ee757 numerical techniques in electromagnetics …mbakr/ece757/lecture3_2016.pdfee757,2017, dr....

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EE757

Numerical Techniques in Electromagnetics

Lecture 3

2 EE757,2017, Dr. Mohamed Bakr

Special Cases in Maxwell’s Equations.

With no sources, Maxwell’s Equations are written as

,

In Cartesian coordinates we have

t

BE)(

t c

DJH )(

t

B

z

E

y

E xyz

t

B

x

E

z

E yzx

t

B

y

E

x

E zxy

t

DJ

z

H

y

H xx

yz

t

DJ

x

H

z

H y

yzx

t

DJ

y

H

x

H zz

xy


Special Cases (Cont’d)

For a 1D case, the fields depend only on one coordinate

(e.g. a uniform plane wave traveling in the x direction)

For example, if /y= /z=0

, (propagation in –ve x)

or

, (propagation in +ve x)

Notice that these two systems are decoupled

t

B

x

E yz

t

DJ

x

H zz

y

t

B

x

E zy

t

DJ

x

H y

yz


Special Cases (Cont’d)

For the 2D problems, the fields depend only on two

coordinates (e.g. TE10 mode in waveguides)

For example, if /y=0

t

B

z

E xy

t

B

x

E

z

E yzx

t

B

x

E zy

t

DJ

z

H xx

y

t

DJ

x

H

z

H y

yzx

t

DJ

x

H zz

y

TEy TMy

Notice that the two systems are decoupled


Transient Response of a Transmission Line

Cd capacitance per unit length F/m

Ld inductance per unit length H/m

u velocity of propagation

After time t, the disturbance traveled a distance x.

A charge Q=Cd x Vs is transferred. It follows that

i=Q/t i=Cd Vs (x/ t)= Cd Vs u

x

C

L


Transient Response of a TL (Cont’d)

The flow of current establishes a flux associated with

line inductance through =Ld x i= Ld x Cd Vs u

Using Faraday’s law we have

Vs=/ t =LdCdVsu2 LdCdu

2=1.0

m/s

It follows that i= Cd Vs u=Vs

LCu

dd

1

L

C

d

d

C

L

i

V Z

d

ds

currentincident

voltageincidento


Open-Ended Transmission Line

After time , the incident voltage and current waves

reach the open end. A reflected current wave of

amplitude is initiated to make the total

current at the open end zero

It follows that for an open-ended transmission line we

have V i= V s , and V r= V s,

ZVI sr

o/

ZVI si

o/ ZVI sr

o/

x

V=Vs uvoltage

current

voltage

current

i=Vs/Zo u

τt 1

τt 1

V=2Vs

Vs/Zo

-Vs/Zo

u

u

l

l/u


Thevenin Equivalent of a Transmission Line

The open circuit voltage is 2V i (open-ended line)

The equivalent resistance is Zo

This Thevenin equivalent is valid only for limited time

period where no reflections takes place on the line

(same V i)

Vi u

Zo

A-A

A

A

2Vi

Zo


Example on Utilizing Thevenin’s Equivalent of a TL

Using Kirchtoff’s voltage law we have

But V=V i+V r V r= V -V i

For open circuit termination R=, =1, V r= V i

For short circuit termination R=0, =-1, V r= -V i

ZR

RVV i

o

2

VZR

RVV

iir

o

2 VVZR

Z-RV

iir

o

o

Vi

Zo

R

2Vi

Zo R


Remember that for a steady-state sinusoidal analysis

we need only the amplitude and phase at each point on

the line

Applying KVL, we get

Leading to

Applying KCL we get

Leading to

Sinusoidal Steady-State Response of a TL

))(Re(),( tjexpVtxV

)()()( xIxLjxxVxV d

)()(

xILjx

xVd

)()()( xxVxCjωxxIxI d

)()(

xVCjx

xId

x xx

xLd

xCd )(xV )( xxV

)(xI )( xxI


Sinusoidal Steady State of a TL (Cont’d)

Differentiating to eliminate the current term we get

Similarly

Using the telegrapher’s differential equations we can

show that and

It follows that

and

)()()( 22

2

2

xVxVCLx

xVdd

)()()( xjexpVxjexpVxV ri

)()()( xjexpIxjexpIxI ri

ZVIii

o/ ZVIrr

o/

)()()( xjexpVxjexpVxV ri

)()()()()( oo xjexpZ/VxjexpZ/VxI ri


ABCD Matrix of a TL Section

At x=0, we have and

Expressing the solutions in terms of the amplitudes at

x=0, we get

VVV ri (0) )()()0( oo Z/VZ/VI ri

)0(

)0(

)cos(/)sin(

)sin()cos(

)(

)(

I

V

xZxj

xZjx

xI

xV

o

o

)(

)(

)cos(/)sin(

)sin()cos(

(0)

(0)

xI

xV

xZxj

xZjx

I

V

o

o

ABCD matrix of a TL section with length x


Discrete Models of Lumped Elements

The lumped element is replaced by a section of a

transmission line

Using these models the voltages and currents are obtained

only at discrete instants of time

Cdl=C, Cd is the capacitance per unit length

u=l/ t=1/ CL dd

Cl

tL

d

d

12

Zc

l

C


Discrete Models of Lumped Elements (Cont’d)

Le = equivalent (parasitic) inductance =

t is selected small enough such that the parasitic

inductance is small

C

t

lC

t

C

LZ

dd

dc

C

tlLd

2

Modeling Steps:

1.Select t, l 2. evaluate Cd, Ld and Zc

3. Assume certain incident voltage at the kth time step

4. Use Thevenin’s equivalent to get the reflected voltages

5. Obtain incident impulses at the (k+1) time step


Discrete Models of Lumped Elements (Cont’d)

Using a similar derivation Cd l=C

u=l/(t/2)=1/ , notice that t is the round-trip time

= error inductance

CL dd

Cl

tL

d

d

1

2

2

lC

t

4

2

C

t

C

LZ

d

dc

2

C

tlLL de

4

2

Zc

l

Open circuit

C


Example

R

C

Vs

Evaluate the transient response of the shown circuit

Evaluate the current

Evaluate the capacitor voltage

Scattering:

Connection:

R)ZVVi cik

S

kk /()2(

iZVV kci

k

C

k 2

VVVi

k

C

k

r

k

VVr

k

i

k

1

Open circuitOpen circuit

ZcV

i

k

Vs

k VC

k

Rik


Example (Cont’d)

Vs

k VC

k

Rik

+

-

Zc

Vi

k2

Open circuit Open circuit

Z c

V

r

k

V

i

k 1


Example (Cont’d)

Try It!


Link Model of an Inductor

L

Using similar derivation to that of the capacitor we have

Ldl=L

Error capacitance Ce=Cdl=(t)2/L

CLtl/u dd/1Ll

tC

d

d

12

t

L

C

LZ

d

dL

Z

l

L


Stub Model of an Inductor

Using similar derivation we have

Ldl=L

Error capacitance Ce=Cdl=(t)2/(4L)

CLt/l/u dd/1)2( Ll

tC

d

d

4

12

t

L

C

LZ

d

dL

2

L

Z

l

L

short circuit


Example

RVs

L

Evaluate the transient response of the shown circuit

short circuit

Vi

k

Vs

k Vk

Rik

ZL

L

Vs

k Vk

Rik

+

-

Z

Vi

k2

L

L

Try It!


Example (Cont’d)

Evaluate the current

Evaluate the inductor voltage

Scattering:

Connection:

R)ZVVi Lik

S

kk /()2(

iZVV kLi

k

C

k 2

VVVi

k

C

k

r

k

VVr

k

i

k

1

V

r

k V

r

k V

r

k V

r

Z L

V V

r k

i

k

1

ee757 numerical techniques in electromagnetics …mbakr/ece757/lecture3_2016.pdfee757,2017, dr....

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