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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