Natural and flat top sampling; Nyquist rate of sin²(400πt)

Sampling

Sampling is the process of converting a continuous-time signal into a discrete-time signal by taking its values at regular intervals $T_s$ (sampling rate $f_s = 1/T_s$). It is the first step of analog-to-digital conversion.

Natural sampling

The signal is multiplied by a train of rectangular pulses of width $\tau$; during each pulse the top follows the signal.

$$s(t) = g(t)c(t), \qquad S(f) = \frac{\tau}{T_s}\sum_n \text{sinc}(nf_s\tau)\,G(f - nf_s)$$

Every spectral copy has the exact shape of $G(f)$, so a LPF recovers $g(t)$ without distortion.

Flat-top sampling

The instantaneous sample value is held constant for width $\tau$ (sample-and-hold):

$$s(t) = \sum_n g(nT_s)h(t - nT_s), \qquad S(f) = f_s H(f)\sum_n G(f - nf_s)$$

with $H(f) = \tau\,\text{sinc}(f\tau)e^{-j\pi f\tau}$. The copies are multiplied by $H(f)$, causing aperture effect (high-frequency loss), corrected by an equalizer $1/H(f)$.

 Signal      .-~~~-.
           .'       '.
 Natural  |‾\ /‾\ /‾‾\|   tops follow curve
 Flat-top |‾| |‾| |‾‾|    tops flat (held)
          Ts  2Ts 3Ts

Nyquist rate and interval of $m(t) = \sin^2(400\pi t)$

Using $\sin^2\theta = \tfrac{1}{2}(1 - \cos 2\theta)$:

$$m(t) = \frac{1}{2} - \frac{1}{2}\cos(800\pi t)$$

It has a dc term and a single tone at $f = 800\pi/(2\pi) = 400$ Hz, so $f_{max} = 400$ Hz.

$$\begin{aligned} f_N &= 2f_{max} = 800\ \text{Hz} \\ T_N &= \frac{1}{800} = 1.25\ \text{ms} \end{aligned}$$

Answer: Nyquist rate = 800 Hz, Nyquist interval = 1.25 ms.