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Fourier Series

Module by: Don Johnson. E-mail the author

Summary: Signals can be composed by a superposition of an infinite number of sine and cosine functions. The coefficients of the superposition depend on the signal being represented and are equivalent to knowing the function itself.

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In Signal Decomposition , we have shown that we could express the square wave as a superposition of pulses. This superposition does not generalize well to other periodic signals: How can a superposition of pulses equal a smooth signal like a sinusoid? Because of the importance of sinusoids to linear systems, you might wonder whether they could be added together to represent a large number of periodic signals. You would be right and in good company as well. Euler and Gauss in particular worried about this problem, and Fourier got the credit even though tough mathematical issues were not settled until later. They worked on what is now known as the Fourier series.

Let st s t have period T T. We want to show that periodic signals, even those that have constant-valued segments like a square wave, can be expressed as sum of harmonically related sine waves.

st= a 0 + k =1 a k cos2πktT+ k =1 b k sin2πktT s t a 0 k 1 a k 2 k t T k 1 b k 2 k t T
(1)
The family of functions called basis functions cos2πktTsin2πktT 2 k t T 2 k t T form the foundation of the Fourier series. No matter what the periodic signal might be, these functions are always present and form the representation's building blocks. They do depend on the signal's period T T, and are indexed by k k. The frequency of each term is kT k T . For k=0 k 0 , the frequency is zero and the corresponding term a 0 a 0 is a constant. The basic frequency 1T 1 T is called the fundamental frequency because all other terms have frequencies that are integer multiples of it. These higher frequency terms are called harmonics: The term at frequency 1T 1 T is the fundamental and the first harmonic, 2T 2 T the second harmonic, etc. Thus, larger values of the series index correspond to higher harmonic-frequency sinusoids. The Fourier coefficients, a k a k and b k b k , depend on the signal's waveform. Because frequency is linked to index, the coefficients implicitly depend on frequency.

Key point:

Assuming we know the period, knowing the Fourier coefficients is equivalent to knowing the signal.

Assume for the moment that the Fourier series works. To find the Fourier coefficients, we note the following orthogonality properties of sinusoids.

Orthogonality

0Tsin2πktTcos2πltTdt=0    for all   k and l t 0 T 2 k t T 2 l t T 0    for all   k and l
(2)

Equation 2

0Tsin2πktTsin2πltTdt={T2  if   (k=l)(k0)(l0) 0  if  (kl)(k=0=l) t 0 T 2 k t T 2 l t T T 2 k l k 0 l 0 0 k l k 0 l
(3)
0Tcos2πktTcos2πltTdt={T2  if   (k=1)(k0)(l0) T  if  k=0=l0  if  kl t 0 T 2 k t T 2 l t T T 2 k 1 k 0 l 0 T k 0 l 0 k l
(4)
To use these, let's multiply the Fourier series for a square wave ( st= a 0 +k=1 a k cos2πktT+k=1 b k sin2πktT s t a 0 k 1 a k 2 k t T k 1 b k 2 k t T ) by cos2πltT 2 l t T and integrate. The idea is that, because integration is linear, the integration will sift out all but the term involving a l a l .
0Tstcos2πltTdt=0T a 0 cos2πltTdt+k=1 a k 0Tcos2πktTcos2πltTdt+k=1 b k 0Tsin2πktTcos2πltTdt t 0 T s t 2 l t T t 0 T a 0 2 l t T k 1 a k t 0 T 2 k t T 2 l t T k 1 b k t 0 T 2 k t T 2 l t T
(5)
The first and third terms are zero; in the second, the only non-zero term in the sum results when the indices k k and l l are equal (but not zero), in which case we obtain a l T2 a l T 2 . If k=0=l k 0 l , we obtain a 0 T a 0 T . Consequently, a l =2T0Tstcos2πltTdt ,   l0 a l 2 T t 0 T s t 2 l t T ,   l 0 All of the Fourier coefficients can be found similarly.
a 0 =1T0Tstdt a 0 1 T t 0 T s t
(6)
a k =2T0Tstcos2ktTdt ,   k0 a k 2 T t 0 T s t 2 k t T ,   k 0
(7)
b k =2T0Tstsin2πktTdt b k 2 T t 0 T s t 2 k t T
(8)

Exercise 1

The expression for a 0 a 0 is referred to as the average value of st s t . Why?

Solution

The average of a set of numbers is the sum divided by the number of terms. Viewing signal integration as the limit of a Riemann sum, the integral corresponds to the average.

Example 1

Let's find the Fourier series representation for the half-wave rectified sinusoid.

st={sin2πtT  if  0t<T20  if  T2t<T s t 2 t T 0 t T 2 0 T 2 t T
(9)
Begin with the sine terms in the series; to find b k b k we must calculate the integral
b k =2T0T2sin2πtTsin2πktTdt b k 2 T t 0 T 2 2 t T 2 k t T
(10)
The key to evaluating such integrals is the classic trigonometric identities.
sinαsinβ=1/2(cosαβcosα+β) α β 12 α β α β
(11)
cosαcosβ=1/2(cosα+β+cosαβ) α β 12 α β α β
(12)
sinαcosβ=1/2(sinα+β+sinαβ) α β 12 α β α β
(13)
Using these identities turns our integral of a product of sinusoids into a sum of integrals of individual sinusoids, which are much easier to evaluate.
0T2sin2πtTsin2πktTdt=1/20T2cos2π(k1)tTcos2π(k+1)tTdt={1/2  if  k=10  if  kk (k2)kN t 0 T 2 2 t T 2 k t T 12 t 0 T 2 2 k 1 t T 2 k 1 t T 12 k 1 0 k k 2 k k
(14)
Thus,
b 1 =1/2 b 1 12
(15)
b 2 = b 3 ==0 b 2 b 3 0
(16)

On to the cosine terms. The average value, which corresponds to a 0 a 0 , equals 1π 1 . The remainder of the cosine coefficients are easy to find, but yield the complicated result

a k ={(2π)1k21  if  k240  if  k is odd a k 2 1 k 2 1 k 2 4 0 k is odd
(17)

Thus, the Fourier series for the half-wave rectified sinusoid has non-zero terms for the average, the fundamental, and the even harmonics. Plotting the Fourier coefficients reveals at what component frequencies the half-wave rectified sinusoid has energy ( Figure 1 ). Furthermore, this figure shows what the Fourier series sum looks like with these coefficients as we add more and more terms. Presumably, you now believe more in the Fourier series.

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