Aperture effect correction; bandpass sampling 80–115 kHz

Aperture effect

In flat-top sampling each sample is held for a finite pulse width $\tau$. This is equivalent to passing ideal samples through a filter with impulse response $h(t) = \text{rect}(t/\tau)$:

$$S(f) = f_sH(f)\sum_n G(f - nf_s), \qquad H(f) = \tau\,\text{sinc}(f\tau)e^{-j\pi f\tau}$$

Since $|H(f)|$ falls as $f$ rises, the higher message frequencies are attenuated (and delayed by $\tau/2$). This amplitude distortion is the aperture effect; the larger $\tau$, the worse it is.

Correction

1. Equalizer: after the reconstruction LPF, use a filter with

$$H_{eq}(f) = \frac{1}{H(f)} = \frac{1}{\tau\,\text{sinc}(f\tau)}, \quad |f| \le W$$

which boosts high frequencies to cancel the sinc droop. 2. Small duty cycle: keep $\tau \ll T_s$; for $\tau/T_s \le 0.1$ the droop at the band edge is under about 0.5%, so equalization is often unnecessary.

Band-pass sampling of 80–115 kHz

$f_L = 80$ kHz, $f_H = 115$ kHz, $B = f_H - f_L = 35$ kHz.

Band-pass sampling condition:

$$\frac{2f_H}{k} \le f_s \le \frac{2f_L}{k - 1}, \qquad k_{max} = \left\lfloor\frac{f_H}{B}\right\rfloor$$ $$\begin{aligned} k_{max} &= \left\lfloor\frac{115}{35}\right\rfloor = \lfloor 3.29 \rfloor = 3 \\ f_{s,min} &= \frac{2f_H}{k} = \frac{2 \times 115}{3} = 76.67\ \text{kHz} \\ \text{upper limit} &= \frac{2f_L}{k - 1} = \frac{2 \times 80}{2} = 80\ \text{kHz} \end{aligned}$$

So any $f_s$ from 76.67 to 80 kHz works, far below the low-pass Nyquist rate $2f_H = 230$ kHz.

Answer: minimum sampling frequency ≈ 76.67 kHz. (If treated as a low-pass signal, 230 kHz would be needed.)