Question

Find the nyquist rate (minimum sampling rate) for each of the following signals.

a) ,x(t) = 2sinc (10t)
b) , x(t) = 3sinc² (100t)
c) , x(t) = 4 rect (10t)
d) . x(t) = -5 sin(20πt)cos(30πt)

Answer 1

a) 20 Hz

b) 200 Hz

c) 20 Hz

d) 100 Hz

The Nyquist rate (minimum sampling rate) for each signal is twice the highest frequency component: 20 Hz, 200 Hz, 20 Hz, and 100 Hz, respectively.

The Nyquist rate (minimum sampling rate) for a signal is twice the highest frequency component in that signal. Let's analyze each signal to find the Nyquist rate:

a)
`\(x(t) = 2\text{sinc}(10t)\)`

- The highest frequency component in
`\(x(t)\) is \(10\) Hz.`

- Nyquist rate
`\(= 2 * 10 = 20\) Hz.`

b)
`\(x(t) = 3\text{sinc}^2(100t)\)`

- The highest frequency component in
`\(x(t)\) is \(100\) Hz.`

- Nyquist rate
`\(= 2 * 100 = 200\) Hz.`

c)
`\(x(t) = 4 \text{ rect}(10t)\)`

- The rectangular pulse function
`\(\text{rect}(t)\)` has infinite bandwidth, but it is typically defined over a certain duration.

- Assuming the duration is finite, the highest frequency component is
`\(10\) Hz.`

- Nyquist rate
`\(= 2 * 10 = 20\) Hz.`

d)
`\(x(t) = -5 \sin(20\pi t)\cos(30\pi t)\)`

-
`The product of \(\sin(20\pi t)\) and \(\cos(30\pi t)\)`results in a signal with frequency components at the sum and difference of the individual frequencies.

- The highest frequency components are
`\(50\) Hz and \(10\) Hz.`

- Nyquist rate
`\(= 2 * 50 = 100\)` Hz.

In summary:

a) 20 Hz

b) 200 Hz

c) 20 Hz

d) 100 Hz