Digital Electronic MCQ Set 1
1. What is the Fourier series representation of a signal x(n) whose period is N?
Answer
Answer: b [Reason:] Here, the frequency F0 of a continuous time signal is divided into 2π/N intervals.
So, the Fourier series representation of a discrete time signal with period N is given as
where ck is the Fourier series coefficient
2. What is the expression for Fourier series coefficient ck in terms of the discrete signal x(n)?
Answer
Answer: d [Reason:] We know that, the Fourier series representation of a discrete signal x(n) is given as 
3. Which of the following represents the phase associated with the frequency component of discrete-time Fourier series(DTFS)?
a) ej2πkn/N
b) e-j2πkn/N
c) ej2πknN
d) None of the mentioned
Answer
Answer: a [Reason:] We know that,
In the above equation, ck represents the amplitude and ej2πkn/N represents the phase associated with the frequency component of DTFS.
4. The Fourier series for the signal x(n)=cos√2πn exists.
a) True
b) False
Answer
Answer: b [Reason:] For ω0=√2π, we have f0=1/√2. Since f0 is not a rational number, the signal is not periodic. Consequently, this signal cannot be expanded in a Fourier series.
5. What are the Fourier series coefficients for the signal x(n)=cosπn/3?
a) c1=c2=c3=c4=0,c1=c5=1/2
b) c0=c1=c2=c3=c4=c5=0
c) c0=c1=c2=c3=c4=c5=1/2
d) None of the mentioned
Answer
Answer: a [Reason:] In this case, f0=1/6 and hence x(n) is periodic with fundamental period N=6.
Given signal is x(n)= cosπn/3=cos2πn/6=1/2 e^(j2πn/6)+1/2 e^(-j2πn/6)
We know that -2π/6=2π-2π/6=10π/6=5(2π/6)
So, we get c1=c2=c3=c4=0 and c1=c5=1/2.
6. What is the Fourier series representation of a signal x(n) whose period is N?
Answer
Answer: b [Reason:] Here, the frequency F0 of a continuous time signal is divided into 2π/N intervals.
So, the Fourier series representation of a discrete time signal with period N is given as
where ck is the Fourier series coefficient
7. What is the average power of the discrete time periodic signal x(n) with period N ?
Answer
Answer: d [Reason:] Let us consider a discrete time periodic signal x(n) with period N.
The average power of that signal is given as
8. What is the equation for average power of discrete time periodic signal x(n) with period N in terms of Fourier series coefficient ck?
Answer
Answer: b [Reason:] We know that
9. What is the Fourier transform X(ω) of a finite energy discrete time signal x(n)?
d) None of the mentioned
Answer
Answer: a [Reason:] If we consider a signal x(n) which is discrete in nature and has finite energy, then the Fourier transform of that signal is given as
10. What is the period of the Fourier transform X(ω) of the signal x(n)?
a) π
b) 1
c) Non-periodic
d) 2π
Answer
Answer: d [Reason:] Let X(ω) be the Fourier transform of a discrete time signal x(n) which is given as
Now
So, the Fourier transform of a discrete time finite energy signal is periodic with period 2π.
11. What is the synthesis equation of the discrete time signal x(n), whose Fourier transform is X(ω)?
d) None of the mentioned
Answer
Answer: c [Reason:] We know that the Fourier transform of the discrete time signal x(n) is
The above equation is known as synthesis equation or inverse transform equation.
12. What is the value of discrete time signal x(n) at n=0 whose Fourier transform is represented as below?
a) ωc.π
b) -ωc/π
c) ωc/π
d) None of the mentioned
Answer
Answer: c [Reason:] We know that, 
Therefore, the value of the signal x(n) at n=0 is ω_c/π.
13. What is the value of discrete time signal x(n) at n≠0 whose Fourier transform is represented as below?
d) None of the mentioned
Answer
Answer: a [Reason:] We know that, 
14. The oscillatory behavior of the approximation of XN(ω) to the function X(ω) at a point of discontinuity of X(ω) is known as Gibbs phenomenon.
a) True
b) False
Answer
Answer: a [Reason:] We note that there is a significant oscillatory overshoot at ω=ωc, independent of the value of N. As N increases, the oscillations become more rapid, but the size of the ripple remains the same. One can show that as N→∞, the oscillations converge to the point of the discontinuity at ω=ωc. The oscillatory behavior of the approximation of XN(ω) to the function X(ω) at a point of discontinuity of X(ω) is known as Gibbs phenomenon.
15. What is the energy of a discrete time signal in terms of X(ω)?
d) None of the mentioned
Answer
Answer: b [Reason:] We know that, 
Digital Electronic MCQ Set 2
1. If the signal to be analyzed is an analog signal, we would pass it through an anti-aliasing filter with B as the bandwidth of the filtered signal and then the signal is sampled at a rate:
a) Fs ≤ 2B
b) Fs ≤ B
c) Fs ≥ 2B
d) Fs = 2B
Answer
Answer: c [Reason:] The filtered signal is sampled at a rate of Fs≥ 2B, where B is the bandwidth of the filtered signal to prevent aliasing.
2. What is the highest frequency that is contained in the sampled signal?
a) 2Fs
b) Fs/2
c) Fs
d) None of the mentioned
Answer
Answer: b [Reason:] We know that, after passing the signal through anti-aliasing filter, the filtered signal is sampled at a rate of Fs≥ 2B=>B≤ Fs/2.Thus the maximum frequency of the sampled signal is Fs/2.
3. The finite observation interval for the signal places a limit on the frequency resolution.
a) True
b) False
Answer
Answer: a [Reason:] After sampling the signal, we limit the duration of the signal to the time interval T0=LT, where L is the number of samples and T is the sample interval. So, it limits our ability to distinguish two frequency components that are separated by less than 1/T0=1/LT in frequency. So, the finite observation interval for the signal places a limit on the frequency resolution.
4. If {x(n)} is the signal to be analyzed, limiting the duration of the sequence to L samples, in the interval 0≤ n≤ L-1, is equivalent to multiplying {x(n)} by:
a) Kaiser window
b) Hamming window
c) Hanning window
d) Rectangular window
Answer
Answer: d [Reason:] The equation of the rectangular window w(n) is given as
w(n)= 1, 0≤ n≤ L-1
=0, otherwise
Thus, we can limit the duration of the signal x(n) to L samples by multiplying it with a rectangular window of length L.
5. What is the Fourier transform of rectangular window of length L?
d) None of the mentioned
Answer
Answer: c [Reason:] We know that the equation for the rectangular window w(n) is given as
w(n)= 1, 0≤ n≤ L-1
=0, otherwise
We know that the Fourier transform of a signal x(n) is given as
6. If x(n)=cosω0n and W(ω) is the Fourier transform of the rectangular signal w(n), then what is the Fourier transform of the signal x(n).w(n)?
a) 1/2[W(ω-ω0)- W(ω+ω0)].
b) 1/2[W(ω-ω0)+ W(ω+ω0)].
c) [W(ω-ω0)+ W(ω+ω0)].
d) [W(ω-ω0)- W(ω+ω0)].
Answer
Answer: b [Reason:] According to the exponential properties of Fourier transform, we get
Fourier transform of x(n).w(n)= 1/2[W(ω-ω0)+ W(ω+ω0)]
7. The characteristic of windowing the signal called “Leakage” is the power that is leaked out into the entire frequency range.
a) True
b) False
Answer
Answer: a [Reason:] We note that the windowed spectrum X ̂(ω) is not localized to a single frequency, but instead it is spread out over the whole frequency range. Thus the power of the original signal sequence x(n) that was concentrated at a single frequency has been spread by the window into the entire frequency range. We say that the power has been leaked out into the entire frequency range and this phenomenon is called as “Leakage”.
8. Which of the following is the advantage of Hanning window over rectangular window?
a) More side lobes
b) Less side lobes
c) More width of main lobe
d) None of the mentioned
Answer
Answer: b [Reason:] The Hanning window has less side lobes and the leakage is less in this windowing technique.
9. Which of the following is the disadvantage of Hanning window over rectangular window?
a) More side lobes
b) Less side lobes
c) More width of main lobe
d) None of the mentioned
Answer
Answer: c [Reason:] In the magnitude response of the signal windowed using Hanning window, the width of the main lobe is more which is the disadvantage of this technique over rectangular windowing technique.
10. The condition with less number of samples L should be avoided.
a) True
b) False
Answer
Answer: a [Reason:] When the number of samples L is small, the window spectrum masks the signal spectrum and, consequently , the DFT of the data reflects the spectral characteristics of the window function. So, this situation should be avoided.
Digital Electronic MCQ Set 3
1. If x(n)=Aejωn is the input of an LTI system and h(n) is the response of the system, then what is the output y(n) of the system?
a) H(-ω)x(n)
b) -H(ω)x(n)
c) H(ω)x(n)
d) None of the mentioned
Answer
Answer: c [Reason:] If x(n)= Aejωn is the input and h(n) is the response o the system, then we know that
2. If the system gives an output y(n)=H(ω)x(n) with x(n)= Aejωnas input signal, then x(n) is said to be Eigen function of the system.
a) True
b) False
Answer
Answer: a [Reason:] An Eigen function of a system is an input signal that produces an output that differs from the input by a constant multiplicative factor known as Eigen value of the system.
3. What is the output sequence of the system with impulse response h(n)=(1/2)nu(n) when the input of the system is the complex exponential sequence x(n)=Aejnπ/2?
Answer
Answer: b [Reason:] First we evaluate the Fourier transform of the impulse response of the system h(n)
4. If the Eigen function of an LTI system is x(n)= Aejnπ and the impulse response of the system is h(n)=(1/2)nu(n), then what is the Eigen value of the system?
a) 3/2
b) -3/2
c) -2/3
d) 2/3
Answer
Answer: d [Reason:] First we evaluate the Fourier transform of the impulse response of the system h(n)
If the input signal is a complex exponential signal, then the input is known as Eigen function and H(ω) is called the Eigen value of the system. So, the Eigen value of the system mentioned above is 2/3.
5. If h(n) is the real valued impulse response sequence of an LTI system, then what is the imaginary part of Fourier transform of the impulse response?
Answer
Answer: a [Reason:] From the definition of H(ω), we have
6. If h(n) is the real valued impulse response sequence of an LTI system, then what is the phase of H(ω) in terms of HR(ω) and HI(ω)?
Answer
Answer: c [Reason:] If h(n) is the real valued impulse response sequence of an LTI system, then H(ω) can be represented as HR(ω)+j HI(ω).
=> 
7. What is the magnitude of H(ω) for the three point moving average system whose output is given by y(n)=1/3[x(n+1)+x(n)+x(n-1)]?
a) 1/[3|1-2cosω|].
b) 1/[3|1+2cosω|].
c) |1-2cosω|.
d) |1+2cosω|.
Answer
Answer: b [Reason:] For a three point moving average system, we can define the output of the system as
8. What is the response of the system with impulse response h(n)=(1/2)nu(n) and the input signal x(n)=10-5sinπn/2+20cosπn?
d) None of the mentioned
Answer
Answer: a [Reason:] The frequency response of the system is
9. What is the magnitude of the frequency response of the system described by the difference equation y(n)=ay(n-1)+bx(n), 0<a<1?
Answer
Answer: d [Reason:] Given y(n)=ay(n-1)+bx(n)
10. If an LTI system is described by the difference equation y(n)=ay(n-1)+bx(n), 0 < a < 1, then what is the parameter ‘b’ so that the maximum value of | H(ω)| is unity?
a) a
b) 1-a
c) 1+a
d) None of the mentioned
Answer
Answer: b [Reason:] We know that,
Since the parameter ‘a’ is positive, the denominator of | H(ω)| becomes minimum at ω=0. So, | H(ω)| attains its maximum value at ω=0. At this frequency we have,
(|b|)/(1-a) =1 =>b=±(1-a).
11. If an LTI system is described by the difference equation y(n)=ay(n-1)+bx(n), 0
Answer
Answer: c [Reason:] From the given difference equation, we obtain
12. The output of the Linear time invariant system cannot contain the frequency components that are not contained in the input signal.
a) True
b) False
Answer
Answer: a [Reason:] If x(n) is the input of an LTI system, then we know that the output of the system y(n) is y(n)= H(ω)x(n) which means the frequency components are just amplified but no new frequency components are added.
13. An LTI system is characterized by its impulse response h(n)=(1/2)nu(n). What is the spectrum of the output signal when the system is excited by the signal x(n)=(1/4)nu(n)?
Answer
Answer: b [Reason:] The frequency response function of the system is
14. What is the frequency response of the system described by the system function H(z)=1/(1-0.8z-1 )?
d) None of the mentioned
Answer
Answer: a [Reason:] Given H(z)=1/(1-0.8z-1)=z/(z-0.8)
Clearly, H(z) has a zero at z=0 and a pole at p=0.8. hence the frequency response of the system is given as
H(ω)= ejω/(ejω-0.8).
Digital Electronic MCQ Set 4
1. If x(n) is a finite duration sequence of length L, then the discrete Fourier transform X(k) of x(n) is given as:
Answer
Answer: a [Reason:] If x(n) is a finite duration sequence of length L, then the Fourier transform of x(n) is given as
If we sample X(ω) at equally spaced frequencies ω=2πk/N, k=0,1,2…N-1 where N>L, the resultant samples are
2. If X(k) discrete Fourier transform of x(n), then the inverse discrete Fourier transform of X(k) is:
Answer
Answer: d [Reason:] If X(k) discrete Fourier transform of x(n), then the inverse discrete Fourier transform of X(k) is given as
3. A finite duration sequence of length L is given as x(n) =1 for 0≤n≤L-1=0 otherwise , then what is the N point DFT of this sequence for N=L?
a) X(k) =L for k=0, 1,2….L-1
b) X(k) =L for k=0
=0 for k=1,2….L-1
c) X(k) =L for k=0
=1 for k=1,2….L-1
d) None of the mentioned
Answer
Answer: b [Reason:] The Fourier transform of this sequence is
If N=L, then X(k)= L for k=0
=0 for k=1,2….L-1
4. The Nth rot of unity WN is given as:
a) ej2πN
b) e -j2πN
c) e-j2π/N
d) ej2π/N
Answer
Answer: c [Reason:] We know that the Discrete Fourier transform of a signal x(n) is given as
Thus we get Nth rot of unity WN= e-j2π/N
5. Which of the following is true regarding the number of computations requires to compute an N-point DFT?
a) N2 complex multiplications and N(N-1) complex additions
b) N2 complex additions and N(N-1) complex multiplications
c) N2 complex multiplications and N(N+1) complex additions
d) N2 complex additions and N(N+1) complex multiplications
Answer
Answer: a [Reason:] The formula for calculating N point DFT is given as
From the formula given at every step of computing we are performing N complex multiplications and N-1 complex additions. So, in a total to perform N-point DFT we perform N2 complex multiplications and N(N-1) complex additions.
6. Which of the following is true?
d) None of the mentioned
Answer
Answer: b [Reason:] If XN represents the N point DFT of the sequence xN in the matrix form, then we know that
7. What is the DFT of the four point sequence x(n)={0,1,2,3}?
a) {6,-2+2j-2,-2-2j}
b) {6,-2-2j,2,-2+2j}
c) {6,-2+2j,-2,-2-2j}
d) {6,-2-2j,-2,-2+2j}
Answer
Answer: c [Reason:] The first step is to determine the matrix W4. By exploiting the periodicity property of W4 and the symmetry property
WNk+N/2= -WNk
The matrix W4 may be expressed as
8. If X(k) is the N point DFT of a sequence whose Fourier series coefficients is given by ck, then which of the following is true?
a) X(k)=Nck
b) X(k)=ck/N
c) X(k)=N/ck
d) None of the mentioned
Answer
Answer: a [Reason:] The Fourier series coefficients are given by the expression
9. What is the DFT of the four point sequence x(n)={0,1,2,3}?
a) {6,-2+2j-2,-2-2j}
b) {6,-2-2j,2,-2+2j}
c) {6,-2-2j,-2,-2+2j}
d) {6,-2+2j,-2,-2-2j}
Answer
Answer: d
Answer: Given x(n)={0,1,2,3}
We know that the 4-point DFT of the above given sequence is given by the expression
In this case N=4
=>X(0)=6,X(1)=-2+2j,X(2)=-2,X(3)=-2-2j.
10. If W4100=Wx200, then what is the value of x?
a) 2
b) 4
c) 8
d) 16
Answer
Answer: c [Reason:] We know that according to the periodicity and symmetry property,
100/4=200/x=>x=8.
Digital Electronic MCQ Set 5
1. If we split the N point data sequence into two N/2 point data sequences f1(n) and f2(n) corresponding to the even numbered and odd numbered samples of x(n), then such an FFT algorithm is known as decimation-in-time algorithm.
a) True
b) False
Answer
Answer: a [Reason:] Let us consider the computation of the N=2v point DFT by the divide and conquer approach. We select M=N/2 and L=2. This selection results in a split of N point data sequence into two N/2 point data sequences f1(n) and f2(n) corresponding to the even numbered and odd numbered samples of x(n), respectively, that is
f1(n)=x(2n)
f2(n)=x(2n+1) ,n=0,1,2…N/2-1
Thus f1(n) and f2(n) are obtained by decimating x(n) by a factor of
2, and hence the resulting FFT algorithm is called a decimation-in-time algorithm.
2. If we split the N point data sequence into two N/2 point data sequences f1(n) and f2(n) corresponding to the even numbered and odd numbered samples of x(n) and F1(k) and F2(k) are the N/2 point DFTs of f1(k) and f2(k) respectively, then what is the N/2 point DFT X(k) of x(n)?
a) F1(k)+F2(k)
b) F1(k)- WNk F2(k)
c) F1(k)+WNkNk F2(k)
d) None of the mentioned
Answer
Answer: c [Reason:] From the question, it is given that
f1(n)=x(2n)
f2(n)=x(2n+1) ,n=0,1,2…N/2-1
3. If X(k) is the N/2 point DFT of the sequence x(n), then what is the value of X(k+N/2)?
a) F1(k)+F2(k)
b) F1(k)- WNk F2(k)
c) F1(k)+WNk F2(k)
d) None of the mentioned
Answer
Answer: b [Reason:] We know that, X(k) = F1(k)+WNk F2(k)
We know that F1(k) and F2(k) are periodic, with period N/2, we have F1(k+N/2)= F1(k) and F2(k+N/2)= F2(k). In addition, the factor WNk+N/2= -WNk.
Thus we get, X(k+N/2)= F1(k)- WNk F2(k).
4. How many complex multiplications are required to compute X(k)?
a) N(N+1)
b) N(N-1)/2
c) N2/2
d) N(N+1)/2
Answer
Answer: d [Reason:] We observe that the direct computation of F1(k) requires (N/2)2 complex multiplications. The same applies to the computation of F2(k). Furthermore, there are N/2 additional complex multiplications required to compute WNk. Hence it requires N(N+1)/2 complex multiplications to compute X(k).
5. The total number of complex multiplications required to compute N point DFT by radix-2 FFT is:
a) (N/2)log2N
b) Nlog2N
c) (N/2)logN
d) None of the mentioned
Answer
Answer: a [Reason:] The decimation of the data sequence should be repeated again and again until the resulting sequences are reduced to one point sequences. For N=2v, this decimation can be performed v=log2N times. Thus the total number of complex multiplications is reduced to (N/2)log2N.
6. The total number of complex additions required to compute N point DFT by radix-2 FFT is:
a) (N/2)log2N
b) Nlog2N
c) (N/2)logN
d) None of the mentioned
Answer
Answer: b [Reason:] The decimation of the data sequence should be repeated again and again until the resulting sequences are reduced to one point sequences. For N=2v, this decimation can be performed v=log2N times. Thus the total number of complex additions is reduced to Nlog2N.
7. The following butterfly diagram is used in the computation of:
a) Decimation-in-time FFT
b) Decimation-in-frequency FFT
c) All of the mentioned
d) None of the mentioned
Answer
Answer: a [Reason:] The above given diagram is the basic butterfly computation in the decimation-in-time FFT algorithm.
8. For a decimation-in-time FFT algorithm, which of the following is true?
a) Both input and output are in order
b) Both input and output are shuffled
c) Input is shuffled and output is in order
d) Input is in order and output is shuffled
Answer
Answer: c [Reason:] In decimation-in-time FFT algorithm, the input is taken in bit reversal order and the output is obtained in the order.
9. The following butterfly diagram is used in the computation of:
a) Decimation-in-time FFT
b) Decimation-in-frequency FFT
c) All of the mentioned
d) None of the mentioned
Answer
Answer: b [Reason:] The above given diagram is the basic butterfly computation in the decimation-in-frequency FFT algorithm.
10. For a decimation-in-time FFT algorithm, which of the following is true?
a) Both input and output are in order
b) Both input and output are shuffled
c) Input is shuffled and output is in order
d) Input is in order and output is shuffled
Answer
Answer: d [Reason:] In decimation-in-frequency FFT algorithm, the input is taken in order and the output is obtained in the bit reversal order.
