Nyquist rate and interval for 6cos50πt + 20sin300πt − 10cos100πt

Nyquist sampling theorem

A band-limited signal with no frequency components above $f_m$ Hz is completely described by, and can be exactly recovered from, its samples taken uniformly at a rate

$$f_s \ge 2f_m$$

The minimum rate $f_s = 2f_m$ is the Nyquist rate and $T_s = 1/(2f_m)$ is the Nyquist interval. Recovery is done by passing the samples through an ideal low-pass filter of cut-off $f_m$. If $f_s < 2f_m$, spectral copies overlap (aliasing) and recovery is impossible.

Nyquist rate and interval of the given signal

$$x(t) = 6\cos 50\pi t + 20\sin 300\pi t - 10\cos 100\pi t$$

Frequencies of the components ($\omega = 2\pi f$):

Highest frequency: $f_m = 150$ Hz.

$$\text{Nyquist rate } f_N = 2f_m = 2 \times 150 = 300\ \text{Hz (samples/s)}$$ $$\text{Nyquist interval } T_N = \frac{1}{f_N} = \frac{1}{300} = 3.33\ \text{ms}$$

Effect of quantizing with 512 levels

$$n = \log_2 512 = 9\ \text{bits/sample}$$

At the Nyquist rate the bit rate is

$$R_b = n f_N = 9 \times 300 = 2700\ \text{bits/s}$$

and the minimum transmission bandwidth is $R_b/2 = 1350$ Hz.

Answer: Nyquist rate = 300 Hz, Nyquist interval = 3.33 ms (9 bits/sample, 2.7 kbps).