Fourier Transforms The complex Fourier series has an important limiting form when the period approaches infinity, i.e., T 0 →∞ or L →∞ . Suppose that in this limit (1) k = nπ L remains large (ranging from −∞ to ∞ ) and (2) c n → 0 since it is proportional to 1/L, but g( k ) = lim L →∞ c n →0 L π c n = 1 2π f ( x )e − ikx −∞ ∞ ∫ dx = finite then we have f ( x ) = c n e ikx n = −∞ ∞ ∑ = lim L →∞ c n →0 1 2π π L g( k )e ikx n = −∞ ∞ ∑ where k = nπ L . The sum over n is in steps of Δn = 1 . Thus, we can write using Δk = π L Δn which becomes infinitesimally small when L becomes large, as a sum over k, which becomes an integral in the limit f ( x ) = lim L →∞ c n →0 1 2π πΔn L g( k )e ikx = n = −∞ ∞ ∑ lim Δk →0 1 2π Δkg( k )e ikx k = −∞ ∞ ∑ = 1 2π g( k )e ikx dk −∞ ∞ ∫ We call g(k) the Fourier Transform of f(x) g( k ) = 1 2π f ( x )e − ikx −∞ ∞ ∫ dx = F( f ) and the last equation is the so-called Fourier inversion formula. We can now obtain an integral representation of the delta- function. This corresponds to the orthogonality condition for the complex exponential Fourier series. We substitute the definition of g(k) into the inversion formula to get Page 1
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Fourier Transforms
The complex Fourier series has an important limiting form when
the period approaches infinity, i.e., T0 →∞ or L→∞ . Suppose
that in this limit
(1) k =nπLremains large (ranging from −∞ to ∞ ) and
(2) cn → 0 since it is proportional to 1/L, but
g(k) = limL→∞cn→0
Lπcn =
12π
f (x)e− ikx−∞
∞
∫ dx = finite
then we have
f (x) = cneikx
n=−∞
∞
∑ = limL→∞cn→0
12π
πLg(k)eikx
n=−∞
∞
∑
where k =nπL. The sum over n is in steps of Δn = 1. Thus, we can
write using
Δk =
πLΔn
which becomes infinitesimally small when L becomes large, as a sum over k, which becomes an integral in the limit
f (x) = limL→∞cn→0
12π
πΔnL
g(k)eikx =n=−∞
∞
∑ limΔk→0
12π
Δkg(k)eikxk=−∞
∞
∑
= 12π
g(k)eikxdk−∞
∞
∫We call g(k) the Fourier Transform of f(x)
g(k) = 1
2πf (x)e− ikx
−∞
∞
∫ dx = F( f )
and the last equation is the so-called Fourier inversion formula.
We can now obtain an integral representation of the delta-function. This corresponds to the orthogonality condition for the complex exponential Fourier series.
We substitute the definition of g(k) into the inversion formula to get
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f (x) = 12π
g(k)eikxdk−∞
∞
∫ = dkeikx1
2πdx '
−∞
∞
∫ e− ikx ' f (x ')−∞
∞
∫
= dx ' f (x ')−∞
∞
∫1
2πdk
−∞
∞
∫ eik (x− x ')⎡
⎣⎢
⎤
⎦⎥ = dx ' f (x ')
−∞
∞
∫ δ (x − x ')
where
δ (x − x ') = 1
2πdk
−∞
∞
∫ eik (x− x ')
Properties
The evaluation of the integrals involved in many Fourier transforms involves complex integration, which we shall learn later. We will just state some properties
Examples:
The Fourier transform of the box function
f (x) =
1 x ≤ a0 x ≥ a
⎧⎨⎩
is
F( f ) = 1
2πdx '
−∞
∞
∫ e− ikx ' f (x ') = 12π
dx '−α
α
∫ e− ikx ' =12π
e− ikx '
ik −α
α
=12π
2sin kαk
The Fourier transform of the derivative of a function is
Fdfdx
⎛⎝⎜
⎞⎠⎟=
12π
dx '−∞
∞
∫ e− ikx ' df (x ')dx '
=12π
f (x ')e− ikx '−∞
∞− (−ik) dx '
−∞
∞
∫ e− ikx ' f (x ')⎡
⎣⎢
⎤
⎦⎥
= ik2π
dx '−∞
∞
∫ e− ikx ' f (x ') = ikF( f )
where we have assumed that f (x)→ 0 as x →∞ . This generalizes to
F
dn fdxn
⎛⎝⎜
⎞⎠⎟= (ik)n F( f )
Other useful properties of the Fourier transform are:
F( f (x)) = g(k) , F( f (x − a)) = e− ikag(k) , F( f (x)eax ) = g(k + ia)
A short table of Fourier Transforms is shown below:
Page 2
f (x) g(k)
δ (x) 12π
0 x>0e−λx x<0
⎧⎨⎩
12π
1a + ik
e−cx2
2 1ce−k2
2c
11+ x2
π2e− k
Convolutions
In general, we define the convolution integral by
h(t) = f (t − τ )g(τ )
0
t
∫ dτ ≡ f ∗ g ≡ convolution integral
The Fourier transform of the product of two functions can be given in terms of the Fourier transforms of the individual function.
FT f (t)g(t)[ ] = 12π
dte− iω t−∞
∞
∫ f (t)g(t)
= 12π
dte− iω t−∞
∞
∫12π
dω 'eiω ' t
−∞
∞
∫ G(ω ') 12π
dω ''eiω '' t
−∞
∞
∫ F(ω '')
FT f (t)g(t)[ ] = 12π
⎛⎝⎜
⎞⎠⎟
3
dω '−∞
∞
∫ dω '' G(ω ')F(ω '')[ ] dt−∞
∞
∫ ei(ω '+ω ''−ω )t
−∞
∞
∫
= 12π
⎛⎝⎜
⎞⎠⎟
3
dω '−∞
∞
∫ dω '' G(ω ')F(ω '')[ ]2πδ (ω '+ω ''−ω )−∞
∞
∫
= 12π
dω '−∞
∞
∫ G(ω ')F(ω −ω ') = G(ω )∗F(ω )
This just the convolution of the Fourier transforms
G(ω ) and F(ω ). It is the fundamental construction needed to solve ODEs using Green’s functions later , as we shall see later.
Clearly, it gives a measure of the overlap of two signals as a function of t.
The symmetry of the Fourier transform and its inverse operation gives the results
Page 3
InvFT F(ω )G(ω )[ ] = 1
2πf (t)⊗ g(t)→ℑ f (t)⊗ g(t)[ ] = 2πF(ω )G(ω )
An example will help us visualize the real complexity of the convolution operation. We consider the convolution of two functions
h(t) = f (t)⊗ g(t) = dτ f (τ )g(t − τ )
−∞
∞
∫Where we choose f(t) = square pulse and g(t) = triangular pulse as shown below.
Now h(t) = area under product integrand f (τ )g(t − τ ) as a function of t.
Procedure(must be done carefully):
(1) plot f (τ ) and g(t − τ ) versus τ ; h(t) is given by this process for all possible t values.
(2) t ≤ 0, say t=-2; In this case it is clear that there is never any overlap of the functions and therefore their product is zero and thus h(t) = 0 for t ≤ 0 .
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Also from the picture below it is clear that h(t) = 0 for t ≤ 1 (say 0.5)
and also for 7 < t (say t = 8) as shown below;
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(4) 1 < t < 2 , say t = 1.5; Here the functions only overlap in the range 1 < t < 2 as shown below.
The red curve is f(tau)*g(tau-t) for t = 1.5. The area under the red curve is h(t).
In this overlap region,
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h(t) = dτ f (τ )g(t − τ )1
t
∫f (τ ) = 2
g(t − τ ) =25
(t − τ ) 0 < (t − τ ) < 5
0 otherwise
⎧⎨⎪
⎩⎪
The range 0 < (t − τ ) < 5 is equivalent to (t − 5) < τ < t . Over this range we have
g(t − τ ) = 2
5(t − τ )
and
h(t) = f (t)⊗ g(t) = dτ 45
(t − τ )1
t
∫ 1 < t < 2
= 25t 2 −
45t +
25
1 < t < 2
(5) 2 < t < 6 , say t = 4 Here the square pulse is entirely inside the triangular pulse as shown below.
Page 7
We have
h(t) = f (t)⊗ g(t) = dτ 45
(t − τ )1
2
∫ 2 < t < 6
= 45t −
65
2 < t < 6
(6) Final region 6 < t < 7 , say t = 6.5; Triangular pulse has moved to the right of the square pulse as shown below.
We have
h(t) = f (t)⊗ g(t) = dτ 45
(t − τ )t−5
2
∫ 6 < t < 7
= −25t 2 +
85t +
425
6 < t < 7
We finally get
h(t) =
0 t < 125t 2 −
45t +
25
1 < t < 2
45t −
65
2 < t < 6
−25t 2 +
85t +
425
6 < t < 7
0 7 < t
⎧
⎨
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
Page 8
which looks like
A very tricky and tedious process to comprehend !!
Correlation
The correlation process gives a measure of the similarity of two signals. One of its most important applications is to pick a known signal out of a sea of noise. The cross-correlation between f(t) and g(t) is defined as
ψ fg (t) = dτ f (τ )g(τ − t)
−∞
∞
∫The cross-correlation of a function with itself
ψ ff (t) = dτ f (τ ) f (τ − t)
−∞
∞
∫is called an autocorrelation.
This operation is similar to the convolution operation except that the second function is not inverted. It is just as tricky as the convolution operation.
We can write
FT dτ f (τ )g(τ − t)
−∞
∞
∫⎡
⎣⎢
⎤
⎦⎥ = 2πF(ω )G(−ω )
Page 9
Examples:
(1) The Square Pulse - Consider the function
f (t) =
1 − T / 2 < t < T / 2 0 otherwise
⎧⎨⎩
f(t) is absolutely integrable so it has a valid Fourier transform. It is given by
F(ω ) = 12π
dte− iω t−T /2
T /2
∫ =2π
sinωT2
ω
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
which looks like (for T=1)
In the limit T →∞ we have (T = 50, in fact, here)
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We get a sharp spike, but the area remains constant. This implies that as T →∞
F(ω )→δ (ω )Formally, we have
F(ω ) = lim
T→∞
12π
dte− iω t =−T /2
T /2
∫12π
dte− iω t =−∞
∞
∫ 2πδ (ω )
(2) Transform of a Delta-Function - Consider the function
Remember this only makes sense inside an integral.
(3) Transform of a Gaussian - Consider the function
f (t) = α
πe−α
2 t2 = normalized Gaussian pulse
Page 11
We choose α = 1. The peak is at α / π . The 1/2 maximum points are
separated by Δt = 1 /α . The area under the curve is = 1.
The Fourier transform is
F(ω ) = 1
2πdt
απ−∞
∞
∫ e−α2 t2 e− iω t =
απ 2
dt−∞
∞
∫ e−(α2 t2 + iω t )
We complete the square to evaluate the integral. We have
α2t2 + iωt = α2t2 + iωt + γ − γ = αt + β( )2 − γ
2αβt = iωt→ β =iω2α
γ = β 2 = −ω 2
4α 2
We thus have
F(ω ) = α
π 2e−ω 2
4α 2 dt−∞
∞
∫ e−(α t+ iω
2α)2
Let
x = αt + iω
2α→ dx = αdt
F(ω ) = 1
π 2e−ω 2
4α 2 dx−∞
∞
∫ e− x2
=12π
e−ω 2
4α 2
which is a different Gaussian.
An important feature is
f (t)→ Δt ≈1α
F(ω )← Δω ≈ 2αΔωΔt ≈ 2
In general for any f(t) we have ΔωΔt ≈ c = constant . In the wave theory of quantum mechanics, this corresponds to the Heisenberg Uncertainty Principle.
The Laplace Transform
Another important integral transform is the Laplace transform.
For a function f(t), we define the Laplace transform by
F(s) = dte− st
0
∞
∫ f (t) = L( f (t))
Page 12
The Laplace transform is a linear operator so that
L(af (t) + bg(t)) = aL( f (t)) + bL(g(t))The Laplace transform has a first shifting property expressed as
if L( f (t)) = F(s) , then L(eat f (t)) = F(s − a)The Laplace transform has a second shifting property expressed as
if L( f (t)) = F(s) , and g(t) =
f (t − a) t>a0 t<a
⎧⎨⎩
then L(g(t)) = e−asF(s)
If we let t = τ / a , then
F(s) = dte− st
0
∞
∫ f (t) = 1a
dτe− sτ /a0
∞
∫ f (τ / a)
and if aσ = s we get
aF(aσ ) = dτe− sτ /a
0
∞
∫ f (τ / a)
Examples:
(1) Heaviside unit step function
L(H (t)) = dte− st
0
∞
∫ H0 (t) = dte− st0
∞
∫ =1s= F(s)
Since integration is only between 0→∞ , this also says that
L(1) = dte− st
0
∞
∫ 1 = dte− st0
∞
∫ =1s
(2) Exponential function f (t) = eat
L(eat ) = dte(a− s )t
0
∞
∫ =1
s − a= F(s − a)
or by first shifting
L(eat f (t)) = F(s − a)
L(1) = 1s
L(eat1) = L(eat ) = 1s − a
(3) Shifted step function
Page 13
Ha (t) =
1 t>a0 t<a
⎧⎨⎩
By second shifting
L(Ha (t)) = L(H0 (t − a)) = e
−asL(H0 (t)) =e−as
s
(4) Euler function eiθ = cosθ + i sinθ . From (2) we have
L(eiω t ) = 1s − iω
= L(cosωt + i sinωt) = L(cosωt) + iL(sinωt)
= 1s − iω
s + iωs + iω
=s
s2 +ω 2 + iω
s2 +ω 2
L(cosωt) = ss2 +ω 2
L(sinωt) = ωs2 +ω 2
(5) Power function t k
L(t k ) = t ke− stdt0
∞
∫ =1sk+1 xke− xdx
0
∞
∫ using x = st
= 1sk+1 Γ(k +1) = k!
sk+1
(6) Power series
f (t) = ant
n
n=0
∞
∑
F(s) = L( f (t)) = an
n=0
∞
∑ L(t n ) = ann=0
∞
∑ n!sn+1
(7) Bessel Function - we will see later that the Bessel function can be written as