Digital Electronic MCQ Set 1
1. Filter parameter optimization technique is used for designing of which of the following?
a) FIR in time domain
b) FIR in frequency domain
c) IIR in time domain
d) IIR in frequency domain
Answer
Answer: d [Reason:] We describe a filter parameter optimization technique carried out in the frequency domain that is representative of frequency domain design methods.
2. In this type of designing, the system function of IIR filter is expressed in which form?
a) Parallel form
b) Cascade form
c) Mixed form
d) Any of the mentioned
Answer
Answer: b [Reason:] The design is most easily carried out with the system function for the IIR filter expressed in the cascade form as
H(z)=G.A(z).
3. It is more convenient to deal with the envelope delay as a function of frequency.
a) True
b) False
Answer
Answer: a [Reason:] Instead of dealing with the phase response ϴ(ω), it is more convenient to deal with the envelope delay as a function of frequency.
4. Which of the following gives the equation for envelope delay?
a) dϴ(ω)/dω
b) ϴ(ω)
c) -dϴ(ω)/dω
d) -ϴ(ω)
Answer
Answer: c [Reason:] Instead of dealing with the phase response ϴ(ω), it is more convenient to deal with the envelope delay as a function of frequency, which is
Tg(ω)= -dϴ(ω)/dω.
5. What is the error in magnitude at the frequency ωk?
a) G.A(ωk) + Ad(ωk)
b) G.A(ωk) – Ad(ωk)
c) G.A(ωk) – A(ωk)
d) None of the mentioned
Answer
Answer: b [Reason:] The error in magnitude at the frequency ωk is G.A(ωk) – Ad(ωk) for 0 ≤ |ω| ≤ π, where Ad(ωk) is the desired magnitude response at ωk.
6. What is the error in delay at the frequency ωk?
a) Tg(ωk)- Td(ωk)
b) Tg(ωk)+ Td(ωk)
c) Td(ωk)
d) None of the mentioned
Answer
Answer: a [Reason:] Similarly as in the previous question, the error in delay at ωk is defined as Tg(ωk)- Td(ωk), where Td(ωk) is the desired delay response.
7. The choice of Td(ωk) for error in delay is complicated.
a) True
b) False
Answer
Answer: a [Reason:] We know that the error in delay is defined as Tg(ωk)- Td(ωk). However, the choice of Td(ωk) for error in delay is complicated by the difficulty in assigning a nominal delay of the filter.
8. If the error in delay is defined as Tg(ωk)- Tg(ω0)- Td(ωkk), then what is Tg(ω0)?
a) Filter delay at nominal frequency in stop band
b) Filter delay at nominal frequency in transition band
c) Filter delay at nominal frequency
d) Filter delay at nominal frequency in pass band
Answer
Answer: d [Reason:] We are led to define the error in delay as Tg(ωk)- Tg(ω0)- Td(ωk), where Tg(ω0) is the filter delay at some nominal centre frequency in the pass band of the filter.
9. We cannot choose any arbitrary function for the errors in magnitude and delay.
a) True
b) False
Answer
Answer: b [Reason:] As a performance index for determining the filter parameters, one can choose any arbitrary function of the errors in magnitude and delay.
10. What does ‘p’ represents in the arbitrary function of error?
a) 2K- dimension vector
b) 3K- dimension vector
c) 4K- dimension vector
d) None of the mentioned
Answer
Answer: c [Reason:] In the error function ‘p’ denotes the 4K dimension vector of the filter coefficients.
11. What should be the value of λ for the error to be placed entirely on delay?
a) 1
b) 1/2
c) 0
d) None of the mentioned
Answer
Answer: a [Reason:] The emphasis on the errors affecting the design may be placed entirely on the delay by taking the value of λ as 1.
12. What should be the value of λ for the error to be placed equally on magnitude and delay?
a) 1
b) 1/2
c) 0
d) None of the mentioned
Answer
Answer: b [Reason:] The emphasis on the errors affecting the design may be equally weighted between magnitude and delay by taking the value of λ as 1/2.
13. Which of the following is true about the squared-error function E(p,G)?
a) Linear function of 4K parameters
b) Linear function of 4K+1 parameters
c) Non-Linear function of 4K parameters
d) Non-Linear function of 4K+1 parameters
Answer
Answer: d [Reason:] The squared error function E(p,G) is a non-linear function of 4K+1 parameters.
14. Minimization of the error function over the remaining 4K parameters is performed by an iterative method.
a) True
b) False
Answer
Answer: a [Reason:] Due to the non-linear nature of E(p,G), its minimization over the remaining 4K parameters is performed by an iterative numerical optimization method.
15. The iterative process may converge to a global minimum.
a) True
b) False
Answer
Answer: b [Reason:] The major difficulty with any iterative procedure that searches for the parameter values that minimize a non-linear function is that the process may converge to a local minimum instead of a global minimum.
Digital Electronic MCQ Set 2
1. Which of the following defines the rectangular window function of length M-1?
a) w(n)= 1, n=0,1,2…M-1
=0, else where
b) w(n)= 1, n=0,1,2…M-1
=-1, else where
c) w(n)= 0, n=0,1,2…M-1
=1, else where
d) None of the mentioned
Answer
Answer: a [Reason:] We know that the rectangular window of length M-1 is defined as
w(n)= 1, n=0,1,2…M-1
=0, else where.
2. The multiplication of the window function w(n) with h(n) is equivalent to the multiplication of H(w) and W(w).
a) True
b) False
Answer
Answer: b [Reason:] According to the basic formula of convolution, the multiplication of two signals w(n) and h(n) in time domain is equivalent to the convolution of their respective Fourier transforms W(w) and H(w).
3. What is the Fourier transform of the rectangular window of length M-1?
Answer
Answer: d [Reason:] We know that the Fourier transform of a function w(n) is defined as
For a rectangular window, w(n)=1 for n=0,1,2….M-1
Thus we get
4. What is the magnitude response |W(ω)| of a rectangular window function?
d) None of the mentioned
Answer
Answer: a [Reason:] We know that for a rectangular window
Thus the window function has a magnitude response
5. What is the width of the main lobe of the frequency response of a rectangular window of length M-1?
a) π/M
b) 2π/M
c) 4π/M
d) 8π/M
Answer
Answer: c [Reason:] The width of the main lobe width is measured to the first zero of W(ω)) is 4π/M.
6. The width of each side lobes decreases with an increase in M.
a) True
b) False
Answer
Answer: a [Reason:] Since the width of the main lobe is inversely proportional to the value of M, if the value of M increases then the main lobe becomes narrower. In fact, the width of each side lobes decreases with an increase in M.
7. With an increase in the value of M, the height of each side lobe:
a) Do not vary
b) Does not depend on value of M
c) Decreases
d) Increases
Answer
Answer: d [Reason:] The height of each side lobes increase with an increase in M such a manner that the area under each side lobe remains invariant to changes in M.
8. As M is increased, W(ω) becomes wider and the smoothening produced by the W(ω) is increased.
a) True
b) False
Answer
Answer: b [Reason:] As M is increased, W(ω) becomes narrower and the smoothening produced by the W(ω) is reduced.
9. Which of the following windows has a time domain sequence 
a) Bartlett window
b) Blackman window
c) Hanning window
d) Hamming window
Answer
Answer: a [Reason:] The Bartlett window which is also called as triangular window has a time domain sequence as
, 0≤n≤M-1.
10. The width of each side lobes decreases with an decrease in M.
a) True
b) False
Answer
Answer: b [Reason:] Since the width of the main lobe is inversely proportional to the value of M, if the value of M increases then the main lobe becomes narrower. In fact, the width of each side lobes decreases with an increase in M.
11. What is the approximate transition width of main lobe of a Hamming window?
a) 4π/M
b) 8π/M
c) 12π/M
d) 2π/M
Answer
Answer: b [Reason:] The transition width of the main lobe in the case of Hamming window is equal to 8π/M where M is the length of the window.
Digital Electronic MCQ Set 3
1. Which of the following is a frequency domain specification?
a) 0≥ 20 log|H(jΩ)|
b) 20 log|H(jΩ)| ≥ KP
c) 20 log|H(jΩ)| ≤ KS
d) All of the mentioned
Answer
Answer: d [Reason:] We are required to design a low pass Butterworth filter to meet the following frequency domain specifications.
KP ≤ 20 log|H(jΩ)| ≤ 0
and 20 log|H(jΩ)| ≤ KS.
2. What is the value of gain at the pass band frequency, i.e., what is the value of KP?
Answer
Answer: b [Reason:] We know that the formula for gain is
3. What is the value of gain at the stop band frequency, i.e., what is the value of KS?
Answer
Answer: a [Reason:] We know that the formula for gain is
4. Which of the following equation is True?
d) None of the mentioned
Answer
Answer: c [Reason:] We know that,
5. Which of the following equation is True?
d) None of the mentioned
Answer
Answer: b [Reason:] We know that,
6. What is the order N of the low pass Butterworth filter in terms of KP and KS?
Answer
Answer: d [Reason:]
7. What is the expression for cutoff frequency in terms of pass band gain?
d) None of the mentioned
Answer
Answer: a [Reason:] We know that,
8. What is the expression for cutoff frequency in terms of stop band gain?
d) None of the mentioned
Answer
Answer: c [Reason:] We know that,
9. The cutoff frequency of the low pass Butterworth filter is the arithmetic mean of the two cutoff frequencies as found above.
a) True
b) False
Answer
Answer: a [Reason:] The arithmetic mean of the two cutoff frequencies as found above is the final cutoff frequency of the low pass Butterworth filter.
10. What is the lowest order of the Butterworth filter with a pass band gain KP= -1 dB at ΩP= 4 rad/sec and stop band attenuation greater than or equal to 20dB at ΩS= 8 rad/sec?
a) 4
b) 5
c) 6
d) 3
Answer
Answer: b [Reason:] We know that the equation for the order of the Butterworth filter is given as
From the given question,
KP= -1 dB, ΩP= 4 rad/sec, KS= -20 dB and ΩS= 8 rad/sec
Upon substituting the values in the above equation, we get
N=4.289
Rounding off to the next largest integer, we get N=5.
Digital Electronic MCQ Set 4
1. If x(n) is a discrete-time signal, then the value of x(n) at non integer value of ‘n’ is:
a) Zero
b) Positive
c) Negative
d) Not defined
Answer
Answer: d [Reason:] For a discrete time signal, the value of x(n) exists only at integral values of n. So, for a non- integer value of ‘n’ the value of x(n) does not exist.
2. The discrete time function defined as u(n)=n for n≥0;=0 for n<0 is an:
a) Unit sample signal
b) Unit step signal
c) Unit ramp signal
d) None of the mentioned
Answer
Answer: c [Reason:] When we plot the graph for the given function, we get a straight line passing through origin with a unit positive slope. So, the function is called as unit ramp signal.
3.The phase function of a discrete time signal x(n)=an, where a=r.ejθ is:
a) tan(nθ)
b) nθ
c) tan-1(nθ)
d) None of the mentioned
Answer
Answer: b [Reason:] Given x(n)=an=(r.ejθ)n =rn.ejnθ
=>x(n)=rn.(cosnθ+jsinnθ)
Phase function is tan-1(cosnθ/sinnθ)=tan-1(tan nθ)=nθ.
4. The signal given by the equation
is known as:
a) Energy signal
b) Power signal
c) Work done signal
d) None of the mentioned
Answer
Answer: a [Reason:] We have used the magnitude-squared values of x(n), so that our definition applies to complex-valued as well as real-valued signals. If the energy of the signal is finite i.e., 0<E<∞ then the given signal is known as Energy signal.
5. x(n)*δ(n-k)=?
a) x(n)
b) x(k)
c) x(k)*δ(n-k)
d) x(k)*δ(k)
Answer
Answer: c [Reason:] The given signal is defined only when n=k by the definition of delta function. So, x(n)*δ(n-k)= x(k)*δ(n-k).
6. A real valued signal x(n) is called as anti-symmetric if:
a) x(n)=x(-n)
b) x(n)=-x(-n)
c) x(n)=-x(n)
d) None of the mentioned
Answer
Answer: b [Reason:] According to the definition of anti-symmetric signal, the signal x(n) should be symmetric over origin. So, for the signal x(n) to be symmetric, it should satisfy the condition x(n)=-x(-n).
7. The odd part of a signal x(t) is:
a) x(t)+x(-t)
b) x(t)-x(-t)
c) (1/2)*(x(t)+x(-t))
d) (1/2)*(x(t)-x(-t))
Answer
Answer: d [Reason:] Let x(t)=xe(t)+xo(t)
=>x(-t)=xe(-t)-xo(-t)
By subtracting the above two equations, we get
xo(t)=(1/2)*(x(t)-x(-t)).
8. Time scaling operation is also known as:
a) Down-sampling
b) Up-sampling
c) Sampling
d) None of the mentioned
Answer
Answer: a [Reason:] If the signal x(n) was originally obtained by sampling a signal xa(t), then x(n)=xa(nT). Now, y(n)=x(2n)(say)=xa(2nT). Hence the time scaling operation is equivalent to changing the sampling rate from 1/T to 1/2T, that is to decrease the rate by a factor of 2. So, time scaling is also called as down-sampling.
9. What is the condition for a signal x(n)=Brn where r=eαT to be called as an decaying exponential signal?
a) 0<r<∞
b) 0<r<1
c) r>1
d) r<0
Answer
Answer: b [Reason:] When the value of ‘r’ lies between 0 and 1 then the value of x(n) goes on decreasing exponentially with increase in value of ‘n’. So, the signal is called as decaying exponential signal.
10. The function given by the equation x(n)=1, for n=0;=0, for n≠0 is a:
a) Step function
b) Ramp function
c) Triangular function
d) Impulse function
Answer
Answer: d [Reason:] According to the definition of the impulse function, it is defined only at n=0 and is not defined elsewhere which is as per the signal given.
Digital Electronic MCQ Set 5
1. The output signal when a signal x(n)=(0,1,2,3) is processed through an ‘Identical’ system is:
a) (3,2,1,0)
b) (1,2,3,0)
c) (0,1,2,3)
d) None of the mentioned
Answer
Answer: c [Reason:] An identical system is a system whose output is same as the input, that is it does not perform any operation on the input and transmits it.
2. If a signal x(n) is passed through a system to get an output signal of y(n)=x(n+1), then the signal is said to be:
a) Delayed
b) Advanced
c) No operation
d) None of the mentioned
Answer
Answer: d [Reason:] For example, the value of the output at the time n=0 is y(0)=x(1), that is the system is advanced by one unit.
3. If the output of the system is 
with an input of x(n) then the system will work as:
a) Accumulator
b) Adder
c) Subtractor
d) Multiplier
Answer
Answer: a [Reason:] From the equation given, y(n)=x(n)+x(n-1)+x(n-2)+…. .This system calculates the running sum of all the past input values till the present time. So, it acts as an accumulator.
4. What is the output y(n) when a signal x(n)=n*u(n)is passed through a accumulator system under the conditions that it is initially relaxed?
a) (n2+n+1)/2
b) (n(n+1))/2
c) (n2+n+2)/2
d) None of the mentioned
Answer
Answer: b [Reason:] Given that the system is initially relaxed, that is y(-1)=0
According to the equation of the accumulator,
5. The block denoted as follows is known as:
a) Delay block
b) Advance block
c) Multiplier block
d) Adder block
Answer
Answer: a [Reason:] If the function to this block is x(n) then the output from the block will be x(n-1). So, the block is called as delay block or delay element.
6. The output signal when a signal x(n)=(0,1,2,3) is processed through an ‘Delay’ system is:
a) (3,2,1,0)
b) (1,2,3,0)
c) (0,1,2,3)
d) None of the mentioned
Answer
Answer: b [Reason:] An delay system is a system whose output is same as the input, but after a delay.
7. The system described by the input-output equation y(n)=nx(n)+bx3(n) is a:
a) Static system
b) Dynamic system
c) Identical system
d) None of the mentioned
Answer
Answer: a [Reason:] Since the output of the system y(n) depends only on the present value of the input x(n) but not on the past or the future values of the input, the system is called as static or memory-less system.
8. Whether the system described by the input-output equations y(n)=x(n)-x(n-1) a Time-variant system?
a) True
b) False
Answer
Answer: b [Reason:] If the input is delayed by k units then the output will be y(n,k)=x(n-k)-x(n-k-1)
If the output is delayed by k units then y(n-k)=x(n-k)-x((n-k)-1)
=>y(n,k)=y(n-k). Hence the system is time-invariant.
9. The system described by the input-output equations y(n)=x2(n) is a Non-linear system.
a) True
b) False
Answer
Answer: a [Reason:] Given equation is y(n)=x2(n)
Let y1(n)=x12(n) and y2(n)=x22(n)
y3(n)=y1(n)+y2(n)= x12(n)+ x22(n)≠(x1(n)+x2(n))2
So the system is non-linear.
10. If the output of the system of the system at any ‘n’ depends only the present or the past values of the inputs then the system is said to be:
a) Linear
b) Non-Linear
c) Causal
d) Non-causal
Answer
Answer: c [Reason:] A system is said to be causal if the output of the system is defined as the function shown below
y(n)=F[x(n),x(n-1),x(n-2),…]
So, according to the conditions given in the question, the system is a causal system.
11. The system described by the input-output equations y(n)=x(-n) is a causal system.
a) True
b) False
Answer
Answer: b [Reason:] For n=-1, y(-1)=x(1)
That is, the output of the system at n=-1 is depending on the future value of the input at n=1. So the system is a non-causal system.
12. If a system do not have a bounded output for bounded input, then the system is said to be:
a) Causal
b) Non-causal
c) Stable
d) Non-stable
Answer
Answer: d [Reason:] An arbitrary relaxed system is said to be BIBO stable if it has a bounded output for every value in the bounded input. So, the system given in the question is a Non-stable system.
