Sampling and Reconstruction of Signals MCQs ( Digital Signal Processing ) MCQs – New Digital Signal Processing MCQs

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Oversampling A/D Converters

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: a
Explanation: The dynamic range of the signal, which is proportional to its standard deviation σ_x, should match the range R of the quantizer, it follows that ∆ is proportional to σ_x. Hence for a given number of bits, the power of 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)?
a) σ2d=2σ2x[1+γxx(1)]
b) σ2d=2σ2x[1−γxx(1)]
c) σ2d=4σ2x[1−γxx(1)]
d) σ2d=3σ2x[1−γxx(1)]
Answer: b
Explanation: σ2d=E[d2(n)]=E[x(n)−x(n−1)]2
= E[x2(n)]−2Ex(n)x(n−1)+E[x2(n−1)]
= 2σ2x[1+γxx(1)].


3. What is the variance of the difference between two successive signal samples, d(n) = x(n)–ax(n-1)?
a) σ2d=2σ2x[1−a2]
b) σ2d=σ2x[1+a2]
c) σ2d=σ2x[1−a2]
d) σ2d=2σ2x[1+a2]
Answer: c
Explanation: 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 σ2d=σ2x[1−a2] .


4. If the difference d(n) = x(n)–ax(n-1), then what is the optimum choice for a = ?
a) γxx(1)σ2x
b) γxx(0)σ2x
c) γxx(0)σ2d
d) γxx(1)σ2d
Answer: a
Explanation: 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 γxx(1)γxx(0)=γxx(1)σ2x.


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: c
Explanation: 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: c
Explanation: 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: a
Explanation: 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: d
Explanation: 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?
a) x^(n)=∑pk=1akx(n+k)
b) x^(n)=∑pk=1akx(n−k)
c) x^(n)=∑pk=0akx(n+k)
d) x^(n)=∑pk=0akx(n−k)
Answer: b
Explanation: The 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 x^(n)=∑pk=1akx(n−k).


10. The simplest form of differential predictive quantization is called?
a) AM
b) BM
c) DM
d) None of the mentioned
Answer: c
Explanation: 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: c
Explanation: 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: b
Explanation: The simplest form of 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: c
Explanation: In DM, the quantizer is a simple 1-bit (two-level) quantizer and the predictor is a first-order predictor.


14. In the equation x_q(n)=ax_q(n-1)+d_q(n), if a = 1 then integrator is called?
a) Leaky integrator
b) Ideal integrator
c) Ideal accumulator
d) Both Ideal integrator & accumulator
Answer: d
Explanation: In the equation x_q(n)=ax_q(n-1)+d_q(n), if a = 1, we have an ideal accumulator (integrator).


15. In the equation x_q(n)=ax_q(n-1)+d_q(n), if a < 1 then integrator is called?
a) Leaky integrator
b) Ideal integrator
c) Ideal accumulator
d) Both Ideal integrator & accumulator
Answer: a
Explanation: In the equation x_q(n)=ax_q(n-1)+ d_q(n), a < 1 results in a ”leaky integrator”.

Sample and Hold

1. What is the main function of (A/D) or ADC converter?
a) Converts Digital to Analog Signal
b) Converts Analog to Digital signal
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: The electronic device that performs this conversion from an analog signal to a digital sequence is called an analog-to-digital (A/D) converter (ADC).


2. What is the main function of (D/A) or DAC converter?
a) Converts Digital to Analog Signal
b) Converts Analog to Digital signal
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: A digital-to-analog (D/A) converter (DAC) takes a digital sequence and produces at its output a voltage or current proportional to the size of the digital word applied to its input.


3. The S/H is a digitally controlled analog circuit that tracks the analog input signal during the sample mode and then holds it fixed during the hold mode to the instantaneous value of the signal at the time the system is switched from the sample to the hold mode.
a) True
b) False
Answer: a
Explanation: The sampling of an analog signal is performed by a sample-and-hold (S/H) circuit. The sampled signal is then quantized and converted to digital form. Usually, the S/H is integrated into the (A/D) converter. The S/H is a digitally controlled analog circuit that tracks the analog input signal during the sample mode and then holds it fixed during the hold mode to the instantaneous value of the signal at the time the system is switched from the sample mode to the hold mode.


4. The time required to complete the conversion of Analog to Digital is ________ the duration of the hold mode of S/H.
a) Greater than
b) Equals to
c) Less than
d) Greater than or Equals to
Answer: c
Explanation: The A/D converter begins the conversion after it receives a convert command. The time required to complete the conversion should be less than the duration of the hold mode of S/H.


5. In A/D converter, what is the time relation between sampling period T and the duration of the sample mode and the hold mode?
a) Should be larger than the duration of sample mode and hold mode
b) Should be smaller than the duration of sample mode and hold mode
c) Should be equal to the duration of sample mode and hold mode
d) Should be larger than or equals to the duration of sample mode and hold mode
Answer: a
Explanation: The A/D converter begins the conversion after it receives a convert command. The sampling period T should be larger than the duration of the sample mode and the hold mode.


6. In the practical A/D converters, what are the distortions and time-related degradations occur during the conversion process?
a) Jitter errors
b) Droops
c) Nonlinear variations in the duration of the sampling aperture
d) All of the mentioned
Answer: d
Explanation: An ideal S/H introduces no distortion in the conversion process and is accurately modeled as an ideal sampler. However, time-related degradations such as errors in the periodicity of the sampling process (“jitter”), nonlinear variations in the duration of the sampling aperture, and changes in the voltage held during conversion (“droop”) do occur in practical devices.


7. In the absence of an S/H, the input signal must change by more than one-half of the quantization step during the conversion, which may be an impractical constraint.
a) True
b) False
Answer: b
Explanation: The use of an S/H allows the A /D converter to operate more slowly compared to the time actually used to acquire the sample. In the absence of an S/H, the input signal must not change by more than one-half of the quantization step during the conversion, which may be an impractical constraint.


8. The noise power σ_n² can be reduced by increasing the sampling rate to spread the quantization noise power over a larger frequency band (-F_s/2, F_s/2).
a) True
b) False
Answer: a
Explanation: The noise power σ_n² can be reduced by increasing the sampling rate to spread the quantization noise power over a larger frequency band (-F_s/2, F_s/2), and then shaping the noise power spectral density by means o f an appropriate filter.


9. What is the process of down sampling called?
a) Decimation
b) Fornication
c) Both Decimation & Fornication
d) None of the mentioned
Answer: a
Explanation: To avoid aliasing, we first filter out the out-of-band (fl, F J 2) noise by processing the wideband signal. The signal is then passed through the low pass filter and re-sampled (down sampled) at the lower rate. The down sampling process is called decimation.


10. If the interpolation factor is I = 256, the A/D converter output can be obtained by averaging successive non-overlapping blocks of 128 bits.
a) True
b) False
Answer: a
Explanation: If the interpolation factor is I = 256, the A/D converter output can be obtained by averaging successive non-overlapping blocks of 128 bits. This averaging would result in a digital signal with a range of values from zero to 256 (b as 8 bits) at the Nyquist rate. The averaging process also provides the required anti-aliasing filtering.


11. The crosshatched areas gives two types of Quantization error in DM, they are?
a) Slope-overload distortion
b) Granular noise
c) Slope-overload distortion & Granular noise
d) None of the mentioned
Answer: c
Explanation: The crosshatched areas illustrate two types of quantization error in DM, slope-overload distortion and granular noise.


12. The slope-overload distortion is avoided, if which of the following conditions satisfy?
a) Min|dx(t)/d(t)| ≤ Δ/T
b) Max|dx(t)/d(t)| ≤ Δ/T
c) |dx(t)/d(t)| ≤Δ/T
d) None of the mentioned
Answer: b
Explanation: The crosshatched areas illustrate two types of quantization error in DM, slope-overload distortion and granular noise. types of quantization error in DM, slope-overload distortion and granular noise. Since the maximum slope A (T in x (n) is limited by the step size, slope-overload distortion can be avoided if max|dx(t)/d(t)|≤Δ/T).


13. In DM, By increasing Δ, reduces the overload distortion but increases the granular noise, and vice versa.
a) True
b) False
Answer: a
Explanation: The granular noise occurs when the DM tracks a relatively flat (slowly changing) input signal. We note that increasing Δ reduces overload distortion but increases the granular noise, and vice versa.


14. Which of the following is the right way to reduce distortion in the DM?
a) By setting up an integrator in front of DM
b) By setting up an integrator behind the DM
c) By setting up an integrator in the middle of DM
d) None of the mentioned
Answer: a
Explanation: We note that increasing Δ reduces overload distortion but increases the granular noise, and vice versa. One way to reduce these two types of distortion is to use an integrator in front of the DM.


15. What are the effects produced by Dm by setting up an integrator at the front of DM?
a) Simplifies the DM decoder
b) Increases correlation of the signal into the DM input
c) Emphasizes the low frequencies of x(t)
d) All of the mentioned
Answer: d
Explanation: One way to reduce these two types of distortion is to use an integrator in front of the DM. This has two effects. First, it emphasizes the low frequencies of x (t) and increases the correlation of the signal into the DM input. Second, it simplifies the DM decoder because the differentiator (inverse system) required at the decoder is canceled by the DM integrator.

Sampling of Band Pass Signals

1. The frequency shift can be achieved by multiplying the band pass signal as given in equation
x(t) = u_c (t) cos⁡2π F_c t-u_s (t) sin⁡2π F_c t by the quadrature carriers cos[2πF_ct] and sin[2πF_ct] and lowpass filtering the products to eliminate the signal components of 2F_c.
a) True
b) False
Answer: a
Explanation: It is certainly advantageous to perform a frequency shift of the band pass signal by and sampling the equivalent low pass signal. Such a frequency shift can be achieved by multiplying the band pass signal as given in the above equation by the quadrature carriers cos[2πF_ct] and sin[2πF_ct] and low pass filtering the products to eliminate the signal components at 2F_c. Clearly, the multiplication and the subsequent filtering are first performed in the analog domain and then the outputs of the filters are sampled.


2. What is the final result obtained by substituting F_c=kB-B/2, T= 1/2B and say n = 2m i.e., for even and n=2m-1 for odd in equation x(nT)= uc(nT)cos2πFcnT−us(nT)sin2πFcnT?
a) (−1)muc(mT1)−us
b) us(mT1−T12)(−1)m+k+1
c) None
d) (−1)muc(mT1)−us(mT1−T12)(−1)m+k+1
Answer: d
Explanation:
x(nT)=uc(nT)cos2πFcnT−us(nT)sin2πFcnT → equ1
=uc(nT)cosπn(2k−1)2−us(nT)sinπn(2k−1)2 → equ2
On substituting the above values in equ1, we get say n=2m, x(2mT)≡xmT(1)=uc(mT1)cosπm(2k−1)=(−1)muc(mT1)
where T1=2T=1B. For n odd, say n=2m-1 in equ2 then we get the result as follows
us(mT1−T12)(−1)m+k+1
Hence proved.


3. Which low pass signal component occurs at the rate of B samples per second with even numbered samples of x(t)?
a) u_c-lowpass signal component
b) u_s-lowpass signal component
c) u_c & u_s-lowpass signal component
d) none of the mentioned
Answer: a
Explanation: With the even-numbered samples of x(t), which occur at the rate of B samples per second, produce samples of the low pass signal component u_c.

Digital Filters Design MCQs

4. Which low pass signal component occurs at the rate of B samples per second with odd numbered samples of x(t)?
a) u_c – lowpass signal component
b) u_s – lowpass signal component
c) u_c & u_s – lowpass signal component
d) none of the mentioned
Answer: b
Explanation: With the odd-numbered samples of x(t), which occur at the rate of B samples per second, produce samples of the low pass signal component u_s.


5. What is the reconstruction formula for the bandpass signal x(t) with samples taken at the rate of 2B samples per second?
a) ∑∞m=−∞x(mT)sin(π/2T)(t−mT)(π/2T)(t−mT)cos2πFc(t−mT)
b) ∑∞m=−∞x(mT)sin(π/2T)(t+mT)(π/2T)(t+mT)cos2πFc(t−mT)
c) ∑∞m=−∞x(mT)sin(π/2T)(t−mT)(π/2T)(t−mT)cos2πFc(t+mT)
d) ∑∞m=−∞x(mT)sin(π/2T)(t+mT)(π/2T)(t+mT)cos2πFc(t+mT)
Answer: a
Explanation: ∑∞m=−∞x(mT)sin(π/2T)(t−mT)(π/2T)(t−mT)cos2πFc(t−mT), where T=1/2B


6. What is the new centre frequency for the increased bandwidth signal?
a) F_c‘= F_c+B/2+B’/2
b) F_c‘= F_c+B/2-B’/2
c) F_c‘= F_c-B/2-B’/2
d) None of the mentioned
Answer: b
Explanation: A new centre frequency for the increased bandwidth signal is F_c‘ = F_c+B/2-B’/2


7. According to the sampling theorem for low pass signals with T₁=1/B, then what is the expression for u_c(t) = ?
a) ∑∞m=−∞uc(mT1)sin(πT1)(t−mT1)(π/T1)(t−mT1)
b) ∑∞m=−∞us(mT1−T12)sin(πT1)(t−mT1+T1/2)(πT1)(t−mT1+T12)
c) ∑∞m=−∞uc(mT1)sin(πT1)(t+mT1)(πT1)(t+mT1)
d) ∑∞m=−∞us(mT1−T12)sin(πT1)(t+mT1+T12)(πT1)(t+mT1+T12)
Answer: a
Explanation: To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T₁=1/B.
uc(t)=∑∞m=−∞uc(mT1)sin(πT1)(t−mT1)(π/T1)(t−mT1).


8. According to the sampling theorem for low pass signals with T₁=1/B, then what is the expression for u_s(t) = ?
a) ∑∞m=−∞uc(mT1)sin(πT1)(t−mT1)(πT1)(t−mT1)
b) ∑∞m=−∞us(mT1−T12)sin(πT1)(t−mT1+T12)(π/T1)(t−mT1+T12)
c) ∑∞m=−∞us(mT1−T12)sin(πT1)(t−mT1−T12)(πT1)(t−mT1−T12)
d) ∑∞m=−∞uc(mT1)sin(πT1)(t+mT1)(πT1)(t+mT1)
Answer: b
Explanation: To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T₁=1/B .
us(t)=∑∞m=−∞us(mT1−T1/2)sin(π/T1)(t−mT1+T1/2)(π/T1)(t−m


9. What is the expression for low pass signal component u_c(t) that can be expressed in terms of samples of the bandpass signal?
a) ∑∞n=−∞(−1)n+r+1x(2nT‘−T‘)sin(π/(2T‘))(t−2nT‘+T‘)(π/(2T‘))(t−2nT‘+T‘)
b) ∑∞n=−∞(−1)nx(2nT‘)sin(π/(2T‘))(t−2nT‘)(π/(2T‘))(t−2nT‘)
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: The low pass signal components u_c(t) can be expressed in terms of samples of the
band pass signal as follows:
uc(t)=∑∞n=−∞(−1)nx(2nT‘)sin(π/(2T‘))(t−2nT‘)(π/(2T‘))(t−2nT‘).


10. What is the expression for low pass signal component u_s(t) that can be expressed in terms of samples of the bandpass signal?
a) ∑∞n=−∞(−1)n+r+1x(2nT‘−T‘)sin(π/(2T‘))(t−2nT‘+T‘)(π/(2T‘))(t−2nT‘+T‘)
b) ∑∞n=−∞(−1)nx(2nT‘)sin(π/(2T‘))(t−2nT‘)(π/(2T‘))(t−2nT‘)
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: The low pass signal components u_s(t) can be expressed in terms of samples of the
band pass signal as follows:
us(t)=∑∞n=−∞(−1)n+r+1x(2nT‘−T‘)sin(π/(2T‘))(t−2nT‘+T‘)(π/(2T‘))(t−2nT‘+T‘)


11. What is the Fourier transform of x(t)?
a) X (F) = 12[Xl(F−Fc)+X∗l(F−Fc)]
b) X (F) = 12[Xl(F−Fc)+X∗l(F+Fc)]
c) X (F) = 12[Xl(F+Fc)+X∗l(F−Fc)]
d) X (F) = 12[Xl(F−Fc)+X∗l(−F−Fc)]
Answer: d
Explanation:
X (F) = ∫∞−∞x(t)e−j2πFtdt
=∫∞−∞{Re[xl(t)ej2πFct]}e−j2πFtdt
Using the identity, R_e(ε)=1/2(ε+ε^*)
X (F) = ∫∞−∞[xl(t)ej2πFct+x∗l(t)e−j2πFct]e−j2πFtdt
=12[Xl(F−Fc)+X∗l(−F−Fc)].


12. What is the basic relationship between the spectrum o f the real band pass signal x(t) and the spectrum of the equivalent low pass signal x_l(t)?
a) X (F) = 12[Xl(F−Fc)+X∗l(F−Fc)]
b) X (F) = 12[Xl(F−Fc)+X∗l(F+Fc)]
c) X (F) = 12[Xl(F+Fc)+X∗l(F−Fc)]
d) X (F) = 12[Xl(F−Fc)+X∗l(−F−Fc)]
Answer: d
Explanation: X(F) = 12[Xl(F−Fc)+X∗l(−F−Fc)], where X_l(F) is the Fourier transform of x_l(t). This is the basic relationship between the spectrum o f the real band pass signal x(t) and the spectrum of the equivalent low pass signal x_l(t).

The Representation of Bandpass Signals

1. Which of the following is the right way of representation of equation that contains only the positive frequencies in a given x(t) signal?
a) X₊(F)=4V(F)X(F)
b) X₊(F)=V(F)X(F)
c) X₊(F)=2V(F)X(F)
d) X₊(F)=8V(F)X(F)
Answer: c
Explanation: In a real valued signal x(t), has a frequency content concentrated in a narrow band of frequencies in the vicinity of a frequency F_c. Such a signal which has only positive frequencies can be expressed as X₊(F)=2V(F)X(F)
Where X₊(F) is a Fourier transform of x(t) and V(F) is unit step function.


2. What is the equivalent time –domain expression of X₊(F)=2V(F)X(F)?
a) F⁽⁺¹⁾[2V(F)]*F⁽⁺¹⁾[X(F)]
b) F^(-1)[4V(F)]*F^(-1)[X(F)]
c) F^(-1)[V(F)]*F^(-1)[X(F)]
d) F^(-1)[2V(F)]*F^(-1)[X(F)]
Answer: d
Explanation: Given Expression, X₊(F)=2V(F)X(F).It can be calculated as follows
x+(t)=∫∞−∞X+(F)ej2πFtdF
=F−1[2V(F)]∗F−1[X(F)]


3. In time-domain expression, x+(t)=F−1[2V(F)]∗F−1[X(F)]. The signal x₊(t) is known as
a) Systematic signal
b) Analytic signal
c) Pre-envelope of x(t)
d) Both Analytic signal & Pre-envelope of x(t)
Answer: d
Explanation: From the given expression, x+(t)=F−1[2V(F)]∗F−1[X(F)].


4. In equation x+(t)=F−1[2V(F)]∗F−1[X(F)], if F−1[2V(F)]=δ(t)+j/πt and F−1[X(F)] = x(t). Then the value of ẋ(t) is?
a) 1π∫∞−∞x(t)t+τdτ
b) 1π∫∞−∞x(t)t−τdτ
c) 1π∫∞−∞2x(t)t−τdτ
d) 1π∫∞−∞4x(t)t−τdτ
Answer: b
Explanation: x+(t)=[δ(t)+j/πt]∗x(t)
x+(t)=x(t)+[j/πt]∗x(t)
ẋ(t)=[j/πt]∗x(t)
=1π∫∞−∞x(t)t−τdτ Hence proved.


5. If the signal ẋ(t) can be viewed as the output of the filter with impulse response h(t) = 1/πt, -∞ < t < ∞ when excited by the input signal x(t) then such a filter is called as __________
a) Analytic transformer
b) Hilbert transformer
c) Both Analytic & Hilbert transformer
d) None of the mentioned
Answer: b
Explanation: The signal ẋ(t) can be viewed as the output of the filter with impulse response h(t) = 1/πt,
-∞ < t < ∞ when excited by the input signal x(t) then such a filter is called as Hilbert transformer.


6. What is the frequency response of a Hilbert transform H(F)=?
a) ⎧⎩⎨−j(F>0)0(F=0)j(F<0)
b) ⎧⎩⎨⎪⎪−j0j(F<0)(F=0)(F>0)
c) ⎧⎩⎨⎪⎪−j0j(F>0)(F=0)(F<0)
d) ⎧⎩⎨⎪⎪j0j(F>0)(F=0)(F<0)
Answer: a
Explanation: H(F) =∫∞−∞h(t)e−j2πFtdt
=1π∫∞−∞1/te−2πFtdt
=⎧⎩⎨⎪⎪−j0j(F>0)(F=0)(F<0)
We Observe that │H (F)│=1 and the phase response ⊙(F) = -1/2π for F > 0 and ⊙(F) = 1/2π for F < 0.


7. What is the equivalent lowpass representation obtained by performing a frequency translation of X₊(F) to X_l(F)= ?
a) X₊(F+F_c)
b) X₊(F-F_c)
c) X₊(F*F_c)
d) X₊(F_c-F)
Answer: a
Explanation: The analytic signal x₊(t) is a bandpass signal. We obtain an equivalent lowpass representation by performing a frequency translation of X₊(F).


8. What is the equivalent time domain relation of x_l(t) i.e., lowpass signal?
a) xl(t)=[x(t)+jẋ(t)]e−j2πFct
b) x(t)+j ẋ(t) = xl(t)ej2πFct
c) xl(t)=[x(t)+jẋ(t)]e−j2πFct & x(t)+j ẋ(t) = xl(t)ej2πFct
d) None of the mentioned
Answer: c
Explanation: xl(t)=x+(t)e−j2πFct
=[x(t)+jẋ(t)]e−j2πFct
Or equivalently, x(t)+j ẋ(t) =xl(t)ej2πFct.


9. If we substitute the equation xl(t)=uc(t)+jus(t) in equation x (t) + j ẋ (t) = x_l(t) e^j2πF_ct and equate real and imaginary parts on side, then what are the relations that we obtain?
a) x(t)=uc(t)cos2πFct+us(t)sin2πFct; ẋ(t)=us(t)cos2πFct−uc(t)sin2πFct
b) x(t)=uc(t)cos2πFct−us(t)sin2πFct; ẋ(t)=us(t)cos2πFct+uc(t)sin2πFct
c) x(t)=uc(t)cos2πFct+us(t)sin2πFct; ẋ(t)=us(t)cos2πFct+uc(t)sin2πFct
d) x(t)=uc(t)cos2πFct−us(t)sin2πFct; ẋ(t)=us(t)cos2πFct−uc(t)sin2πFct
Answer: b
Explanation: If we substitute the given equation in other, then we get the required result


10. In the relation, x(t) = uc(t)cos2πFct−us(t)sin2πFct the low frequency components u_c and u_s are called _____________ of the bandpass signal x(t).
a) Quadratic components
b) Quadrature components
c) Triplet components
d) None of the mentioned
Answer: b
Explanation: The low frequency signal components u_c(t) and u_s(t) can be viewed as amplitude modulations impressed on the carrier components cos2πF_ct and sin2πF_ct, respectively. Since these carrier components are in phase quadrature, u_c(t) and u_s(t) are called the Quadrature components of the bandpass signal x (t).


11. What is the other way of representation of bandpass signal x(t)?
a) x(t) = R_e[xl(t)ej2πFct]
b) x(t) = R_e[xl(t)ejπFct]
c) x(t) = R_e[xl(t)ej4πFct]
d) x(t) = R_e[xl(t)ej0πFct]
Answer: a
Explanation: The above signal is formed from quadrature components, x(t) = R_e[xl(t)ej2πFct] where R_e denotes the real part of complex valued quantity.


12. In the equation x(t) = R_e[xl(t)ej2πFct], What is the lowpass signal x_l (t) is usually called the ___ of the real signal x(t).
a) Mediature envelope
b) Complex envelope
c) Equivalent envelope
d) All of the mentioned
Answer: b
Explanation: In the equation x(t) = R_e[x_l(t)e^(j2πF_ct)], R_e denotes the real part of the complex valued quantity in the brackets following. The lowpass signal x_l (t) is usually called the Complex envelope of the real signal x(t), and is basically the equivalent low pass signal.


13. If a possible representation of a band pass signal is obtained by expressing x_l (t) as xl(t)=a(t)ejθ(t) then what are the equations of a(t) and θ(t)?
a) a(t) = u2c(t)+u2s(t)−−−−−−−−−−√ and θ(t)=tan−1us(t)uc(t)
b) a(t) = u2c(t)−u2s(t)−−−−−−−−−−√ and θ(t)=tan−1us(t)uc(t)
c) a(t) = u2c(t)+u2s(t)−−−−−−−−−−√ and θ(t)=tan−1uc(t)us(t)
d) a(t) = u2s(t)−u2c(t)−−−−−−−−−−√ and θ(t)=tan−1us(t)uc(t)
Answer: a
Explanation: A third possible representation of a band pass signal is obtained by expressing xl(t)=a(t)ejθ(t) where a(t) = u2c(t)+u2s(t)−−−−−−−−−−√ and θ(t)=tan−1us(t)uc(t).


14. What is the possible representation of x(t) if x_l(t)=a(t)e^(jθ(t))?
a) x(t) = a(t) cos[2πF_ct – θ(t)]
b) x(t) = a(t) cos[2πF_ct + θ(t)]
c) x(t) = a(t) sin[2πF_ct + θ(t)]
d) x(t) = a(t) sin[2πF_ct – θ(t)]
Answer: b
Explanation: x(t) = R_e[xl(t)ej2πFct]
= R_e[a(t)ej[2πFct+θ(t)]]
= a(t)cos[2πFct+θ(t)]
Hence proved.


15. In the equation x(t) = a(t)cos[2πF_ct+θ(t)], Which of the following relations between a(t) and x(t), θ(t) and x(t) are true?
a) a(t), θ(t) are called the Phases of x(t)
b) a(t) is the Phase of x(t), θ(t) is called the Envelope of x(t)
c) a(t) is the Envelope of x(t), θ(t) is called the Phase of x(t)
d) none of the mentioned
Answer: c
Explanation: In the equation x(t) = a(t) cos[2πF_ct+θ(t)], the signal a(t) is called the Envelope of x(t), and θ(t) is called the phase of x(t).

Quantization and Coding

1. The basic task of the A/D converter is to convert a discrete set of digital code words into a continuous range of input amplitudes.
a) True
b) False
Answer: b
Explanation: The basic task of the A/D converter is to convert a continuous range of input amplitude into a discrete set of digital code words. This conversion involves the processes of Quantization and Coding.


2. What is the type of quantizer, if a Zero is assigned a quantization level?
a) Midrise type
b) Mid tread type
c) Mistreat type
d) None of the mentioned
Answer: b
Explanation: If a zero is assigned a quantization level, the quantizer is of the mid treat type.


3. What is the type of quantizer, if a Zero is assigned a decision level?
a) Midrise type
b) Mid tread type
c) Mistreat type
d) None of the mentioned
Answer: a
Explanation: If a zero is assigned a decision level, the quantizer is of the midrise type.


4. What is the term used to describe the range of an A/D converter for bipolar signals?
a) Full scale
b) FSR
c) Full-scale region
d) FS
Answer: b
Explanation: The term Full-scale range (FSR) is used to describe the range of an A/D converter for bipolar signals (i.e., signals with both positive and negative amplitudes).


5. What is the term used to describe the range of an A/D converter for uni-polar signals?
a) Full scale
b) FSR
c) Full-scale region
d) FSS
Answer: a
Explanation: The term Full scale (FS) is used for uni-polar signals.


6. What is the fixed range of the quantization error e_q(n)?
a) –Δ6 < e_q(n) ≤ Δ6
b) –Δ4 < e_q(n) ≤ Δ4
c) –Δ2 < e_q(n) ≤ Δ2
d) –Δ16 < e_q(n) ≤ Δ16
Answer: c
Explanation: The quantization error e_q(n) is always in the range – Δ2 < e_q(n) ≤ Δ2, where Δ is quantizer step size.


7. If the dynamic range of the signal is smaller than the range of quantizer, the samples that exceed the quantizer are clipped, resulting in large quantization error.
a) True
b) False
Answer: b
Explanation: If the dynamic range of the signal, defined as x_max-x_min, is larger than the range of the quantizer, the samples that exceed the quantizer range are clipped, resulting in a large (greater than Δ2) quantization error.


8. What is the relation defined by the operation of quantizer?
a) x_q(n) ≡ Q[x(n)] = x^k
b) x_q(n) = Q[x(n)] = x^k, if x(n) ∈ I_k
c) x_q(k) ≡ Q[x(k)] = x^k
d) none of the mentioned
Answer: b
Explanation: The possible outputs of the quantizer (i.e., the quantization levels) are denoted as x^1,x^2,…x^L. The operation of the quantizer is defined by the relation, x_q(n) ≡ Q[x(n)]= x^k, if x(n) ∈ I_k.


9. What is the step size or the resolution of an A/D converter?
a) Δ = (R)/2^(b+1)
b) Δ = (R)/2^(b-1)
c) Δ = (R)/3^(b+1)
d) Δ = (R)/2
Answer: a
Explanation: The coding process in an A/D converter assigns a unique binary number to each quantization level. If we have L levels, we need at least L different binary numbers. With a word length of b + 1 bits we can represent 2^b+1 distinct binary numbers. Hence we should have 2^(b+1) > L or, equivalently, b + 1 > log2 L. Then the step size or the resolution of the A/D converter is given by
Δ = (R)/2^(b+1), where R is the range of the quantizer.


10. In the practical A/D converters, if the first transition may not occur at exactly + 1/2 LSB, then such kind of error is known as ____________
a) Scale-factor error
b) Offset error
c) Linearity error
d) All of the mentioned
Answer: b
Explanation: We note that practical A/D converters may have offset error (the first transition may not occur at exactly + 1/2 LSB).


11. In the practical A/D converters, if the difference between the values at which the first transition and the last transition occur is not equal to FS – 2LSB, then such error is known as _________
a) Scale-factor error
b) Offset error
c) Linearity error
d) All of the mentioned
Answer: a
Explanation: We note that practical A/D converters scale-factor (or gain) error (the difference between the values at which the first transition and the last transition occur is not equal to F_S — 2LSB).


12. In the practical A/D converters, if the differences between transition values are not all equal or uniformly changing, then such error is known as?
a) Scale-factor error
b) Offset error
c) Linearity error
d) All of the mentioned
Answer: c
Explanation: We note that practical A/D converters, linearity error (the differences between transition values are not all equal or uniformly changing).

Digital to Analog Conversion Sample and Hold

1. What is the ideal reconstruction formula or ideal interpolation formula for x(t) = _________
a) ∑∞−∞x(nT)sin(π/T)(t−nT)(π/T)(t−nT)
b) ∑∞−∞x(nT)sin(π/T)(t+nT)π/T)(t+nT
c) ∑∞−∞x(nT)sin(2π/T)(t−nT)2π/T)(t−nT
d) ∑∞−∞x(nT)sin(4π/T)(t−nT)(4π/T)(t−nT)
Answer: a
Explanation: x(t) = ∑∞−∞x(nT)sin(π/T)(t−nT)(π/T)(t−nT) where the sampling interval T = 1/F_s=1/2B, F_s 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?
a) g(t)=sin(2πt/T)(πt/T)
b) g(t)=sin(πt/T)(πt/T)
c) g(t)=sin(6πt/T)(πt/T)
d) g(t)=sin(3πt/T)(πt/T)
Answer: b
Explanation: The reconstruction of the signal x (t) from its samples as an interpolation problem and have described the function:g(t)=sin(πt/T)(πt/T).


3. What is the frequency response of the analog filter corresponding to the ideal interpolator?
a) H(F)={T,|F|≤12T=Fs/20,|F|>14T
b) H(F)={T,|F|≥12T=Fs/20,|F|>14T
c) H(F)={T,|F|≤12T=Fs/20,|F|>12T
d) H(F)={T,|F|≤14T=Fs/20,|F|>14T
Answer: c
Explanation: The analog filter corresponding to the ideal interpolator has a frequency response:
H(F)={T,|F|≤12T=Fs/20,|F|>12T, H(F) is the Fourier transform of the interpolation function g(t).


4. The reconstruction of 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: a
Explanation: The reconstruction of 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: a
Explanation: 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: a
Explanation: 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. The 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 of the binary word.


7. The time required for the output of 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: b
Explanation: An important parameter of a D/A converter is its settling time, which is defined as the time required for the output of 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: a
Explanation: The application of 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: b
Explanation: 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 of 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?
a) h(t)={1,0≤t≤T0,otherwise
b) h(t)={1,0≥t≥T0,otherwise
c) h(t)={1,0<t≤T0,otherwise
d) None of the mentioned
Answer: a
Explanation: W hen viewed as a linear filter, the S/H has an impulse response:
h(t)={1,0≤t≤T0,otherwise

Sampling and Reconstruction of Signals MCQs ( Digital Signal Processing ) MCQs – New Digital Signal Processing MCQs