Calculate the Nyquist rate and Nyquist interval from the given signal,
$x(t)=5cos50\pi
t+20sin300\pi t+5cos100\pi t$
Solution:
$x(t)=5cos50\pi
t+20sin300\pi t+5cos100\pi t$
This
is like below equation
$x(t)=5cos\omega_1
t+20sin\omega_2 t+5cos\omega_3 t$
$\omega_1=50\pi$
$\omega_2=300\pi$
$\omega_3=100\pi$
To
calculate the maximum frequency from above values is
$\omega=2\pi
f$
Therefore,
$f_1=25Hz$
$f_2=150Hz$
$f_3=50Hz$
Maximum frequency is nothing but highest frequency in a given signal
maximum
frequency $f_{max}=150Hz$
To
calculate the Nyquist rate or sampling rate
$f_{Nr}=2f_{max}$
$f_{Nr}=2\times
150$
$f_{Nr}=300
Hz$
To calculate the Nyquist interval, $T_{Ni}$ is
$T_{Ni}=\frac{1}{f_{Nr}}$
or $T_{Ni}=\frac{1}{2\times f_{max}}$
$T_{Ni}=\frac{1}{300}$
$T_{Ni}=0.0033
s$
$T_{Ni}=3.3 ms$
Reference material
ReplyDeletehttps://electronicspost.com/what-is-nyquist-rate-and-nyquist-interval/
Deletehttps://ccsuniversity.ac.in/bridge-library/pdf/ENGG-EI-4th-sem-Signal-and-Systems-Code-BT-403-the-Sampling-theorem-and-its-implications.pdf
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