This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “A2D and D2A Converters”.

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

View Answer

Explanation: The process of converting a continuous-time signal into a discrete-time signal by taking samples of 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

View Answer

Explanation: 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

View Answer

Explanation: 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

c) Quadratic interpolation

d) All of the mentioned

View Answer

Explanation: 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 as Quadratic interpolation.

a) F=f*T(where T is sampling period)

b) f=F*T

c) No relation

d) None of the mentioned

View Answer

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

6. What is 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)

View Answer

Explanation: 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

View Answer

Explanation: According to Nyquist rate, to avoid aliasing the sampling frequency should be equal to twice of 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

View Answer

Explanation: 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 5000samples/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

View Answer

Explanation: 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)

a) eq(t)=eq(n)

b) eq(t)

d) Not related

View Answer

Explanation: If it obeys sampling theorem, then the only error in A/D conversion is quantization error. So, the error is same for both analog and discrete-time signal.

11. The quality of output signal from a A/D converter is measured in terms of:

a) Quantization error

b) Quantization to signal noise ratio

c) Signal to quantization noise ratio

d) Conversion constant

View Answer

Explanation: The quality is measured by taking the ratio of noises of input signal and the quantized signal i.e., SQNR and is measured in terms of dB.

12. Which bit coder is required to code a signal with 16 levels?

a) 8 bit

b) 4 bit

c) 2 bit

d) 1 bit

View Answer

Explanation: To code a signal with L number of levels, we require a coder with (log L/log 2) number of bits. So, log16/log2=4 bit coder is required.

**Sanfoundry Global Education & Learning Series – Digital Signal Processing.**

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