# Signals & Systems Questions and Answers – Symmetry Properties of the Fourier Series

This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Symmetry Properties of the Fourier Series”.

1. How can fourier series calculations be made easy?
a) Using symmetry conditions
b) Using formula
c) Using integration
d) Calculations are easy anyways

Explanation: Fourier series calculations are made easy because the series consists of sine and cosine functions and if they are in symmetry they can be easily done. Some integration is always even or odd, hence, we can calculate.

2. What is the function of an even signal?
a) x(t) = -x(t)
b) x(t) = x(-t)
c) x(t) = -x(-t)
d) x(t) = x(t+1)

Explanation: An even signal is one in which the functional values of the signal in t and –t is same. Hence, even signal is one in which x(t) and x(-t) is same.

3. What is the function of an odd signal?
a) x(t) = -x(t)
b) x(t) = x(-t)
c) x(t) = -x(-t)
d) x(t) = x(t+1)

Explanation: An even signal is one in which the functional values of the signal in t is negative of the value of -t . Hence, even signal is one in which x(-t) is –x(t).

4. What is the product of an even signal and odd signal?
a) Even signal
b) Odd signal
c) Mixture of even and odd
d) Odd signals sometimes

Explanation: The product of an even and odd signal is an odd signal. But product or sum of two even signals is even signal.

5. What are the values of an and bn when the signal is even?
a) an=0 and bn=0
b) an=0 and bn=4/T∫x(t)cos(nwt)dt
c) an=4/T∫x(t)cos(nwt)dt and bn=0
d) an=4/T∫x(t)sin(nwt)dt and bn=4/T∫x(t)cos(nwt)dt

Explanation: In an even signal the summation of 0 to T/2 is in sine series is zero.
Hence, 4/T∫x(t)sin(nwt)dt=0, hence bn=0. So, a0= 2/T∫x(t)dt
And an=4/T∫x(t)cos(nwt)dt .
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6. What are the values of an and bn when the signal is even?
a) an=0 and bn=0
b) an=0 and bn=4/T∫x(t)sin(nwt)dt
c) an=4/T∫x(t)cos(nwt)dt and bn=0
d) an=4/T∫x(t)sin(nwt)dt and bn=4/T∫x(t)cos(nwt)dt

Explanation: In an odd signal the summation of 0 to T/2 is in cosine series is zero.
Hence, 4/T∫x(t)cos(nwt)dt=0, hence an=0. So, a0= 0
And bn=4/T∫x(t)sin(nwt)dt .

7. How can we define half wave symmetry?
a) x(t) = -x(t±T)
b) x(t) = x(t±T/2)
c) x(t) = x(t±T)
d) x(t) = -x(t±T/2)

Explanation: x(t) = -x(t±T/2) is how we define a half wave symmetry. In this case, the waveform is neither even nor odd, it must be both.

7. How can we define the coefficients half wave symmetry when n is even?
a) an=0 and bn=0 and a0=0
b) an=4/T∫x(t)cos(nwt)dt and bn=0= a0
c) an=4/T∫x(t)sin(nwt)dt and bn=4/T∫x(t)cos(nwt)dt and a0=0
d) an=4/T∫x(t)sin(nwt)dt and bn=4/T and a0=0

Explanation: If it is half wave symmetry then we define the fourier coefficients as-
an=0 and bn=0 and a0=0 if n is an even number.

8. How can we define the coefficients a half wave symmetry when n is even?
a) an=0 and bn=0 and a0=0
b) an=4/T∫x(t)cos(nwt)dt and bn=0= a0
c) an=4/T∫x(t)sin(nwt)dt and bn=4/T∫x(t)cos(nwt)dt and a0=0
d) an=4/T∫x(t)sin(nwt)dt and bn=4/T and a0=0

Explanation: If it is a half wave symmetry then we define the fourier coefficients as-
an=0 and an=4/T∫x(t)cos(nwt)dt and bn=4/T∫x(t)sin(nwt)dt.

9. When does a wave posses a quarter wave symmetry?
a) It has either even or odd symmetry
b) It has half wave symmetry
c) even/odd symmetry and half wave symmetry
d) It is even in one quarter and odd in the other

Explanation: A signal is said to posses quarter wave symmetry when it is either odd or even symmetry and has half wave symmetry. These signals are shifted to left or right by T/4 with a half wave symmetry but the even symmetry of the original unshifted towards the odd symmetry it is said to be quarter wave symmetry.

10. What are the types of symmetry shown by signals?
a) Even symmetry and odd symmetry
b) Even, odd and quarter wave symmetry
c) Even, odd, half-wave and quarter wave symmetry
d) Half wave symmetry

Explanation: The types of symmetry shown by signals are- Even, odd, half-wave and quarter wave symmetry.

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