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What is Nyquist Rate and Nyquist Interval?
Nyquist Rate of Sampling
The theoretical minimum sampling rate at which a signal can be sampled and still can be reconstructed from its samples without any distortion is called the Nyquist rate of sampling.
Mathematically,
$$\mathrm{Nyquist\: Rate,\mathit{f_{N}}\mathrm{=}2\mathit{f_{m}}}$$
Where, $\mathit{f_{m}}$is the maximum frequency component present in the signal.
If the signal is sampled at the rate greater than the Nyquist rate, then the signal is called over sampled.
If the signal is sampled at the rate less than its Nyquist rate, then it is said to be under sampled.
Nyquist Interval
When the rate of sampling is equal to the Nyquist rate, then the time interval between any two adjacent samples is called the Nyquist interval.
Mathematically, the Nyquist interval is given by,
$$\mathrm{Nyquist \:interval\:\mathrm{=}\:\frac{1}{\mathit{f_{N}}}\mathrm{=}\frac{1}{\mathit{\mathrm{2}f_{m}}}}$$
Numerical Example
Determine the Nyquist rate and Nyquist interval corresponding to signal given by,
$$\mathrm{\mathit{x\mathrm{\left(\mathit{t}\right )}\mathrm{=}\mathrm{1}\:\mathrm{+}\:\mathrm{sin\: 3000}\mathit{\pi t}\:\mathrm{+}\:\mathrm{cos\: 5000}\mathit{\pi t}}}$$
Solution
The given signal is,
$$\mathrm{\mathit{x\mathrm{\left(\mathit{t}\right )}\mathrm{=}\mathrm{1}\:\mathrm{+}\:\mathrm{sin\: 3000}\mathit{\pi t}\:\mathrm{+}\:\mathrm{cos\: 5000}\mathit{\pi t}}}$$
For this signal, we have,
Highest frequency component in 1 is 0.
Highest frequency component in term $\mathrm{sin\: 3000}\mathit{\pi t}\:\mathrm{=}\:\mathrm{sin\: \mathit{\omega _{m\mathrm{1}}\mathit{t}}\: is\: 3000\:\mathit{\pi}}$.
Highest frequency component in term $\mathrm{cos\: 5000}\mathit{\pi t}\:\mathrm{=}\:\mathrm{cos\: \mathit{\omega _{m\mathrm{2}}\mathit{t}}\: is\: 5000\: \mathit{\pi}}$.
Therefore, the maximum frequency component in the signal $\mathit{x\mathrm{\left ( \mathit{t} \right )}}$ is $\mathit{\omega_{m}\:\mathrm{=}\:\mathit{\omega _{m\mathrm{2}}}\mathrm{=}\mathrm{5000}\mathit{\pi}}$,i.e., highest of 0, 3000$\mathit{\pi}$ and 4000 $\mathit{\pi}$.
Thus, the maximum frequency component present in the signal is given by,
$$\mathrm{\mathit{f_{m}\mathrm{=}\frac{\mathit{\omega _{m}}}{\mathrm{2}\pi }}\mathrm{=}\frac{5000}{2\mathit{\pi}}\:\mathrm{=}\:2500\:\mathrm{HZ}}$$
Hence, the Nyquist rate of sampling is,
$$\mathrm{\mathrm{Nyquist \:rate,\mathit{f_{N}}\mathrm{=}2\mathit{f_{m}}}}$$
$$\mathrm{\Rightarrow \mathit{f_{N}\:\mathrm{=}\:\mathrm{2\times 2500}\:\mathrm{=}\:\mathrm{5000\:HZ}}}$$
And the Nyquist interval is,
$$\mathrm{\mathrm{Nyquist \:interval\:\mathrm{=}\:\frac{1}{\mathit{f_{N}}}\mathrm{=}\frac{1}{2\mathit{f_{m}}}}}$$
$$\mathrm{\Rightarrow \mathrm{Nyquist\: interval\:\mathrm{=}\:\frac{1}{5000}}\:\mathrm{=}\:0.2\:\mathrm{ms}}$$
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