HOW DELTA V/2=100? Example 6.2 A signal m(t) band-limited to 3 kHz is sampled at
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HOW DELTA V/2=100?
Example 6.2 A signal m(t) band-limited to 3 kHz is sampled at a rate 33 % higher than the Nyquist rate. The maximum acceptable erroin the sample amplitude (the maximum quantizaon eror) is 0.5% of the peak amplitude my. The quantized samples are binary coded. Find the minimum bandwidth of a channel required to transmit the encoded binary signal. If 24 such signals are time-division-multiplexed. determine the minimum transmission bandwidth required to transmit the muhiplexed signal The Nyquist sampling rate is R 2x 3000 6000 Hz (samples per second). The actual sampling rate is R-6000 x (H) 8000 Hz. The quantization step is , and the maximum quantization error is Av/2. 6.2 Fulse Code Modulation (PCM) 279 Therefore, from Eq. (6.31), For binary coding. L must be a power of 2. Hece, the next higher value of L that is a power of 2 is 256. From Eq. (6.37). we need n = log, 256 = 8 bits per sample. We require to transmit a total of C = 8 x 8000 = 64,000 bit/s. Because we can transmit up to 2 bits per hertz of bandwidth, we require a minimum transmission bandwidth BT = 2 = 32 kHz. The multiplexed signal has a lotal of CM = 24 x 64,000 = 1.536 Mbits, whichExplanation / Answer
Some part of this question is missing, amplitude of message signal should be given
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