from fourier series to fourier transforms. recall that where now let t become large... and so ω...
TRANSCRIPT
From Fourier Series to
Fourier Transforms
Recall that
n
Txninecxf /2)(
where ,...2,1,0,1 /2
2/
2/
ndxexf
Tc Txni
T
T
n
Now let T become large ...
andT
nπω
2
so ω becomes small ...
deFxf xi
2
1)(
dxexfF xi )(
Fourier Transform
of f(x)
Inverse Fourier Transform of F(ω).
Example 1Determine the Fourier Transform of
113 tututf
113 tututf
1
1
3
tiei
dtetfF tiωω )(
1
1
3 dteF ti
sin6
ii eei
3
ii eei
3
sin23
ii
sinc6)( F)(sinc
sin
113 tututf ω
ωω
sin6F
Note: F(ω) is REAL in this example.
These are the graphs of f(t) and F(ω):
Example 2Determine the Fourier Transform of
tetutf 2
i
2
1
dtetfF tiωω )( tetutf 2
0
2 dteeF tit
0
2 dte ti
0
2
2
1 tiei
102
1
i
Note: F(ω) is COMPLEX in this example.
Draw the graph of the modulus of F(ω) (the amplitude spectrum).
tetutf 2 ω
ωi
F
2
1
24
1
2
1
2
1
iiF
tetutf 2 F
Even Functions
If f is an even function, then tftf
0
cos2)( dtttfdtetfF ti ωω ω
This result arises because cosine is even ...
... and so is even ... ttf ωcos
Example 3Determine the Fourier Transform of
1111 tututtututtf
Even function!
1
0
cos12 dtttF
1cos22
1
0
1
0
sin12
sin12 dt
ttt
1
0
sin2
dtt
002
1
0
cos2
t AAA 22 sincos2cos
A2sin21AA 2sin22cos1
2
2)(
F 2/sin2 2 2/sin
)2/(
1 22
2/sinc2
Odd Functions
If f is an odd function, then tftf
0
sin2)( dtttfidtetfF ti ωω ω
This result arises because sine is odd ...
... and so is even ... ttf ωsin
Example 4Determine the Fourier Transform of
2424 tututututf
Odd function!
2
0
sin42 dttiF
2424 tututututf
0
sin2 dtttfiF ωω
2sin16i
2
0
cos8
t
i
12cos8
i
2sin28
i
FIm
Summary:
Example f(t) F(ω)
1 Even Real
2 Neither odd nor even
Complex
3 Even Real
4 Odd Imaginary
113 tutuω
ωsin6
tetu 2
ωi2
1
...11 tutut
2sinc2 ω
...24 tutuω
ω2sin16i
Special Case 0,2
aetf at
Use this known result:
πdxe x2
Substitute yax
πdyea ay2
Now use:a
ity
2
ω
atiat
a
itaay
42
22
22
a
dte atiat πωω
4
22
Hence:
atiat ea
dte 4
2
2ω
ω πor
set at iofTransformFourier2 ae
a4
2
Now look at Tutorial 1