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A telecommunication system monitors 10 substations, each containing 5 signal channels. Each channel has a signal bandwidth of 100 Hz. The signals are quantized and encoded. The sampling rate is twice the Nyquist rate. The maximum quantization error is 0.25% of the peak amplitude mp. The required overall bit rate of the system.
Bandwidth per channel = 100 Hz
Sampling rate = 2 Nyquist = 2 2 100 = 400 samples/sec
Quantization error = 0.25% of mp ⇒
For uniform quantizer : = L = 400 levels
L = 2n
n = log2(400) ≈ 8.64
⇒ 9 bits/sample
Bit rate per channel = 400 9 = 3600 bps
Total channels = 10 substations 5 channels = 50
Total bit rate = 50 3600 = 180,000 bps = 180 kbp
A telecommunication system monitors 10 substations, each containing 5 signal channels. Each channel has a signal bandwidth of 100 Hz. The signals are quantized and encoded. The sampling rate is twice the Nyquist rate. The maximum quantization error is 0.25% of the peak amplitude mp. The required overall bit rate of the system.
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