Sampling 3cos200πt + 2sin300πt − 3cos500πt

Frequencies in $x(t)$ ($\omega = 2\pi f$):

(i) Minimum sampling rate

$f_m = 250$ Hz, so

$$F_{s,\min} = 2f_m = 2\times250 = 500\ \text{Hz}$$

Answer: 500 samples/s (Nyquist rate).

(ii) Discrete-time signal for $F_s = 400$ Hz

Put $t = nT = n/F_s = n/400$:

$$\begin{aligned} x[n] &= 3\cos\left(\frac{200\pi n}{400}\right) + 2\sin\left(\frac{300\pi n}{400}\right) - 3\cos\left(\frac{500\pi n}{400}\right) \\ &= 3\cos\left(\frac{\pi n}{2}\right) + 2\sin\left(\frac{3\pi n}{4}\right) - 3\cos\left(\frac{5\pi n}{4}\right) \end{aligned}$$

The last term has digital frequency $5\pi/4 > \pi$, so it is aliased. Since $\cos\left(\frac{5\pi n}{4}\right) = \cos\left(2\pi n - \frac{3\pi n}{4}\right) = \cos\left(\frac{3\pi n}{4}\right)$:

$$x[n] = 3\cos\left(\frac{\pi n}{2}\right) + 2\sin\left(\frac{3\pi n}{4}\right) - 3\cos\left(\frac{3\pi n}{4}\right)$$

In normalized frequency: $\frac{100}{400} = \frac14$, $\frac{150}{400} = \frac38$, and $\frac{250}{400} = \frac58$, which aliases to $1 - \frac58 = \frac38$ (i.e. 150 Hz).

(iii) Reconstructed signal

An ideal reconstruction filter passes only $|f| \le F_s/2 = 200$ Hz, so each digital frequency $\omega$ in $[0, \pi]$ maps back to $f = \frac{\omega F_s}{2\pi}$:

$$x_r(t) = 3\cos(200\pi t) + 2\sin(300\pi t) - 3\cos(300\pi t)$$

Answer: $x_r(t) = 3\cos 200\pi t + 2\sin 300\pi t - 3\cos 300\pi t$.

Because $F_s = 400$ Hz is below the Nyquist rate (500 Hz), the 250 Hz component has been aliased to 150 Hz, so $x_r(t) \ne x(t)$.