Aperture effect, DPCM and DM slope overload numerical

Aperture effect in flat-top sampling

In flat-top sampling each sample is held for a pulse width $T$. The flat-top signal is the ideally sampled signal convolved with a rectangular pulse $h(t)$ of width $T$:

$$S(f) = f_s\sum_k M(f - kf_s)\,H(f), \qquad H(f) = T\,\text{sinc}(fT)\,e^{-j\pi fT}$$

The message spectrum is multiplied by $\text{sinc}(fT)$, so high frequencies are attenuated. This amplitude distortion (plus a delay $T/2$) is the aperture effect. It is reduced by keeping $T \ll T_s$ (duty ratio ≤ 0.1) or by an equalizer after the reconstruction filter with response $1/H(f)$.

DPCM scheme

PCM sends full samples although adjacent samples are highly correlated. DPCM sends only the quantized prediction error:

m[n]->(+)-e[n]->[Quant.]--+--eq[n]-->[Encoder]--> out
       ^-                 |
     m^[n]               (+)<-+
       |                  |   |
       +--[Predictor]<-mq[n]  |
              +---------------+
Receiver: eq[n]->(+)->mq[n]->[LPF]; feedback via
          the same predictor gives m^[n]

$e[n] = m[n] - \hat m[n]$, $m_q[n] = \hat m[n] + e_q[n] = m[n] + q[n]$. Since $e[n]$ is small, fewer bits are needed, or SQNR improves by the prediction gain.

Numerical: maximum amplitude without slope overload

Given: $W = 3$ kHz, $f_s = 5 \times$ Nyquist rate, $f_m = 2$ kHz, $\Delta = 250$ mV.

$$f_s = 5 \times 2W = 5 \times 6\ \text{kHz} = 30\ \text{kHz}$$

No slope overload requires $2\pi f_m A_m \le \Delta f_s$:

$$A_{m,max} = \frac{\Delta f_s}{2\pi f_m} = \frac{0.25 \times 30000}{2\pi \times 2000} = \frac{7500}{12566.4} = 0.597\ \text{V}$$

Answer: maximum amplitude $\approx 0.597$ V (597 mV).