250+ TOP MCQs on Fourier Series Coefficients – 2 and Answers

Signals & Systems Objective Questions on “Fourier Series Coefficients – 2”.

1. The Fourier series coefficient for the signal 10δ(t) is ___________
A. 1
B. Cos ((frac{π}{2}) k)
C. sin ((frac{π}{2}) k)
D. 2

Answer: D
Clarification: (X[k] = frac{1}{T} displaystyleint_{-frac{T}{2}}^{frac{T}{2}} Aδ(t)e^{-jkωt} ,dt)
= (frac{A}{2})
Here, A=10, T=5
∴ X[k] = 2.

2. The Fourier series coefficient for the periodic rectangular pulses of height 2A is ____________
A. (frac{2A}{jkπ} ,sin,⁡ frac{π}{2} k)
B. (frac{2A}{jkπ} ,cos⁡, frac{π}{2} k)
C. (frac{2A}{kπ} ,sin⁡, frac{π}{2} k)
D. (frac{2A}{kπ} ,cos⁡, frac{π}{2} k)

Answer: C
Clarification: (X[k] = frac{1}{T} displaystyleint_{-frac{T}{2}}^{frac{T}{2}} x(t)e^{-jkωt} ,dt)
(= frac{1}{T} displaystyleint_{-frac{T}{4}}^{frac{T}{4}} Ae^{-jkωt} ,dt)
(= frac{2A}{T}[frac{e^{-jkωt}}{-jkω}]) (from –(frac{T}{4}) to (frac{T}{4}))
(=frac{2A}{kπ} ,sin⁡, frac{π}{2} k).

3. The Fourier series coefficient for the periodic signal x(t) = sin²t is _____________
A. –(frac{1}{4}) δ[k-1] + (frac{1}{2}) δ[k] – (frac{1}{4}) δ[k+1]
B. –(frac{1}{4}) δ[k-2] + (frac{1}{2}) δ[k] – (frac{1}{4}) δ[k+2]

C. –(frac{1}{2}) δ[k-1] + δ[k] – (frac{1}{2}) δ[k+1]
D. –(frac{1}{2}) δ[k-2] + δ[k] – (frac{1}{2}) δ[k+2]

Answer: A
Clarification: sin²t = ((frac{e^{jt} – e^{-jt}}{2j})^2)
= –(frac{1}{4}) (e^2jt – 2 + e^-2jt)
The fundamental period of sin2t is π and ω = (frac{2π}{π}) = 2
∴ X[k] = –(frac{1}{4}) δ[k-1] + (frac{1}{2}) δ[k] – (frac{1}{4}) δ[k+1].

4. The Fourier series coefficient of time domain signal x (t) is X[k] = jδ[k-1] – jδ[k+1] + δ[k+3] + δ[k-3], the fundamental frequency of the signal is ω=2π. The signal is ___________
A. 2(cos 3πt – sin πt)
B. -2(cos 3πt – sin πt)
C. 2(cos 6πt – sin 2πt)
D. -2(cos 6πt – sin 2πt)

Answer: C
Clarification: (x (t) = ∑_{k=-∞}^∞ X[k]e^{j2πkt})
= je^j2πt – je^-j2πt + e^j6πt + e^-j6πt
= 2(cos 6πt – sin 2πt).

5. The Fourier series coefficient of time domain signal x (t) is X[k] = ((-frac{1}{3})^{|k|}). The fundamental frequency of signal is ω=1. The signal is _____________
A. (frac{4}{5 + 3 sin⁡})
B. (frac{5}{4 + 3 sin⁡t})
C. (frac{5}{4 + 3 cos⁡t})
D. (frac{4}{5 + 3 sin⁡t})

Answer: D
Clarification: (x (t) = ∑_{k=-∞}^∞ X[k]e^{jkt})
Or, x (t) = (∑_{k=-∞}^{-1} (-frac{1}{3})^{-k} e^{jk} + ∑_{k=0}^∞ (-frac{1}{3})^k e^{jkt})
= (frac{frac{-1}{3} e^{-jt}}{1+frac{1}{3} e{-jt}} + frac{1}{1 + frac{1}{3} e^{jt}})
= (frac{4}{5 + 3 sin⁡t}).

6. The Fourier series coefficient of the signal y(t) = x(t-t₀) + x(t+t₀) is _____________
A. 2 cos ((frac{2π}{t}) kt₀) X[k]
B. 2 sin ((frac{2π}{t}) kt₀) X[k]
C. 2 cos ((frac{2π}{t}) kt₀)
D. 2 sin ((frac{2π}{t}) kt₀)

Answer: A
Clarification: x (t-t₀) is periodic with period T. the Fourier series coefficient of x (t-t₀) is X₁[k] = (frac{1}{T}) ∫ x (t-t₀)e^-jkωt dt
= e^-jkωt₀ X[k]
Similarly, the Fourier series coefficient of x (t+t₀) is X₂[k] = e^jkωt₀ X[k]
The Fourier series coefficient of x (t-t₀) + x (t+t₀) is
Y[k] = X₁[k] + X₂[k]
= e^-jkωt₀ X[k] + e^jkωt₀ X[k]
= 2 cos ((frac{2π}{t}) kt₀) X[k].

7. The Fourier series coefficient of the signal y(t) = Even{x(t)} is ___________
A. (frac{X[k]+X[-k]}{2})
B. (frac{X[k]-X[-k]}{2})
C. (frac{X[k]+X^* [-k]}{2})
D. (frac{X[k]-X^* [-k]}{2})

Answer: A
Clarification: even {x (t)} = (frac{x(t)+x(-t)}{2})
The Fourier series coefficient transform of x (t) is
X₁[k] = (frac{1}{T}) ∫ x (-t)e^-jkωt dt
= (frac{1}{T}) ∫ x (α)e^jkωα dα
= X [-k]
∴ The Fourier coefficient of Even{x(t)} = Y[k] = (frac{X[k]+X[-k]}{2}).

8. The Fourier series coefficient of the signal y(t) = Re{x(t)} is ____________
A. (frac{X[k]+X[-k]}{2})
B. (frac{X[k]-X[-k]}{2})
C. (frac{X[k]+X^* [-k]}{2})
D. (frac{X[k]-X^* [-k]}{2})

Answer: C
Clarification: Re{x (t)} = (frac{x(t)+x^* (-t)}{2})
The Fourier coefficient of x* (t) is
X₁[k] = (frac{1}{T}) ∫ x* (t)e^-jkωt dt = (X_1^*) [-k]
Or, (X_1^*) [k] = (frac{1}{T}) ∫ x(t)e^jkωt dt = X [-k]
So, X₁[k] = (X_1^*) [-k]
∴ Y[k] = (frac{X[k]+X^* [-k]}{2}).

9. The Fourier series coefficient of the signal y(t) = (frac{d^2 x(t)}{dt^2}) is _____________
A. ((frac{2πk}{T})^2 X[k])
B. –((frac{2πk}{T})^2 X[k])
C. j((frac{2πk}{T})^2 X[k])
D. -j((frac{2πk}{T})^2 X[k])

Answer: B
Clarification: (x (t) = ∑_{k=-∞}^∞ X[k]e^{j frac{2π}{T} kt})
Now, (frac{dx(t)}{dt} = -j (frac{2π}{T})k ∑_{k=-∞}^∞ X[k]e^{j frac{2π}{T} kt})
And, (frac{d^2 x(t)}{dt^2} = -(frac{2π}{T})^2 k^2 ∑_{k=-∞}^∞ X[k]e^{j frac{2π}{T} kt})
∴ Y[k] = – ((frac{2πk}{T})^2 X[k]).

10. The Fourier series coefficient of the signal y(t) = x(4t-1) is ______________
A. (frac{8π}{T} X[k])
B. (frac{4π}{T}X[k])
C. (e^{-jk frac{8π}{T}} X[k])
D. (e^{jk frac{8π}{T}} X[k])

Answer: C
Clarification: The period of x (4t) is a fourth of the period of x (t). The Fourier series coefficient of x (4t) is still X[k]. Hence, the coefficient of x (4t-1) is (e^{-jk frac{8π}{T}} X[k]).

11. The discrete time Fourier coefficients of (∑_{m=-∞}^∞ δ[n-4m]) is ____________
A. –(frac{1}{4}) for all k
B. (frac{1}{4}) for all k
C. –(frac{1}{2}) for all k
D. (frac{1}{2}) for all k

Answer: B
Clarification: N=4, ω = (frac{2π}{4} = frac{π}{2})
(X[k] = frac{1}{4} ∑_{n=4}^3 x[n]e^{-j(frac{π}{2})nk})
= (frac{1}{4}) x[0]
= (frac{1}{4}) for all k.

12. The discrete time Fourier coefficient of cos²((frac{π}{8}) n) is ______________
A. (frac{π}{2})(δ(k+1] + 2δ[k] + δ[k-1])
B. (frac{1}{4})j(δ(k+1] + 2δ[k] + δ[k-1])
C. (frac{1}{4})(δ(k+1] + 2δ[k] + δ[k-1])
D. (frac{π}{4})(δ(k+1] + 2δ[k] + δ[k-1])

Answer: C
Clarification: N=8, ω = (frac{2π}{8} = frac{π}{4})
X[n] = cos² ((frac{π}{8}) n) = (frac{1}{4}(e^{j(frac{π}{8})n} + e^{-j(frac{π}{8})n})^2)
(= frac{1}{4}(e^{j(frac{π}{8})n} + 2 + e^{-j(frac{π}{8}n})^2)
Or, X[k] = (frac{1}{4})(δ(k+1] + 2δ[k] + δ[k-1]).

13. V(t) = 5,    0≤t<1;
t,           t≥1;
The Laplace transform of V (t) is ___________
A. (frac{5}{s} + frac{e^{-s}}{s^2} + frac{4e^{-s}}{s})
B. (frac{5}{s} + frac{e^{-s}}{s^2} – frac{4e^{-s}}{s})
C. (frac{5}{s} – frac{e^{-s}}{s^2} – frac{4e^{-s}}{s})
D. (frac{5}{s} – frac{e^{-s}}{s^2} + frac{4e^{-s}}{s})

Answer: B
Clarification: V (t) = 5 + u (t) (t-5)
L {5 + u (t) (t-5)} = (frac{5}{s}) + L {u (t) (t-5)}
= (frac{5}{s} + e^{-s}) L {t-4}
= (frac{5}{s} + e^{-s} (frac{1}{s^2} – frac{4}{s}))
= (frac{5}{s} + frac{e^{-s}}{s^2} – frac{4e^{-s}}{s}).

14. W(t) = 2,   0≤t<4;
t²,          t≥4;
The Laplace transform of W (t) is ___________
A. (frac{2}{s} – e^{-4s} (frac{2}{s^3} – frac{8}{s^2} – frac{14}{s}))
B. (frac{2}{s} + e^{-4s} (frac{2}{s^3} – frac{8}{s^2} – frac{14}{s}))
C. (frac{2}{s} – e^{-4s} (frac{2}{s^3} + frac{8}{s^2} + frac{14}{s}))
D. (frac{2}{s} + e^{-4s} (frac{2}{s^3} + frac{8}{s^2} + frac{14}{s}))

Answer: D
Clarification: W (t) = 2 + u (t) (t²-2)
L {2 + u (t) (t²-2)} = (frac{2}{s}) + L {u (t) (t²-2)}
= (frac{2}{s} + e^{-4s}) L {(t+4)² -2}
= (frac{2}{s} + e^{-4s}) L {t² + 8t + 14}
= (frac{2}{s} + e^{-4s} (frac{2}{s^3} + frac{8}{s^2} + frac{14}{s})).

15. U(t) = 0,   0≤t<7;
(t-7)³,          t≥7;
The Laplace transform of U (t) is ___________
A. (frac{6e^{-7s}}{s^4})
B. (frac{e^{-7s}}{s^4})
C. (frac{6e^{-7s}}{s^3})
D. (frac{3e^{-7s}}{s^3})

Answer: A
Clarification: U (t) = u (t) (t-7)³
L {u (t) (t-7)³} = e^-7s L {t³}
= (frac{3!e^{-7s}}{s^4} = frac{6e^{-7s}}{s^4}).