FOURIER SERIES

Notation: A is a vector (boldface); Ax 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  =  Axx + Ayy + Azz  =  SAixi   where   x1=x,  x2=y, and  x3=z ;

where Ax = A· x ,  Ay = A· y , and  Az = A· z ; or  Ai = A· xi ;

since  x· x = 1;  x· y = 0; or  xi· xi = 1 and  xi· xj = 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 Ancos(nq ) + S Bnsin(nq )

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

                         B0  =  (1/2p) qoò qo+2p f(q) sin(0) dq    =   0   (so we don’t have to bother with this term)

                         An  =  (1/p) qoò qo+2p f(q) cos(nq) dq

                         Bn  =  (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.  This is often called spectral analysis.

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 to to to+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 xo to xo+l .

 

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

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

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

            cos(nq)  =  ½(einq + e-inq) , and  sin(nq)  =  (1/i) ½(einq - e-inq) .

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

                        f(q)  =  S an cos(nq) + S bn sin(nq)  =

                        S an {½(einq + e-inq)} + S ibn {-½ (einq - e-inq)}  =

                        S ½ (an - ibn) einq + S ½ (an + ibn) e-inq   (where n is always positive).

Check:  ½ (an - ibn) einq + ½ (an + ibn) e-inq  =  ½ (an - ibn)*[cos(nq) + i sin(nq)]  + ½ (an + ibn)*[cos(nq) - i sin(nq)]  =  an cos(nq) + bn sin(nq).

If we now let  cn>0 = ½ (an - ibn) , and let  cn<0 = ½ (an + ibn)   (where n is +),

and c0 = ½ a0 , then we can write:

                        f(q ) = S cn einq   for n both + and - and 0, where

            for n>0:            cn  =  ½ (an - ibn)  =  (1/2p) qoò qo+2p f(q) {cos(nq) - sin(nq)} dq ,

            or         cn  =  (1/2p ) qoò qo+2p f(q) e-inq dq   (for n>0)  , and

            for n<0:            cn  =  ½ (an + ibn)  =  (1/2p) qoò qo+2p f(q) {cos(nq) + sin(nq)} dq ,

            or         cn  =  (1/2p ) qoò qo+2p f(q) einq dq   (for n<0 with n always +).

This can be generalized to:

                        cn  =  (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 cn einq      for n both + and -,    where

                        cn  =  (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  l < x < ½l

 

2. The general form for the Fourier series: 

f(q )  =  ½ao + n=1S¥ an cos(nq) + n=1S¥ bn sin(nq) , where

am  =  (1/p ) qoò qo+2p f(q) cos(mq) dq , and

bm  =  (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)  =  1/2ao + n=1S¥ an cos(n2px/l) + n=1S¥ bn sin(n2px/l) , where

                        am  =  (1/p ) -p ò +p f(x) cos(m2px/l) d(2px/l) and

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

            or

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

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

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

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

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

 

4. Evaluate am and bm :

            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 am = 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

            bm  =  (-1)m+1 2A/(mp) .

 

5. Put am and bm 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 sine wave: Y(x,t)  =  A sin(kx-wt)  =  A sin(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   (wt is then an angle);

for crest of wave  (phase angle of crest is constant at 90° = ½p rad for the sine function):

            qcrest  =  ½p  =  kxcrest- wtcrest ,   or   xcrest  =  (½p + wtcrest)/k  ;

for speed of crest of wave: vphase  =  vcrest  =  dxcrest/dtcrest ,   so

            vphase  =  d([½p + wt]/k)/dt   =  w/k   (here t = tcrest for short);

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

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

                        vphase  =  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).

                        Ygroup(x,t)  =  A(x) sin(Kx - wgt)  

            where A(x) is the shape of the group, K is the 2p/l for the group, wg is the angular velocity of the group, and vg = wg/K is the group velocity.  Can we get some expression for K, wg, and/or vg in terms of the waves that make up the group?

            At t=0 sec, Ygroup(x,0)  =  A(x) sin(Kx)  where A(x)  =  nS an cos(knx)

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

                        Ygroup(x,0)  =  nS an cos(knx) sin(Kx) .

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

                        to get sinq cosf  =  ½[sin(q+f) + sin(q-f)]  ,

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

                        Ygroup(x,0)  =  nS ½ an { sin[(K+kn)x] + sin[(K-kn)x] }

                         =   nS ½ an { sin[(kn+K)x] - sin[(kn-K)x]    since sin(-q) = -sin(q).

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

                        Ygroup(x,t)  =  nS ½ an { sin[(kn+K)x-w+t] - sin[(kn-K)x-w-t] }

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

[Recall that vphase = w/k, and in some cases vphase may not be constant but may depend on (i.e., vary with) w.  Recall from the previous section that for materials, vphase = c/n where n depended on frequency, i.e., n(w).]

            Since w is a function of k  [v = w/k, or w(k) = vphasek, and vphase may depend on k], we can expand w(k) in a Taylor Series about kn:

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

            now let's let vg º (dw/dk)  so that w± = w(kn±K) » w(kn) ± vgK , so

            Ygroup(x,t)  =  nS ½ an {sin[(kn+K)x -(wn+vgK)t] - sin[(kn-K)x -(wn-vgK)t]}

                        or re-grouping terms:

            Ygroup(x,t)  =  nS ½ an {sin[(knx-wnt)+K(x-vgt)] - sin[(knx-wnt)-K(x-vgt)]}.

 

            We can again use our trig identity:

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

                                    where q = (knx-wnt) and f = K(x-vgt), to get:

            Ygroup  =  nS an cos[(knx-wnt)] sin(K[x-vgt])   ;

                        but here the sin(K[x-vgt]) can come out of the summation, so

            Ygroup  =  { nS an cos(knx-wnt)} sin(K[x-vgt]) = A(x,t) sin(K[x-vgt]) ,

                        where we identify the function A(x,t) as the original Fourier series; that is, the shape moves through space with a speed of vg, hence the name:  group velocity.

 

3.  Review:

                        vphase = 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]) ;

 

                        vgroup = dw/dk .

 

4.  Special case:   If vphase = constant, then w = vphasek , and so

                        vgroup = dw/dk = d[vphasek]/dk = vphase .

5.  Note:  We made an approximation when we expanded w(k) in a Taylor Series about k=kn and kept only the zero order and first order terms.  Also, the amplitude is a function of both space (x) and time (t), so the shape may change with time if vphase is not constant, and this is called dispersion.

To play with Fourier Series and to see the group velocity, see this excel spreadsheet.

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