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A sinusoidal signal with amplitude 20 V and bandwidth 35 kHz is sampled at 2.5 times its Nyquist rate and then quantized using a 128-level quantizer. Determine:

i) The resulting data bit rate?

The Bit rate= 2Wm=2(35K)(7)(2.5)=1.225M bit/s

ii) The Signal to Quantization Noise Ratio (SQNR) in dBs of the resulting sampled signal assuming the sinusoidal signal amplitude covers all levels of the quantizer?

I know the power of the signal , But how to get the noise power.?

iii) The minimum required baseband channel bandwidth needed for transmitting the digital bits assuming that Manchester line coding is used.

The required bandwidth is 2R= 2(1.225M)

Can someone help me solving this problem ?

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  • \$\begingroup\$ A sine wave has a band width of zero theoretically. It may have a frequency of 35 kHz of course. \$\endgroup\$ – Andy aka Dec 12 '16 at 11:21
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I know the power of the signal, but how to get the noise power?

The quantization noise can be approximated as a sawtooth wave that has a peak-to-peak amplitude equal to the step size of the quantizer. You just need to calculate the RMS voltage of this sawtooth wave and compare it to the RMS value of the sinewave signal.

BTW, this part of the problem really has nothing to do with Nyquist, so your title is a bit misleading.

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  • \$\begingroup\$ The effective SNR of in-band signal is decreased as the sample rate exceeds Nyquist so it does improve SNR. Just offhand I'd guess about 4 dB. \$\endgroup\$ – doug Dec 12 '16 at 6:13
  • \$\begingroup\$ @doug: That improvement occurs only after a subsequent filtering stage (in either the digital or analog domain) -- a step not mentioned in the problem. \$\endgroup\$ – Dave Tweed Dec 12 '16 at 11:49
  • \$\begingroup\$ of course, but it is an important aspect of oversampling, trivial in cost, and would be a part of any actual design unless one already had significant excess SNR. I would expect this would be discussed by the instructor when reviewing the problem. \$\endgroup\$ – doug Dec 12 '16 at 15:27

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