Nyquist criteria, aliasing; 30sin100πt + …

Nyquist criterion

For a signal band-limited to $f_m$ Hz, the sampling frequency must satisfy

$$f_s \ge 2f_m$$

The minimum value $f_N = 2f_m$ is the Nyquist rate and $T_N = 1/(2f_m)$ is the Nyquist interval. When this holds, the spectral copies produced by sampling, $X_s(f) = f_s\sum_k X(f - kf_s)$, do not overlap, and an ideal low-pass filter of cutoff $f_m$ to $f_s - f_m$ recovers the original signal exactly.

Aliasing effect

If $f_s < 2f_m$, the copies at $0$ and $\pm f_s$ overlap. A component of frequency $f > f_s/2$ then appears at the alias frequency $|f - kf_s|$ (the nearest one in $0$ to $f_s/2$). Example: 900 Hz sampled at 1200 Hz looks like 300 Hz. The overlap cannot be undone after sampling. It is prevented by an anti-aliasing low-pass filter before the sampler and by choosing $f_s > 2f_m$.

 fs >= 2fm:  /\     /\     /\        separate copies
           -+--+---+--+---+--+--> f
              0       fs     2fs
 fs < 2fm:   /\/\/\/\/\              copies overlap
           -+----+----+---> f
            0    fs   2fs

Nyquist rate of $x(t) = 30\sin(100\pi t) + 40\cos(400\pi t)\sin(300\pi t)$

Convert the product using $\sin A\cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$ with $A = 300\pi t$, $B = 400\pi t$:

$$\begin{aligned} 40\cos(400\pi t)\sin(300\pi t) &= 20\left[\sin(700\pi t) + \sin(-100\pi t)\right] \\ &= 20\sin(700\pi t) - 20\sin(100\pi t) \end{aligned}$$

So

$$x(t) = 10\sin(100\pi t) + 20\sin(700\pi t)$$

Highest frequency $f_m = 350$ Hz.

$$f_N = 2f_m = 700\ \text{Hz}, \qquad T_N = \frac{1}{700} = 1.43\ \text{ms}$$

Answer: Nyquist rate $= 700$ samples/s ($\omega_N = 1400\pi$ rad/s); Nyquist interval $\approx 1.43$ ms.