow sa(frac{ωτ}{2})) is not a fourier transform pair.
(G(t)leftrightarrow τsa(frac{ωτ}{2})) (or) (G(t)leftrightarrow G(t) τ sinc(τf)).
10. Find the fourier transform of the unit step function.
A. πδ(ω) + (frac{1}{ω})
B. πδ(ω) + (frac{1}{jω})
C. πδ(ω) – (frac{1}{jω})
D. δ(ω) + (frac{1}{jω})
Answer: B
Clarification: We know that sgn(t) = 2u(t) – 1.
u(t) = (frac{1}{2})[sgn(t)+1]
Its Fourier transform is F[u(t)] = (frac{1}{2}) F[sgn(t)] + (frac{1}{2}) F[1]
As the Fourier transforms F[1] = 2πδ(ω) and [sgn(t)] = (frac{2}{jω}), hence
F[u(t)] = πδ(ω) + (frac{1}{jω}).