Signals & Systems Multiple Choice Questions 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
Answer: A
Clarification: 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)
Answer: B
Clarification: 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)
Answer: A
Clarification: 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
Answer: B
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
Answer: C
Clarification: 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 .
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
Answer: B
Clarification: 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)
Answer: D
Clarification: 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
Answer: A
Clarification: 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
Answer: C
Clarification: 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