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Classic 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.

Note: You are viewing an old version of this document. The latest version is available here.

The classic Fourier series as derived originally expressed a periodic signal (period TT) in terms of harmonically related sines and cosines.

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 complex Fourier series and the sine-cosine series are identical, each representing a signal's spectrum. The Fourier coefficients, ak ak and bk bk, express the real and imaginary parts respectively of the spectrum while the coefficients ck ck of the complex Fourier series express the spectrum as a magnitude and phase. Equating the classic Fourier series to the complex Fourier series, an extra factor of two and complex conjugate become necessary to relate the Fourier coefficients in each. ck=12(akibk) ck 1 2 ak bk

Exercise 1

Derive this relationship between the coefficients of the two Fourier series.

Solution

Write the coefficients of the complex Fourier series in Cartesian form as ck=Ak+iBk ck Ak Bk and substitute into the expression for the complex Fourier series. k=ckei2πktT=k=(Ak+iBk)ei2πktT k ck 2kt T k Ak Bk 2kt T Simplifying each term in the sum using Euler's formula, (Ak+iBk)ei2πktT = (Ak+iBk)(cos2πktT+isin2πktT) = Akcos2πktTBksin2πktT+i(Aksin2πktT+Bkcos2πktT) Ak Bk 2kt T = Ak Bk 2kt T 2kt T = Ak 2kt T Bk 2kt T Ak 2kt T Bk 2kt T We now combine terms that have the same frequency index in magnitude. Because the signal is real-valued, the coefficients of the complex Fourier series have conjugate symmetry: ck=ck¯ ck ck or Ak=Ak Ak Ak and Bk=Bk Bk Bk . After we add the positive-indexed and negative-indexed terms, each term in the Fourier series becomes 2Akcos2πktT2Bksin2πktT 2 Ak 2kt T 2 Bk 2kt T . To obtain the classic Fourier series, we must have 2Ak=ak 2 Ak ak and 2Bk=bk 2 Bk bk .

Just as with the complex Fourier series, we can find the Fourier coefficients using the orthogonality properties of sinusoids. Note that the cosine and sine of harmonically related frequencies, even the same frequency, are orthogonal.

k,l,kZlZ:0Tsin2πktTcos2πltTdt=0 k l k l t 0 T 2 k t T 2 l t T 0
(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 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) cosαcosβ=12(cosα+β+cosαβ) sinαcosβ=12(sinα+β+sinαβ) α β 1 2 α β α β α β 1 2 α β α β α β 1 2 α β α β
(3)
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, for example, multiply the Fourier series for a signal by the cosine of the lth lth harmonic 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
(4)
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 k,k0: a k =2T0Tstcos2πktTdt b k =2T0Tstsin2πktTdt a 0 1 T t 0 T s t k k 0 a k 2 T t 0 T s t 2 k t T b k 2 T t 0 T s t 2 k t T
(5)

Exercise 2

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.

Exercise 3

What is the Fourier series for a unit-amplitude square wave?

Solution

We found that the complex Fourier series coefficients are given by ck=2iπk ck 2 k . The coefficients are pure imaginary, which means ak=0 ak 0 . The coefficients of the sine terms are given by bk=(2ck) bk 2 ck so that bk={4πk  if   k  odd 0  if   k  even bk 4 k k  odd 0 k  even Thus, the Fourier series for the square wave is

sqt=k134πksin2πktT sq t k k 1 3 4 k 2 k t T
(6)

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
(7)
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
(8)
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
(9)
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 ={(2π1k21)  if  k240  if  k odd a k 2 1 k 2 1 k 2 4 0 k odd
(10)

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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