---
title: "PAM, PWM, PPM; PCM of 4.8 MHz TV signal with 512 levels"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:pam-pwm-ppm-tv-4-8mhz"
source: https://www.bctengineer.com/q/semester_6/communication_systems/pam-pwm-ppm-tv-4-8mhz
site: BCT Engineering
---

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

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2070 Chaitra (CS II), 1.5+1+1.5+1.5+1.5 marks
- Video: [PCM Examples | Pulse Code Modulation Solved Problems | Digital Communication | Engineering Funda](https://www.youtube.com/watch?v=MMbu5CXVxIY) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/pam-pwm-ppm-tv-4-8mhz
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

Define PAM, PWM and PPM with corresponding waveforms. A Television signal having a bandwidth of 4.8 MHz is transmitted using binary PCM system. Given that the number of quantization levels is 512. Determine:
i) Code word length
ii) Transmission bandwidth
iii) Final bit rate
iv) Output signal to quantization noise ratio

## Short answer

PAM varies pulse amplitude, PWM varies pulse width, PPM varies pulse position. For TV signal: code word length 9 bits, transmission bandwidth 43.2 MHz, bit rate 86.4 Mbps, SQNR 55.95 dB.

## Answer

#### 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.

```text
      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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In PWM the quantity that varies with the sample is:
   A. Pulse amplitude
   B. Pulse width
   C. Pulse position
   D. Pulse count
   Answer: B. PAM varies amplitude, PWM width, PPM position. (Key point: Pulse modulation types)

2. For 512 levels the code word length is:
   A. 8
   B. 9
   C. 10
   D. 512
   Answer: B. n = log2 512 = 9 bits. (Key point: Code word length)

3. With W = 4.8 MHz and n = 9, the bit rate is:
   A. 43.2 Mbps
   B. 86.4 Mbps
   C. 9.6 Mbps
   D. 172.8 Mbps
   Answer: B. fs = 9.6 MHz, Rb = 9 x 9.6 = 86.4 Mbps. (Key point: Bit rate)

4. Minimum transmission bandwidth is:
   A. 86.4 MHz
   B. 4.8 MHz
   C. 43.2 MHz
   D. 9.6 MHz
   Answer: C. BT = Rb/2 = nW = 43.2 MHz. (Key point: Bandwidth)

## Related in Communication Systems

- [E1 hierarchy; PCM of 4.2 MHz TV signal with 512 levels](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ae1-pcm-tv-4-2mhz-512) — Communication Systems, chapter 4
- [PCM bit rate for 4.5 MHz TV signal with 1024 levels](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-tv-4-5mhz-1024) — Communication Systems, chapter 4
- [Short note: PAM, PWM and PPM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashort-note-pam-pwm-ppm) — Communication Systems, chapter 4