Lecture 3 Transmission Line Junctions Time Harmonic Exitation

8/2/2019 Lecture 3 Transmission Line Junctions Time Harmonic Exitation

1/23

EECS 117

Lecture 3: Transmission Line Junctions / Time

Harmonic Excitation

Prof. Niknejad

University of California, Berkeley

University of California, Berkeley EECS 117 Lecture 3 p. 1/


2/23

Transmission Line Menagerie

striplinemicrostripline coplanar

rectangular

waveguide

coaxialtwo wires

T-Lines come in many shapes and sizes

Coaxial usually 75 or 50 (cable TV, Internet)Microstrip lines are common on printed circuit boards(PCB) and integrated circuit (ICs)

Coplanar also common on PCB and ICsTwisted pairs is almost a T-line, ubiquitous forphones/Ethernet



3/23

Waveguides and Transmission Lines

The transmission lines weve been considering havebeen propagating the TEM mode or TransverseElectro-Magnetic. Later well see that they can also

propagation other modesWaveguides cannot propagate TEM but propagationTM (Transverse Magnetic) and TE (Transverse Electric)

In general, any set of more than one losslessconductors with uniform cross-section can transmitTEM waves. Low loss conductors are commonly

approximated as lossless.



4/23

Cascade of T-Lines (I)

Z01 Z02

z = 0

i1 i2

v1 v2

Consider the junction between two transmission linesZ01 and Z02

At the interface z = 0, the boundary conditions are thatthe voltage/current has to be continuous

v+1

+ v1

= v+2

(v

+

1 v

1 )/Z01 = v

+

2 /Z02



5/23

Cascade of T-Lines (II)

Solve these equations in terms of v+1

The reflection coefficient has the same form (easy to

remember) =

v1

v+1

=Z02 Z01Z01 + Z02

The second line looks like a load impedance of valueZ02

Z01 Z02

z = 0

i1

+v1



6/23

Transmission Coefficient

The wave launched on the new transmission line at theinterface is given by

v+

2 = v+

1 + v

1 = v+

1 (1 + ) = v+

1

This transmitted wave has a coefficient

= 1 + = 2Z02Z01 + Z02

Note the incoming wave carries a power

Pin =|v+1|2

2Z01



7/23

Conservation of Energy

The reflected and transmitted waves likewise carry apower of

Pref = |v

1 |22Z01

= ||2 |v+1 |22Z01

Ptran = |v+2 |22Z02

= ||2 |v+1 |22Z02

By conservation of energy, it follows that

Pin = Pref + Ptran

1

Z02 2

+

1

Z012

=

1

Z01

You can verify that this relation holds!



8/23

Bounce Diagram

Consider the bouncediagram for the follow-ing arrangement

Z01 Z02

Rs

RL

1 2

Time

Space

td

2td

3td

4td

5td

6td

v+1

1v+1

L1v+

1

L12v

+

1

sL12v+1

jv+1

jsv+1

s2jv

+1

1 1 + 2

td1



9/23

Junction of Parallel T-Lines

Z01Z02

z = 0

Z03

Again invoke voltage/current continuity at the interface

v+1

+ v1

= v+2

= v+3

v+1 v

1

Z01=

v+2

Z02+

v+3

Z02

But v+2

= v+3

, so the interface just looks like the case oftwo transmission lines Z01 and a new line with char.impedance Z01

||Z02.



10/23

Reactive Terminations (I)

Rs

Vs

Z0, td L

Lets analyze the problem intuitively first

When a pulse first sees the inductance at the load, itlooks like an open so 0 = +1

As time progresses, the inductor looks more and more

like a short! So = 1



11/23

Reactive Terminations (II)

So intuitively we might expect the reflection coefficientto look like this:

1 2 3 4 5

-1

-0.5

0.5

1

t/

The graph starts at +1 and ends at 1. In between wellsee that it goes through exponential decay (1st order

ODE)



12/23

Reactive Terminations (III)

Do equations confirm our intuition?

vL = Ldi

dt= L

d

dtv

+

Z0 v

Z0

And the voltage at the load is given by v+ + v

v + LZ0

dv

dt= L

Z0dv+

dt v+

The right hand side is known, its the incomingwaveform



13/23

Solution for Reactive Term

For the step response, the derivative term on the RHSis zero at the load

v+ =Z0

Z0 + RsVs

So we have a simpler case dv+

dt = 0

We must solve the following equation

v +L

Z0

dv

dt=

v+

For simplicity, assume at t = 0 the wave v+ arrives atload



14/23

Laplace Domain Solution I

In the Laplace domain

V(s) +sL

Z0V(s)

L

Z0v(0) =

v+/s

Solve for reflection V(s)

V(s) = v

(0)L/Z01 + sL/Z0

v+

s(1 + sL/Z0)

Break this into basic terms using partial fraction

expansion

1

s(1 + sL/Z0)

=1

1 + sL/Z0+

L/Z0

1 + sL/Z0


L l D i S l i (II)


15/23

Laplace Domain Solution (II)

Invert the equations to get back to time domain t > 0

v(t) = (v(0) + v+)et/ v+

Note that v(0) = v+ since initially the inductor is anopen

So the reflection coefficient is(t) = 2et/ 1

The reflection coefficient decays with time constantL/Z0


Ti H i S d S


16/23

Time Harmonic Steady-State

Compared with general transient case, sinusoidal case

is very easy t j

Sinusoidal steady state has many importantapplications for RF/microwave circuits

At high frequency, T-lines are like interconnect fordistances on the order of

Shorted or open T-lines are good resonators

T-lines are useful for impedance matching


Wh Si id l St d St t ?


17/23

Why Sinusoidal Steady-State?

Typical RF system modulates a sinusoidal carrier(either frequency or phase)

If the modulation bandwidth is much smaller than the

carrier, the system looks like its excited by a puresinusoid

Cell phones are a good example. The carrier frequency

is about 1 GHz and the voice digital modulation is about200 kHz(GSM) or 1.25 MHz(CDMA), less than a 0.1% ofthe bandwidth/carrier


G li d Di ib d Ci i M d l


18/23

Generalized Distributed Circuit Model

Z Z Z Z Z

Y Y Y Y Y

Z: impedance per unit length (e.g. Z = jL + R)

Y

: admittance per unit length (e.g. Y

= jC

+ G

)A lossy T-line might have the following form (but wellanalyze the general case)

L

C

R

G

L

C

R

G

L

C

R

G

L

C

R

G


Ti H i T l h E ti


19/23

Time Harmonic Telegraphers Equations

Applying KCL and KVL to a infinitesimal section

v(z + z) v(z) = Zzi(z)

i(z + z) i(z) = Yzv(z)Taking the limit as before (z 0)

dv

dz= Zi(z)

didz = Y v(z)


Si St d St t (SSS) V lt /C t


20/23

Sin. Steady-State (SSS) Voltage/Current

Taking derivatives (notice z is the only variable) wearrive at

d

2

vdz2 = Zdidz = Y Zv(z) = 2v(z)

d2i

dz2 = Ydv

dz = Y Zi(z) =

2

i(z)Where the propagation constant is a complex function

= + j =

(R

+ jL

)(G

+ jC

)

The general solution to D2G 2G = 0 is ez


L l Li f SSS


21/23

Lossless Line for SSS

The voltage and current are related (just as before, butnow easier to derive)

v(z) = V+

ez

+ V

ez

i(z) =V+

Z0ez V

Z0ez

Where Z0 =

Z

Y is the characteristic impedance of the

line (function of frequency with loss)

For a lossless line we discussed before, Z = jL andY = jC

Propagation constant is imaginary

=

jLjC = j

LCUniversity of California, Berkeley EECS 117 Lecture 3 p. 21/

B k t Ti D i


22/23

Back to Time-Domain

Recall that the real voltages and currents are the and parts of

v(z, t) = ezejt = ejtz

Thus the voltage/current waveforms are sinusoidal inspace and time

Sinusoidal source voltage is transmitted unaltered ontoT-line (with delay)

If there is loss, then has a real part , and the wavedecays or grows on the T-line

ez = ezejz

The first term represents amplitude response of theT-line


Passive T Line/Wave Speed


23/23

Passive T-Line/Wave Speed

For a passive line, we expect the amplitude to decaydue to loss on the line

The speed of the wave is derived as before. In order to

follow a constant point on the wavefront, you have tomove with velocity

d

dt (t z = constant)

Or, v = dzdt =

=

1

LC


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Lecture 3 Transmission Line Junctions Time Harmonic Exitation