Mathematica for Fourier Series and Transforms

Mathematica for Fourier Series and Transforms

Fourier Series

Periodic odd step function

Use built-in function "UnitStep" to define."Mod" allows one to make the function periodic, with the "-Pi" shiftingthe fundamental region of the Mod to -Pi to Pi (rather than 0 to 2Pi).The period is taken to be 2 Pi, symmetric around the origin, so the function iseven.“Exclusions->None” makes the plot include the steps.

ff0@x_D = -1 ê 2 + UnitStep@Mod@x, 2 Pi, -PiDD;

ff0plot = Plot@ff0@xD, 8x, -3 Pi, 3 Pi<, Ticks Ø 88-3 Pi, -2 Pi, -Pi, 0, Pi, 2 Pi, 3 Pi<<,PlotStyle Ø 8Thick, Red<, Exclusions Ø NoneD

-3 p -2 p -p p 2 p 3 p

-0.4

-0.2

0.2

0.4

Truncated Fourier Series---since odd can use "FourierSinSeries"

In[170]:= ft6ff0 = FourierSinSeries@ff0@xD, x, 6D

Out[170]=2 Sin@xD

p+2 Sin@3 xD

3 p+2 Sin@5 xD

5 p

Faster for evaluation to define Fourier Series directly (given that we’ve worked it out by hand in class notes):

ftff0@x_, nmax_D := Sum@Sin@H2 n + 1L xD ê H2 n + 1L, 8n, 0, nmax<D 2 ê Pi

ftff0@x, 2D

2 JSin@xD + 13Sin@3 xD + 1

5Sin@5 xDN

p

Comparing step to truncated Fourier series:

In[156]:= stepplot@n_D :=Plot@8ff0@xD, ftff0@x, nD<, 8x, -Pi, Pi<, Ticks Ø 88-Pi, -Pi ê 2, 0, Pi ê 2, Pi<<,PlotStyle Ø 88Thick, Red<, 8Thick, Blue<<, Exclusions Ø NoneD

In[160]:= stepplot@0D

Out[160]=-p -

p

2

p

2p

-0.6

-0.4

-0.2

0.2

0.4

0.6

Make animation showing convergence:

In[163]:= Manipulate@stepplot@nD, 8n, 0, 20, 1<D

Out[163]=

n

1

-p -p

2

p

2p

-0.6

-0.4

-0.2

0.2

0.4

0.6

Comparing 3, 21 and 101 term Fourier Series to original function

2 Fourier.nb

Plot@8ff0@xD, ftff0@x, 2D, ftff0@x, 20D, ftff0@x, 100D<,8x, -Pi, Pi<, Ticks Ø 88-Pi, -Pi ê 2, 0, Pi ê 2, Pi<<, PlotStyle Ø88Thick, Red<, 8Thick, Blue<, 8Thick, Green<, 8Thick, Black<<, Exclusions Ø NoneD

-p -p

2

p

2p

-0.6

-0.4

-0.2

0.2

0.4

0.6

Zoom in to see Gibbs phenomenon: lack of convergence near step.Here use 21, 101 and 501 term series.

Plot@8ff0@xD, ftff0@x, 20D, ftff0@x, 100D, ftff0@x, 500D<,8x, -0.1, +0.1<, PlotRange Ø 88-.1, 0.1<, 80.4, .6<<, PlotStyle Ø88Thick, Red<, 8Thick, Green<, 8Thick, Black<, 8Thick, Orange<<, Exclusions Ø NoneD

-0.10 -0.05 0.05 0.10

0.45

0.50

0.55

0.60

Using complex fourier series

In[174]:= ft6ff0complex = FourierSeries@ff0@xD, x, 6D

Out[174]=Â ‰-Â x

p- ‰Â x

p+Â ‰-3 Â x

3 p-Â ‰3 Â x

3 p+Â ‰-5 Â x

5 p-Â ‰5 Â x

5 p

Simplify doesn’t help, but FullSimplify brings result back into Sin form found above

In[178]:= FullSimplify@ft6ff0complexD

Out[178]=2 H15 Sin@xD + 5 Sin@3 xD + 3 Sin@5 xDL

15 p

Fourier.nb 3

Even "barrier" function

Use built-in function "UnitStep" to define."Mod" allows one to make the function periodic, with the "-Pi" shiftingthe fundamental region of the Mod to -Pi to Pi (rather than 0 to 2Pi).The period is taken to be 2 Pi, symmetric around the origin, so the function iseven.

ff1@x_D = UnitStep@Mod@x + Pi ê 2, 2 Pi, -PiDD UnitStep@Mod@Pi ê 2 - x, 2 Pi, -PiDD;

ff1plot = Plot@ff1@xD, 8x, -3 Pi, 3 Pi<, Ticks Ø 88-3 Pi, -2 Pi, -Pi, 0, Pi, 2 Pi, 3 Pi<<,PlotStyle Ø 8Thick, Red<, Exclusions Ø NoneD

-3 p -2 p -p p 2 p 3 p

0.2

0.4

0.6

0.8

1.0

For an even function can use Cosine Fourier Series, here up to Cos[6x].We will call this a 4 term series.

ft6ff1@x_D = FourierCosSeries@ff1@xD, x, 6D

1

2+2 Cos@xD

p-2 Cos@3 xD

3 p+2 Cos@5 xD

5 p

The function does a pretty poor job of representing the "barrier" at this order.

Plot@8ff1@xD, ft6ff1@xD<, 8x, -3 Pi, 3 Pi<,Ticks Ø 88-3 Pi, -2 Pi, -Pi, 0, Pi, 2 Pi, 3 Pi<<,PlotStyle Ø 88Thick, Red<, 8Thick, Blue<<, Exclusions Ø NoneD

-3 p -2 p -p p 2 p 3 p

0.2

0.4

0.6

0.8

1.0

4 Fourier.nb

If one uses "FourierSeries" one gets the complex form, which is equivalent to that found above:

FourierSeries@ff1@xD, x, 6D

1

2+‰-Â x

p+‰Â x

p-‰-3 Â x

3 p-‰3 Â x

3 p+‰-5 Â x

5 p+‰5 Â x

5 p

Since the general form is known, it is faster (for evaluation) to write out the Fourier Transform explicitly.The number of terms is 2 nmax+2 (including the constant).

In[164]:= ftff1@x_, nmax_D := 1 ê 2 + Sum@H-1L^n Cos@H2 n + 1L xD ê H2 n + 1L, 8n, 0, nmax<D 2 ê Pi

Making and manipulating plots:

In[165]:= barrierplot@n_D :=Plot@8ff1@xD, ftff1@x, nD<, 8x, -Pi, Pi<, Ticks Ø 88-Pi, -Pi ê 2, 0, Pi ê 2, Pi<<,PlotStyle Ø 88Thick, Red<, 8Thick, Blue<<, Exclusions Ø NoneD

In[166]:= barrierplot@1D

Out[166]=

-p -p

2

p

2p

0.2

0.4

0.6

0.8

1.0

Fourier.nb 5

In[167]:= Manipulate@barrierplot@nD, 8n, 0, 20, 1<D

Out[167]=

n

3

-p -p

2

p

2p

0.2

0.4

0.6

0.8

1.0

Comparing the 4, 22 and 102 term Cosine Series to the function itself (which is no longer visible):

Plot@8ff1@xD, ftff1@x, 2D, ftff1@x, 20D, ftff1@x, 100D<,8x, -Pi, Pi<, Ticks Ø 88-Pi, -Pi ê 2, 0, Pi ê 2, Pi<<, PlotStyle Ø88Thick, Red<, 8Thick, Blue<, 8Thick, Green<, 8Thick, Black<<, Exclusions Ø NoneD

-p -p

2

p

2p

0.2

0.4

0.6

0.8

1.0

Zooming in on the edge of the step, showing the 22, 102 and 502 term series.(Colors match those in the previous plot.)Note the oscillation near the edge with the fixed 9% overshoot (called the Gibbs phenomenon).

6 Fourier.nb

Plot@8ff1@xD, ftff1@x, 20D, ftff1@x, 100D, ftff1@x, 500D<, 8x, Pi ê 2 - 0.1, Pi ê 2 + 0.1<,PlotRange Ø 88Pi ê 2 - .1, Pi ê 2 + 0.1<, 80.9, 1.1<<, PlotStyle Ø88Thick, Red<, 8Thick, Green<, 8Thick, Black<, 8Thick, Orange<<, Exclusions Ø NoneD

1.55 1.60 1.65

0.95

1.00

1.05

1.10

Using Complex Fourier Series

In[179]:= ft6ff1complex = FourierSeries@ff1@xD, x, 6D

Out[179]=1

2+‰-Â x

p+‰Â x

p-‰-3 Â x

3 p-‰3 Â x

3 p+‰-5 Â x

5 p+‰5 Â x

5 p

In[180]:= FullSimplify@ft6ff1complexD

Out[180]=15 p + 60 Cos@xD - 20 Cos@3 xD + 12 Cos@5 xD

30 p

General real function (neither odd nor even)

In[185]:= ff4@x_D := Piecewise@883 HMod@x, 2 Pi, -PiD + PiL ê Pi - 1, Mod@x, 2 Pi, -PiD < -Pi ê 2<,8-Mod@x, 2 Pi, -PiD ê Pi, Mod@x, 2 Pi, -PiD ¥ -Pi ê 2<<D

In[186]:= ff4plot = Plot@ff4@xD, 8x, -3 Pi, 3 Pi<, Ticks Ø 88-3 Pi, -2 Pi, -Pi, 0, Pi, 2 Pi, 3 Pi<<,PlotStyle Ø 8Thick, Red<, Exclusions Ø NoneD

Out[186]=

-3 p -2 p -p p 2 p 3 p

-1.0

-0.8

-0.6

-0.4

-0.2

0.2

0.4

Use “FourierTrigSeries” to get Cosine/Sine seriesHere is the result up to 6th order

Fourier.nb 7

Use “FourierTrigSeries” to get Cosine/Sine seriesHere is the result up to 6th order

In[188]:= ft6ff4@x_D = FourierTrigSeries@ff4@xD, x, 6D

Out[188]= -1

4+4 Cos@xD

p2-2 Cos@2 xD

p2+4 Cos@3 xD

9 p2+

4 Cos@5 xD

25 p2-2 Cos@6 xD

9 p2-4 Sin@xD

p2+4 Sin@3 xD

9 p2-4 Sin@5 xD

25 p2

In[189]:= Plot@8ff4@xD, ft6ff4@xD<, 8x, -3 Pi, 3 Pi<,Ticks Ø 88-3 Pi, -2 Pi, -Pi, 0, Pi, 2 Pi, 3 Pi<<,PlotStyle Ø 88Thick, Red<, 8Thick, Blue<<, Exclusions Ø NoneD

Out[189]=

-3 p -2 p -p p 2 p 3 p

-1.0

-0.8

-0.6

-0.4

-0.2

0.2

0.4

Here’s the complex version:

In[191]:= ft6ff4complex = FourierSeries@ff4@xD, x, 6D

Out[191]= -1

4+H2 - 2 ÂL ‰-Â x

p2+H2 + 2 ÂL ‰Â x

p2-‰-2 Â x

p2-‰2 Â x

p2+J29+ 2 Â

9N ‰-3 Â x

p2+

J29- 2 Â

9N ‰3 Â x

p2+J

225

- 2 Â

25N ‰-5 Â x

p2+J

225

+ 2 Â

25N ‰5 Â x

p2-‰-6 Â x

9 p2-‰6 Â x

9 p2

In[192]:= FullSimplify@ft6ff4complexD

Out[192]=1

900 p2I3600 Cos@xD - 1800 Cos@2 xD + 400 Cos@3 xD + 144 Cos@5 xD -

25 I9 p2 + 8 Cos@6 xD + 144 Sin@xD - 16 Sin@3 xDM - 144 Sin@5 xDM

Even function with discontinuity in derivative

Here we consider an example with a discontinuity in derivative but not in the function itself.

ff3@x_D = Piecewise@88Cos@xD, -Pi ê 2 <= Mod@x, 2 Pi, -PiD < Pi ê 2<<D

Cos@xD - p

2§ Mod@x, 2 p, -pD < p

2

0 True

8 Fourier.nb

ff3plot = Plot@ff3@xD, 8x, -3 Pi, 3 Pi<, Ticks Ø 88-3 Pi, -2 Pi, -Pi, 0, Pi, 2 Pi, 3 Pi<<,PlotStyle Ø 8Thick, Red<, Exclusions Ø NoneD

-3 p -2 p -p p 2 p 3 p

0.2

0.4

0.6

0.8

1.0

Starting with the Cos[2x] term, the number in the denominator is n^2-1.So the coefficients fall like 1/n^2, compared to the 1/n we saw with a discontinuity in the function itself. This is general.

ft6ff3@x_D = FourierCosSeries@ff3@xD, x, 6D

1

p+Cos@xD

2+2 Cos@2 xD

3 p-2 Cos@4 xD

15 p+2 Cos@6 xD

35 p

ft20ff3@x_D = FourierCosSeries@ff3@xD, x, 20D

1

p+Cos@xD

2+2 Cos@2 xD

3 p-2 Cos@4 xD

15 p+2 Cos@6 xD

35 p-2 Cos@8 xD

63 p+2 Cos@10 xD

99 p-

2 Cos@12 xD

143 p+2 Cos@14 xD

195 p-2 Cos@16 xD

255 p+2 Cos@18 xD

323 p-2 Cos@20 xD

399 p

ft100ff3@x_D = FourierCosSeries@ff3@xD, x, 100D;

From this view, 20 terms is enough to do a good job, with 6 showing oscillations:

ff3plot = Plot@8ff3@xD, ft6ff3@xD, ft20ff3@xD<,8x, -3 Pi, 3 Pi<, Ticks Ø 88-3 Pi, -2 Pi, -Pi, 0, Pi, 2 Pi, 3 Pi<<,PlotStyle Ø 88Thick, Red<, 8Thick, Blue<, 8Thick, Green<<, Exclusions Ø NoneD

-3 p -2 p -p p 2 p 3 p

0.2

0.4

0.6

0.8

1.0

Zooming in, we need > 100 terms to approach the discontinuity in derivative.But there is no Gibbs phenomenon here.

Fourier.nb 9

Zooming in, we need > 100 terms to approach the discontinuity in derivative.But there is no Gibbs phenomenon here.

ff3plot = Plot@8ff3@xD, ft6ff3@xD, ft20ff3@xD, ft100ff3@xD<,8x, Pi ê 2 - 0.1, Pi ê 2 + 0.2<, PlotStyle Ø88Thick, Red<, 8Thick, Blue<, 8Thick, Green<, 8Thick, Black<<, Exclusions Ø NoneD

1.55 1.60 1.65 1.70 1.75

0.02

0.04

0.06

0.08

0.10

Integrals needed in discussion of Gibbs phenomenon (see lecture notes)Integrate@Sin@xD ê x, 8x, 0, Infinity<D ê Pi

1

2

Integrate@Sin@xD ê x, 8x, 0, Pi<D ê Pi êê N

0.58949

Showing that the sum indeed asymptotes to the integral above:

ss@N_D := Sum@Sin@H2 n + 1L Pi ê H2 HN + 1LLD ê H2 n + 1L,8n, 0, N<, Assumptions Ø 8N œ Integers, N > 0<D 2 ê Pi

ListPlot@Table@ss@nD, 8n, 5, 500<DD

100 200 300 400 500

0.589492

0.589494

0.589496

0.589498

0.589500

0.589502

Fourier transforms

10 Fourier.nb

Fourier transforms

Section 7.12 problem number 10

This is one way of writing the function in Mathematica

In[193]:= f10@t_D = Piecewise@88-2 Ha + tL, -a § t < 0<, 82 Ha - tL, 0 < t < a<<D;

Plotting: note that the "/. a->1" command tells Mathematica to set the parameter a to 1 in the function.

In[194]:= f10plot = Plot@f10@tD ê. a Ø 1, 8t, -2, 2<, PlotStyle Ø 8Thick, Red<D

Out[194]=-2 -1 1 2

-2

-1

1

2

Obtaining the Fourier Transform. The "FourierParameters" must be set to match our conventions (see the Documentation for their defini-tions).The “Assumptions -> a>0” allows Mathematica to ignore the case where a<0 in which case, given the definitions above, the function vanishes.“FullSimplify” leads to a more compact answer than “Simplify”

In[200]:= ft10@alp_D = FullSimplify@FourierTransform@f10@tD, t, alp,FourierParameters Ø 8-1, -1<, Assumptions Ø a > 0DD

Out[200]=2 Â H-a alp + Sin@a alpDL

alp2 p

Since our function is odd, we can also use the "FourierSinTransform", which gives the imaginary part of the previous result

In[196]:= img10@alp_D = FourierSinTransform@f10@tD,t, alp, FourierParameters Ø 8-1, -1<, Assumptions Ø a > 0D

Out[196]=2 H-a alp + Sin@a alpDL

alp2 p

Fourier.nb 11

In[197]:= Plot@img10@alpD ê. a Ø 1, 8alp, -15, 15<, PlotStyle Ø 8Thick, Blue<D

Out[197]=-15 -10 -5 5 10 15

-0.2

-0.1

0.1

0.2

Here is the inverse Fourier Transform which brings us back to the original function (expressed in a different way).

In[212]:= invft10@t_D = FullSimplify@InverseFourierTransform@ft10@alpD, alp, t, FourierParameters Ø 8-1, -1<DD

Out[212]= Ha - tL Sign@a - tD + 2 a Sign@tD - Ha + tL Sign@a + tD

In[213]:= Plot@invft10@tD ê. a Ø 1, 8t, -2, 2<, PlotStyle Ø 8Thick, Red<D

Out[213]=-2 -1 1 2

-2

-1

1

2

12 Fourier.nb