Post on 06-Dec-2020
Evaluation of Definite Integrals via the Residue Theorem
ECE 6382
Notes are from D. R. Wilton, Dept. of ECE
1
David R. Jackson
Fall 2020
Notes 11
Review of Singular Integrals
Logarithmic singularities are examples of integrable singularities:
( )1 1 1
0 00 0Ln Ln Ln 1 lim Ln 0lim lim
x xx dx x dx x x x x x
εεε ε = →→ →= = − = − =∫ ∫ since
1x
Ln x
2
Note: There might be numerical trouble if one integrates this function numerically!
π θ π− < <( ) ( )Ln lnz r iθ= +
Singularities like 1/x are non-integrable.
3
Review of Singular Integrals (cont.)
x
1x
1ε
( )1 1 1
0 0 0
1 1 Lnlim limx
dx dx xx x εεε ε =→ →
= = = −∞∫ ∫
Review of Cauchy Principal Value Integrals
2 0 2 0 2
1 01 1 0Ln Ln
x x
dx dx dxI x xx x x =− =− −
= = + = +∫ ∫ ∫
Consider the following integral:
A finite result is obtained if the integral interpreted as
( )2 2 2
11 10 0
0
lim lim Ln Ln
lim Ln
x x
dx dx dxI x xx x x
ε ε
εεε ε
εε
− −
=− =+− −→ →
→
= = + = +
=
∫ ∫ ∫Ln1 Ln2 Lnε− + −( ) Ln2=
The infinite contributions from the two symmetrical shadedparts shown exactly cancel in this limit. Integrals evaluated in this way are said to be (Cauchy) principal value (PV) integrals:
2 2
1 1PV dx dxI
x x− −= =∫ ∫
or
4
x
1/xε−
ε1−
2
Notation:
1/x singularities are examples of singularities integrable only in the principal value (PV) sense.
Principal value integrals must not start or end at the singularity, but must pass through them to permit cancellation of infinities
x
1x
5
Cauchy Principal Value Integrals (cont.)
Singularities like 1/x2 are non- integrable (even in the PV sense).
2 2
2
2
2
1
1
1 1 1 1 1 1
, 0sgn( ) , 0
b
a
x
x
dx dxx x a b
xxx x
ε
ε ε ε− + = + + − → ∞
> = − <
∫ ∫
but note that has a PV integral
6
x
2
1x
a bεε−
Cauchy Principal Value Integrals (cont.)
Summary of some results:
• Ln x is integrable at x = 0
• 1/xα is integrable at x = 0 for 0 < α < 1
• 1/xα is non-integrable at x = 0 for α ≥ 1
• f (x)sgn(x)/|x|α has a PV integral at x = 0 for α < 2 if f(x) is smooth
(The above results translate to singularities at a point a via the transformation x → x-a.)
7
Singular Integral Examples
Integrals Along the Real Axis
( )f x dx∞
−∞∫
On the large semicircle we then have
0
lim ( ) lim ( ) 0R
i i
R RC
f z dz f Re iRe dπ
θ θ θ→∞ →∞
= =∫ ∫
( )( ) 2 ( )C
I f z dz f x dx i f zπ∞
−∞
= = = ∑∫ ∫
residues of in the UHP
f is analytic in the UHP except for a finite number of poles (can be extended to handle poles on the real axis via PV integrals).
f is , i.e. in the UHP.( )1 /o z ( )lim 0z
z f z→∞
=
8
Assumptions:
Hence
R−
: ,iR
i
C z R edz iR e d
θ
θ θ
=
=
Rx
y×
××
×
Note:If the function is analytic in the LHP except for poles, then we would close the
contour in the LHP.
UHP: Upper Half Plane
Example:
( )( ) ( )( )2 2
2 22 2 2 20
4( )
1 12 29 4 9 4
f z
z dx zI dxz z z z
z
∞ ∞
−∞
−=
= = =+ + + +
∫ ∫
2 ( ) ( )
( )3
Res 3 + Res 2
3limz i
i f i f i
z ii
π
π→
−=
( ) ( )
2
3 3
z
z i z i+ − ( )( )2
2 22
2 + lim
4 z i
z iddzz →
−
+ ( ) ( )
2
22 9 2
z
z z i+ − ( )
( )( ) ( ) ( )( )( )( ) ( )
2
2 22 2 2
2 42 2
2
9 2 2 2 2 2 2 93 + lim50 9 2
3 13 50 200 200
z i
z i
z z i z z z z i z i ziiz z i
i ii
π
ππ
→
+ + + − + + + + = + + = − =
9
( ) ( )2
22 20 9 4
x dxIx x
∞
=+ +
∫
( )1
0 01
0
1Res ( ) ( ) ( )1 !
mm
mdf z z z f z
m dz z z
−
− = − − =
( ) ( )2
22 20 2009 4
x dxIx x
π∞
= =+ +
∫
z = 2iz = 3i
x
y
2z i=(Note the double pole at !)
×
×
Integrals Along the Real Axis (cont.)
( ) ( ) ( ) ( )
( ) ( )
1
2 2,10
2 00
lim1 1 1 1
lim lim1 1
R
R
R
C RC R C C
i
RC C
dz dzIz z z z
i edzI Iz z
ρ
ρ
ρ
ρρ
θ
ρρ
ρ
− −
→∞− − +→
→∞ →→
= = + + + + + + +
= + + = + + +
∫ ∫ ∫ ∫ ∫
∫ ∫
( )21 1i i
d
e eθ θ
θ
ρ ρ − + + ( ) ( )
2
2 20
2 2 2
2 20 0
lim1 1
lim lim 02 2 2
i
i iR
i i
R R
iR e dR e R e
id ie d i ie d iI I IR R
π π φ
φ φπ
π π πφ φ
π
φ
θ φ π φ π
→∞
− −
→∞ →∞
++ +
= + + = + + = + +
∫ ∫
∫ ∫ ∫
Consider the following integral:I
10
Cauchy Principal Value IntegralsExample
( ) ( )2 1 1dxI
x x∞
−∞=
+ +∫
2CiI I π
= +
Hence,
Evaluate:z i=
z i= −
: ,iR
i
C z R edz iR e d
φ
φ φ
=
=
: 1 ,i
i
C z edz i e d
θρ
θ
ρρ θ
+ ==
RR−1z = −
×
×
×
C
We could have also used the formula from Notes 10 for going
halfway around a simple pole on a small semicircle.
[ ]( )
2 Res ( ) Res ( 1) 2 limC z i
z iI i f z i f z iπ π
→
−= = + = − =
( )z i− ( )( )( )
1
1lim
1 z
z
z i z → −
++
+ + ( ) ( )2 1 1z z+ +
( )1 1 1 1 1 12 2 2
2 1 2 4 2 4 2 2 2i ii i i i
i i iπ ππ π π
− − − = + = + = + = + +
Next, evaluate the integral ICusing the residue theorem:
11
( ) ( )2 1 1dxI
x x∞
−∞=
+ +∫
2I π
=
Hence:
Cauchy Principal Value Integrals (cont.)
2 2CI iπ π = + 2C
iI I π= +
Recall:Thus:
z i=
z i= −
: ,iR
i
C z R edz iR e d
φ
φ φ
=
=
: 1 ,i
i
C z edz i e d
θρ
θ
ρρ θ
+ ==
RR−1z = −
×
×
×
C
Choosing the contour shown, the contribution from the semicircular arc vanishes as R → ∞ by Jordan’s lemma.
Fourier Integrals( ) i axf x e dx
∞
−∞∫
f is analytic in the UHP except for a finite number of poles (can easily be extended to handle poles on the real axis via PV integrals),
lim ( ) 0, 0 arg ( )z
f z z zπ→∞
= ≤ ≤ in UHP
0a >Assume (close the path in the UHP)
12
Assumptions:
(See next slide.)Fourier
R−
: ,iR
i
C z R edz iR e d
θ
θ θ
=
=
Rx
y
××
×
Jordan’s lemma
( )
( )
/2cos sin sin
0 0/2 2
0
( ) ( ) 2 max ( )
2 max ( ) 2
R
iaz i iaR aR i i aR
CaR
i
f z e dz f Re e e iRe d R f Re e d
R f Re e d R
π πθ θ θ θ θ θ
π θθ π
θ θ
θ
− −
−
= ≤
≤ =
∫ ∫ ∫
∫ ( ) max ( )2
if Rea R
θ π ( )1 0aRe−− →
2sin20 / 2 : sin
aRaRe e
θθ πθθ π θ
π−−⇒≤ ≤ ≥ ≤
13θ
1
2π
sinθ
2θπ
Fourier Integrals (cont.)
lim ( ) 0, 0 arg ( )z
f z z zπ→∞
= ≤ ≤ in UHP
Assume
( ) 0R
iaz
C
f z e dz →∫Then
Proof:
R−
: ,iR
i
C z R edz iR e d
θ
θ θ
=
=
Rx
y
××
×
Fourier Integrals (cont.)
( ) i axI f x e dx∞
−∞
= ∫
( ) ( ) 2 ( )iax iaz iaz
C
I f x dx f z dz i f ze e eπ∞
−∞
= = = ∑∫ ∫ residues of in the UHP
14
We then have
0a >
Question: What would change if a < 0 ?
R−
: ,iR
i
C z R edz iR e d
θ
θ θ
=
=
Rx
y
××
×
Use the symmetries of cosλx and sinλx and the Euler formula, we have:
Example:
( )2 20
cos , , 0.xI dx ax a
λ λ∞
= >+∫
cos sini xe x i xλ λ λ= +Note :
2 212
i xeI dxx a
λ∞
−∞
=+∫ (imaginary part vanishes by symmetry!)
( )2 2 2 2
1 1 12 Res = 2 lim2 2 2
i z i z
z iaz ia
z iae eI dz i iz a z a
λ λ
π π∞
→−∞ =
− = = + +
∫ ( ) ( )
i ze
z ia z ia
λ
+ −
= 2 iπ 12 2
aei
λ−
2
aeaa
λπ −
=
15
Fourier Integrals (cont.)
2 20
cos2
ax eIx a a
λλ π∞ −
= =+∫
z = ia
x
y
×
Rational Functions of sin and cos
( )2
0
sin ,cosf dπ
θ θ θ∫ f is finite within the interval. f is a rational function (ratio of polynomials) of sinθ, cosθ.
1 1
,
sin , cos2 2
i i dzz e dz ie d d iz
z z z zi
θ θ θ θ
θ θ− −
⇒= = = −
− += =
unit circle
Let
( )1 1
1
1 1
2
0
,2 2
12 ,2 2
sin ,cosz
z z z z dzI fi z
z z z zfz i
f d iπ
π
θ θ θ− −
=
− −
− +=
− +=
= − ∫
∑
∫
Residues of inside the unit circle
1 11 ,2 2
z z z zf zz i
− − − +
will be a rational function ofNote : 16
Assumptions:
Example:
( ) ( )( )
( )( )( )
1
12
2
5 504 4 21
2 121 1
12
121
1sin
4 42 5 2 2 2
2 2 lim2
z ziz
z z
z iz
d dzI iz
i dzidzz iz z i z i
z idz iz i z i
π θθ
π
−−=
= =
→ −=
= = − + +
= − =+ − + +
+= =
+ +
∫ ∫
∫ ∫
∫
( ) ( )12
2
2z i z i+ +83π
=
17
2
504 sin
dIπ θ
θ=
+∫
Multiply top and bottom by 4i.
2
504
8sin 3
dIπ θ π
θ= =
+∫
12z i= −
2z i= −
1z =
×
×
x
y
Rational Functions of sin and cos (cont.)
Exponential Integrals There is no general rule for choosing the contour of integration; if the integral
can be done by contour integration and the residue theorem, the contour is usually specific to the problem.
, 0 11
ax
xeI dx a
e
∞
−∞
= < <+∫
Consider the contour integral over the path shown in the figure:
1 2 3 41 1
az az
C z zC
e eI dz dze eγ γ γ γ
= = + + + + +
∫ ∫ ∫ ∫ ∫
The contribution from each contour segment in the limit must be separately evaluated (next slide).
R → ∞
(2 1) , 0, 1, 2,z n i nπ= + = ± ±
The integrand has simple poles at
18
Example:
4γ
3γ
2γ1γ
R− R
2z iπ=
z iπ=x
y
×
× 3z iπ=
Exponential Integrals (cont.)
1
1 : , ,
lim1
az
zR
z x dz dxe dz I
eγ
γ
→∞
= =
=+∫
3
3
2 2
: 2 , ,
lim lim1 1
Raz axia ia
z xR RR
z x i dz dx
e edz e dx e Ie e
π π
γ
γ π−
→∞ →∞
= + =
= = −+ +∫ ∫
( )
2
2
2
0
2 2 1
0 0
1
: ,
1 1
0,1 1
az aR iay
z R iy
a RaR
R R
z R iy dz idy
e e edz i dye e e
e edy dye e
a
π
γ
π π
γ
−
−
= + =
=+ +
≤− −
<= →
∫ ∫
∫ ∫19
R → ∞
y π=Smallest for :
1 1R iy Re e e+ ≥ −
1Re y
1 R iye e+
4γ
3γ
2γ1γ
R− R
2z iπ=
z iπ=x
y
×
× 3z iπ=
Exponential Integrals (cont.)
4
4
0
2
2
0
: ,
0
,
1 1
0,1
az aR iay
z R iy
aR
R
z R iy dz idy
e e edz i dye e e
e d aye
γ π
π
γ−
−
−
−
= − + =
=+ +
≤ →−
>
∫ ∫
∫
Hence
( ) ( )
( ) ( ) ( ) ( )
21 2 Res
Res lim lim lim1 1 1
ia
azaz az
z z iz i z i z i
e I i f z i
z i ee ef z i z i z ie e
π
ππ π π
π π
ππ π π −→ → →
− = =
−= = − = − =
+ − 1− ( ) ( )
( )
212
12
lim1
azia
z i
z i z i
e ez i
π
π
π π
π→
+ − + − +
= = −− − − −
( )( )( ) ( )0
0
0
0
( )( ) , 0( )
Res ( ) lim lim1
1
az
z z z i z
a i a ia i
i
g zf z h zh z
g z ef z i dh z edz
e e ee
π
π ππ
π
π→ →
= =
⇒ = = =′ +
= = = −−
Alternatively,
20
4γ
3γ
2γ1γ
R− R
2z iπ=
z iπ=x
y
×
Exponential Integrals (cont.)
Therefore
22 2
1 1
ax ia ia
x iae ie i eI dx
e e
π π
π
π π∞
−∞
− −= = =
+ −∫ iae π ( ), 0 1
sinia iaa
ae eπ π
ππ−
= < <−
( )21 2ia iae I ieπ ππ− = −
21
We thus have
, 0 11 sin
ax
xeI dx a
e aπ
π
∞
−∞
= = < <+∫
Hence
Integration Around a Branch Cut A given contour of integration is chosen: usually problem specific, usually must
not cross a branch cut.
We take advantage of the fact that the integrand changes across the branch cut.
Usually an evaluation of the contribution from the branch point is required.
0
, 0 11
kxI dx kx
∞ −
= < <+∫
22
Example:
Assume the branch 0 ≤ θ < 2π
( kx− = positive real)
1z = −x
y
×
Note: We choose the branch cut on the positive real axis (the axis of integration).
Integration Around a Branch Cut (cont.)
0
, 0 11
kxI dx kx
∞ −
= < <+∫
First, note the integral exists since the integral of the asymptotic forms of the integrand at both limits exists:
0
1
0, 11
, 01
kk
x
kk
x
x x x kxx x x kx
−−
→
−− −
→∞
→ = <+
→ = ∞ >+
which is integrable at
which is integrable at
23
Integration Around a Branch Cut
0
, 0 11
kxI dx kx
∞ −
= < <+∫
We’ll evaluate the integral using the contour shown.
24
( ) ( ) ( )lnln ln ln ln , 0 2k kk r ik z k z k r ik r ik k ikz e e e e e e e r eθ θ θ θ θ π
− −− +− − − − − − −= = = = = = < <
0 2θ π< <
iz re θ=
RC
1z = −
0C
R
εε
1L
2Lx
y
2ε
0ε →
×0r
Now consider the various contributions to the contour integral:
( ) ( )
( )1 2 0
2 Res 1 ,
1
RL L C Ck
f z dz i f
zf zz
π+ + +
−
= −
=+
∫
where
0 0 0 0: , ,i i k k ikC z r e dz ir e d z r eθ θ θθ − − −= = =
( )0 0
0
010 0
00, 002 20
lim lim 01
k ik ik i ik
ir rC
r e ir e df z dz ir e dr e
ε θ θθ θ
θπ ε πε
θ θ− −
− −
→ →−→
= = →+∫ ∫ ∫
: , ,i i k k ikRC z Re dz iRe d z R eθ θ θφ − − −= = =
( )2 2
,00
lim lim 01
R
k ik ik ik
iR RC
R e iRe df z dz iR e dRe
π ε πθ θθ
θεε
φ φ− − −
− −
→∞ →∞→
= = →+∫ ∫ ∫
25
Integration Around a Branch Cut (cont.)
RC
1z = −
0C
R
εε
1L
2Lx
y
2ε
0ε →
×0r
1 : , ,i i k k ikL z re dz e dr z r eε ε ε− − −= = =
( )1 00
( )
000
lim1 1
R k ki k
iRL rr
r dr r drf z dz e Ire r
ε εε
ε
∞− −−
→∞→→
= = =+ +∫ ∫ ∫
( ) ( )2 22 : , ,i iki i k kL z re re dz e dr z r eπ ε π εε ε− − −− − − −= = = =
( ) [ ]0
2 0
(2 ) 2 2
000
lim1 1
r k ki k i k i k
iRL Rr
r dr r drf z dz e e e Ire r
ε π ε π πε
ε
∞− −− + − − −
−→∞→→
= = − = −+ +∫ ∫ ∫
26
Integration Around a Branch Cut (cont.)
For the path L2:
For the path L1:( )
1
kzf zz
−
=+
RC
1z = −
0C
R
εε
1L
2Lx
y
2ε
0ε →
×0r
( ) ( )
( )
2
1
1 2 Res 1
2 lim 1
i k
z
e I i f z
i z
π π
π
−
→−
− = = −
= +1
kzz
−
+
( )
( )2 1
2
2
k
ki
ik
i
i e
i e
π
π
π
π
π
−
−
−
= −
=
=
( )2
2 21
ik ik
i k
i e i eIe
π π
π
π π− −
−= =
− ike π− ( ), 0 1
sinik i kk
ke eπ π
ππ−
= < <−
Hence
Therefore, we have
27
Integration Around a Branch Cut (cont.)
Note: Arg (-1) = π here.
0
, 0 11 sin
kxI dx kx k
ππ
∞ −
= = < <+∫
Hence
“Dispersion” Relations
28
Assumptions:
( )( ) 0 , z
f z u iv
f z
= +
→
(including the real axis)is analytic in the UHP
in the UHP
0iz x e θε− =Let
( ) ( ) ( )0
0
0
0 00
limix R
iRC R x
f x ef z f xdz dx
z x x x e
θε
θεε
ε
ε
−
→∞− +→
+= + + − −
∫ ∫ ∫
ii e θε ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2
0
0 0 00
0 0 00 0 0
2
2
1 1( ) ( )
d i f x
f xdx i f x i f x i f x
x x
f x v x u xif x u x iv x dx dx dxi x x x x x x
π
π
θ π
π π π
π π π
∞
−∞
∞ ∞ ∞
−∞ −∞ −∞
=
= − =−
= + = = −− −
⇒
−⇒
∫
∫
∫ ∫ ∫
(From the residue theorem or the Cauchy integral theorem.)
CPV ( )0i f xπ
UHP: Upper Half Plane
y
RR−x0x
ε×
0Rε
→ ∞→
C
“Dispersion” Relations (cont.)
29
( ) ( ) ( )0 0 0
0 0
1( ) ( )v x u xif x u x iv x dx dxx x x xπ π
∞ ∞
−∞ −∞
= + = −− −∫ ∫
Equate the real and imaginary parts:
( )
( )
00
00
1( )
1( )
v xu x dx
x x
u xv x dx
x x
π
π
∞
−∞
∞
−∞
=−
= −−
∫
∫
Next, relabel: 0,x x x x′→ →
“Dispersion” Relations (cont.)
( ), ( )u x v x⇒ are of one anotherHilbert Transforms
30
Hence, we have
( )
( )
1( )
1( )
v xu x dx
x x
u xv x dx
x x
π
π
∞
−∞
∞
−∞
′′=
′ −
′′= −
′ −
∫
∫ Hilbert
( )( ) 0 ,
f z u iv
f z z
= +
→
(including the real axis)
Assumptions :is analytic in the UHP
in the UHP
“Dispersion” Relations: Circuit Theory
31
Note 1: A pole in the LHP would corresponds to a nonphysical growing response:ir ti ti t
r ii e e e ωωωω ω ω −= + ⇒ =
Note 2: The system is assumed to be unable to respond to a signal at very high frequency.
( ) ( ) ( ) ( ) ( ) ( )* ;R R I IH H H H H Hω ω ω ω ω ω− = ⇒ − = − = −
(see Appendix)
Symmetry property:
Reω
Imω
LHP: Lower Half Plane
( )( )
( )
1
2
exp(
( )
( ) 0
)
,
i
H
t
H ω
ω
ω
ω→ → ∞
time convention
is an
Assumptio
alytic in the LHP
for in the
ns
LHP
:
( ) ( ) ( )R IH H i Hω ω ω= +( )inV ω+
−( )outV ω
+
−
( ) ( )( ) /out inH V Vω ω ω≡
Transfer function
“Dispersion” Relations: Circuit Theory (cont.)
32
Use the path shown below:
( )0
( ) ( )limiR
iRC R
H eH Hd de
θω ε
θω εε
ω εω ωω ωω ω ω ω ε
−
→∞− +→
+ ′ ′′ ′= + + ′ ′− −
∫ ∫ ∫
CPV
ii e θε
( )
( )
( ) ( )
( ) ( )
0
2
1
i H
d i H
Hd i H
HH d
i
π ω
π
θ π ω
ωω π ω
ω ω
ωω ω
π ω ω
∞
−∞
−
∞
−∞
= −
′′ = −
′ −
′
⇒
⇒′
′= −−
∫
∫
∫
ReRR−
Im
ε
ω
ω′
C
×
(From the residue theorem or the Cauchy integral theorem.)
33
( ) ( )1 HH d
iω
ω ωπ ω ω
∞
−∞
′′= −
′ −∫
( ) ( ) ( ) ( ) ( )1 I RR I
H HiH H iH d dω ω
ω ω ω ω ωπ ω ω π ω ω
∞ ∞
−∞ −∞
′ ′′ ′= + = − +
′ ′ −⇒
−∫ ∫
( ) ( )
( ) ( )
1
1
IR
RI
HH d
HH d
ωω ω
π ω ω
ωω ω
π ω ω
∞
−∞
∞
−∞
′′= −
′ −
′′=
′ −
∫
∫
We thus have
“Dispersion” Relations: Circuit Theory (cont.)
The real and imaginary parts of the transfer functionare Hilbert transforms of each other.
34
We can also derive an alternative form:
( ) ( ) ( )( )( )2 2
0
21 1R RI
H HH d d
ω ω ωω ω ω
π ω ω π ω ω
∞ ∞
−∞
′ ′′ ′= = −
′ − ′ −∫ ∫
Similarly, we have
“Dispersion” Relations: Circuit Theory (cont.)
( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( )( )( )( )
( )( )( )
0
00
0
0 0
0
2 20
1 1 1
1 1
1 1
1
21
I I IR
I I
I I
I
I
H H HH d d d
H Hd d
H Hd d
Hd
Hd
ω ω ωω ω ω ω
π ω ω π ω ω π ω ω
ω ωω ω
π ω ω π ω ω
ω ωω ω
π ω ω π ω ω
ω ω ω ω ωω
π ω ω ω ω
ω ωω
π ω ω
∞ ∞
−∞ −∞
∞
∞
∞ ∞
∞
∞
′ ′ ′′ ′ ′= − = − −
′ ′ ′− − −
′ ′−′ ′= −
′ ′− − −
′ ′′ ′= − −
′ ′+ −
′ ′ ′− + +′= −
′ ′+ −
′ ′′= −
′ −
∫ ∫ ∫
∫ ∫
∫ ∫
∫
∫
ω ω′ ′→ −First one :Use
This integral starts at zero.
( ) ( )I IH Hω ω− = −First one :Use
35
Summarizing, we have
( ) ( ) ( )( )( )
( ) ( ) ( )( )( )
2 20
2 20
21 1
21 1
I IR
R RI
H HH d d
H HH d d
ω ω ωω ω ω
π ω ω π ω ω
ω ω ωω ω ω
π ω ω π ω ω
∞ ∞
−∞
∞ ∞
−∞
′ ′ ′′ ′= − = −
′ − ′ −
′ ′′ ′= =
′ − ′ −
∫ ∫
∫ ∫
Dispersion Relation: Circuit Theory (cont.)
( )( ) 0,
HH
ωω ω→ → ∞
is analytic in the LHPfo
Assumption
r in t
s:
he LHP
Kramers-Kronig Relations
( ) ( ) 1,r rε ω ε ω ω→The is analytic in the LHP and in LHPrelative permittivity
( ) ( ) 1
r
e
e r
εχχ ω ε ω
≡≡
≡ −
relative permittivityelectric susceptibility
Material parameters :
36
( ) ( )
( ) ( )
2 20
2 20
Im ( ) 12Re ( ) 1
Re ( ) 12Im ( ) 1
rr
rr
d
d
ω ε ωε ω ω
π ω ω
ω ε ωε ω ω
π ω ω
∞
∞
′ ′ −′− = −
′ −
′ −′− =
′ −
∫
∫
( ) 0eχ ω ω→ → ∞as in LHP
Assumption:
Similar to the transfer - function analysis, one obtains the disperson relations :Kronig - Kramers
0 eP Eε χ= (polarization per unit volume)
Kramers-Kronig Relations (cont.)
Kramers
37
( ) ( )
( ) ( )( )
2 20
2 20
21
12
rr
rr
d
d
ω ε ωε ω ω
π ω ω
ω ε ωε ω ω
π ω ω
∞
∞
′ ′′ ′′ ′= +
′ −
′ ′ −′′ ′= −
′ −
∫
∫
The final form of the Kramers-Kronig relations is then:
( )( ) ( ) ( )r r rj j iε ω ε ω ε ω′ ′′= −Denote using instead of
Note: This shows that if there is no loss (εr′′ = 0), then the relative permittivity must be 1. Hence, practical materials will always have some loss.
Laplace Transform
38
( ) ( )0
stF s f t e dt∞
−≡ ∫
The Laplace transform is defined as:
Assume
( ) 0Re s γ>
( ) 0tf t Aeγ<
Then F(s) is analytic in the region ( )Re s
( )Im s
0γ
Analytic××
×
( ) ( ) ( ) ( )00
Re >stF s t f t e dt s γ∞
−′ = −∫ valid forNote :
Laplace Transform (cont.)
39
( ) ( )12
i tg t g e dωω ωπ
∞
−∞
= ∫
Define a function g(t) (using any γ > γ0):
We then have, from the inverse Fourier transform,
( ) ( ), 00, 0
te f t tg t
t
γ− >≡
<
Then, from the relation between f and g:
The Fourier transform of g(t) exists.(The function g(t) stays finite along the entire real axis
and tends to zero.)
( ) ( )1 , 02
t i tf t g e e d tγ ωω ωπ
∞
−∞
= >∫
s iγ ω≡ +
( ) ( )( )1 , 02
ist
i
f t g i s e ds ti
γ
γ
γπ
+ ∞
− ∞
= − − >∫
Let
( )Re s
( )Im s
γ
C××
×
( ) ( ) i tg g t e dωω ω∞
−
−∞
≡
∫
40
( ) ( )( )1 , 02
ist
i
f t g i s e ds ti
γ
γ
γπ
+ ∞
− ∞
= − − >∫
( )( ) ( ) ( )( )
( ) ( )( )
( ) ( )
( )
( )
0
0
0
i i s t
i i s t
s t
st
g i s g t e dt
g t e dt
g t e dt
f t e dt
F s
γ
γ
γ
γ∞
− − −
−∞
∞− − −
∞− −
∞−
− − =
=
=
=
=
∫
∫
∫
∫
We have for the integrand term:
( )Re s
( )Im s
γ
C××
×
Laplace Transform (cont.)
41
( ) ( )1 , 02
ist
i
f t F s e ds ti
γ
γπ
+ ∞
− ∞
= >∫
“Bromwich integral”
( )Re s
( )Im s
γ
C××
×
Laplace Transform (cont.)
The Bromwich contour C is chosen to the right of all singularities of the function F(s).
“Inverse Laplace transform”
Hence we have:
42
( ) ( )12
R
st
C
f t F s e dsiπ
= ∫
0t <
Close the contour to the right.
( ) 0f t =
Laplace Transform (cont.)
Consider the case:
( )Re s
( )Im s
γ
RC×
×
×
∞C
( ) ( )12
ist
i
f t F s e dsi
γ
γπ
+ ∞
− ∞
= ∫ RCThe integrand is analytic inside .
( )
:
00 ( )
R
st
C C
eF s
→ ∞
→
→ assumption
Only the vertical path contributes as
Right :Top and bottom :
(by Cauchy’s theorem)
43
Laplace Transform (cont.)
( ) ( ) , 012 0, 0
ist
i
f t tF s e ds
i t
γ
γπ
+ ∞
− ∞
>=
<∫
Summary
( )Re s
( )Im s
γ
C××
×
Note: This inverse Laplace transform integral gives us zero for t < 0, no matter how the original
function f(t) was defined for t < 0.
44
Close the contour to the left.
( ) ( )12
L
st
C
f t F s e dsiπ
= ∫
( ) Cf t = ∑ residues at poles to the left of
(This assumes that there are only pole singularities.)
Laplace Transform (cont.)Evaluation of the Bromwich integral for the case of poles:
( )Re s
( )Im s
γ
××
×
LC
−∞C
( ) ( )12
ist
i
f t F s e dsi
γ
γπ
+ ∞
− ∞
= ∫0t >
( )
:
00 ( )
L
st
C C
eF s
→ ∞
→
→ assumption
Only the vertical path contributes as
Left :Top and bottom :
Poles only
Laplace Transform (cont.)
45
Example ( ) atf t e=
( ) ( ) ( )
00 0
1s a t s a tat stF s e e dt e dt es a
∞∞ ∞− − − −− −
= = =−∫ ∫
( ) ( )1 , ReF s s as a
= >−
( ) 1 1 ,2
ist
i
f t e ds ai s a
γ
γ
γπ
+ ∞
− ∞
= >−∫
( )Re s
( )Im s
×
LC−∞
γa
( ) 1 12 Res2
st
s a
f t i ei s a
ππ =
= −
( ) , 0atf t e t⇒ = >
For t > 0:
Appendix
46
( ) ( )*H Hω ω− =Proof of symmetry property:
( ) ( )12
i th t H e dωω ωπ
∞
−∞
= =∫ real
( ) ( ) ( ) ( ) ( ) ( )* * * * *1 1 1 12 2 2 2
i t i t i t i th t h t H e d H e d H e d H e dω ω ω ωω ω ω ω ω ω ω ωπ π π π
∞ −∞ ∞ ∞′ ′−
−∞ ∞ −∞ −∞
′ ′ ′ ′= = = − − = − = −∫ ∫ ∫ ∫
ω ω′ = −Use
( ) ( )*H Hω ω− =
so
( ) ( )*1 12 2
i t i tH e d H e dω ωω ω ω ωπ π
∞ ∞
−∞ −∞
= −∫ ∫
Hence:
Impulse response:
( ) ( ) i tH h t e dtωω∞
−
−∞
= ∫
ω ω′ →Relabel
( ) ( )1 1 *F H F Hω ω− − = −
( ) ( ) ( )R IH H i Hω ω ω= +( )inV ω+
−( )outV ω
+
−
( ) ( )( ) /out inH V Vω ω ω≡
Transfer function