Arrange the Nyquist sampling interval of the signal in descending order

(A) sin c(300t)

(B) sin c(300t) + sin c^{2}(300t)

(C) sin c(200t)

(D) sin c(200t) + sinc^{2}(200t)

(E) sin c(200t) + sin c(500t)

Choose the correct answer from the options given below:

(1) (A), (C), (B), (D), (E)

(2) (C), (D), (E), (A), (B)

(3) (C), (A), (D), (E), (B)

(4) (A), (E), (D), (C), (B)

This question was previously asked in

UGC NET Paper 2: Electronic Science Nov 2020 Official Paper

Option 3 : 3

Official Paper 1: Held on 24 Sep 2020 Shift 1

16726

50 Questions
100 Marks
60 Mins

__Concept:__

Nyquist Sampling interval is defined as the maximum time that:

1) is between regularly spaced samples that will permit the signal waveform to be completely determined.

2) and is equal to reciprocal to twice of the highest frequency in the sampling signal.

i.e., if signal frequency is f_{m} then :

**Nyquist rate f _{s} ≥ 2f_{m}**

And Nyquist interval:

\({T_s} \le \frac{1}{{2{f_m}}}\)

** sinc function:** It is the Fourier transform of rectangular function with no scaling.

\(\sin c\left( x \right) = \frac{{\sin \left( x \right)}}{x}\)

\(\sin c\left( x \right) = \left\{ {\begin{array}{*{20}{c}} 1&{x = 0}\\ {\frac{{\sin x}}{x}}&{otherwise} \end{array}} \right.\)

__Analysis__**:**

If two or more than two signals are added with signal frequency then sampling frequency will be:

\({f_s} = 2\max \left( {{f_{{m_1}}},\;{f_{{m_2}}},\;{f_m}} \right)\)

And \({T_s} = \frac{1}{{{F_s}}}\)

__For A:__

\({\omega _m} = 300\;\pi \)

\(2\pi {f_m} = 300\pi \)

\({T_s} = \frac{1}{{2{f_m}}} = \frac{1}{{2 \times 150}}\)

\({T_s} = \frac{1}{{300}}sec\)

f_{m} = 150 Hz

__For B:__

ω_{1} = 300π and ω_{2}= 2 × 300π = 600π

fm_{1} = 150, fm_{2} = 300

max (fm_{1}, fm_{2}) = 300 Hz

\({T_s} = \frac{1}{{2 \times 300}} = \frac{1}{{600}}sec\)

__For C:__

ω_{m} = 200π ⇒ f_{m} = 100 Hz

\({T_s} = \frac{1}{{2 \times 100}} = \frac{1}{{200}}sec\)

__For D:__

ω_{1} = 200π, ω_{2} = 2 × 200π = 400 π

fm_{1 }= 100, fm_{2} = 200

max (fm_{1}, fm_{2}) = 200 Hz

\({T_s} = \frac{1}{{2 \times 200}} = \frac{1}{{400}}sec\)

__For E:__

ω_{1} = 200π, ω_{2} = 500π

f_{1 }= 100, f_{2} = 250

max (f_{1}, f_{2}) = 250

\({T_s} = \frac{1}{{2 \times 250}} = \frac{1}{{500}}\)

Thus T_{c} > T_{A} > T_{D} > T_{E} > T_{B}

**So, Option 3 correct.**