FOURIER SERIES

Notation: A is a vector (boldface); A_x is a component of a vector;

x is a unit vector (boldface + italic); i,j,n,m are all integers.

1. The idea of Fourier Series is that of breaking a vector into components:

A = A_xx + A_yy + A_zz = SA_ix_i where x₁=x, x₂=y, and x₃=z ;

where A_x = A· x , A_y = A· y , and A_z = A· z ; or A_i = A· x_i ;

since x· x = 1; x· y = 0; or x_i· x_i = 1 and x_i· x_j = 0 if i ¹ j.

2. The basis of Fourier Series are the results (with n and m integers):

_qoò ^qo+2p sin(nq) sin(nq) dq = p (n¹0) ; _qoò ^qo+2p sin(nq) sin(mq) dq = 0 (n¹m)

_qoò ^qo+2p cos(nq) cos(nq) dq = p (n¹0) ; _qoò ^qo+2p cos(nq) cos(mq) dq = 0 (n¹m)

_qoò ^qo+2p sin(nq) cos(nq) dq = 0 (n¹0) ; _qoò ^qo+2p sin(nq) cos(mq) dq = 0 (n¹m)

3. The method of Fourier Series is:

f(q ) = S A_ncos(nq ) + S B_nsin(nq )

where A₀ = (1/2p) _qoò ^qo+2p f(q) cos(0) dq

A_n = (1/p) _qoò ^qo+2p f(q) cos(nq) dq

B_n = (1/p ) _qoò ^qo+2p f(q) sin(nq) dq

4. The use of Fourier Series is for breaking up waves (whether in space or in time [or both]) into sine and cosine functions.

a) For functions that oscillate in time, q must be replaced by wt (where w is the angular frequency in radians/sec), dq by w dt [recall that w = 2p/T (where T is the period of oscillation)], and the limits of the integral by t_o to t_o+T.

b) For functions that oscillate in space, q must be replaced by kx (where k is the wavevector in radians/meter), dq by k dx [recall that k = 2p/l (where l is the wavelength)], and the limits of the integral by x_o to x_o+l .

5. A Fourier Series can also be expressed in terms of exponentials due to the following identity (provable by Taylor Series Expansion):

e^inq = cos(nq ) + i sin(nq ) , and e^-inq = cos(nq ) - i sin(nq )

where n is considered positive. Therefore, we can write:

cos(nq ) = 1/2(e^inq + e^-inq) , and sin(nq) = (1/i) 1/2(e^inq - e^-inq) .

Therefore, we can write the Fourier Series as: (recall 1/i = i/i² = -i)

f(q ) = S a_n cos(nq) + S b_n sin(nq) =

S a_n {1/2(e^inq + e^-inq)} + S b_n {-1/2(e^inq - e^-inq)} =

S 1/2(a_n -b_n) e^inq + S 1/2(a_n +b_n) e^-inq (where n is always positive).

If we now let c_n>0 = 1/2(a_n - b_n) , and let c_n<0 = 1/2(a_n + b_n) (where n is +),

and c₀ = 1/2 a₀ , then we can write:

f(q ) = S c_n e^inq for n both + and - and 0, where

for n>0: c_n = 1/2(a_n - b_n) = (1/2p) _qoò ^qo+2p f(q) {cos(nq) - sin(nq)} dq ,

or c_n = (1/2p ) _qoò ^qo+2p f(q) e^-inq dq (for n>0) and

for n<0: c_n = 1/2(a_n + b_n) = (1/2p) _qoò ^qo+2p f(q) {cos(nq) + sin(nq)} dq ,

or c_n = (1/2p ) _qoò ^qo+2p f(q) e^inq dq (for n<0 with n always +).

This can be generalized to:

c_n = (1/2p ) _qoò ^qo+2p f(q) e^-inq dq (for all n, both + and -).

Result: we have the exponential form for Fourier Series:

f(q ) = S c_n e^inq for n both + and -, where

c_n = (1/2p ) _qoò ^qo+2p f(q) e^-inq dq (for all n, both + and -).

FOURIER SERIES: an example - the SAWTOOTH WAVE 

1. Define the repeating function:

f(x) = 2Ax/l for -1/2l < x < 1/2l

2. The general form for the Fourier series: 

f(q ) = 1/2a_o + _n=1S^¥ a_n cos(nq) + _n=1S^¥ b_n sin(nq) , where

a_m = (1/p ) _qoò ^qo+2p f(q) cos(mq) dq , and

b_m = (1/p) _qoò ^qo+2p f(q) sin(mq) dq .

3. Convert from q to x:

In our case, x repeats over a distance l

(just as q repeats over an angle of 2p ).

Therefore, we make the substitution q = 2p x/l

(you should see that as x goes from 0 to l , q goes from 0 to 2p ): 

f(x) = 1/2a_o + _n=1S^¥ a_n cos(n2px/l) + _n=1S^¥ b_n sin(n2px/l) , where

a_m = (1/p ) _{-p ò} ^+p f(x) cos(m2px/l) d(2px/l) and

b_m = (1/p ) _-p ò ^+p f(x) sin(m2px/l) d(2px/l) ;

or

a_m = (1/p )(2p /l ) _-1/2l ò ^+1/2l f(x) cos(m2px/l) dx and

b_m = (1/p )(2p /l ) _-1/2l ò ^+1/2l f(x) sin(m2px/l) dx ;

or substituting in f(x) = 2Ax/l :

a_m = (2/l ) _-1/2l ò ^+1/2l (2A/l ) x cos(m2px/l) dx and

b_m = (2/l ) _-1/2l ò ^+1/2l (2A/l ) x sin(m2px/l) dx .

4. Evaluate a_m and b_m :

Since cosine is an even function [cos(-q) = cos(q)] and f(x) = (2A/l)x is an odd function [ f(-x) = -f(x) ], the integral about an interval symmetric about the origin will be zero. Hence a_m = 0 for all m. [You can also integrate by parts to get this result.]

We can integrate ò x sin(ax) dx by parts (here a=m2p /l ) so that

b_m = (-1)^m+1 2A/(mp) .

5. Put a_m and b_m back into the Fourier series expression to get: 

f(x) = (2A/p) sin(2px/l) + (-2A/2p) sin(4px/l) + (2A/3p) sin(6px/l) + ...

Homework Problem: Problem #13: Fourier Series: Find the expressions for the coefficients for the Fourier Series for a square wave.

GROUP VELOCITY

We now consider a group of waves with possibly different phase velocities:

1. Pure cosine wave: Y(x,t) = A cos(kx-wt) = A cos(q)

where q = q(x,t) = phase angle = kx-wt

where k = 2p radians/l (kx is then an angle)

and w = 2p radians/T (w t is then an angle);

for crest of wave (phase angle of crest is constant at 90° = 1/2p rad):

q_crest = 1/2p = kx_crest- wt_crest , or x_crest = (1/2p + wt_crest)/k ;

for speed of crest of wave: v_phase = v_crest = dx_crest/dt_crest , so

v_phase = d([1/2p + wt]/k)/dt = w/k (here t = t_crest for short);

recall that k=2p/l and w =2p/T so that:

v_phase = (2p/T) / (2p/l ) = l/T , and since f = (1/T)

v_phase = w/k = lf .

2. Group of waves: A group of sine waves will add together to form some pattern that also repeats (this is the Fourier Series in reverse).

Y_group(x,t) = A(x) cos(Kx - w_gt)

where A(x) is the shape of the group, K is the 2p/l for the group, and w_g is the angular velocity of the group.

At t=0 sec, Y_group(x,0) = A(x) cos(Kx) where A(x) = _nS b_n sin(k_nx)

(here A(x) is expressed as a Fourier Series) , so

Y_group(x,0) = _nS b_n sin(k_nx) cos(Kx) .

We can now use two trig identities [sin(q ±f ) = sinq cosf ± cosq sinf ]

to get sinq cosf = 1/2[sin(q+f) + sin(q-f)] ,

and with q =k_nx and f =Kx, we get

Y_group(x,0) = _nS 1/2 b_n { sin[(k_n+K)x] + sin[(k_n-K)x] } ,

and since sin(-q) = -sin(+q) , we can write:

Y_group(x,0) = _nS 1/2 b_n { sin[(K+k_n)x] - sin[(K-k_n)x] } .

Now put in the time dependence such that wherever we had a Kx, we put in an additional -wt:

Y_group(x,t) = _nS 1/2 b_n { sin[(K+k_n)x-w₊t] - sin[(K-k_n)x-w_-t] }

where we use w_± to indicate that w depends on k=(K±k_n) .

[Recall that v_phase = w/k, and in some cases v_phase may not be constant but may depend on (vary with) w .]

Since w is a function of k [w(k) = v_phasek], we can expand w(k) in a Taylor Series about k=K:

w(K±k_n) = w(K) ± (dw/dk)_K k_n + higher order terms which we neglect ;

now let's let v_g º (dw/dk)_K so that w_± = w(K±k_n) » w(K) ± v_gk_n , so

Y_group(x,t) = _nS 1/2 b_n {sin[(K+k_n)x - (w+v_gk_n)t] - sin[(K-k_n)x - (w-v_gk_n)t] }

or re-grouping terms:

Y_group(x,t) = _nS 1/2 b_n { sin[(Kx-wt)+k_n(x-v_gt)] - sin[(Kx-wt)-k_n(x-v_gt)] }.

We can again use our trig identity:

sin(q +f) + sin(q -f) = 2 sinq sinf ,

where q = (Kx-wt) and f = k_n(x-v_gt) , to get:

Y_group = _nS b_n sin(Kx-wt) cos[k_n(x-v_gt)] ;

but here the sin(Kx-wt) can come out of the summation, so

Y_group = { _nS b_n cos[k_n(x-v_gt) } sin(Kx-wt) = A(x-v_gt) sin(Kx-wt) ,

where we identify the function A(x-v_gt) as the original Fourier series with x replaced by (x-v_gt); that is, the shape moves through space with a speed of v_g, hence the name group velocity.

3. Review:

v_phase = w/k = lf

(good for any pure sine wave [or cosine wave] of wavelength l and frequency f [or wavevector k and angular speed w]) ;

v_group = dw /dk .

4. Special case: If v_phase = constant, then w = v_phasek , and so

v_group = dw /dk = d[v_phasek]/dk = v_phase .

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