Communication Math

Khaled Mahmud

Source Coding, PCM, Sampling

Q1.

An analog signal source is sampled at a rate of 16000 samples/sec and the sampled values are quantized and encoded with 12 bits.

(a) How many bytes will be required to store 2 seconds of signal?

(b) If you need to transmit the sampled data continuously, what is the required data rate?

Solution

Given,

$ R_{Sampling}=16000\ sample/sec \\ \frac{Bit}{Sample}=\ 12 $

(a)

$ \begin{aligned} Recorded\ data,\ D\\ &=(Time)\ x\ (Sampling\ rate)\ x\ \frac{bit}{sample}\\ &=2x16000x12\ bit \\ &=384000\ bit\\ &=48000\ byte \end{aligned} $

(b)

$ \begin{aligned} R_b&=(Sampling\ rate)x(Bit\ per\ sample)\\ &=16000x12\\ &=192\ kbps \end{aligned} $

Q2.

After analyzing the spectrum of a signal you determine that the frequency contents of the signal lies between 2.5 kHz and 25 kHz. (a) What sampling rate would you suggest for A/D conversion of this signal? (b) What is the bandwidth of the signal?

Solution

Given,

$ Lowest\ frequency,\ F_L=2.5\ kHz $

$ Highest\ frequency,\ F_H=25\ kHz $

(a)

By Nyquist theorem,

$ \begin{aligned} R_{sampling}\\ &=2xF_H\\ &=2x25\\ &=50\ ksample/sec \end{aligned} $

(b)

$ \begin{aligned} Bandwidth,\ B\\ &=(F_H-F_L)\\ &=25-2.5\\ &=22.5\ kHz \end{aligned} $

Q3.

In circuit switched telephony, one toll quality voice channel is carried by 64 kbps data channel. Explain how this number is derived.

Solution

Assumtions of basic circuit switched telephony:

By Nyquest theorem,

Required sampling rate,

$ \begin{aligned} R_{sampling} &= 2x 4000\\ &= 8000\ sample/sec \end{aligned} $

Required data rate,

$ \begin{aligned} R &= (sampling\ rate)x(bit\ per\ sample)\\ &=8000 \frac{sample}{sec}\ x\ 8 \frac{bit}{sample}\\ &= 64000\ \frac{bit}{sec}\\ &= 64\ kbps \end{aligned} $

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