Modulation

1. Modulation

2. Low-frequency messages cannot be sent over an airwave medium. • The low-frequency messages can be converted into high-frequency signals for the purpose of airwave communication. • The “conversion” process of a low-frequency message to a high-frequency signal is usually accomplished by modulatingor changing a high-frequency signal using a low-frequency message. • The high frequency signal can be modulated by changing its amplitude, its frequency or its phase.

3. Amplitude Modulation • Amplitude Modulation (AM) is simply the changing of the amplitude of a high-frequency carrier by a low-frequency signal. • This changing of the amplitude is accomplished by direct multiplication of the high frequency carrier by a modulating signal.

4. Suppose we take the product of two signals: X x1(t) x(t) x2(t)

5. What is the spectrum of the product?

7. The resultant spectrum of the product is the convolution of the spectra of the factors. Example: Find the Fourier transform of the product x(t)=x1(t)x2(t), where x1(t) = 1. Solution:

8. Since x(t)=x2(t), we must have In general, for any function X(f), we have

9. Example: Find the Fourier transform of the product x(t)=x1(t)x2(t), where x2(t) = ej2pf2t. Since x(t)= x1(t)ej2pf2t, we also have

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

11. Example: Find the Fourier transform of the product x(t)=x1(t)x2(t), where x2(t) = cos2pf2t. thus,

12. As a more concrete example, suppose X1(f) f -½ ½ Further suppose that f2 = 3. We have

13. X(f) f -2½ 2½ 3½ -3½

14. As another concrete example, suppose X1(f) f -fm fm

15. If fm=1 and f2 = 3. We have X(f) f -4 -2 2 4

16. These last examples were examples of Double-Sideband, Suppressed Carrier (DSB-SC) Amplitude Modulation. In the very last example the “carrier” frequency f2 is 3 and the modulating frequency fm is 1. Suppressed Carrier X(f) Lower Sideband Upper Sideband f -4 -2 2 3 4

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

18. Exercise: Let m(t) = cos 2pfm1t + cos 2pfm2t, where fm1=1, fm2=2 and fc = 10. Sketch the resultant spectrum. Indicate frequencies and relative amplitudes.

19. Exercise: Suppose we replaced m(t) with 1 + m(t). Let m(t) = cos 2pfmt. Sketch the resultant spectrum. Indicate frequencies and relative amplitudes. Would it be appropriate for this modulation system to say that the carrier is suppressed?

20. Exercise: Given X m(t) xc(t) coswct How would we modify xc(t) to generate a Single-Sideband spectrum?

21. Phase Modulation • Up until this point we have used a modulating signal m(t) to alter or modulate the amplitudeof a carrier. [In the case of DSB-SC, m(t)was the amplitude.] • Suppose that we use m(t) to modulate the phase of the carrier.

22. Suppose that our modulated carrier is expressed as where qi(t) is the instantaneous phase. In the simplest case, for an unmodulated carrier.

23. With our instantaneous phase notation, we can define the instantaneous frequency: If qi(t) = wct, as in our simplest case, we must have wi=wc.

24. Now suppose that our instantaneous phase is not so simple: The function m(t) is our modulating signal. The modulating signal m(t) directly effects thephaseof the carrier.

25. Let m(t) be a simple sinewave: where

26. Our instantaneous phase becomes

27. Our modulated carrier becomes Expanding this expression using a simple trigonometric identity, we have

28. The complex parts of this expression are the compositetrigonometric functions: Can we find the spectra for these expressions? If we can, we can find the spectrum of xc(t).

29. Composite trigonometric functions can be evaluated using Bessel functions. Bessel functions are denoted by Jn(kp) and can implemented using the generating function for Bessel functions:

30. Let a = jejwmt = ej(wmt+p/2), we have Thus,

31. Plugging this into the generating function, we have

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

33. We may now insert these expressions back into the original expression:

34. What we have is spectral components at wc± nwmwhose magnitudes are J±n(kp). The magnitude of J+n(kp) is the same as that of J-n(kp).

35. 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

36. Exercise: Using the Bessel function table in Ziemer on page 131, sketch the spectrum of the modulated carrier Xc(f) for fc = 100, fm = 10, and kp=1.0 and 2.0.

37. Frequency Modulation • Can we use m(t) to modulate the frequency rather than the phase of the carrier. • The difference between phase modulation and frequency modulation deals only with the instantaneous phase:

38. Instead of we will have

39. 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).

40. Let m(t) be a simple sinewave: A plot of wi(t) versus time is shown on the following slide for wc = 1 and fm = 1.

42. We may now compute the instantaneous phase:

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

44. The modulated carrier for frequency modulation is Evaluating the composite trigonometric functions

45. can be performed by using the generating function for Bessel functions using a = ejwmt

46. Proceeding as before, we have Equating real and imaginary components we have

47. Inserting these expressions into the modulated carrier expression, we have

48. The magnitude spectrum is the same as that for phase modulation.

49. The bessel function can be evaluated in MATLAB using the function >> bessel (n, beta);

50. Exercise: Suppose fc = 1000, kf = 20 and fm = 5. Sketch the spectrum of the frequency-modulated carrier.