Reconstruction methods; Nyquist of product

Methods of reconstruction

Reconstruction (interpolation) means rebuilding the continuous signal $x(t)$ from its samples $x(nT)$.

1. Ideal (band-limited) interpolation: pass the impulse-sampled signal through an ideal LPF with gain $T$ and cutoff $\omega_c = \omega_s/2$:

$$x_r(t) = \sum_{n=-\infty}^{\infty} x(nT)\,\frac{\sin\!\big(\omega_c(t-nT)\big)}{\omega_c(t-nT)}\cdot\frac{\omega_c T}{\pi}$$

Each sample is replaced by a sinc pulse; the result equals $x(t)$ exactly when $\omega_s > 2\omega_M$. It is not realizable because the ideal LPF is non-causal. 2. Zero-order hold (ZOH): each sample is held constant until the next sample, giving a staircase. Impulse response $h_0(t) = 1$ for $0 \le t < T$. Simple (used in DACs) but adds distortion, so a smoothing (reconstruction) filter follows it. 3. First-order hold (linear interpolation): adjacent samples are joined by straight lines. Impulse response is a triangle of width $2T$. Smoother than ZOH but still approximate.

ZOH:   _    __          FOH:    /\
      | |__|  |__              /  \__/\
      staircase                straight lines

Nyquist rate of $x(t) = \frac{1}{2\pi}\cos(200\pi t)\cos(300\pi t)$

Use $\cos A\cos B = \tfrac12[\cos(A-B) + \cos(A+B)]$:

$$\begin{aligned} x(t) &= \frac{1}{2\pi}\cdot\frac12\big[\cos(100\pi t) + \cos(500\pi t)\big] \\ &= \frac{1}{4\pi}\cos(100\pi t) + \frac{1}{4\pi}\cos(500\pi t) \end{aligned}$$

$f_M = 250$ Hz.

$$f_N = 2 f_M = 500\ \text{Hz}$$

Answer: Nyquist rate = 500 Hz (500 samples/s, $\omega_N = 1000\pi$ rad/s).