Digital Electronic MCQ Set 1
1. Which of the following is true in the case of Butterworth filters?
a) Smooth pass band
b) Wide transition band
c) Not so smooth stop band
d) All of the mentioned
Answer
Answer: d [Reason:] Butterworth filters have a very smooth pass band, which we pay for with a relatively wide transmission region.
2. What is the magnitude frequency response of a Butterworth filter of order N and cutoff frequency ΩC?
d) None of the mentioned
Answer
Answer: a [Reason:] A Butterworth is characterized by the magnitude frequency response
where N is the order of the filter and ΩC is defined as the cutoff frequency.
3. What is the factor to be multiplied to the dc gain of the filter to obtain filter magnitude at cutoff frequency?
a) 1
b) √2
c) 1/√2
d) 1/2
Answer
Answer: c [Reason:] The dc gain of the filter is the filter magnitude at Ω=0.
We know that the filter magnitude is given by the equation
4. What is the value of magnitude frequency response of a Butterworth low pass filter at Ω=0?
a) 0
b) 1
c) 1/√2
d) None of the mentioned
Answer
Answer: b [Reason:] The magnitude frequency response of a Butterworth low pass filter is given as
At Ω=0 => |H(jΩ)|=1 for all N.
5. As the value of the frequency Ω tends to ∞, then |H(jΩ)| tends to:
a) 0
b) 1
c) ∞
d) None of the mentioned
Answer
Answer: a [Reason:] We know that the magnitude frequency response of a Butterworth filter of order N is given by the expression
In the above equation, if Ω→∞ then |H(jΩ)|→0.
6. |H(jΩ)| is a monotonically increasing function of frequency.
a) True
b) False
Answer
Answer: b [Reason:] |H(jΩ)| is a monotonically decreasing function of frequency, i.e., |H(jΩ2)| < |H(jΩ1)| for any values of Ω1 and Ω2 such that 0 ≤ Ω1 < Ω2.
7. What is the magnitude squared response of the normalized low pass Butterworth filter?
a) 1/(1+Ω-2N)
b) 1+Ω-2N
c) 1+Ω2N
d) 1/(1+Ω^2N)
Answer
Answer: d [Reason:] We know that the magnitude response of a low pass Butterworth filter of order N is given as
8. What is the transfer function of magnitude squared frequency response of the normalized low pass Butterworth filter?
Answer
Answer: a [Reason:] We know that the magnitude squared frequency response of a normalized low pass Butterworth filter is given as
9. Where does the poles of the transfer function of normalized low pass Butterworth filter exists?
a) Inside unit circle
b) Outside unit circle
c) On unit circle
d) None of the mentioned
Answer
Answer: c [Reason:] The transfer function of normalized low pass Butterworth filter is given as
The poles are therefore on a circle with radius unity.
10. What is the general formula that represent the phase of the poles of transfer function of normalized low pass Butterworth filter of order N?
a) π/N k+π/2N k=0,1,2…N-1
b) π/N k+π/2N+π/2 k=0,1,2…2N-1
c) π/N k+π/2N+π/2 k=0,1,2…N-1
d) π/N k+π/2N k=0,1,2…2N-1
Answer
Answer: d [Reason:] The transfer function of normalized low pass Butterworth filter is given as
11. What is the Butterworth polynomial of order 3?
Answer
Answer: d [Reason:] Given that the order of the Butterworth low pass filter is 3.
Therefore, for N=3 Butterworth polynomial is given as B3(s)=(s-s0) (s-s1) (s-s2)
12. What is the Butterworth polynomial of order 1?
a) s-1
b) s+1
c) s
d) None of the mentioned
Answer
Answer: b [Reason:] Given that the order of the Butterworth low pass filter is 1.
Therefore, for N=1 Butterworth polynomial is given as B3(s)=(s-s0).
We know that, sk=e(jπ((2k+1)/2N)) e(jπ/2)
=>s0= -1
=> B1(s)=s-(-1)=s+1.
13. What is the transfer function of Butterworth low pass filter of order 2?
a) 1/(s2+√2 s+1)
b) 1/(s2-√2 s+1)
c) s2-√2 s+1
d) s2+√2 s+1
Answer
Answer: a [Reason:] We know that the Butterworth polynomial of a 2nd order low pass filter is
B2(s)= s2+√2 s+1
Thus the transfer function is given as 1/(s2+√2 s+1).
Digital Electronic MCQ Set 2
1. What is the ideal reconstruction formula or ideal interpolation formula for x(t) = ___
Answer
Answer: a [Reason:] 
where the sampling interval T = 1/Fs=1/2B,Fs is the sampling frequency and B is the bandwidth of the analog signal.
2. What is the new ideal interpolation formula described after few problems with previous one?
Answer
Answer: b [Reason:] The reconstruction of the signal x ( t) from its samples as an interpolation problem and have described the function:
3. What is the frequency response of the analog filter corresponding to the ideal interpolator?
Answer
Answer: c [Reason:] The analog filter corresponding to the ideal interpolator has a frequency response:
, H(F) is the Fourier transform of the interpolation function g(t).
4. The reconstruction o f the signal from its samples as a linear filtering process in which a discrete-time sequence of short pulses (ideally impulses) with amplitudes equal to the signal samples, excites an analog filter.
a) True
b) False
Answer
Answer: a [Reason:] The reconstruction o f the signal from its samples as a linear filtering process in which a discrete-time sequence of short pulses (ideally impulses) with amplitudes equal to the signal samples, excites an analog filter.
5. The ideal reconstruction filter is an ideal low pass filter and its impulse response extends for all time.
a) True
b) False
Answer
Answer: a [Reason:] The ideal reconstruction filter is an ideal low pass filter and its impulse response extends for all time. Hence the filter is noncausal and physically nonrealizable. Although the interpolation filter with impulse response given can be approximated closely with some delay, the resulting function is still impractical for most applications where D /A conversion are required.
6. D /A conversion is usually performed by combining a D /A converter with a sample-and-hold (S/H ) and followed by a low pass (smoothing) filter.
a) True
b) False
Answer
Answer: a [Reason:] D /A conversion is usually performed by combining a D /A converter with a sample-and hold (S/H) and followed by a low pass (smoothing) filter. T he D /A converter accepts at its input, electrical signals that correspond to a binary word, and produces an output voltage or current that is proportional to the value o f the binary word.
7. The time required for the output o f the D /A converter to reach and remain within a given fraction of the final value, after application of the input code word is called?
a) Converting time
b) Setting time
c) Both Converting & Setting time
d) None of the mentioned
Answer
Answer: b [Reason:] An important parameter o f a D /A converter is its settling time, which is defined as the time required for the output o f the D /A converter to reach and remain within a given fraction (usually,±1/2 LSB) of the final value, after application of the input code word.
8. In D/A converter, the application of the input code word results in a high-amplitude transient, called?
a) Glitch
b) Deglitch
c) Glitter
d) None of the mentioned
Answer
Answer: a [Reason:] The application o f the input code word results in a high-amplitude transient, called a “glitch.” This is especially the case when two consecutive code words to the A /D differ by several bits.
9. In a D/A converter, the usual way to solve the glitch is to use deglitcher. How is the Deglitcher designed?
a) By using a low pass filter
b) By using a S/H circuit
c) By using a low pass filter & S/H circuit
d) None of the mentioned
Answer
Answer: b [Reason:] The usual way to remedy this problem is to use an S/H circuit designed to serve as a “deglitcher”. Hence the basic task of the S/H is to hold the output of the D /A converter constant at the previous output value until the new sample at the output o f the D /A reaches steady state, and then it samples and holds the new value in the next sampling interval. Thus the S/H approximates the analog signal by a series of rectangular pulses whose height is equal to the corresponding value of the signal pulse.
10. What is the impulse response of an S/H, when viewed as a linear filter?
d) None
Answer
Answer: a [Reason:] W hen viewed as a linear filter, the S/H has an impulse response:
Digital Electronic MCQ Set 3
1. What is the pass band edge frequency of an analog low pass normalized filter?
a) 0 rad/sec
b) 0.5 rad/sec
c) 1 rad/sec
d) 1.5 rad/sec
Answer
Answer: c [Reason:] Let H(s) denote the transfer function of a low pass analog filter with a pass band edge frequency ΩP equal to 1 rad/sec. This filter is known as analog low pass normalized prototype.
2. If H(s) is the transfer function of a analog low pass normalized filter and Ωu is the desired pass band edge frequency of new low pass filter, then which of the following transformation has to be performed?
a) s→ s / Ωu
b) s→ s .Ωu
c) s→ Ωu/s
d) None of the mentioned
Answer
Answer: a [Reason:] If Ωu is the desired pass band edge frequency of new low pass filter, then the transfer function of this new low pass filter is obtained by using the transformation s→ s / Ωu.
3. Which of the following is a low pass-to-high pass transformation?
a) s→ s / Ωu
b) s→ Ωu / s
c) s→ Ωu.s
d) none of the mentioned
Answer
Answer: b [Reason:] The low pass-to-high pass transformation is simply achieved by replacing s by 1/s. If the desired high pass filter has the pass band edge frequency Ωu, then the transformation is
s→ Ωu / s.
4. Which of the following is the backward design equation for a low pass-to-low pass transformation?
Answer
Answer: d [Reason:] If Ωu is the desired pass band edge frequency of new low pass filter, then the transfer function of this new low pass filter is obtained by using the transformation s→ s / Ωu. If ΩS and Ω’S are the stop band frequencies of prototype and transformed filters respectively, then the backward design equation is given by
.
5. Which of the following is a low pass-to-band pass transformation?
Answer
Answer: c [Reason:] If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band pass filter, then the transformation to be performed on the normalized low pass filter is
6. Which of the following is the backward design equation for a low pass-to-high pass transformation?
Answer
Answer: b [Reason:] If Ωu is the desired pass band edge frequency of new high pass filter, then the transfer function of this new high pass filter is obtained by using the transformation s→ Ωu /s. If ΩS and Ω’S are the stop band frequencies of prototype and transformed filters respectively, then the backward design equation is given by
.
7. Which of the following is a low pass-to-band stop transformation?
d) None of the mentioned
Answer
Answer: c [Reason:] If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band stop filter, then the transformation to be performed on the normalized low pass filter is
8. If 
then which of the following is the backward design equation for a low pass-to-band pass transformation?
a) ΩS= |B|
b) ΩS= |A|
c) ΩS= Max{|A|,|B|}
d) ΩS= Min{|A|,|B|}
Answer
Answer: d [Reason:] If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band pass filter and Ω1 and Ω2 are the lower and upper cutoff stop band frequencies of the desired band pass filter, then the backward design equation is
ΩS= Min{|A|,|B|}
where,.
9. If 
then which of the following is the backward design equation for a low pass-to-band stop transformation?
a) ΩS= Max{|A|,|B|}
b) ΩS= Min{|A|,|B|}
c) ΩS= |B|
d) ΩS= |A|
Answer
Answer: b [Reason:] If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band stop filter and Ω1 and Ω2 are the lower and upper cutoff stop band frequencies of the desired band stop filter, then the backward design equation is
ΩS= Min{|A|,|B|}
where, .
10. Which of the following is a low pass-to-high pass transformation?
a) s→ s / Ωu
b) s→ Ωu / s
c) s→ Ωu.s
d) None of the mentioned
Answer
Answer: b [Reason:] The low pass-to-high pass transformation is simply achieved by replacing s by 1/s. If the desired high pass filter has the pass band edge frequency Ωu, then the transformation is
s→ Ωu / s.
Digital Electronic MCQ Set 4
1. Which of the following operation has to be performed to increase the sampling rate by an integer factor I?
a) Interpolating I+1 new samples
b) Interpolating I-1 new samples
c) Extrapolating I+1 new samples
d) Extrapolating I-1 new samples
Answer
Answer: b [Reason:] An increase in the sampling rate by an integer factor of I can be accomplished by interpolating I-1 new samples between successive values of the signal.
2. In one of the interpolation process, we can preserve the spectral shape of the signal sequence x(n).
a) True
b) False
Answer
Answer: a [Reason:] The interpolation process can be accomplished in a variety of ways. Among them there is a process that preserves the spectral shape of the signal sequence x(n).
3. If v(m) denote a sequence with a rate Fy=I.Fx which is obtained from x(n), then which of the following is the correct definition for v(m)?
a) x(mI), m=0,±I,±2I….
0, otherwise
a) x(mI), m=0,±I,±2I….
x(m/I), otherwise
c) x(m/I), m=0,±I,±2I….
0, otherwise
d) None of the mentioned
Answer
Answer: c [Reason:] If v(m) denote a sequence with a rate Fy=I.Fx which is obtained from x(n) by adding I-1 zeros between successive values of x(n). Thus
v(m)= x(m/I), m=0,±I,±2I….
0, otherwise.
4. If X(z) is the z-transform of x(n), then what is the z-transform of interpolated signal v(m)?
a) X(zI)
b) X(z+I)
c) X(z/I)
d) X(zI)
Answer
Answer: d [Reason:] By taking the z-transform of the signal v(m), we get
5. If x(m) and v(m) are the original and interpolated signals and ωy denotes the frequency variable relative to the new sampling rate, then V(ωy)= X(ωyI).
a) True
b) False
Answer
Answer: a [Reason:] The spectrum of v(m) is obtained by evaluating V(z)= X(zI) on the unit circle. Thus V(ωy)= X(ωyI), where ωy denotes the frequency variable relative to the new sampling rate.
6. What is the relationship between ωx and ωy?
a) ωy= ωx.I
b) ωy= ωx/I
c) ωy= ωx+I
d) None of the mentioned
Answer
Answer: b [Reason:] We know that the relationship between sampling rates is Fy=IFx and hence the frequency variables ωx and ωy are related according to the formula
ωy= ωx/I.
7. The following sampling rate conversion technique is interpolation by a factor I.
a) True
b) False
Answer
Answer: a [Reason:] From the diagram, the values are interpolated between two successive values of x(n), thus it is called as sampling rate conversion using interpolation by a factor I.
8. The following sampling rate conversion technique is interpolation by a factor I.
a) True
b) False
Answer
Answer: b [Reason:] The sampling rate conversion technique given in the diagram is decimation by a factor D.
9. Which of the following is true about the interpolated signal whose spectrum is V(ωy)?
a) (I-1)-fold non-periodic
b) (I-1)-fold periodic repetition
c) I-fold non periodic
d) I-fold periodic repetition
Answer
Answer: d [Reason:] We observe that the sampling rate increase, obtained by the addition of I-1 zero samples between successive values of x(n), results in a signal whose spectrum is a I-fold periodic repetition of the input signal spectrum.
10. C=I is the desired normalization factor.
a) True
b) False
Answer
Answer: a [Reason:] The amplitude of the sampling rate converted signal should be multiplied by a factor C, whose value when equal to I is called as desired normalization factor.
Digital Electronic MCQ Set 5
1. For a given number of bits, the power of quantization noise is proportional to the variance of the signal to be quantized.
a) True
b) False
Answer
Answer: a [Reason:] The dynamic range of the signal, which is proportional to its standard deviation σx , should match the range R of the quantizer, it follows that ∆ is proportional to σx. Hence for a given number o f bits, the power o f the quantization noise is proportional to the variance of the signal to be quantized.
2. What is the variance of the difference between two successive signal samples, d(n) = x(n) – x(n-1) ?
Answer
Answer: b [Reason:] 
3.What is the variance of the difference between two successive signal samples, d(n) = x(n) –ax(n-1) ?
Answer
Answer: c [Reason:] An even better approach is to quantize the difference, d(n) = x(n) –ax(n-1), w here a is a parameter selected to minimize the variance in d(n). Therefore 
4. If the difference d(n) = x(n) –ax(n-1), then what is the optimum choice for a = ?
Answer
Answer: a [Reason:] An even better approach is to quantize the difference, d(n) = x(n) –ax(n-1), w here a is a parameter selected to minimize the variance in d(n). This leads to the result that the optimum choice of a is 
5. What is the quantity ax(n-1) is called?
a) Second-order predictor of x(n)
b) Zero-order predictor of x(n)
c) First-order predictor of x(n)
d) Third-order predictor of x(n)
Answer
Answer: c [Reason:] In the equation d(n) = x(n) –ax(n-1), the quantity ax(n-1) is called a First-order predictor of x(n).
6. The differential predictive signal quantizer system is known as?
a) DCPM
b) DMPC
c) DPCM
d) None of the mentioned
Answer
Answer: c [Reason:] A differential predictive signal quantizer system. This system is used in speech encoding and transmission over telephone channels and is known as differential pulse code modulation (DPCM ).
7. What is the expansion of DPCM?
a) Differential Pulse Code Modulation
b) Differential Plus Code Modulation
c) Different Pulse Code Modulation
d) None of the mentioned
Answer
Answer: a [Reason:] A differential predictive signal quantizer system. This system is used in speech encoding and transmission over telephone channels and is known as differential pulse code modulation (DPCM ).
8. What are the main uses of DPCM?
a) Speech Decoding and Transmission over mobiles
b) Speech Encoding and Transmission over mobiles
c) Speech Decoding and Transmission over telephone channels
d) Speech Encoding and Transmission over telephone channels
Answer
Answer: d [Reason:] A differential predictive signal quantizer system. This system is used in speech encoding and transmission over telephone channels and is known as differential pulse code modulation (DPCM ).
9. To reduce the dynamic range of the difference signal d(n) = x(n) – x ̂(n), thus a predictor of order p has the form?
Answer
Answer: b [Reason:] T he goal of the predictor is to provide an estimate x ̂(n) of x(n) from a linear combination of past values of x(n), so as to reduce the dynamic range of the difference signal d(n) = x(n) – x ̂(n). Thus a predictor of order p has the form 
10. The simplest form of differential predictive quantization is called?
a) AM
b) BM
c) DM
d) None of the mentioned
Answer
Answer: c [Reason:] The simplest form of differential predictive quantization is called delta modulation (DM ).
11. What is the abbreviation of DM?
a) Diameter Modulation
b) Distance Modulation
c) Delta Modulation
d) None of the mentioned
Answer
Answer: c [Reason:] The simplest form of differential predictive quantization is called delta modulation (DM ).
12. In DM, the quantizer is a simple ________ bit and ______ level quantizer?
a) 2-bit, one-level
b) 1-bit, two-level
c) 2-bit, two level
d) 1-bit, one level
Answer
Answer: b [Reason:] T he simplest form o f differential predictive quantization is called delta modulation (DM ). In DM, the quantizer is a simple 1-bit (two -level) quantizer.
13. In DM, What is the order of predictor is used?
a) Zero-order predictor
b) Second-order predictor
c) First-order predictor
d) Third-order predictor
Answer
Answer: c [Reason:] In DM, the quantizer is a simple 1-bit (two -level) quantizer and the predictor is a first-order predictor.
14. In the equation xq(n)=axq(n-1)+ dq(n), if a = 1 then integrator is called?
a) Leaky integrator
b) Ideal integrator
c) Ideal accumulator
d) Both Ideal integrator & accumulator
Answer
Answer: d [Reason:] In the equation xq(n)=axq(n-1)+ dq(n), if a = 1, we have an ideal accumulator (integrator).
15. In the equation xq(n)=axq(n-1)+ dq(n), if a < 1 then integrator is called?
a) Leaky integrator
b) Ideal integrator
c) Ideal accumulator
d) Both Ideal integrator & accumulator
Answer
Answer: a [Reason:] In the equation xq(n)=axq(n-1)+ dq(n), a < 1 results in a ”leaky integrator”.
