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21.4.2  Continuous Fourier Transform

The Fourier transform of a function f is defined by

  F(ω)=
+∞


−∞
ei ω t f(tdt =F

f(t)

(ω),

where ω∈ℝ. The fourier command computes the Fourier transform by table lookup and application of transform properties.

The inverse Fourier transform, as its name implies, takes a Fourier transform F(ω) and returns the original function f(t). It is given by:

  f(t)=
1
2 π
 
+∞


−∞
ei ω t F(ω)  dω=F−1

F(ω)

(t)=
1
F

F(ω)

(−t).

The ifourier command computes the inverse Fourier transform by using the above identity.

The Fourier command is the inert form of fourier. It is used by fourier and ifourier when returning unknown transforms.

Transforming rational functions.

An arbitrary rational function can be transformed as long as its full partial fraction decomposition can be found.

Examples

Find the Fourier transform of f(t)=t/t3−19 t+30.

F:=fourier(t/(t^3-19t+30))
     
π  
i ei ω−21 i e−3 i ω+16 i e−2 i ω
sign
ω
56
          

Indeed:

ifourier(F)
     
t
t3−19 t+30
          

Find the transform of f(t)=t2+1/t2−1.

F:=fourier((t^2+1)/(t^2-1))
     
2 π  
δ
ω
sign
ω
sinω
          

Indeed:

ifourier(F)
     
t2+1
t2−1
          

Find the transform of f(t)=2/1+4t4.

F:=fourier(2/(1+4t^4))
     
π  


sin



ω
2



+cos



ω
2






e

ω
2
 
          

Indeed:

ifourier(F)
     
2
t4+1
          

As a special case, you can transform all poylnomials. For example:

fourier(3x^2+2x+1,x,s)
     
2 π  
δ
s
+2 i δ
s,1
−3 δ
s,2

          
Transforming general functions.

A range of other (generalized) functions and distributions can be transformed, as demonstrated in the following examples.

Examples

fourier(Dirac(x-1)+Dirac(x+1),x)
     
cosω           
fourier(exp(-2*abs(x-1)),x,s)
     
ei s
s2+4
          
fourier(BesselJ(3,x),x,s)
     
s 
s2−3

i sign
s+1
+i sign
s−1

s2+1
          
F:=fourier(sin(x)*sign(x),x,s)
     
2
s2−1
          
ifourier(F,s,x)
     
sign
x
sinx
          
fourier(log(abs(x)),x,s)
     
π 

s
−2 π  γ  δ
s
          
F:=fourier(abs(2t-3)^(-3/4))
     
Γ


1
4



2
+2
 


ω
1
4
 


3


 
 e
i ω
2
 


2
1
4
 


3


 
 
ω
          
ifourier(F)
     
1



−2 t+3
1
4
 


3


 
          
fourier(rect(x),x,s)
     
sin


s
2



s
          
fourier(exp(-abs(x))*sinc(x),x,s)
     
arctan
s+1
arctan
s−1
          
fourier(1/sqrt(abs(x)),x,s)
     
2
 
π 

s
          
F:=fourier(1/cosh(2x),x,s)
     
π 
e
1
4
 π  s
 
+e
1
4
 π  s
 
          
ifourier(F,s,x)
     
2
e−2 x+ex
          
F:=fourier(Gamma(1+i*x/3),x,s)
     
6 π  e

s+e−3 s
 
          
ifourier(F,s,x)
     
Γ


1
3
 i x+1


          
fourier(atan(x/4)/x,x,s)
     
π  ugamma
0,4 
s

          
assume(a>0):; fourier(exp(-a*x^2+b),x,s)
     
a
 
π 
 e
s2
a
+b
 
a
          

Piecewise functions can be transformed if defined as

piecewise(t<a1,f1,x<a2,f2,…,t<an,fn,f0)

where f0,…,fn are expressions and a1,a2,…,an are real numbers such that a1<a2<⋯<an . Inequalities may be strict or non-strict. For example:

f:=piecewise(t<-1,exp(t+1),t<1,1,exp(2-2t)):; F:=fourier(f)
     
i ω sinω+3 ω cosω+4 sinω
ω 
ω−2 i

ω+i
          
ifourier(F)
     
θ
t+1
θ
t−1
+θ
t−1
e−2 t+2+θ
t−1
et+1
          

You can verify that the above expression coincides with f(t) by plotting them together with the plot command. Alternatively, you can apply piecewise to the result (see Section 5.1.3):

piecewise(ans())
     






et+1,t<−1
1,t<1
e

t−2
 
,
otherwise
          
Convolution Theorem and applications.

The Fourier transform behaves nicely when applied to convolutions. Recall that the convolution fg (see Section 21.2.5) of two functions f and g is defined by

  (f∗ g)(t)=
+∞


−∞
f(ug(tudu.

The Convolution Theorem states that F(fg)=F(fF(g).

As an example, we use this result for computing the convolution of f(t)=e−|t| with itself. Note that Xcas is unable to compute ff from the definition:

f(t):=exp(-abs(t)):; int(f(u)*f(t-u),u=-inf..inf)
     
undef           

Using Fourier transform:

F:=fourier(f(t))
     
2
ω2+1
          
collect(ifourier(F^2))
     


t
+1
e

t
 
          

The above result is the desired convolution (ff)(t)=∫−∞+∞f(uf(tudu.


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