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Expected GATE-2022 Geophysics Solution (10)

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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]

 

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