Sample and Hold MCQ [Free PDF] - Objective Question Answer for Sample and Hold Quiz

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

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

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

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

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 the 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 equal to the duration of sample mode and hold mode

Answer: A

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 that 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

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

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 σn2 can be reduced by increasing the sampling rate to spread the quantization noise power over a larger frequency band (-Fs/2, Fs/2).

A. True
B. False

Answer: A

The noise power σn2 can be reduced by increasing the sampling rate to spread the quantization noise power over a larger frequency band (-Fs/2, Fs/2), and then shaping the noise power spectral density by means of 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

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

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 give 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
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?

Answer: B

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

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

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

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 to the DM input. Second, it simplifies the DM decoder because the differentiator (inverse system) required at the decoder is canceled by the DM integrator.

16. The frequency shift can be achieved by multiplying the bandpass signal as given in the equation x(t) = \(u_c (t) cos⁡2π F_c t-u_s (t) sin⁡2π F_c t\) by the quadrature carriers cos[2πFct] and sin[2πFct] and lowpass filtering the products to eliminate the signal components of 2Fc.

A. True
B. False

Answer: A

It is certainly advantageous to perform a frequency shift of the bandpass signal by sampling the equivalent low pass signal. Such a frequency shift can be achieved by multiplying the bandpass signal as given in the above equation by the quadrature carriers cos[2πFct] and sin[2πFct] and low pass filtering the products to eliminate the signal components at 2Fc.

Clearly, the multiplication and the subsequent filtering are first performed in the analog domain and then the outputs of the filters are sampled.

17. What is the final result obtained by substituting Fc=kB-B/2, T= 1/2B and say n = 2m i.e., for even and n=2m-1 for odd in equation x(nT)= \(u_c (nT)cos⁡2πF_c nT-u_s (nT)sin⁡ 2πF_c nT\)?

A. \((-1)^m u_c (mT_1)-u_s\)

B. \(u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}\)

C. \((-1)^m u_c (mT_1)- u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}\)

D. None

Answer: C

\(x(nT)=u_c (nT)cos⁡ 2πF_c nT-u_s (nT)sin⁡2πF_c nT\) → equ1

=\(u_c (nT)cos⁡\frac{πn(2k-1)}{2}-u_s(nT)sin⁡\frac{πn(2k-1)}{2}\) → equ2

On substituting the above values in equ1, we get say n=2m, \(x(2mT) ≡ xmT_{(1)} = u_c (mT_1)cos⁡πm(2k-1)=(-1)^m u_c (mT_1)\)

where \(T_1=2T=\frac{1}{B}\). For n odd, say n=2m-1 in equ2 then we get the result as follows

\(u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}\)

Hence proved.

18. Which low pass signal component occurs at the rate of B samples per second with even-numbered samples of x(t)?

A. uc-lowpass signal component
B. us-lowpass signal component
C. uc & us-lowpass signal component
D. none of the mentioned

Answer: A

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 uc.

19. Which low pass signal component occurs at the rate of B samples per second with odd-numbered samples of x(t)?

A. uc – lowpass signal component
B. us – lowpass signal component
C. uc & us – lowpass signal component
D. none of the mentioned

Answer: B

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 us.

20. What is the reconstruction formula for the bandpass signal x(t) with samples taken at the rate of 2B samples per second?

A. \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t-mT)\)

B. \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t-mT)\)

C. \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t+mT)\)

D. \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t+mT)\)

Answer: A

\(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t-mT)\), where T=1/2B

21. What is the new center frequency for the increased bandwidth signal?

A. Fc‘= Fc+B/2+B’/2
B. Fc‘= Fc+B/2-B’/2
C. Fc‘= Fc-B/2-B’/2
D. None of the mentioned

Answer: B

A new center frequency for the increased bandwidth signal is

Fc‘ = Fc+B/2-B’/2

22. According to the sampling theorem for low pass signals with T1=1/B, then what is the expression for uc(t) = ?

A. \(\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(π/T_1)(t-mT_1)}\)

B. \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1+T_1/2)}{(\frac{π}{T_1})(t-mT_1+\frac{T_1}{2})}\)

C. \(\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t+mT_1)}{(\frac{π}{T_1})(t+mT_1)}\)

D. \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t+mT_1+\frac{T_1}{2})}{(\frac{π}{T_1})(t+mT_1+\frac{T_1}{2})}\)

Answer: A

To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T1=1/B.
\(u_c (t)=\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(π/T_1)(t-mT_1)}\).

23. According to the sampling theorem for low pass signals with T1=1/B, then what is the expression for us(t) = ?

A. \(\sum_{m=-∞}^∞ u_c (mT_1) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(\frac{π}{T_1})(t-mT_1)}\)

B. \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1+\frac{T_1}{2})}{(π/T_1)(t-mT_1+\frac{T_1}{2})}\)

C. \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1-\frac{T_1}{2})}{(\frac{π}{T_1})(t-mT_1-\frac{T_1}{2})}\)

D. \(\sum_{m=-∞}^∞ u_c (mT_1) \frac{sin⁡(\frac{π}{T_1}) (t+mT_1)}{(\frac{π}{T_1})(t+mT_1)}\)

Answer: B

To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T1=1/B .

\(u_s (t)=\sum_{m=-∞}^∞ u_s (mT_1-T_1/2) \frac{sin⁡(π/T_1) (t-mT_1+T_1/2)}{(π/T_1)(t-mT_1+T_1/2)}\)

24. What is the expression for low pass signal component uc(t) that can be expressed in terms of samples of the bandpass signal?

A. \(\sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(π/(2T^{‘}))(t-2nT^{‘}+T^{‘})}\)

B. \(\sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘})}{(π/(2T^{‘}))(t-2nT^{‘})}\)

C. All of the mentioned

D. None of the mentioned

Answer: B

The low pass signal components uc(t) can be expressed in terms of samples of the
band pass signal as follows:

\(u_c (t) = \sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘})}{(π/(2T^{‘}))(t-2nT^{‘})}\).

25. What is the expression for low pass signal component us(t) that can be expressed in terms of samples of the bandpass signal?

A. \(\sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(π/(2T^{‘}))(t-2nT^{‘}+T^{‘})}\)

B. \(\sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘})}{(π/(2T^{‘}))(t-2nT^{‘})}\)

C. All of the mentioned

D. None of the mentioned

Answer: A

The low pass signal components us(t) can be expressed in terms of samples of the band pass signal as follows:

\(u_s (t) = \sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(π/(2T^{‘}))(t-2nT^{‘}+T^{‘})}\)

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