32 lines
1.1 KiB
TeX
32 lines
1.1 KiB
TeX
\hsize=4in
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\nopagenumbers
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\parindent=0pt
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3.4 A black and white television signal has a bandwidth of about 4.2 MHz.
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What bit rate is required if this signal is to be digitized with uniform
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PCM at an SQR of 30 dB?
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Use a sampling-rate to Nyquist-rate ratio comparable to that used for PCM
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voice encoding.
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\bigskip
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The Nyquist criterion tells us that the sampling rate must be greater
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than twice the bandwidth. The excess sampling factor for voice PCM is 4000/3400
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(see page 107).
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Therefore the sampling rate is
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$$f_s=2\times4.2\,{\rm MHz}\times{4000\over3400}=9.88\,{\rm MHz}$$
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Equation 3.4 gives us the signal to noise ratio.
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$${\rm SQR}=7.78+20\log_{10}(A/q)$$
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Solve for $A/q$ for an SQR of 30 dB.
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$$\log_{10}(A/q)={30-7.78\over20}=1.11$$
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$$A/q=10^{1.11}=12.9$$
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We see that the quantization interval $q$ must divide the input amplitude
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$A$ into at least 13 intervals.
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Therefore four bits are required to encode the sample.
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In addition, another bit is required to represent the sign of the amplitude.
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(The signal swings between $+A$ and $-A$.)
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The overall bit rate is
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$$5\,{\rm bits}\times9.88\,{\rm MHz}
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=49.4\,{\rm Mbits/sec}$$
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\end
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