q2.md

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• Name: Mark Jia
• NSID: mij623
• StuN: 11271998

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Q2 ( 4 )

What is the minimum bandwidth needed to achieve a data rate of B bits/sec if the signal is transmitted using

• NRZ,
• MLT-3, and
• Manchester encoding, respectively? Explain your answer.

A2.1

use $w$ to represent bandwidth to avoid confusion (since $B$ is used as the data rate (in bits/sec))

Nyquist: B = $2w log_2 V$ bits/sec w: bandwidth V: discrete levels

NRZ

binary -> 2 states: V=2

$$ \begin{align} B &= 2w log_2 V \\ &= 2w \times 1 \\ &= 2w\\ w &= \frac{B}{2} \end{align} $$

minimum bandwidth needs to be $\frac{B}{2}$ Hz

MLT-3

Has three possible states => V=3

To encode any 1 bit (information is 1 or 0), it could be one of three states (-1, 0, 1).

Hence, to encoding B bits (2^B of information), it would have (3^B) different encodings in MLT-3.

So B bits as the data would mean $\frac{3^B}{2^B}\times B$ information sent through physical layer, which, $log_2{3^B}=B\times log_2 3$ bits $$ \begin{align} & log_2 3 \times B &= 2w log_2 3 \ & w &= \frac{B}{2} \end{align} $$

The minimum bandwidth needs to be $\frac{B}{2}$ Hz

Manchester

Has 50% coding effieiency (the other half is for clocking info) Hence, to send B bits $=2^B$ information, requires an encoding of 2B bits of data sent through physical.

$$ \begin{align} & 2\times B &= 2w log_2 2\\ & w &= \frac{B}{log_2 2}\\ & w &= B \end{align} $$

Hence, the minimum bandwidth needs to be $B$ Hz.

A2.2 FDM

Ten signals, each requiring 4000 Hz, are multplexed onto a channel using FDM.

What is the minimum bandwidth required for the multiplexed channel? Assume that the guard bands are 400 Hz wide.

FDM

Each chan-> 4 kHz, each guard-> 0.4 kHz

10 signals would need 9 guard bands.

$$ \begin{align} 10 \times 4 \text{ kHz} + 9 \times 0.4 \text{ kHz}\\ &=43.6 \text{ kHz} \end{align} $$

The minimum bandwidth for this multiplexed channel is 43.6 kHz.