Previous Year Questions- Fourier Series | Signals and Systems - Electrical Engineering (EE) PDF Download

Q1: The discrete time Fourier series representation of a signal x[n] with period N is written as  A discrete time periodic signal with period N = 3, has the non-zero Fourier series coefficients: a₋₃ = 2 and a₄ = 1. The signal is     (2022)
(a)
(b)
(c)
(d)
Ans: (b)
Sol: We have,
Put n = 0 in eq. (1)
x(0) = a₀+a₁ = 2+1 = 3
Put n = 1 in eq. (1)
These two conditions satisfy by the option (B). Hence, option (B) will be correct.

Q2: A periodic function f(t), with a period of 2π, is represented as its Fourier series,  
the Fourier series coefficients a₁ and b₁ of f(t) are      (2019)
(a)
(b)
(c)
(d)
Ans: (d)
Sol: 
Q3: Let the signal  be passed through an LTI system with frequency response H(ω) , as given in the figure below
The Fourier series representation of the output is given as      (SET-1 (2017))
(a) $4000+4000𝑐𝑜𝑠\left(2000𝜋𝑡\right)$4000 + 4000cos(2000πt) + 4000cos(4000πt)
(b) 2000+2000cos(2000πt) + 2000cos(4000πt)
(c) $4000𝑐𝑜𝑠\left(2000𝜋𝑡\right)$4000cos (2000πt)
(d) $2000𝑐𝑜𝑠\left(2000𝜋𝑡\right)$2000cos (2000πt)
Ans: (c)
Sol: Time period:
Since, x(t) is even-halt wave symmetric. So, its expansion will contain only odd harmonics of cos. Therefore, coefficient of fundamental harmonic is
Now, frequency components available in expansion are
As, LTI system given in the question will pass upto 5000 π rad/sec frequency component of input.So, output will have only one component of sfrequency 2000π rad/sec
Thus, y(t) = expansion of output = a₁cosω₀t = 4000cos 2000πt  

Q4: Consider  
Here, ⌊t⌋ represents the largest integer less than or equal to t and ⌈t⌉ denotes the smallest integer greater than or equal to t. The coefficient of the second harmonic component of the Fourier series representing g(t) is _________.       (SET-1   (2017))
(a) 0
(b) 1
(c) 2
(d) 3
Ans: (a)
Sol: Given that,  
where,
⌊t⌋= greatest integer less than or equal to 't'.
⌈t⌉= smallest integer greater than or equal to 't'.
Now,
Since, g(t) is nonperiodic. So, there is no Fourier series expansion of this signal and hence no need to calculate harmonic here.

Q5: Let f(x) be a real, periodic function satisfying f(-x) = -f(x). The general form of its Fourier series representation would be      (SET-2  (2016))
(a)
(b)
(c)
(d)
Ans: (b)
Sol: Given that,
 f(−x) = −f(x)
So, function is an odd function.
So, the fourier series will have sine term only. So,

Q6: The signum function is given by
The Fourier series expansion of sgn(cos(t)) has      (SET-1(2015))
(a) only sine terms with all harmonics
(b) only cosine terms with all harmonics.
(c) only sine terms with even numbered harmonics.
(d) only cosine terms with odd numbered harmonics.
Ans: (d)
Sol: So, cos(t) is
So, sgn(cos t) is a rectangular signal which is even and has half wave symmetry.
So, Fourier series will have only cosine terms with add harmonics only.

Q7: Let g : [0, ∞) → [0, ∞) be a function defined by g(x) = x-[x], where [x] represents the integer part of x . (That is, it is the largest integer which is less than or equal to x ). The value of the constant term in the Fourier series expansion of g(x) is____.       (SET-1 (2014))
(a) 0
(b) 0.5
(c) 0.75
(d) 1
Ans: (b)
Sol: Given function g(t) = x−[x]
where, [x] is a integer part of x
Then function g(x) will be
The value of the constant term (or) dc term in the Fourier series expansion of g(x) is

Q8: For a periodic square wave, which one of the following statements is TRUE ?     (SET-1 (2014))
(a) The Fourier series coefficients do not exist
(b) The Fourier series coefficients exist but the reconstruction converges at no point
(c) The Fourier series coefficients exist but the reconstruction converges at most point
(d) The Fourier series coefficients exist and the reconstruction converges at every point.
Ans: (c)
Sol: 

Reconstruction of signal by its Fourier series coefficient is not possible at those points where signal is discontinuous.
In the above figure, at integer multiples of 'T/2', signal recovery is not possible by using its coefficient.
Therefore, reconstruction of x(t) by using its coefficient is possible at most of the points except those instants where x(t) is discontinous.

Q9: For a periodic signal
v(t) = 30sin100t + 10cos300t + 6sin(500t + π/4),
the fundamental frequency in rad/s       (2013)
(a) 100
(b) 300
(c) 500
(d) 1500
Ans: (a)

Q10: The fourier series expansion  of the periodic signal shown below will contain the following nonzero terms       (2011)
(a) a₀ and b_n, n = 1, 3, 5,... ∞
(b) $𝑎0𝑎𝑛𝑑𝑎𝑛,𝑛=1,2,3,...\infty$a₀ and a_n, n = 1, 2, 3 ,... ∞
(c) $𝑎0,𝑎𝑛𝑎𝑛𝑑𝑏𝑛,𝑛=1,3,5,...\infty$a₀, a_n and b_n, n = 1, 3, 5,... ∞
(d) a₀ and a_n, n = 1, 3, 5 ,... ∞
Ans: (d)
Sol: Let, x(t) = Even and Hws
Fourier series expansion of x(t) contains cos terms with odd harmonics.
Now, f(t) = 1 + x(t)  
Fourier series of f(t) contains dc and cos terms with odd harmonics.

Q11: The second harmonic component of the periodic waveform given in the figure has an amplitude of
(a) 0
(b) 1
(c) 2/π
(d) √5
Ans: (a)
Sol: The given signal is odd as wel as having half wave symmetry.
So, it has only sine terms with odd harmonics. So, for second harmonic term amplitude = 0.

Q12: The Fourier Series coefficients of a periodic signal x(t), expressed as 𝑥(𝑡) =  are given by a₋₂ = 2−j1, 𝑎₋₁ = 0.5 + 𝑗0.2, a₀ = j2, a₁ = 0.5− j0.2,  𝑎₂ = 2+𝑗1 and  a_k = 0 for ∣k∣ > 2 Which of the following is true ?      (2009)
(a) x(t) has finite energy because only finitely many coefficients are non-zero
(b) x(t) has zero average value because it is periodic
(c) The imaginary part of x(t) is constant
(d) The real part of x(t) is even
Ans: (c)
Sol: 
Q13: Let x(t) be a periodic signal with time period T, Let 𝑦(𝑡) = x(t − t₀) + x(t + t₀) for some t₀. The Fourier Series coefficients of y(t) are denoted by b_k. If b_k = 0 for all odd k , then t₀ can be equal to      (2008 )
(a) T/8
(b) T/4
(c) T/2
(d) 2T
Ans: (b)
Sol: y(t) = x(t − t₀) + x(t + x₀)
Since, x(t) is periodic with period T.
Therefore, x(t − t₀) and x(t + t₀) will also be periodic with period T.  
a_k is Fourier series coefficient of signal x(t)
therefore,

Q14: A signal x(t) is given by  
Which among the following gives the fundamental fourier term of x(t)?     (2007)
(a)
(b)
(c)
(d)
Ans: (a)
Sol: According to defination of signal given in question the x(t) will be as
So it is periodic with period,
Therefore, fundamental angular frequency
Now,
Comparing it with

Q15: x(t) is a real valued function of a real variable with period T. Its trigonometric Fourier Series expansion contains no terms of frequency ω = 2π(2k)/T; k = 1, 2... Also, no sine terms are present. Then x(t) satisfies the equation        (2006)
(a) $𝑥\left(𝑡\right)=-𝑥(𝑡-𝑇)$x(t) = −x(t − T)
(b) x(t) = x(T − t) = −x(−t)
(c) $𝑥\left(𝑡\right)=𝑥(𝑇-𝑡)=-𝑥(𝑡-𝑇/2)$x(t) = x(T − t) = −x(t − T/2)
(d) x(t) = x(t − T) = x(t − T/2)
Ans: (c)
Sol: Since trigonometric fourier series has no sine terms and has only cosine terms therefore this will be an even signal i.e. it will satisfy.
x(t) = x(−t)
or, we can write,
 x(t − T) = x(−t + T)
but signal is periodic with period T.  
therefore x(t − T) = x(t)
therefore, x(t) = x(T − t)...(i)
Now, since signal contains only odd harmonics i.e. no terms of frequency
i.e. no even harmonics.
This means signal contains half wave symmetry
this implies that,

Q16: The Fourier series for the function f(x) = sin²x is    (2005)
(a) $\sin 𝑥+\sin 2𝑥$sinx + sin2x
(b) $1-\cos 2𝑥$1 − cos2x  
(c) sin2x + cos2x
(d) $0.5-0.5\cos 2𝑥$0.5 − 0.5cos2x  
Ans: (d)
Sol: f(n) = sin²x
for finding the fourier series expansion

Q17: For the triangular wave from shown in the figure, the RMS value of the voltage is equal to    (2005)
(a)  
(b)
(c) 1/3
(d)
Ans: (a)
Sol: From the wave symmetry,

Q18: The rms value of the periodic waveform given in figure is
(a) 2√6 A
(b) 6√2 A
(c) √4/3 A
(d) 1.5 A
Ans: (a)
Sol: 
Q19: Fourier Series for the waveform, f(t) shown in figure is    (2002)
(a)
(b)
(c)
(d)
Ans: (c)
Sol: ∵ f(t) is an even function with half waves symmetry,
∴ dc term as well as sine terms = 0
Only the cosine terms with odd harmonics will be present.