Analog to Digital (A2D) Converter MCQ Quiz – Objective Question with Answer for Analog to Digital (A2D) Converter

1. Which of the following should be done in order to convert a continuous-time signal to a discrete-time signal?

A. Sampling
B. Differentiating
C. Integrating
D. None of the mentioned

The process of converting a continuous-time signal into a discrete-time signal by taking samples of the continuous-time signal at discrete time instants is known as ‘sampling’.

2. The process of converting discrete-time continuous valued signal into discrete-time discrete valued (digital) signal is known as ____________

A. Sampling
B. Quantization
C. Coding
D. None of the mentioned

In this process, the value of each signal sample is represented by a value selected from a finite set of possible values. Hence this process is known as ‘quantization’

3. The difference between the unquantized x(n) and quantized xq(n) is known as ___________

A. Quantization coefficient
B. Quantization ratio
C. Quantization factor
D. Quantization error

Quantization error is the difference in the signal obtained after sampling i.e., x(n), and the signal obtained after quantization i.e., xq(n) at any instant of time.

4. Which of the following is a digital-to-analog conversion process?

A. Staircase approximation
B. Linear interpolation
D. All of the mentioned

The process of joining in terms of steps is known as staircase approximation, connecting two samples by a straight line is known as Linear interpolation, connecting three samples by fitting a quadratic curve is called Quadratic interpolation.

5. The relation between analog frequency ‘F’ and digital frequency ‘f’ is?

A. F=f*T(where T is sampling perioD.
B. f=F*T
C. No relation
D. None of the mentioned

Consider an analog signal of frequency ‘F’, which when sampled periodically at a rate of Fs=1/T samples per second yields a frequency of f=F/Fs=>f=F*T.

6. What is the output signal when a signal x(t)=cos(2*pi*40*t) is sampled with a sampling frequency of 20Hz?

A. cos(pi*n)
B. cos(2*pi*n)
C. cos(4*pi*n)
D. cos(8*pi*n)

From the question F=40Hz, Fs=20Hz
=>f=F/Fs
=>f=40/20
=>f=2Hz
=>x(n)=cos(4*pi*n).

7. If ‘F’ is the frequency of the analog signal, then what is the minimum sampling rate required to avoid aliasing?

A. F
B. 2F
C. 3F
D. 4F

According to the Nyquist rate, to avoid aliasing the sampling frequency should be equal to twice the analog frequency.

8. What is the Nyquist rate of the signal x(t)=3cos(50*pi*t)+10sin(300*pi*t)-cos(100*pi*t)?

A. 50Hz
B. 100Hz
C. 200Hz
D. 300Hz

The frequencies present in the given signal are F1=25Hz, F2=150Hz, F3=50Hz
Thus Fmax=150Hz and from the sampling theorem,
Nyquist rate=2*Fmax
Therefore, Fs=2*150=300Hz.

9. What is the discrete-time signal obtained after sampling the analog signal x(t)=cos(2000*pi*t)+sin(5000*pi*t) at a sampling rate of 5000 samples/sec?

A. cos(2.5*pi*n)+sin(pi*n)
B. cos(0.4*pi*n)+sin(pi*n)
C. cos(2000*pi*n)+sin(5000*pi*n)
D. none of the mentioned

From the given analog signal, F1=1000Hz F2=2500Hz and Fs=5000Hz
=>f1=F1/Fs and f2=F2/Fs
=>f1=0.2 and f2=0.5
=>x(n)=cos(0.4*pi*n)+sin(pi*n).

10. If the sampling rate Fs satisfies the sampling theorem, then the relation between quantization errors of the analog signal(eq(t)) and discrete-time signal(eq(n)) is?

A. eq(t)=eq(n)
B. eq(t)<eq(n)
C. eq(t)>eq(n)
D. not related