Expected GATE-2022 Geophysics Solution (10)

Determine the Nyquist rate and Nyquist interval of given below signal,

$x(t)=sin500\pi t \times cos300\pi t$

Solution:

Given signal,

$x(t)=sin500\pi t \times cos300\pi t---(1)$

Important Trigonometric formulas,

$2 sinA cosB=\left[sin(A+B)+sin(A-B) \right]---(2)$

$2 cosA sinB=\left[sin(A+B)-sin(A-B) \right]$

From given equation similar to equation (2)

Then,

$sin 500t \times cos300\pi t=\frac{1}{2}[sin(500\pi +300\pi)t+sin(500\pi-300\pi)t]$

$sin 500t \times cos300\pi t=\frac{1}{2}[sin800 \pi t+sin200 \pi t]$

This is like below equation

$x(t)=\frac{1}{2}[sin\omega_1 t+sin\omega_2 t]$

$\omega_1=800\pi$

$\omega_2=200\pi$

To calculate the maximum frequency from above values is

$\omega=2\pi f$

Therefore,

$f_1=400Hz$

$f_2=100Hz$

Maximum frequency is nothing but highest frequency in a given signal

maximum frequency $f_{max}=400Hz$

To calculate the Nyquist rate or sampling rate

$f_{Nr}=2f_{max}$

$f_{Nr}=2\times 400$

$f_{Nr}=800Hz$

To calculate the Nyquist interval, $T_{Ni}$ is

$T_{Ni}=\frac{1}{f_{Ni}}$ or $T_{Ni}=\frac{1}{2\times f_{max}}$

$T_{Ni}=\frac{1}{800}$

$T_{Ni}=0.00125 s$

$T_{Ni}=1.25 ms$

Extra information:

Important Trigonometric formulas,
$2 cosA cosB=\left[cos(A+B)+cos(A-B) \right]$
$2 sinA sinB=\left[cos(A-B)-cos(A+B) \right]$

GEOPHYSICS