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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:
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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
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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.
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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
⎧
⎨
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
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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 )
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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
f (t) = δ (t). The transform is
F(ω ) = 1
2πdte− iω tδ (t) =
−∞
∞
∫12π
The inverse transform is
12π
dte− iω t12π
=−∞
∞
∫ δ (t)
Now
FT
dfdt
⎛⎝⎜
⎞⎠⎟= iωFT ( f ) = iωF(ω )
Therefore for the square pulse we have
dfdt
= δ (t + T / 2) − δ (t − T / 2)
ℑdfdt
⎛⎝⎜
⎞⎠⎟= ℑ δ (t + T / 2) − δ (t − T / 2)( )
But
FT ( f (t − t0 )) = e− iω t0FT ( f (t))
Thus,
FTdfdt
⎛⎝⎜
⎞⎠⎟= e− iω (−T /2) − e− iω (T /2)( )FT δ (t)( ) = i 2
πsinωT
2= iωF(ω )
F(ω ) = 2π
sinωT2
ω
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
for the square pulse ( as before)
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
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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))
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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
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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
J0 (t) = (−1)k
22k (k!)2 t2k
k=0
∞
∑ = Bessel function of order 0
We have
L(J0 (t)) =
(−1)k
22k (k!)2L(t 2k
k=0
∞
∑ ) = (−1)k
22k (k!)2(2k)!s2k+1k=0
∞
∑Now
(2k)!= 1 ⋅2 ⋅ 3 ⋅ ...........2k = 2 ⋅ 4 ⋅6 ⋅ ......... ⋅2k ⋅1 ⋅ 3 ⋅5 ⋅ ....... ⋅ (2k −1) = 2k k!⋅1 ⋅ 3 ⋅5 ⋅ ....... ⋅ (2k −1)
so
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L(J0 (t)) =
(−1)k
22k (k!)2(2k)!s2k+1k=0
∞
∑ =1s1+ (−1)k
22k k!1 ⋅ 3 ⋅5 ⋅ ....... ⋅ (2k −1)
s2kk=1
∞
∑⎡
⎣⎢
⎤
⎦⎥
Now using the binomial theorem
(1+ x)n = 1+ n1!x +
n(n −1)2!
x2 + ....+ n(n −1)......(n − k +1)k!
xk + ....
1+ 1s2
⎛⎝⎜
⎞⎠⎟−1/2
= 1+(−1) 1
2⎛⎝⎜
⎞⎠⎟
1!1s2
⎛⎝⎜
⎞⎠⎟+(−1)2 1
2⎛⎝⎜
⎞⎠⎟32
⎛⎝⎜
⎞⎠⎟
2!1s2
⎛⎝⎜
⎞⎠⎟2
+ ...+(−1)k 1
2⎛⎝⎜
⎞⎠⎟32
⎛⎝⎜
⎞⎠⎟ ...
2k −12
⎛⎝⎜
⎞⎠⎟
2!1s2
⎛⎝⎜
⎞⎠⎟k
or
L(J0 (t)) =
1s1+ 1
s2⎛⎝⎜
⎞⎠⎟−1/2
=1s2 +1
Wow!
(7) Dirac Delta Function
L δ (t − t0 )[ ] = δ (t − t0 )e
− stdt0
∞
∫ = e− st0
There is a tables of Laplace transforms.
More examples - typical electric circuit input functions):
(1) Consider the square pulse
f (t) =0 for t<a A for a<t<b 0 for t>b
⎧⎨⎪
⎩⎪
We rewrite this function as
f (t) = AHa (t) − AHb (t)and then using linearity and (3) we have
L( f (t)) = A e
−as
s− A
e−bs
s=Ase−as − e−bs⎡⎣ ⎤⎦
(2) Consider the time-dependent pulse
f (t) =0 for 0<t<1 t 2 for 1<t<2 0 for 2<t
⎧⎨⎪
⎩⎪
We rewrite this function as
f (t) = H1(t)t2 − H2 (t)t
2
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g(t) = t 2 is not in the proper form to use the second shift property. We fix it by the following algebra
f (t) = H1(t) (t −1)2 + 2(t −1) +1⎡⎣ ⎤⎦ − H2 (t) (t − 2)
2 + 4(t − 2) + 4⎡⎣ ⎤⎦Now using the second shift property and linearity we have
L( f (t)) = L H1(t) (t −1)2 + 2(t −1) +1⎡⎣ ⎤⎦( ) − L H2 (t) (t − 2)2 + 4(t − 2) + 4⎡⎣ ⎤⎦( ) = (t −1)2 + 2(t −1) +1⎡⎣ ⎤⎦e
− stdt1
∞
∫ − (t − 2)2 + 4(t − 2) + 4⎡⎣ ⎤⎦e− stdt
2
∞
∫
= e− s x2 + 2x +1⎡⎣ ⎤⎦e− sxdx
0
∞
∫ − e−2s x2 + 4x + 4⎡⎣ ⎤⎦e− sxdx
0
∞
∫
= e− s 2s3 +
2s2 +
1s
⎡⎣⎢
⎤⎦⎥− e−2s 2
s3 +4s2 +
4s
⎡⎣⎢
⎤⎦⎥
Finally, let us use the table to find the inverse Laplace transform.
(1) Consider
F(s) = 2s
s2 + 4Now
L(cosωt) = s
s2 +ω 2 → L(2cos2t) = 2ss2 + 4
→ f (t) = 2cos2t = L−1 ss2 +ω 2
⎛⎝⎜
⎞⎠⎟
(2) Consider
F(s) = 6s
s2 + 4s +13Now
F(s) = 6s
s2 + 4s +13=
6s(s + 2)2 + 9
=6(s + 2) −12(s + 2)2 + 9
=6(s + 2)(s + 2)2 + 9
−12
(s + 2)2 + 9which is a form where we can use the first shift property. We have
L(eat f (t)) = F(s − a) and L(cosωt) = s
s2 +ω 2
or
L(e−2t cos3t) = L(cos3(t + 2)) = (s + 2)
(s + 2)2 + 9and
L(eat f (t)) = F(s − a) and L(sinωt) = ω
s2 +ω 2
or
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L(e−2t sin 3t) = L(sin 3(t + 2)) = 3
(s + 2)2 + 9Therefore,
F(s) = 6(s + 2)(s + 2)2 + 9
−12
(s + 2)2 + 9= 6L(e−2t cos 3t) − 4L(e−2t sin 3t)
= L(6e−2t cos 3t − 4e−2t sin 3t)or
f (t) = L−1 6s
s2 + 4s +13⎛⎝⎜
⎞⎠⎟= 2e−2t (3cos3t − 2sin 3t)
(3) Consider
F(s) = 4e−2s
s2 −16= e−2s
4s2 −16
⎡⎣⎢
⎤⎦⎥
Now
L(sinh 4t) = 4
s2 −16→ L−1
4s2 −16
⎛⎝⎜
⎞⎠⎟= sinh 4t
The second shift property then gives
L(sinh 4(t − 2)) = e−2sL(sinh 4t) = e−2s 4
s2 −16or
f (t) = L−1 4e−2s
s2 −16⎛⎝⎜
⎞⎠⎟=
sinh 4(t − 2) for t>20 for t<2
⎧⎨⎩
Other Properties of the Laplace Transform
Laplace Transform of Derivatives and Integrals
I will just quote some results here without proof.
L( f ') = sL( f ) − f (0)
L( f (n) ) = snL( f ) − sn−1 f (0) − sn−2 f '(0) − .......− f (n−1)(0)
L f (τ )dτ
0
t
∫⎛
⎝⎜⎞
⎠⎟=1sL( f )
Examples:
(1) Consider f (t) = t sin t . We have
f '(t) = t cos t − sin t and f ''(t) = −t sin t + 2cos tNow
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L( f '' ) = s2L( f ) − sf (0) − f '(0) = s2L( f )
L(t sin t) = 1s2L 2cos t − t sin t( ) = 1
s2L 2cos t( ) − L t sin t( )⎡⎣ ⎤⎦
1+ 1s2
⎛⎝⎜
⎞⎠⎟L(t sin t) = 1
s2L 2cos t( ) = 2
s2L cos t( ) = 2
s2s
s2 +1
L(t sin t) = 2s(s2 +1)2
(2) If
L(t sin2t) = 2s
(s2 + 4)2
F(s) = 8
(s2 + 4)2
L( f ) = F(s) = 8(s2 + 4)2
=1s
8s(s2 + 4)2
L f (τ )dτ0
t
∫⎛
⎝⎜⎞
⎠⎟=1sL( f )→ L τ sin2τdτ
0
t
∫⎛
⎝⎜⎞
⎠⎟=1sL(t sin2t) = 1
s2s
(s2 + 4)2
or
L( f ) = 4L τ sin2τdτ
0
t
∫⎛
⎝⎜⎞
⎠⎟→ f (t) = 4 τ sin2τdτ
0
t
∫ = −2t cos2t + sin2t
Derivatives and Integrals of the Laplace Transform
L(t n f (t)) = (−1)n F (n) (s)
Lf (t)t
⎛⎝⎜
⎞⎠⎟= F(σ )dσ
s
∞
∫Laplace Transforms of Periodic Functions
We now consider functions that are periodic with period a, i.e.,
f (t) = f (t + na) n=1,2,3,.........The transform is
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L( f (t)) = f (t)e− stdt0
∞
∫ = f (t)e− stdt0
a
∫ + f (t)e− stdta
2a
∫ + f (t)e− stdt2a
3a
∫ + ......
= f (t)e− stdt0
a
∫ + f (t + a)e− s(t+a)dt0
a
∫ + f (t + 2a)e− s(t+2a)dt0
a
∫ + ......
= f (t)e− stdt0
a
∫ + f (t)e− s(t+a)dt0
a
∫ + f (t)e− s(t+2a)dt0
a
∫ + ......
= 1+ e− sa + e−2sa + ....( ) f (t)e− stdt0
a
∫ =1
1− e− saf (t)e− stdt
0
a
∫Inverse Laplace Transforms
Partial Fractions
(1) Consider two polynomials P(s) and Q(s) such that
degree(Q) > degree(P)
and Q(s) = (s − a1)(s − a2 )(s − a3)............(s − an ) with all the roots ai distinct.We saw earlier that we can then write
F(s) = P(s)
Q(s)=
A1(s − a1)
+A2
(s − a2 )+
A3(s − a3)
+ ......+ An(s − an )
We then have
lims→ak
P(s)Q(s)
(s − ak ) = Ak
= lims→ak
(s − ak )Q(s)
P(s) = P(ak ) lims→ak
(s − ak )Q(s)
= P(ak ) lims→ak
dds
(s − ak )
ddsQ(s)
= P(ak ) lims→ak
1Q '(s)
=P(ak )Q '(ak )
Example:
Suppose we know
F(s) = s3 + 3s2 − 2s + 4
s(s −1)(s − 2)(s2 + 4s + 3)then what is f(t) or how do we find the inverse Laplace transform? We have
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F(s) = s3 + 3s2 − 2s + 4s(s −1)(s − 2)(s2 + 4s + 3)
=s3 + 3s2 − 2s + 4
s(s −1)(s − 2)(s +1)(s + 3)
= A1
s+
A2
(s −1)+
A3
(s − 2)+
A4
(s +1)+
A5
(s + 3)Using
Ak =
P(ak )[Q(s) / (s − ak )]s=ak
we get
A1 = 2 / 3 , A2 = −3 / 4 , A3 = 2 / 3 , A4 = −2 / 3 , A5 = 1 /12so that
F(s) = 2 / 3
s−3 / 4(s −1)
+2 / 3(s − 2)
−2 / 3(s +1)
+1 /12(s + 3)
Using
L(eat ) = 1
s − awe have
f (t) = 2
3−34et +
23e2t −
23e− t +
112e − 3t
When the roots of Q(s) are not all distinct we get a different
result. Suppose there is a repeated factor (s − a1)m , i.e.,
Q(s) = (s − a1)m (s − a2 )(s − a3)............
We then get
F(s) = P(s)
Q(s)=
Bm(s − a1)
m +Bm−1
(s − a1)m−1 + .....+
B2(s − a1)
2 +B1
(s − a1)+
A2(s − a2 )
+A3
(s − a3)+ ......
where
Ak =
P(ak )[Q(s) / (s − ak )]s=ak
and
Bi =1
(m −1)!d (m− i )
ds(m− i )
P(s)Q(s) / (s − a1)
m
⎡
⎣⎢
⎤
⎦⎥s=a1
Bm =P(a1)
Q(s) / (s − a1)m⎡⎣ ⎤⎦s=a1
Two final cases are:
Q(s) = ((s − a)2 + b2 )(s − a1)(s − a2 )(s − a3)............
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F(s) = P(s)
Q(s)=
B1s + B2(s − a)2 + b2
+A1
(s − a1)+
A2(s − a2 )
+ ......
Ak =
P(ak )[Q(s) / (s − ak )]s=ak
B1(a + ib) + B2 =P(s)
Q(s) / (s − a)2 + b2⎡⎣ ⎤⎦
⎡
⎣⎢⎢
⎤
⎦⎥⎥s=a+ ib
and
Q(s) = ((s − a)2 + b2 )2 (s − a1)(s − a2 )(s − a3)............
F(s) = P(s)
Q(s)=
C1s + C2
((s − a)2 + b2 )2+
B1s + B2(s − a)2 + b2
+A1
(s − a1)+
A2(s − a2 )
+ ......
Ak =
P(ak )[Q(s) / (s − ak )]s=ak
c1(a + ib) + c2 =P(s)
Q(s) / (s − a)2 + b2⎡⎣ ⎤⎦2
⎡
⎣⎢⎢
⎤
⎦⎥⎥s=a+ ib
B1(a + ib) + B2 =dds
P(s)Q(s) / (s − a)2 + b2⎡⎣ ⎤⎦
2
⎡
⎣⎢⎢
⎤
⎦⎥⎥s=a+ ib
A Convolution Theorem
Suppose that we have two functions f(t) and g(t). We define the integral
h(t) = f (t − τ )g(τ )
0
t
∫ dτ ≡ f ∗ g ≡ convolution integral
If
L( f (t)) = F(s) and L(g(t)) = G(s)then we have
H (s) = dte− st
0
∞
∫ h(t) = F(s)G(s) = L( f ∗ g)
The Error Function as an Example of Convolution
As we defined earlier, the error function is defined by
erf (x) = 2
πe− t
2
dt0
x
∫From the table
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L
e− t
t⎛⎝⎜
⎞⎠⎟=
πs +1
→ L−11s +1
⎛⎝⎜
⎞⎠⎟=e− t
πtNow the convolution theorem says that
L−1
1s
1s +1
⎛⎝⎜
⎞⎠⎟= f ∗ g = f (t − τ )g(τ )dτ = (1) e
−τ
πτdτ
0
t
∫0
t
∫where
f (t) = 1 , g(t) = e− t
πt→ F(s) = 1
s,G(s) = 1
s +1
Changing variables with x = τ we get
L−1
1s
1s +1
⎛⎝⎜
⎞⎠⎟=
e−τ
πτdτ =
2π0
t
∫ e− x2
0
t
∫ dx = erf ( t )
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