PAM, PWM, PPM; PCM of 4.8 MHz TV signal with 512 levels

PAM, PWM and PPM

• PAM (Pulse Amplitude Modulation): the amplitude of regularly spaced pulses varies with the sample value; width and position are fixed.
• PWM (Pulse Width Modulation): the width (duration) of each pulse varies with the sample value; amplitude and leading edge are fixed.
• PPM (Pulse Position Modulation): the position of each pulse shifts from its nominal time in proportion to the sample; amplitude and width are fixed. It is usually derived from the trailing edge of PWM.

      small       medium      large       medium

PAM                           +--+
                  +--+        |  |        +--+
      +--+        |  |        |  |        |  |
______|  |________|  |________|  |________|  |__________

PWM   +-+         +---+       +------+    +---+
      | |         |   |       |      |    |   |
______| |_________|   |_______|      |____|   |_________

PPM     ++            ++             ++       ++
        ||            ||             ||       ||
________||____________||_____________||_______||________
      0           Ts          2Ts         3Ts

For the same four samples: PAM pulse heights follow the samples, PWM pulses get wider as the sample grows, and each PPM pulse sits at the trailing edge of the PWM pulse, so it moves later as the sample grows.

Numerical

Given: $W = 4.8$ MHz, $L = 512$, sampling at the Nyquist rate.

i) Code word length

$$n = \log_2 512 = 9\ \text{bits}$$

ii) Transmission bandwidth (minimum, Nyquist)

$$\begin{aligned} f_s &= 2W = 9.6\ \text{MHz} \\ B_T &= \frac{nf_s}{2} = nW = 9 \times 4.8 = 43.2\ \text{MHz} \end{aligned}$$

iii) Final bit rate

$$R_b = nf_s = 9 \times 9.6 \times 10^6 = 86.4\ \text{Mbps}$$

iv) Output SQNR (full-load sinusoid)

$$\begin{aligned} \text{SQNR} &= 1.5L^2 = 1.5 \times 512^2 = 393\,216 \\ \text{SQNR}_{dB} &= 1.76 + 6.02 \times 9 = 55.95\ \text{dB} \end{aligned}$$

(Using $3L^2$: 58.96 dB.)

Answer: $n = 9$ bits; $B_T = 43.2$ MHz; $R_b = 86.4$ Mbps; SQNR ≈ 55.95 dB.