2071 Chaitra(Q) State Nyquist Sampling theorem. Determine the Nyquist rate and Nyquist interval for a continuous-time signal x(t)=6cos50πt+20sin300πt-10cos100πt is to be sampled and quantized using 512 levels.

The Statement of Nyquist Sampling theorem given below :

“A continuous-time signal may be completely represented in its samples and recovered back if the sampling frequency is fs >= 2fm”. Here f_s is the sampling frequency and f_s is the maximum frequency present in signal.

Numerical Part:

given, x(t)=6cos50πt+20sin300πt-10cos100πt ———-(a)

let the three frequencies present in eq(a) be ω1, ω2, and ω3

so that, the new equation for signal eq(a) is:

x(t)=6cosω₁t+20sinω₂t-10cosω₃t ———————–(b)

comparing eq(a) and eq(b) we get,,

ω₁t=50πt -> ω₁ =50π

2π f₁ =50π, therefore f₁ =25 Hz

Similarly,

f₂ =150 Hz

and f3= 50 Hz

also , signal (a) is sampled and quantized using 512 levels.

so, q= 2ⁿ

or, 512=2ⁿ

on solving we get

n = 9 bits.

i.e f₂ >f₃ >f₁

Therefore f_m =f₂ = 150

Nyquist rate (f_s )= 2 fm

=2*150

=300 Hz

Nyquist interval (Ts)= 1/fs=1/2fm

=1/(2*150)

=0.037 sec

=3.7*10^-3 Sec