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Amplitude Modulation

Module by: Don Johnson. E-mail the author

Summary: Understanding and performing amplitude modulation through Fourier methods.

Example 1

In communications, a very important operation on a signal s⁢t s t is to amplitude modulate it. Using this operation more as an example rather than elaborating the communications aspects here, we want to compute the Fourier transform—the spectrum—of

(1+s⁢t)⁢cos2⁢π⁢ f c ⁢t 1 s t 2 f c t
(1)
Thus,
s⁢t=(1+s⁢t)⁢cos2⁢π⁢ f c ⁢t=cos2⁢π⁢ f c ⁢t+s⁢t⁢cos2⁢π⁢ f c ⁢t s t 1 s t 2 f c t 2 f c t s t 2 f c t
(2)
For the spectrum of cos2⁢π⁢ f c ⁢t 2 f c t , we use the Fourier series. Its period is 1 f c 1 f c , and its only nonzero Fourier coefficients are c ± 1 =12 c ± 1 1 2 . 1 The second term is not periodic unless s⁢t s t has the same period as the sinusoid. Using Euler's relation, the spectrum of the second term can be derived as
s⁢t⁢cos2⁢π⁢ f c ⁢t=∫−∞∞S⁢f⁢ei⁢2⁢π⁢f⁢tdf⁢cos2⁢π⁢ f c ⁢t=12⁢∫−∞∞S⁢f⁢ei⁢2⁢π⁢(f+ f c )⁢tdf+12⁢∫−∞∞S⁢f⁢ei⁢2⁢π⁢(f− f c )⁢tdf=12⁢∫−∞∞S⁢f− f c ⁢ei⁢2⁢π⁢f⁢tdf+12⁢∫−∞∞S⁢f+ f c ⁢ei⁢2⁢π⁢f⁢tdf=∫−∞∞S⁢f− f c +S⁢f+ f c 2⁢ei⁢2⁢π⁢f⁢tdf s t 2 f c t f S f 2 f t 2 f c t 1 2 f S f 2 f f c t 1 2 f S f 2 f f c t 1 2 f S f f c 2 f t 1 2 f S f f c 2 f t f S f f c S f f c 2 2 f t
(3)
Exploiting the uniqueness property of the Fourier transform, we have
F⁢s⁢t⁢cos2⁢π⁢ f c ⁢t=S⁢f− f c +S⁢f+ f c 2 F s t 2 f c t S f f c S f f c 2
(4)
This component of the spectrum consists of the original signal's spectrum delayed and advanced in frequency. The spectrum of the amplitude modulated signal is shown in Figure 1.

Figure 1: A signal which has a triangular shaped spectrum is shown in the top plot. Its highest frequency—the largest frequency containing power—is W W  Hz. Once amplitude modulated, the resulting spectrum has "lines" corresponding to the Fourier series components at ±⁢ f c ± f c and the original triangular spectrum shifted to components at ±⁢ f c ± f c and scaled by 12 1 2 .
modulated
modulated (spectrum5.png)

Note how in this figure the signal s⁢t s t is defined in the frequency domain. To find its time domain representation, we simply use the inverse Fourier transform.

Exercise 1

What is the signal s⁢t s t that corresponds to the spectrum shown in the upper panel of Figure 1?

Solution

The signal is the inverse Fourier transform of the triangularly shaped spectrum, and equals s⁢t=W⁢sinπ⁢W⁢tπ⁢W⁢t2 s t W W t W t 2

Exercise 2

What is the power in x⁢t x t , the amplitude-modulated signal? Try the calculation in both the time and frequency domains.

Solution

The result is most easily found in the spectrum's formula: the power in the signal-related part of x⁢t x t is half the power of the signal s⁢t s t .

In this example, we call the signal s⁢t s t a baseband signal because its power is contained at low frequencies. Signals such as speech and the Dow Jones averages are baseband signals. The baseband signal's bandwidth equals W W , the highest frequency at which it has power. Since x⁢t x t 's spectrum is confined to a frequency band not close to the origin (we assume f c ≫W ≫ f c W ), we have a bandpass signal. The bandwidth of a bandpass signal is not its highest frequency, but the range of positive frequencies where the signal has power. Thus, in this example, the bandwidth is 2⁢W Hz 2 W Hz. Why a signal's bandwidth should depend on its spectral shape will become clear once we develop communications systems.

Footnotes

  1. From Euler's relation, cos2⁢π⁢ f c ⁢t=12⁢e−(2⁢π⁢ f c ⁢t)+12⁢e2⁢π⁢ f c ⁢t 2 f c t 1 2 2 f c t 1 2 2 f c t . This is the Fourier series for cosine!

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