Five television signals (video and audio) with a bandwidth of 4.5 MHz each, are
ID: 2078956 • Letter: F
Question
Five television signals (video and audio) with a bandwidth of 4.5 MHz each, are to be transmitted by binary PCM. The signals are sampled at 25% above the Nyquist rate. Samples are quantized into 1024 levels. In every frame, framing and synchronization requires an additional 4% extra bits. A PCM encoder is used to convert these signals before they are time-multiplexed into a single data stream. Determine the sampling rate and the sampling interval for each channel. At the quantizer, how many bits are used to represent each sample. In every frame, how many bits are there? Determine the data rate of the multiplexed signal. Determine the minimum required bandwidth of baseband communication.Explanation / Answer
Solution:-
a) Sampling rate: =1.25*Nyquist rate=1.25*2*4.5M= 11.25 M samples /sec
Sampling interval: =1/sampling rate= 1/(11.25 M)= 88.88889 n sec
b) Log2 1024=10
c) 5*(10)*1.04=52 bits
d) Date rate (bits/sec)= 52 bits/sampling interval(sec)=52 bits* sampling rate
= (52)(11.25M)=585 M bits/sec
e) Min baseband BW= Rate/2=585 M /2 =292.5 M HZ
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