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Modulation

Modulation. Suppose we take the product of two signals:. In general, for any function X(f) , we have. In general, for DSB-SC, we have. X. m(t). x c (t). cos w c t. Phase Modulation. Equating the real and the imaginary parts, we have. A general sketch of the spectrum is shown below.

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Modulation

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  1. Modulation

  2. Suppose we take the product of two signals:

  3. In general, for any function X(f), we have

  4. In general, for DSB-SC, we have X m(t) xc(t) coswct

  5. Phase Modulation

  6. Equating the real and the imaginary parts, we have

  7. A general sketch of the spectrum is shown below. Xc(f) J0(kp) J1(kp) J-1(kp) J-2(kp) J2(kp) f fc-2fm fc-fm fc fc+fm fc+2fm

  8. Frequency Modulation

  9. Instead of we will have

  10. The reason for this change can be seen by finding the corresponding instantaneous frequency: We see that the frequency varies about a nominal value wc by kfm(t).

  11. We may now compute the instantaneous phase:

  12. The quantity kf/wm is called the modulation index and is denoted by the letter b.

  13. Noise

  14. The power spectral density of the noise is the power per Hzof the noise. Dimensionally, Power = (Power Spectral Density) x (Bandwidth). In general, when the power spectral density is not constant, we can find the power from the power spectral density using

  15. Parseval’s Theorem for Fourier Series Given a signal described by a Fourier series, the power can be found from its Fourier series coefficients by using something called Parseval’s Theorem for Fourier series.

  16. It can be shown that (exercise for the student): This function acts like a selector function sifting out those terms for which n=m. So,

  17. Parseval’s Theorem for Fourier Transforms Parseval’s theorem for Fourier transforms uses Fourier transformsto calculate the energy in a signal. Energy is equal to power times time. If the power is not constant, energy is equal to the integral of the power over time.

  18. The Quadrature Decomposition of Noise Very often noise is added to a modulated signal, that is a signal whose frequency components are centered about some high frequency fc. The corresponding noise is usually passed through a band-pass filter in the demodulation process. The noise of interest is a band-pass signal.

  19. A bandpass signal can be represented as a modulated form of lowpass signals. The lowpass signals are, effectively, upconvertedto bandpass signals. If we have a bandpass noise signal n(t), it can be represented in terms of lowpass signals nc(t)and ns(t):

  20. The power in each the in-phase and quadrature components,nc(t), ns(t), are the same as that of the noise itself, n(t).

  21. The power spectral density of n(t) is proportional to the square of its spectrum. Sn(f) A2/2 f B The power in n(t) can be calculated by integrating the power spectral density.

  22. P = 2 [ B•A2/2] = BA2. If we compute the power spectral density and the power of nc(t)and ns(t) we get the same value: Snc(f), Sns(f) A2 f B

  23. Noise in Modulation

  24. The calculation of the signal-to-noise improvement (SNRI) is shown in the following diagram. so(t) si(t) Demodulator ni(t) no(t)

  25. The notation < > means time average.

  26. Baseband Digital Transmission

  27. Let s(t) be a noiseless digital signal (the transmitted signal). Let n(t) be additive Gaussian noise. The signal n(t) is a random process as a function of time. At each instant in time, n(t) is equal to a is a random variable n which has a Gaussian distribution. The received signal (with noise) is

  28. The noise random variable n has a Gaussian probability density function with a variance of s2. [We have (n/s) in place of n and an extra s term in the denominator.]

  29. Example: Suppose we were transmitting TTL with Gaussian noise whose variance is unity. Plot the distribution functions for the received signal. Solution:The probability density functions would be

  30. The probability that the noise is greater than 2.5 volts can be found by integrating the probability density function for the noise.

  31. We can formally solve this unequal logic level likelihood problem by starting with the following: In words, if the probability of a logic one is greater than the probability of a logic zero (given received voltage values, logic level likelihoods, et al.), then we decide in favor of a logic zero. Likewise for logic one.

  32. Inserting these expressions into our original decision rule, we have Or,

  33. To find-out the BER, we use the theorem of total probability:

  34. Modulated Digital Transmission

  35. 1 0 1 1 0 1

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

  37. 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 - - - -

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

  39. 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 00 10

  40. 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.

  41. 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.

  42. Now, if what would ds(t) and s be?

  43. So,

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

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