Q.1 A 30 channel TDM PCM telephone system transmits voice (band-limited 4KHz) wi
ID: 2265666 • Letter: Q
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Q.1 A 30 channel TDM PCM telephone system transmits voice (band-limited 4KHz) with an encoding of 8 bits per sample. Two framing bits are added to each frame. (a) Determine number of pulses per frame and bits per frame. (b) What are its bit rate and the bit interval? (c) What is the step size of the encoder if full-scale voltage is 5 V? What is its quantization error? (d) What is the minimum bandwidth requirement of the channel? Q.2 Given the message signal 01001011 with a bit rate of 1 Mbps. Sketch the data signal using AMI line coding. What's its minimum transmissiorn bandwidth? Sketch the signal with Manchester coding. Find bandwidth. Sketch the signal with dibit signaling. Find bandwidth.Explanation / Answer
Answer:-1)a) Number of pulse per frame = Number of bits per frame = (30 x 8) + 2 bits = 242 bit/frame
b) Signal is band limited to 4kHz so sampling must be done at least 8 kHz. So frame/sec = 8000, bits/frame = 242, hence bit rate = 242 x 8000 bps = 1.936 Mbps. Thus bit interval = 0.52 micro-s.
c) Step size = 5/28 = 19.53 mV. Quantization Error = step size/2 = 9.765 mV.
d) Minimum bandwidth required is equal to the half of the bit rate. So BW = 1.936/2 MHz = 0.968 MHz.
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