notes 11 evaluation of definite integrals via the residue ...courses.egr.uh.edu/ece/ece6382/class...

Evaluation of Definite Integrals via the Residue Theorem

ECE 6382

Notes are from D. R. Wilton, Dept. of ECE

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

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.

Review of Singular Integrals (cont.)

( )1 1 1

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

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

lim lim Ln Ln

lim Ln

dx dx dxI x xx x x

εεε ε

− −

=− =+− −→ →

= = + = +

∫ ∫ ∫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:

1 1PV dx dxI

x x− −= =∫ ∫

1/xε−

ε1−

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

Cauchy Principal Value Integrals (cont.)

Singularities like 1/x2 are non- integrable (even in the PV sense).

1 1 1 1 1 1

, 0sgn( ) , 0

dx dxx x a b

ε ε ε− + = + + − → ∞

> = − <

∫ ∫

but note that has a PV integral

a bεε−

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.)

Singular Integral Examples

Integrals Along the Real Axis

( )f x dx∞

−∞∫

On the large semicircle we then have

lim ( ) lim ( ) 0R

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→∞

Assumptions:

C z R edz iR e d

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

1 12 29 4 9 4

z dx zI dxz z z z

∞ ∞

−∞

= = =+ + + +

∫ ∫

2 ( ) ( )

Res 3 + Res 2

3limz i

i f i f i

( ) ( )

z i z i+ − ( )( )2

2 + lim

z iddzz →

+ ( ) ( )

22 9 2

z z i+ − ( )

( )( ) ( ) ( )( )( )( ) ( )

2 22 2 2

2 42 2

9 2 2 2 2 2 2 93 + lim50 9 2

3 13 50 200 200

z z i z z z z i z i ziiz z i

+ + + − + + + + = + + = − =

( ) ( )2

22 20 9 4

x dxIx x

1Res ( ) ( ) ( )1 !

mdf z z z f z

m dz z z

− = − − =

( ) ( )2

22 20 2009 4

x dxIx x

= =+ +

z = 2iz = 3i

2z i=(Note the double pole at !)

Integrals Along the Real Axis (cont.)

( ) ( ) ( ) ( )

( ) ( )

2 2,10

lim1 1 1 1

lim lim1 1

C RC R C C

dz dzIz z z z

i edzI Iz z

− −

→∞− − +→

→∞ →→

= = + + + + + + +

= + + = + + +

∫ ∫ ∫ ∫ ∫

∫ ∫

( )21 1i i

e eθ θ

ρ ρ − + + ( ) ( )

2 20 0

lim1 1

lim lim 02 2 2

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

θ φ π φ π

→∞

− −

→∞ →∞

= + + = + + = + +

∫ ∫

∫ ∫ ∫

Consider the following integral:I

Cauchy Principal Value IntegralsExample

( ) ( )2 1 1dxI

x x∞

−∞=

+ +∫

2CiI I π

Hence,

Evaluate:z i=

z i= −

C z R edz iR e d

: 1 ,i

C z edz i e d

ρρ θ

RR−1z = −

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− ( )( )( )

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:

( ) ( )2 1 1dxI

x x∞

−∞=

+ +∫

Hence:

2 2CI iπ π = + 2C

iI I π= +

Recall:Thus:

z i= −

C z R edz iR e d

: 1 ,i

C z edz i e d

ρρ θ

RR−1z = −

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)

Assumptions:

(See next slide.)Fourier

C z R edz iR e d

Jordan’s lemma

/2cos sin sin

0 0/2 2

( ) ( ) 2 max ( )

2 max ( ) 2

iaz i iaR aR i i aR

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

− −

∫ ∫ ∫

∫ ( ) max ( )2

if Rea R

θ π ( )1 0aRe−− →

2sin20 / 2 : sin

aRaRe e

θθ πθθ π θ

π−−⇒≤ ≤ ≥ ≤

Fourier Integrals (cont.)

lim ( ) 0, 0 arg ( )z

f z z zπ→∞

= ≤ ≤ in UHP

Assume

( ) 0R

f z e dz →∫Then

Proof:

C z R edz iR e d

( ) i axI f x e dx∞

−∞

( ) ( ) 2 ( )iax iaz iaz

I f x dx f z dz i f ze e eπ∞

−∞

= = = ∑∫ ∫ residues of in the UHP

We then have

Question: What would change if a < 0 ?

C z R edz iR e d

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 :

i xeI dxx a

−∞

=+∫ (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

π π∞

→−∞ =

− = = + +

∫ ( ) ( )

z ia z ia

= 2 iπ 12 2

λπ −

ax eIx a a

λλ π∞ −

= =+∫

z = ia

Rational Functions of sin and cos

sin ,cosf dπ

θ θ θ∫ f is finite within the interval. f is a rational function (ratio of polynomials) of sinθ, cosθ.

sin , cos2 2

i i dzz e dz ie d d iz

z z z zi

θ θ θ θ

θ θ− −

⇒= = = −

− += =

unit circle

( )1 1

12 ,2 2

sin ,cosz

z z z z dzI fi z

z z z zfz i

f d iπ

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:

( ) ( )( )

( )( )( )

5 504 4 21

2 121 1

4 42 5 2 2 2

2 2 lim2

d dzI iz

i dzidzz iz z i z i

z idz iz i z i

π θθ

−−=

→ −=

= = − + +

= − =+ − + +

∫ ∫

( ) ( )12

2z i z i+ +83π

504 sin

dIπ θ

Multiply top and bottom by 4i.

8sin 3

dIπ θ π

12z i= −

2z i= −

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

xeI dx a

−∞

= < <+∫

Consider the contour integral over the path shown in the figure:

1 2 3 41 1

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

Example:

2γ1γ

R− R

2z iπ=

z iπ=x

× 3z iπ=

Exponential Integrals (cont.)

1 : , ,

z x dz dxe dz I

→∞

: 2 , ,

lim lim1 1

Raz axia ia

z xR RR

z x i dz dx

e edz e dx e Ie e

γ π−

→∞ →∞

= = −+ +∫ ∫

az aR iay

z R iy

z R iy dz idy

e e edz i dye e e

e edy dye e

≤− −

<= →

∫ ∫

∫ ∫19

R → ∞

y π=Smallest for :

1 1R iy Re e e+ ≥ −

1 R iye e+

2γ1γ

R− R

2z iπ=

z iπ=x

× 3z iπ=

az aR iay

z R iy

z R iy dz idy

e e edz i dye e e

e d aye

= − + =

≤ →−

∫ ∫

( ) ( )

( ) ( ) ( ) ( )

21 2 Res

Res lim lim lim1 1 1

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

ππ π π

ππ π π −→ → →

− = =

−= = − = − =

+ − 1− ( ) ( )

z i z i

e ez i

+ − + − +

= = −− − − −

( )( )( ) ( )0

( )( ) , 0( )

Res ( ) lim lim1

z z z i z

a i a ia i

g zf z h zh z

g z ef z i dh z edz

e e ee

π ππ

π→ →

⇒ = = =′ +

= = = −−

Alternatively,

2γ1γ

R− R

2z iπ=

z iπ=x

Therefore

ax ia ia

x iae ie i eI dx

π π∞

−∞

− −= = =

+ −∫ iae π ( ), 0 1

sinia iaa

ae eπ π

ππ−

= < <−

( )21 2ia iae I ieπ ππ− = −

We thus have

, 0 11 sin

xeI dx a

−∞

= = < <+∫

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 11

kxI dx kx

∞ −

= < <+∫

Example:

Assume the branch 0 ≤ θ < 2π

( kx− = positive real)

1z = −x

Note: We choose the branch cut on the positive real axis (the axis of integration).

Integration Around a Branch Cut (cont.)

, 0 11

kxI dx kx

∞ −

= < <+∫

First, note the integral exists since the integral of the asymptotic forms of the integrand at both limits exists:

x x x kxx x x kx

−−

−− −

→∞

→ = <+

→ = ∞ >+

which is integrable at

Integration Around a Branch Cut

, 0 11

kxI dx kx

∞ −

= < <+∫

We’ll evaluate the integral using the contour shown.

( ) ( ) ( )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=

1z = −

0ε →

Now consider the various contributions to the contour integral:

( ) ( )

( )1 2 0

2 Res 1 ,

RL L C Ck

f z dz i f

π+ + +

0 0 0 0: , ,i i k k ikC z r e dz ir e d z r eθ θ θθ − − −= = =

( )0 0

00, 002 20

lim lim 01

k ik ik i ik

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

lim lim 01

k ik ik ik

R e iRe df z dz iR e dRe

π ε πθ θθ

θεε

φ φ− − −

− −

→∞ →∞→

= = →+∫ ∫ ∫

1z = −

0ε →

1 : , ,i i k k ikL z re dz e dr z r eε ε ε− − −= = =

( )1 00

lim1 1

R k ki k

iRL rr

r dr r drf z dz e Ire r

ε εε

∞− −−

→∞→→

= = =+ +∫ ∫ ∫

( ) ( )2 22 : , ,i iki i k kL z re re dz e dr z r eπ ε π εε ε− − −− − − −= = = =

( ) [ ]0

(2 ) 2 2

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

∞− −− + − − −

−→∞→→

= = − = −+ +∫ ∫ ∫

For the path L2:

For the path L1:( )

kzf zz

1z = −

0ε →

( ) ( )

1 2 Res 1

2 lim 1

e I i f z

→−

− = = −

( )2 1

i e i eIe

π π− −

−= =

− ike π− ( ), 0 1

sinik i kk

ke eπ π

ππ−

= < <−

Therefore, we have

Note: Arg (-1) = π here.

, 0 11 sin

kxI dx kx k

∞ −

= = < <+∫

“Dispersion” Relations

Assumptions:

( )( ) 0 , z

f z u iv

(including the real axis)is analytic in the UHP

in the UHP

0iz x e θε− =Let

( ) ( ) ( )0

limix R

iRC R x

f x ef z f xdz dx

z x x x e

θεε

→∞− +→

+= + + − −

∫ ∫ ∫

ii e θε ( )

( ) ( ) ( ) ( )

0 0 00

0 0 00 0 0

1 1( ) ( )

d i f x

f xdx i f x i f x i f x

f x v x u xif x u x iv x dx dx dxi x x x x x x

π π π

−∞

∞ ∞ ∞

−∞ −∞ −∞

= − =−

= + = = −− −

−⇒

∫ ∫ ∫

(From the residue theorem or the Cauchy integral theorem.)

CPV ( )0i f xπ

UHP: Upper Half Plane

RR−x0x

→ ∞→

“Dispersion” Relations (cont.)

( ) ( ) ( )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:

v xu x dx

u xv x dx

−∞

= −−

Next, relabel: 0,x x x x′→ →

“Dispersion” Relations (cont.)

( ), ( )u x v x⇒ are of one anotherHilbert Transforms

Hence, we have

v xu x dx

u xv x dx

−∞

′′=

′ −

′′= −

′ −

∫ Hilbert

( )( ) 0 ,

f z u iv

(including the real axis)

Assumptions :is analytic in the UHP

in the UHP

“Dispersion” Relations: Circuit Theory

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:

LHP: Lower Half Plane

( )( )

ω→ → ∞

time convention

Assumptio

alytic in the LHP

for in the

( ) ( ) ( )R IH H i Hω ω ω= +( )inV ω+

−( )outV ω

( ) ( )( ) /out inH V Vω ω ω≡

Transfer function

“Dispersion” Relations: Circuit Theory (cont.)

Use the path shown below:

( ) ( )limiR

H eH Hd de

θω ε

θω εε

ω εω ωω ωω ω ω ω ε

→∞− +→

+ ′ ′′ ′= + + ′ ′− −

∫ ∫ ∫

ii e θε

( ) ( )

Hd i H

θ π ω

ωω π ω

ωω ω

π ω ω

−∞

′′ = −

′ −

⇒′

′= −−

ReRR−

(From the residue theorem or the Cauchy integral theorem.)

( ) ( )1 HH d

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

∞ ∞

−∞ −∞

′ ′′ ′= + = − +

′ ′ −⇒

−∫ ∫

( ) ( )

ωω ω

π ω ω

ωω ω

π ω ω

−∞

′′= −

′ −

′′=

′ −

We thus have

The real and imaginary parts of the transfer functionare Hilbert transforms of each other.

We can also derive an alternative form:

( ) ( ) ( )( )( )2 2

21 1R RI

H HH d d

ω ω ωω ω ω

π ω ω π ω ω

∞ ∞

−∞

′ ′′ ′= = −

′ − ′ −∫ ∫

Similarly, we have

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( )( )( )( )

( )( )( )

I I IR

H H HH d d d

H Hd d

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

Summarizing, we have

( ) ( ) ( )( )( )

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,

ωω ω→ → ∞

is analytic in the LHPfo

Assumption

r in t

he LHP

Kramers-Kronig Relations

( ) ( ) 1,r rε ω ε ω ω→The is analytic in the LHP and in LHPrelative permittivity

( ) ( ) 1

εχχ ω ε ω

≡≡

≡ −

relative permittivityelectric susceptibility

Material parameters :

( ) ( )

Im ( ) 12Re ( ) 1

Re ( ) 12Im ( ) 1

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

( ) ( )

( ) ( )( )

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

( ) ( )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

Analytic××

( ) ( ) ( ) ( )00

Re >stF s t f t e dt s γ∞

−′ = −∫ valid forNote :

Laplace Transform (cont.)

( ) ( )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

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

f t g i s e ds ti

− ∞

= − − >∫

( )Re s

( )Im s

( ) ( ) i tg g t e dωω ω∞

−∞

( ) ( )( )1 , 02

f t g i s e ds ti

− ∞

= − − >∫

( )( ) ( ) ( )( )

( ) ( )( )

( ) ( )

i i s t

g i s g t e dt

g t e dt

f t e dt

− − −

−∞

∞− − −

∞− −

∞−

− − =

We have for the integrand term:

( )Re s

( )Im s

( ) ( )1 , 02

f t F s e ds ti

− ∞

= >∫

“Bromwich integral”

( )Re s

( )Im s

The Bromwich contour C is chosen to the right of all singularities of the function F(s).

“Inverse Laplace transform”

Hence we have:

( ) ( )12

f t F s e dsiπ

Close the contour to the right.

( ) 0f t =

Consider the case:

( )Re s

( )Im s

( ) ( )12

f t F s e dsi

− ∞

= ∫ RCThe integrand is analytic inside .

00 ( )

→ ∞

→ assumption

Only the vertical path contributes as

Right :Top and bottom :

(by Cauchy’s theorem)

( ) ( ) , 012 0, 0

f t tF s e ds

− ∞

Summary

( )Re s

( )Im s

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.

Close the contour to the left.

( ) ( )12

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

−∞C

( ) ( )12

f t F s e dsi

− ∞

= ∫0t >

00 ( )

→ ∞

→ assumption

Only the vertical path contributes as

Left :Top and bottom :

Poles only

Example ( ) atf t e=

( ) ( ) ( )

1s a t s a tat stF s e e dt e dt es a

∞∞ ∞− − − −− −

= = =−∫ ∫

( ) ( )1 , ReF s s as a

= >−

( ) 1 1 ,2

f t e ds ai s a

− ∞

= >−∫

( )Re s

( )Im s

LC−∞

( ) 1 12 Res2

f t i ei s a

ππ =

( ) , 0atf t e t⇒ = >

For t > 0:

Appendix

( ) ( )*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− =

( ) ( )*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

notes 11 evaluation of definite integrals via the residue ...courses.egr.uh.edu/ece/ece6382/class...

Documents

adapted from notes by ece 5317-6351 prof. jeffery t...

shannon's sampling theorem, incongruent residue classes and...

conserved cys residue influences catalytic properties of...

residue formulas for meromorphic functions on surfaces

· introduction the original aim of these notes is to...

univerza v ljubljani, fakulteta za farmacijo...when...

liquefied residue of kenaf core wood produced at different

7. tillage and residue management - amazon web services

equivariant intersection theory and bott's residue

notes notes - zgravweb.net · follow your bliss —joseph...

fall 2021 david r. jackson - courses.egr.uh.edu

the residue of nic and super - 東北学院大学sacl/the...

stima aree residue - comune.macerata.it fileooggetto: stima...

outline notes: diffusive flux notes: advection-diffusion...

amyloid fibril formation by aβ , a seven-residue fragment...

adapted from notes by ece 5317-6351 prof. jeffery...

national residue plan brazil (2009) mongolia

ece 2201 f18 exam 1 solutions - courses.egr.uh.edu

residue reviews / ruckstands-berichte: residue of pesticides...

utilization of dried malt residue as dairy cow feed