Ideal sampling and reconstruction; Nyquist rate of ½π cos(400πt)cos(1000πt)

Ideal sampling

In ideal (impulse) sampling the message $g(t)$, band-limited to $W$, is multiplied by an impulse train of period $T_s$:

$$g_\delta(t) = g(t)\sum_n \delta(t - nT_s) = \sum_n g(nT_s)\,\delta(t - nT_s)$$

The impulse train has the Fourier series $\delta_{T_s}(t) = f_s\sum_n e^{j2\pi nf_st}$, so

$$G_\delta(f) = f_s\sum_{n=-\infty}^{\infty} G(f - nf_s)$$

The spectrum is $G(f)$ repeated every $f_s$ (scaled by $f_s$).

 G(f)        /\
        ____/  \____
           -W   W
 Gd(f) /\    /\    /\
  ____/  \__/  \__/  \___   (fs > 2W)
     -fs    0     fs
          |<-LPF->|

Reconstruction

• If $f_s \ge 2W$, the copies do not overlap. An ideal LPF of gain $T_s$ and cut-off $B$ ($W \le B \le f_s - W$) passes only the $n = 0$ copy, giving back $G(f)$.
• In the time domain, the LPF impulse response is $h(t) = 2BT_s\,\text{sinc}(2\pi Bt)$, where here $\text{sinc}\,x = \sin x/x$. With $f_s = 2W$ and $B = W$, $h(t) = \text{sinc}(2\pi Wt)$, and

$$g(t) = \sum_n g(nT_s)\,\text{sinc}\big(2\pi W(t - nT_s)\big)$$

This is the interpolation formula: each sample is replaced by a sinc pulse, and the sum is exactly $g(t)$.

• If $f_s < 2W$, copies overlap (aliasing) and $g(t)$ cannot be recovered.
• In practice a guard band ($f_s > 2W$) is used since ideal LPFs do not exist.

Nyquist rate of $\tfrac{1}{2\pi}\cos(400\pi t)\cos(1000\pi t)$

(The constant in front only scales amplitude, so it does not affect the answer.)

Using $\cos A\cos B = \tfrac{1}{2}[\cos(A - B) + \cos(A + B)]$:

$$\frac{1}{2\pi}\cos(400\pi t)\cos(1000\pi t) = \frac{1}{4\pi}\big[\cos(600\pi t) + \cos(1400\pi t)\big]$$

Frequencies: $300$ Hz and $700$ Hz, so $f_{max} = 700$ Hz.

$$\begin{aligned} f_N &= 2 \times 700 = 1400\ \text{Hz} \\ T_N &= \frac{1}{1400} = 0.714\ \text{ms} \end{aligned}$$

Answer: Nyquist rate = 1400 Hz, Nyquist interval = 0.714 ms.