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Modulated Digital Transmission

Modulated Digital Transmission. Digital modulation is the process of using digital information to alter or modulate the amplitude , phase or frequency of a sinewave.

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Modulated Digital Transmission

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  1. Modulated Digital Transmission

  2. Digital modulation is the process of using digital information to alter or modulate the amplitude, phase or frequency of a sinewave. As an example, suppose we wished to use digital information to modulate the amplitude of a sinewave. The result is called On-Off Keying (OOK) and is shown on the following page.

  3. 1 0 1 1 0 1

  4. To generate an OOK signal, we simply multiply the digital (baseband) signal by the unmodulated carrier. X

  5. In a variation of OOK, we multiply the sinewave by a bipolar or antipodal version of the digital waveform: 1 0 1 1 0 1 + + + + 0 volts - - - -

  6. The resultant modulated waveform is actually a form of phase-modulation called BPSK. X

  7. BPSK has two phases: 0° and 180° corresponding to logic one and logic zero respectively. logic one 0° logic zero 180°

  8. Just as we can have two phases with BPSK, we can have four phases with QPSK. 90° 0° 180° 270°

  9. Since we have four phases, we cannot simply assign these phases to logic one and logic zero. Instead, we assign each of the four phases to pairs of bits. 01 11 11 10

  10. Thus, we modulate the following digital waveform using QPSK: 1 0 1 1 0 1

  11. The digital modulation processes are fairly simple: simply multiply the bits by sinewaves (or cosinewaves). In the case of QPSK, we multiply the odd bits by sinewaves and the even bits by cosinewaves and add them together. The demodulation processes are very similar to that of DSB-SC AM. In DSB-SC AM, we multiplied the modulated carrier by a sinewave (at the carrier frequency) and then low-pass filtered the product.

  12. LPF X xc(t) cos wct The digital demodulation process is very similar, except that, instead of a low-pass filter, we use an integrator.

  13. digitally-modulated signal X We must also interpret the output of the integrator appropriately. The value of the output will determine if input signal corresponds to a one or a zero.

  14. Let s(t) be the modulated signal. For OOK, we have The digital demodulator would look like the following

  15. ds(t) X s(t) s cos wct The products(t) cos wct is denoted by ds(t) We denote the output of the integrator by s.

  16. Now, if what would dn(t) and s be?

  17. Now the integral is taken over a period T corresponding to the time to transmit a single bit. The bit period T is also an integral multiple of periods of wc. T

  18. If this is true, we have

  19. So,

  20. Now, we need to interpretthis integrator output. To do this interpretation, all we need to do is look at what the demodulatordoes: it converts a modulated sinewave to one of two values: {T/2, 0}. The resultant output is just like a baseband digital signal. To interpret this digital signal, we simply set a threshold (typically at the halfway point): anything above this threshold is considered to be a logic one, and anything below this threshold is considered to be a logic zero.

  21. So, for OOK output s, we perform the following comparison:

  22. We can use the same demodulator for BPSK: ds(t) X s(t) s cos wct where

  23. The product and the output of the integrator become

  24. Or,,

  25. The interpretation or comparison of s for BPSK becomes

  26. We now have detection thresholds for the outputs of digital demodulators. The next step is to determine the bit error-rates (BER’s) for these demodulators/detectors. We can determine the bit error-rates in much the same way that we determined the bit error-rates for baseband digital transmission, reception and detection.

  27. For baseband digital transmission, we took the bit error-rate to be where d is the distance between either logic level and the detection threshold (e.g, 2.5 volts), and where sn is the standard deviation of the noise (the square-root of the variance of the noise).

  28. For modulated digital transmission and reception the formula is much the same: where d is the distance between detected logic level (e.g., T/2) and the threshold (e.g, T/4), and where snd is the standard deviation of the detected noise.

  29. All we need to do is find the detected noise variancesnd. In order to best find the detected noise variance (and determine the BER), we will first make a small change in the modulation and demodulation of BPSK and QPSK: we will change the carrier from to

  30. The modulated signal for BPSK becomes We will also change the signal for the local oscillator in the demodulator as well.

  31. ds(t) X s(t) s (2/T) cos wct Given the new values for s(t), and the new demodulator, the new values for ds(t) and s become

  32. Since we now have The detection threshold is now at ½.

  33. For BPSK and the same demodulator, we have The detection threshold remains at zero.

  34. Now what happens when we pass noise through such a demodulator. dn(t) X n(t) n (2/T) cos wct

  35. The output of the demodulator is The variance of n is the average of the square of n.

  36. Now let us examine

  37. This quantity, the average of the product of a function with itself at a different time, is called the autocorrelation of n(t). We denote the autocorrelation of n(t) by the letter R. If the noise is something called wide-sense stationary, the autocorrelation is only dependent upon the time difference between t and s.

  38. As it turns out, the Fourier transform of the autocorrelation is the power spectral density (this is the Wiener Khinchine theorem). When we take the Fourier transform of Rn(t), we get the power spectral density Sn(f).

  39. The nature of depends upon the nature of the power spectral density of n(t). Suppose n(t) is Gaussian white noise with power spectral density

  40. The inverse Fourier transform of a constant is an impulse function: So,

  41. So the variance of the detected noise becomes Using the sifting property of delta functions, we have

  42. We can now calculate the bit error rate for OOK and BPSK: OOK BPSK

  43. Now, let us do another variation of OOK and BPSK: OOK BPSK

  44. We have introduced a factor A into the amplitude of the signals. We shall update the demodulator appropriately: ds(t) X s(t) s A(2/T) cos wct

  45. The outputs of the demodulator become OOK BPSK

  46. The bit error rates are now OOK BPSK

  47. Finally, let us let A = Es. OOK BPSK

  48. The bit error rates are now OOK BPSK

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