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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
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.

k,l,kZlZ:0Tsin2πktTcos2πltTdt=0 k l k l t 0 T 2 k t T 2 l t T 0
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 0Tcos2πktTcos2πltTdt={T2  if   (k=l)(k0)(l0) T  if  k=0=l0  if  kl t 0 T 2 k t T 2 l t T T 2 k l k 0 l 0 T k 0 l 0 k l These orthogonality relations follow from the following important trigonometric identities.
sinαsinβ=12(cosαβcosα+β) α β 1 2 α β α β
cosαcosβ=12(cosα+β+cosαβ) α β 1 2 α β α β sinαcosβ=12(sinα+β+sinαβ) α β 1 2 α β α β These identities allow you to substitute a sum of sines and/or cosines for a product of them. Each term in the sum can be integrating by noticing one of two important properties of sinusoids.
  • The integral of a sinusoid over an integer number of periods equals zero.
  • The integral of the square of a unit-amplitude sinusoid over a period TT equals T2 T 2 .

To use these, let's multiply the Fourier series for a signal ( 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
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, l,l0: a l =2T0Tstcos2πltTdt l l 0 a l 2 T t 0 T s t 2 l t T All of the Fourier coefficients can be found similarly.
a 0 =1T0Tstdt a 0 1 T t 0 T s t
k,k0: a k =2T0Tstcos2πktTdt k k 0 a k 2 T t 0 T s t 2 k t T b k =2T0Tstsin2πktTdt b k 2 T t 0 T s t 2 k t T

Exercise 1

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


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

Finding the Fourier series coefficients for the square wave is very simple. sqTt sqT t . Mathematically, this signal can be expressed as sqTt={1  if  0t<T21  if  T2t<T sqT t 1 0 t T2 1 T2 t T The expressions for the Fourier coefficients have the common form

a k b k =2T0T2cos2πktTsin2πktTdt2TT2Tcos2πktTsin2πktTdt a k b k 2 T t 0 T 2 2 k t T 2 k t T 2 T t T 2 T 2 k t T 2 k t T
The cosine coefficients a k a k are all zero, and the sine coefficients are
b k ={4πk  if  k is odd0  if  k is even b k 4 k k is odd 0 k is even
Thus, the Fourier series for the square wave is
sqt=k134πksin2πktT sq t k k 1 3 4 k 2 k t T
Consequently, the square wave equals a sum of sinusoids, but only those having frequencies equal to odd multiples of the fundamental frequency 1T 1 T . The coefficients decay slowly as the frequency index kk increases. This index corresponds to the kk-th harmonic of the signal's period.

Example 2

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
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
Using our trigonometric identities turns our integral of a product of sinusoids into a sum of integrals of individual sinusoids, which are much easier to evaluate.
0T2sin2πtTsin2πktTdt=120T2cos2π(k1)tTcos2π(k+1)tTdt={12  if  k=10  otherwise   t 0 T 2 2 t T 2 k t T 1 2 t 0 T 2 2 k 1 t T 2 k 1 t T 1 2 k 1 0
Thus, b 1 =12 b 1 1 2 b 2 = b 3 ==0 b 2 b 3 0

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

Thus, the Fourier series for the half-wave rectified sinusoid has non-zero terms for the average, the fundamental, and the even harmonics.

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