Question

In a telemetry system, eight message signals having 2 kHz bandwidth, each are time division multiplexed using a binary PCM. The sampling rate is 25% above the Nyquist rate and the error in sampling amplitude cannot be greater than 1% of the peak to peak amplitude. If raised cosine pulses with roll-off factor $a=0.05$ are used then

• A. the value of minimum number of bits 'n' is equal to 6
• B. the bandwidth of the signal is equal to 240 kHz
• C. the sampling rate is equal to 4 kHz
• D. the minimum bit rate the can be achieved is equal to 0.24 Mbps

Answer 1

Answer: (A), (D)

the minimum bit rate the can be achieved is equal to 0.24 Mbps

Error $=\frac{\Delta }{2}=\frac{V_{PP}}{2L}=0.01V$
Sicne error $\le \frac{V_{PP}}{100}$
So, $\frac{V_{PP}}{{2.2}^n}\le \frac{V_{PP}}{100}$
$\Rightarrow n\ge 5.64$ and $n_{min}=6$
$f_s=1.25f_n=1.25\times 2\times 2\times {10}^3=5kHz$
Thus, the bit are $R_b=nN{f}_s=8\times 6\times 5\times {10}^3=240kbps$
Therefore, the bandwidth is given by
$BW=(1+\alpha ).\frac{R_b}{2}=(1+0.05)\times \frac{240}{2}kbps=126kHz$