Microwave Circuit Design

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Microwave Circuit
Design
ECE 5250/4250 Lecture Notes
Fall 2009
© 1989, 1991, 1993, 1995, & 2009
Mark A. Wickert
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Review of
Transmission Line
Theory
Chapter
1
• Transmission lines and waveguides are used to transport
electromagnetic energy at microwave frequencies from one
point in a system to another
• The desirable features of a transmission line or waveguide
are:
– Single-mode propagation over a wide band of frequencies
– Small attenuation
• The transmission line structures of primary interest for this
course are those for which the dominant mode of propagation
is a transverse electromagnetic (TEM) wave
• Recall that for TEM waves the components of electric and
magnetic fields in the direction of wave propagation are zero
• We wish to consider transmission lines which consist of two
or more parallel conductors which have axial uniformity
• That is to say their cross-sectional shape and electrical properties do not vary along the axis of propagation
ECE 5250/4250 Microwave Circuits
1–1
Review of Transmission Line Theory
Figure 1.1: Transmission lines with axial uniformity
• The TEM wave solution for axially uniform transmission
lines can be obtained using:
– Field Analysis - (obtain electric and magnetic field waves
analogous to uniform plane waves)
– Distributed-Circuit Analysis - (obtain voltage and current
waves)
• Distributed circuit analysis will be at the forefront of all analysis in this course, in particular consider Pozar1, “Modern
microwave engineering involves predominantly distributed
circuit analysis and design, in contrast to the waveguide and
field theory orientation of earlier generations”
1. David Pozar, Microwave Engineering, 3rd edition, John Wiley, New York,
2005.
1–2
ECE 5250/4250 Microwave Circuits
Distributed Circuit Analysis
Distributed Circuit Analysis
Figure 1.2: Distributed element circuit model
Parameters:
L – series inductance per unit length due to energy storage in
the magnetic field
C – shunt capacitance per unit length due to energy storage in
the electric field
R – series resistance per unit length due to power loss in the
conductors
G – shunt conductance per unit length due to power loss in
the dielectric. (i.e. ! = !" – j!#% !# $ 0 )
• Using KCL, KVL and letting &z ' 0 it can be shown that
(i ) z% t *
– (v ) z% t *
--------------------- = Ri ) z% t * + L ----------------(t
(z
(1.1)
) z% t *
–------------------( i ) z% t *- = Gv ) z% t * + C (v
-----------------(t
(z
(1.2)
and
ECE 5250/4250 Microwave Circuits
1–3
Review of Transmission Line Theory
• For now assume the line is lossless, that is R = 0 and
G = 0 , so we have:
(v = L (i
---– ----(t
(z
(v
(i- = C ----– ---(t
(z
(1.3)
• Now differentiate the first equation with respect to z and the
second with respect to time t
2
2
( i(-------v- = – L ---------2
(t(z
(z
2
2
(-------v(
i
----------- = C 2
(z(t
(t
(1.4)
• Combine the two resulting equations to get
2
2
(-------v- = LC (-------v- (voltage eqn.)
(t 2
(z 2
(1.5)
similarly obtain
2
2
( i
( i
-------2- = LC ------2- (current eqn.)
(t
(z
(1.6)
• These are in the form of the classical one-dimensional wave
equation, often seen in the form
2
2
1- ------( y(-------y- = ---+ 2p (t 2
(z 2
(1.7)
where + p has dimension and significance of velocity
1–4
ECE 5250/4250 Microwave Circuits
Distributed Circuit Analysis
• A well known solution to the wave equation is
+
z-/ + y – . t + ---z-/
y = y . t – ---, + , + p
p
(1.8)
– y + propagates in the positive z direction
– y – propagates in the negative z direction
• This solution can be checked by noting that
0
0
0
2
(x
1
z
(y
(y = (y
1
---------- ----- = ------ ----------% x = t 0 ------------+
(x
(z
(x
+
(z
p
p
(1.9)
• A solution for the voltage wave equation is thus
+
z-/ + v – . t + ---z-/
v ) z% t * = v . t – ---, + p, + p-
(1.10)
• The current equation can be written in a similar manner, but
+
–
it can also be written in terms of V and V since
(i
– (v
--------- = L ---(t
(z
(1.11)
+
–
(i
(v
(v
– -------- + -------- = L ---(t
(z
(z
(1.12)
or
Letting x = t 0 z 3 + p and using the chain rule the above equation becomes
+
–
1 (v (i
– 1- -------(v
-------- ------– ----+
=
L
+ p (x + p (x
(t
ECE 5250/4250 Microwave Circuits
(1.13)
1–5
Review of Transmission Line Theory
• This implies that
+
–
1
z
z
i ) z% t * = --------- v . t – -----/ – v . t + -----/
, + , + L+ p
p
p
(1.14)
1 +
z-/ – v – . t + ---z-/
i ) z% t * = ----- v . t – ---, + , + Z0
p
p
(1.15)
or
where
1
+ p = -----------% (velocity of propagation)
LC
Z0 =
L-% (characteristic impedance)
--C
(1.16)
• At this point the general lossless line solution is incomplete.
The functions v+ and v- are unknown, but must satisfy the
boundary conditions imposed by a specific problem
• The time domain solution for a lossless line, in particular the
analysis of transients, can most effectively be handled by
using Laplace transforms
• If the source and load impedances are pure resistances and
the source voltage consists of step functions or rectangular
pulses, then time domain analysis is most convenient
• In the following we will first consider resistive load and
source impedances, later the analysis will be extended to
complex impedance loads using Laplace transforms
1–6
ECE 5250/4250 Microwave Circuits
Transient Analysis with Resistive Loads
• The Laplace transform technique will in theory allow for a
generalized time-domain analysis of transmission lines
• In Section 2 sinusoidal steady-state analysis will be introduced. This approach offers greatly reduced analysis complexity
Transient Analysis with Resistive Loads
Infinite Length Line
Figure 1.3: Infinite length line circuit diagram
• Assume that the line is initially uncharged, i.e. for z 4 0 and
t50
+
z-/ + v – . t + ---z-/ = 0
v ) z% t * = v . t – ---, + , + p
(1.17)
p
and
1 +
z-/ – v – . t + ---z-/
i ) z% t * = ----- v . t – ---, + , + Z0
p
p
= 0
(1.18)
the above equations imply that
+
z
v . t – -----/ = 0 for t – z 3 + p 6 0
, + -
(1.19)
p
ECE 5250/4250 Microwave Circuits
1–7
Review of Transmission Line Theory
and
–
z
v . t + -----/ = 0 for all t
, + -
(1.20)
p
+
Note: For the given initial conditions only v ) t – z 3 + p * can
exist on the line.
• We thus conclude that
:
8
8
z
p
----- ; 0
9 for all t – +
1 +
p
z-/ 8
i ) z% t * = ----- v . t – ---Z 0 , + p- 8
7
+
z-/
v ) z% t * = v . t – ---, + -
• Suppose that at t = 0
+
(1.21)
a voltage source v g ) t * is
applied through a source resistance R g , at z = 0
• Apply Ohm's law at z = 0 for t > 0 and we obtain
v g ) t * – v ) 0% t * = i ) 0% t *R g
(1.22)
Rg +
v g ) t * – v ) t * = ------ v ) t *
Zg
(1.23)
or
+
which implies
Z0
+
v ) t * = ------------------ v g ) t *
Z0 + Rg
1–8
(1.24)
ECE 5250/4250 Microwave Circuits
Transient Analysis with Resistive Loads
• The final result is that under the infinite line length assumption for any z we can write
Z0
z/
----------------- v g . t – ---v ) z% t * =
Z 0 + R g , + p-
(1.25)
1
z-/
i ) t% z * = ------------------ v g . t – ---Z 0 + R g , + p-
(1.26)
– Note: That the infinite length of line appears as a voltage
divider to the source
– Voltages and currents along the line appear as replicas of
the input values except for the time delay z 3 + p
Terminated Line
Figure 1.4: Terminated line circuit diagram
• Note: As a matter of convenience the reference point z = 0
has been shifted to the load end of the line
• Suppose that a wave traveling in the z + direction is incident
upon the load, R L , at z = 0
• Thus,
1 +.
+
z-/ and i ) z% t * = ----z-/
v t – ---v ) z% t * = v . t – ---, + Z 0 , + pp
ECE 5250/4250 Microwave Circuits
(1.27)
1–9
Review of Transmission Line Theory
• By applying Ohm’s law at the load, we must have
v ) 0% t * = R L i ) 0% t *
(1.28)
• This condition cannot, in general, be met by the incident
wave alone since
v ) z% t * = Z 0 i ) z% t *
(1.29)
• Since the line was initially discharged, it is reasonable to
assume that a fraction, < L , of the incident wave is reflected
from the load resistance, i.e.,
–
+
v ) t * = <L v ) t *
(1.30)
• The load voltage is now
+
–
+
v ) 0% t * = v ) t * + v ) t * = ) 1 + < L *v ) t *
(1.31)
• Similarly the load current is
+
+
–
) 1 + < L *v ) t *
v )t* v )t*
i ) 0% t * = ------------ – ------------ = --------------------------------Z0
Z0
Z0
(1.32)
• To satisfy Kirchoff’s laws,
1 + <L
Net
load
voltage
--------------------------------------- = R L = Z 0 --------------1 – <L
Net load current
(1.33)
or the more familiar result
RL – Z0
< L = ----------------RL + Z0
1–10
(1.34)
ECE 5250/4250 Microwave Circuits
Transient Analysis with Resistive Loads
– Note: < L is real by assumption
– Note: To terminate the line without reflection use
RL = Z0 .
General Transmission Line Problem
Figure 1.5: Circuit for general transmission line problem
• From earlier analysis, we know that for 0 5 t 5 l 3 + p = T l ,
:
Z0
z
.
/
v ) z% t * = ------------------ v g t – ----- 8
Z 0 + R g , + p- 8
9 0 5 t 5 Tl
1
z-/ 8
i ) z% t * = ------------------ v g . t – ---Z 0 + R g , + p- 87
(1.35)
• When t = T l the leading edge of v g ) t * has traveled to the
load end of the line ( z = l )
• Assuming R L $ Z 0 a reflected wave now returns to the
source during the interval T l 5 t 6 2T l
=
8
8
8
8
>
8
8
8
8
?
Z0
2l z
+ ------------------ < L v g . t – ----- + -----/
, +
Z0 + Rg
p +p
=
8
8
8
>
8
8
8
?
v ) z% t * =
Z0
z/
------------------ v g . t – ---Z 0 + R g , + pv ) t – z 3 +p *
v ) t + z 3 +p *
+
ECE 5250/4250 Microwave Circuits
–
(1.36)
1–11
Review of Transmission Line Theory
and
1
1
z
2l z
i ) z% t * = ------------------ v g . t – -----/ – ------------------ < L v g . t – ----- + -----/ (1.37)
, +
Z 0 + R g , + p- Z 0 + R g
p +p
–
• When the load reflected wave v ) t + z 3 + p * arrives at the
source ( z = 0 ), a portion of it will be reflected towards the
load provided R g $ Z 0
• The reflection that takes place is independent of the source
voltage
• The wave traveling in the positive Z direction after
2l 3 + p = 2T l seconds has elapsed, can be found by applying
Ohm's law for z = 0 and t = 2T l :
v g ) 2T l * – v ) 0% 2T l * = R g i ) 0% 2T l *
(1.38)
• Now substitute
+
–
v ) 0% 2T l * = v ) 2T l * + v ) 2T l *
1 +
–
i ) 0% 2T l * = ----- @ v ) 2T l * – v ) 2T l * A
Z0
(1.39)
+
and solve for v ) 2T l *
• The results is
Z0
Rg –Z0
–
v ) 2T l * = v g ) 2T l * ------------------ + v ) 2T l * -----------------Rg + Z0
Z0 + Rg
=
8
8
>
8
8
?
=
8
8
>
8
8
?
+
–
incident part of v g ) t * reflected portion of v ) 2T l *
Z0
Z0
= v g ) 2T l * ------------------ + v g ) 0 * ------------------ < L < g
Z0 + Rg
Z0 + Rg
1–12
(1.40)
ECE 5250/4250 Microwave Circuits
Transient Analysis with Resistive Loads
where < L is the source reflection coefficient defined as
Rg – Z0
< g = ----------------Rg + Z0
(1.41)
Note: < g is real by assumption in this case
• The wave incident on the load during the interval
2T l 5 t 6 3T l , thus consists of the original source signal plus
a reflected component due to mismatches at both the load and
source ends of the line
Z0
z-/ + < v . t + z-----------– 2l-/
v ) z% t * = ------------------ v g . t – ---L g,
Z 0 + R g , + p+p z + 2l
+ < g < L v g . t – -------------/ % 2T l 5 t 6 3T l
,
+p -
(1.42)
and
1
z-/ – < v . t + z-----------– 2l-/
i ) z% t * = ------------------ v g . t – ---Z 0 + R g , + p- L g ,
+p z + 2l-/ % 2T 5 t 6 3T
+ < g < L v g . t – -----------l
l
,
+ -
(1.43)
p
• The process of reflections occurring at both the source and
load ends continues in such a way that in general over the n th
time interval, ) n – 1 *T l 5 t 5 nT l , the v ) z% t * and i ) z% t * solutions each require n terms involving v g ) t *
ECE 5250/4250 Microwave Circuits
1–13
Review of Transmission Line Theory
Example: Consider the following circuit.
Figure 1.6: Transmission line circuit
Here we have
Z0
1
1
------------------ = ---% < L = – 1% < g = ---% T l = 2Bs
2
4
Z0 + Rg
(1.44)
a) Find V ) 0% t * and I ) 0% t * for T p = 1Bs
b) Find V ) 0% t * and I ) 0% t * for T p = 6Bs
• For both parts (a) and (b) the basic circuit operation is as follows:
i) An 8v pulse will propagate toward the load, reaching the
load in 2 ms.
ii) A -8v pulse will be reflected from the load, requiring 2 ms
to reach the source.
iii) A -4v pulse will be reflected at the source. It will take 2
ms to reach the load.
iv) The process continues.
• The +z and -z direction propagating pulses can be displayed
on a distance-time plot or bounce diagram (see Figure 1.7)
1–14
ECE 5250/4250 Microwave Circuits
Transient Analysis with Resistive Loads
– In the bounce diagram the boundaries at z = 0 and z = 400
m are represented as surfaces with reflection coefficient of
1/2 and -1 respectively.
– To obtain the voltage waveform at say z = 380 m as a function of time, you sum the contributions from the various +z
and -z traveling waves
– For pulses that are short in comparison with the one-way
delay time of the line, only at most two wave terms need to
be included at a time.
– For long pulses (in the limit say a step function) all wave
terms need to be included
Figure 1.7: Bounce diagram showing pulse propagation
ECE 5250/4250 Microwave Circuits
1–15
Review of Transmission Line Theory
a) T p = 1Bs : at t = 4Bs we have
v ) 0% 4Bs * = – 4 + ) – 8 * = – 12v.
1
i ) 0% 4Bs * = ------ @ – 4 – ) – 8 * A = 0.08A.
50
b) T p = 6Bs : at t = 4Bs we have
v ) 0% 4Bs * = ) 8 + ) – 4 * * + ) – 8 * = – 4v.
1
i ) 0% 4Bs * = ------ @ ) 8 + ) – 4 * * – ) – 8 * A = 0.24A.
50
• We can simulate this result using the Agilent Advanced
Design System (ADS) software
• In this example we will use ideal transmission line elements
Current
measurement
means
A short circuit
Source
R
R1
R=150 Ohm
t
VtPulse
SRC1
Vlow=0 V
Vhigh=32 V
Delay=0 nsec
Edge=linear
Rise=50 nsec
Fall=50 nsec
Width=1 usec
Period=20 usec
I_Probe
I_Probe1
TLIND
TLD1
Z=50.0 Ohm
Delay=2 usec
R
R2
R=0 Ohm
‘TLIND’ allows the line
to be spec’d in terms of
propagation delay
TRANSIENT
Tran
Tran1
StopTime=20.0 usec
MaxTimeStep=50 nsec
Figure 1.8: Circuit schematic (Example1.dsn)
1–16
ECE 5250/4250 Microwave Circuits
Transient Analysis with Resistive Loads
• The source end voltage, v ) 0% t *
10
T p = 1Bs
Source (v)
5
0
-5
-10
-15
0
2
4
6
8
10
12
14
16
18
20
time, usec
• The source end current entering the line, i ) 0% t *
0.20
T p = 1Bs
I_Probe1.i (A)
0.15
0.10
0.05
0.00
-0.05
0
2
4
6
8
10
12
14
16
18
20
time, usec
• Modify the schematic by increasing the pulse width to 6 Bs
ECE 5250/4250 Microwave Circuits
1–17
Review of Transmission Line Theory
• The source end voltage, v ) 0% t *
10
T p = 6Bs
Source (v)
5
0
-5
-10
-15
0
2
4
6
8
10
12
14
16
18
20
time, usec
• The source end current entering the line, i ) 0% t *
0.25
T p = 6Bs
I_Probe1.i (A)
0.20
0.15
0.10
0.05
0.00
-0.05
0
2
4
6
8
10
12
14
16
18
20
time, usec
1–18
ECE 5250/4250 Microwave Circuits
Transient Analysis with Resistive Loads
Transient Analysis using Laplace Transforms
Figure 1.9: Lossless line with arbitrary terminations
• In the time domain we know that for a lossless line
+
–
z
z
v ) z% t * = v . t – -----/ + v . t + -----/
, + , + -
(1.45)
1 +
z-/ – v – . t + ---z-/
i ) z% t * = ----- v . t – ---, + , + Z0
p
p
(1.46)
p
p
and
+
–
where v ) t – z 3 + p * and v ) t + z 3 + p * are determined by the
boundary conditions imposed by the source and load
• Laplace transform each side of the above equations with
respect to the time variable, using the time shift theorem
which is given by
L C f ) t – c * D = F ) s *e
– sc
(1.47)
where F ) s * is the laplace transform of f ) t *
• The result is
+
v ) z% s * = v ) s *e
– sz 3 + p
–
+ v ) s *e
sz 3 + p
(1.48)
and
ECE 5250/4250 Microwave Circuits
1–19
Review of Transmission Line Theory
– sz 3 + p
sz 3 + p
–
1 +
i ) z% s * = ----- @ v ) s *e
– v ) s *e
A
Z0
+
+
–
(1.49)
–
where v ) s * = L C v ) t * D and v ) s * = L C v ) t * D
Case 1: Matched Source
• For the special case of Z g ) s * = Z 0 , the source is matched to
the transmission line which eliminates multiple reflections
• Thus, we can write
Z0
+
- = 1--- V g ) s *
V ) s * = V g ) s * ----------------Z0 + Z0
2
(1.50)
at z = l , the incident wave is reflected with the reflection
coefficient
ZL ) s * – Z0
< L ) s * = ------------------------ZL ) s * + Z0
(1.51)
so,
+
V ) l% s * = V ) s *e
– sl 3 + p
@ 1 + <L ) s * A
(1.52)
which implies that
–
V ) s * = < L ) s *e
– s2l 3 + p +
V )s*
(1.53)
• Finally, for 0 5 z 5 l we can write
– sz 3 + p
– s ) 2l – z * 3 + p
1
V ) z% s * = --- V g ) s * @ e
+ < L ) s *e
A
2
– sz 3 + p 1
– s2l 3 + p
sz 3 + p (1.54)
1
= --- V g ) s *e
+ --- < L ) s *e
V g ) s *e
2
2
1–20
ECE 5250/4250 Microwave Circuits
Transient Analysis with Resistive Loads
• To obtain V ) z% t * inverse transform:
–1
1
z
v ) z% t * = --- v g . t – -----/ + L @ < L ) s *V g ) s * A
2 , + p-
t ' t + ) z – 2l * 3 + p
(1.55)
Example:
• Let v g ) t * = v 0 u ) t * and Z L be a parallel RC connection
Figure 1.10: Parallel RC circuit
Find: v ) z% t *
• To begin with in the s-domain we can write
1
R -----Cs = ------------------R Z L ) s * = ---------------11 + RCs
R + ----Cs
(1.56)
R -–Z
------------------0
ZL ) s * – Z0
1
+
RCs
< L ) s * = -------------------------- = -------------------------------ZL ) s * + Z0
R - Z
------------------+ 0
1 + RCs
R – Z0
--------------- – s
R + Z0
R–Z
RCZ 0
b–s
= ----------------------- = -----------% a = ---------------% b = --------------0RCZ 0
a+s
RCZ 0
R – Z0
--------------- + s
RCZ 0
(1.57)
and
ECE 5250/4250 Microwave Circuits
1–21
Review of Transmission Line Theory
• Now since V g ) s * = v 0 L C u ) t * D = v 0 3 s
v 0 –sz 3 + p b – s – s ) 2l – z * 3 + p
---V ) z% s * = - e
+ ----------- e
2s
a+s
(1.58)
• To inverse transform first apply partial fractions to
K1
K2
b
–
s
------------------- = ------ + ----------s)a + s*
s s+a
(1.59)
Clearly,
R – Z0
b
K 1 = --- = --------------R + Z0
a
–) a + b *
– 2R
K 2 = -------------------- = --------------a
R + Z0
(1.60)
so
v 0 1 –sz 3 +p
---V ) z% s * = - --- e
2 s
= R – Z0 1
2R
1 :
+ > --------------- E --- – --------------- E ----------- 9e
? R + Z0 s R + Z0 s + a 7
s ) 2l – z *
– -------------------+p
(1.61)
and
v0
R – Z0
–1
z-/ + = --------------v ) z% t * = L C V ) z% s * D = ----- u . t – ---2 , + p- >? R + Z 0
– 2R
--------------- e
R + Z0
1–22
2l – z R + Z
– . t – -------------/ --------------0,
+ p - RCZ 0 :
(1.62)
. 2l – z-/
9u , t – -----------+p 7
ECE 5250/4250 Microwave Circuits
Transient Analysis with Resistive Loads
• As a special case consider z = l
v0
R – Z0
2R
---v ) z% t * =
1 + --------------- – --------------- e
2
R + Z0 R + Z0
R
= v 0 --------------- 1 – e
R + Z0
l R+Z
– . t – -----/ --------------0, + p- RCZ 0
R + Z0
l-/ -------------– . t – ---, + p- RCZ 0
l-/
u . t – ---, + p
(1.63)
l
u . t – -----/
, + p
Figure 1.11: Sketch of the general v ) l% t * waveform
Example: The results of an ADS simulation when using
v 0 = 2 v, Z 0 = R = 50 ohms, C = 50 pf, and T l = 5 ns is
shown below
Figure 1.12: ADS Circuit diagram with nodes numbered
• The ADS schematic is shown below
ECE 5250/4250 Microwave Circuits
1–23
Review of Transmission Line Theory
Source
Load
R
R1
R=50 Ohm
t
TLIND
TLD1
Z=50.0 Ohm
Delay=5 nsec
VtPulse
SRC1
Vlow=0 V
Vhigh=2 V
Delay=0 nsec
Edge=linear
Rise=0.1 nsec
Fall=0.1 nsec
Width=1 usec
Period=1 usec
R
R2
R=50 Ohm
C
C1
C=50 pF
TRANSIENT
Tran
Tran1
StopTime=20.0 nsec
MaxTimeStep=0.1 nsec
• Next we plot the source and load end waveforms
1.0
load end
source end
Load
Source
0.8
0.6
0.4
0.2
0.0
0
2
4
6
8
10
12
14
16
18
20
time, nsec
1–24
ECE 5250/4250 Microwave Circuits
Transient Analysis with Resistive Loads
Case 2: Mismatched Source Impedance
• For the general case where Z g ) s * $ Z 0 to reflections occur at
the source end of the line as well as at the load end
• The expression for V ) z% s * now will consist of an infinite
number of terms as shown below:
Z0
– sz 3 + p
– s ) 2l – z * 3 + p
V ) z% s * = V g ) s * ------------------------- @ e
+ < L ) s *e
Z0 + Zg ) s *
+ < L ) s *< g ) s *e
– s ) 2l + z * 3 + p
+ < L ) s *< g ) s *< L ) s *e
(1.64)
– s ) 4l – z * 3 + p
+ < L ) s *< g ) s *< L ) s *< g ) s *e
– s ) 4l + z * 3 + p
+ FA
where
ZL ) s * – Z0
Zg ) s * – Z0
------------------------<L ) s * =
and < g ) s * = ------------------------ZL ) s * + Z0
Zg ) s * + Z0
(1.65)
• In terms of +z and -z propagating waves we can write
V g ) s *Z 0 – sz 3 +p
------------------------ e
V ) z% s * =
Z0 + Zg ) s *
+e
sz 3 + p G
H
G
H
n
n
< L ) s *< g ) s *e
– s ) 2n *l 3 + p
n=0
(1.66)
n+1
<L
n
) s *< g ) s *e
– s ) 2n + 2 *l 3 + p
n=0
ECE 5250/4250 Microwave Circuits
1–25
Review of Transmission Line Theory
Example: Consider a circuit with v 0 = 2 v, Z 0 = 50 ohms,
T l = 5 ns, Z g = R g = 100 ohms, and Z L a parallel RC circuit
with R = 100 ohms and C = 20 pf, as shown below in Figure
1.13.
Figure 1.13: Circuit diagram with Spice nodes indicated
• In the s-domain the solution is of the form
2 –sz 3 +p
V ) z% s * = --- e
3
+e
n
G
) b – s * . 1---/ n –s ) 2n *l 3 + p
--------------------e
n , 3n = 0 s)s + a*
H
sz 3 + p G
n+1
n – s ) 2n + 2 *l 3 +
)
b
–
s
*
1
p
.
/
----------------------------- --- e
n + 1 , 3s
)
s
+
a
*
n=0
H
(1.67)
• To inverse transform V ) z% s * note that each series term consists of the product of a constant, a ratio of polynomials in s ,
– sI
and a time shift exponential (i.e. e )
• In the time-domain each series term to within a constant is of
the form
–1 =
n
)b – s* :
L > ---------------------n 9
? s)s + a* 7
1–26
% n = 0% 1% 2% F
(1.68)
t ' t – In
ECE 5250/4250 Microwave Circuits
Transient Analysis with Resistive Loads
• A partial fraction expansion of the ratio of polynomials in
(1.68) is
n
K 1 K 12
K 2n
K 22
)
b
–
s
*
F
--------------------- = ------ + ----------- + ------------------- + + ------------------n
n
s s + a ) s + a *2
s)s + a*
)s + a*
(1.69)
where
n
K 1 = b----n
a
(1.70)
and
)n – k*
K 2k
n
1
d
)b – s*
- ----------------= ------------------ ----------------) n – k *! ds ) n – k *
s
% k = 1% 2% F% n (1.71)
s = –a
• To obtain a partial solution for comparison with a Spice simulation we will solve (1.69) for n = 0 , 1, and 2.
– Case n = 0 :
1
--- J u ) t *
s
(1.72)
b–s
b 3 a )a + b* 3 a
b a + b – at
u)t*
------------------- = ---------- – ------------------------ J --- – ------------ e
a
s)s + a*
s
s+a
a
(1.73)
– Case n = 1
ECE 5250/4250 Microwave Circuits
1–27
Review of Transmission Line Theory
– n=2
2
2
2
2
2
2
) b – s * = b-------------3 a - + )---------------------------1 – b 3 a -* – -------------------------)b + a* 3 a
--------------------2
2
s
s+a
s)s + a*
)s + a*
2/
2
.
– at ) b + a *
– at
b
J ----- + K 1 – b-----L e – ------------------- te
u)t*
2 ,
2a
a
a
2
(1.74)
• Using Mathematica the analytical solution valid for t up to
25 ns was obtained
1–28
ECE 5250/4250 Microwave Circuits
Transient Analysis with Resistive Loads
• This required the use of +z and -z wave solutions from the
series for n = 0 and 1
• The theoretical voltage waveforms at z = 0 and z = 1 are
shown below
1.2
Line Voltage (v)
1.0
Load
0.8
Source
0.6
0.4
0.2
0.0
5
10
15
20
t (ns)
25
!0.2
• A circuit simulation using ADS was also run, the results
compare favorably as expected
ECE 5250/4250 Microwave Circuits
1–29
Review of Transmission Line Theory
Source
Load
R
R1
R=100 Ohm
t
C
C1
C=20 pF
TLIND
TLD1
Z=50.0 Ohm
Delay=5 nsec
VtStep
SRC1
Vlow=0 V
Vhigh=2 V
Delay=0 nsec
Rise=0.05 nsec
R
R2
R=100 Ohm
TRANSIENT
Tran
Tran1
StopTime=25.0 nsec
MaxTimeStep=0.05 nsec
• Plots of v ) 0% t * * and v ) l% t *
1.2
(v)
0.8
Load
Source
1.0
0.6
m1
Load
Source
m3
m2
m1
indep(m1)=1.148E-8
plot_vs(Load, time)=0.889
m2
indep(m2)=1.608E-8
plot_vs(Source, time)=0.963
0.4
0.2
m3
indep(m3)=2.443E-8
plot_vs(Load, time)=0.988
0.0
-0.2
0
2
4
6
8
10
12
14
16
18
20
22
24
26
time, nsec
1–30
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
Sinusoidal Signals on Transmission Lines
Voltage and Current Relationships
• Consider an axially uniform transmission line operating in
the TEM or Quasi-TEM mode
Figure 1.14: Voltage and current at z on an axially uniform line
• Let the sinusoidal voltage and current at z be given by
v ) t% z * = Re C V ) z *e
i ) t% z * = Re C I ) z *e
jMt
jMt
D = V ) z * cos @ Mt + NV ) z * A
D = I ) z * cos @ Mt + NI ) z * A
(1.75)
where the complex phasors V and I represent the voltage and
current without the time dependence e jMt
• Since a sinusoidal steady-state solution is desired, phasor
notation may be used to solve the transmission line equations
• Note that with e
jMt
suppressed it is implied that
n
(-------V- = ) jM * n V
n
(t
(1.76)
• By writing the basic line equations in phasor notation we
obtain
(V
------ = – ) R + jML *I
(z
ECE 5250/4250 Microwave Circuits
(1.77)
1–31
Review of Transmission Line Theory
and
(I
----- = – ) G + jMC *V
(z
(1.78)
• The voltage wave equation becomes
2
(-------V- – ) RG – M 2 LC *V – jM ) RC + LG *V = 0
2
(z
(1.79)
where R, G, L, and C are the primary transmission line
parameters defined earlier
• The general solution is
+ – Oz
V) z* = V e
– Oz
+V e
(1.80)
where
O = P + jQ
=
) R + jML * ) G + jMC *
=
– M LC + RG + jM ) RC + LG *
(1.81)
2
with P being the line attenuation constant in nepers per meter
and Q is the line phase constant in radians per meter
+
– V is the constant associated with the wave propagating in
–
the +z direction and V is the constant associated with the
wave propagating in the -z direction
• The current I ) z * can be shown to have general solution of the
form
+ – Oz
I)z* = I e
1–32
– Oz
+I e
(1.82)
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
• By substituting (1.77) into (1.82) we obtain
–1
(V ) z *
I ) z * = . -------------------/ -------------, R + jML- (z
+ – Oz
– Oz
–1
= ------------------- ) – OV e + OV e *
R + jML
=
(1.83)
G
+ jMC- ) V + e –Oz – V – e Oz *
-------------------R + jML
• The line characteristic impedance is defined as
+ jML =
Z0 = R
------------------O
R + jML-------------------G + jMC
(1.84)
which then relates the voltage and current on the line as
+
–
also
VZ0 = V
------ = –-------+
–
I
I
(1.85)
• Finally we can write I ) z * as
+
–
V –Oz V Oz
I ) z * = ------ e – ------ e
Z0
Z0
(1.86)
• In the time domain the steady state solution voltage waveform is
+
+
– Pz
–
–
Pz
v ) z% t * = V cos ) Mt – Qz + NV *e
+ V cos ) Mt + Qz + NV *e
ECE 5250/4250 Microwave Circuits
(1.87)
1–33
Review of Transmission Line Theory
Lossless Line
• Special Case: For an ideal lossless line R = G = 0 , O and
Z 0 reduce to
O = jQ = jM LC ) P = 0 *
LZ 0 = --C
(1.88)
• Since Q = M LC the phase velocity, + p , on the line is
1
+ p = ----------LC
(1.89)
which allows Z 0 to also be written as
1
Z 0 = --------+p C
(1.90)
• This result implies that Z 0 can be obtained by knowing the
velocity of propagation in the medium and the capacitance
per unit length of the transmission line structure
• In free space
8
1
+ p = c = --------------- S 3 R10 m/s
B0 !0
(1.91)
where ! 0 and B 0 are the free space permittivity and permeability respectively
• For a dielectrically filled structure where it is assumed
that B = B 0 , ! = !" – j!"" S ! 0 ! r and T = 0 , thus
+p = c 3 !r
1–34
(1.92)
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
where ! r is the relative permittivity of the medium
• For partially filled transmission line structures such as
microstrip, the velocity of propagation can be written as
c + p = ----------! eff
(1.93)
where ! eff is the effective value of relative permittivity.
(Note: ! eff 5 ! r ).
Low Loss Line
• Special Case: For most practical transmission line structures
the losses are small, which is to say, R « ML and G « MC
• This low-loss assumption allows the expression for O to be
simplified
• To begin with write
RG + jM ) RC + LG *O = jM LC 1 + ----------------------------------------------2
– M LC
132
(1.94)
using the binomial expansion
)1 + x*
132
1
S 1 + --- x for x « 1
2
• Thus
(1.95)
=
>
?
=
:
1 – RG
j
O S jM LC > 1 + --- ------------- – ---- . R
--- + G
----/ 9
2 M 2 LC M , L C- 7
?
U0
ECE 5250/4250 Microwave Circuits
1–35
Review of Transmission Line Theory
or
1
R G
O = P + jQ S --- LC . --- + ----/ + jM LC
, L C2
(1.96)
Note that Q did not change from the lossless case
• Using a similar approach on Z 0 results in
Z0 =
R + jML
--------------------- =
G + jMC
) R + jML * ) G – jMC *
-------------------------------------------------2
2 2
G +M C
2
+ jM ) GL – RC * + M LCS RG
-------------------------------------------------------------------2 2
M C
=
(1.97)
L- 1 RG + jM ) LG – RC *
--+ -----------------------------------------------2
C
M LC
L
1 G
LR / U --S ---- 1 + --- j . ------- – ------C
2 , MC MLC
Note also that under the low-loss assumption Z 0 is still
approximately real
1–36
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
Terminated Lossless Line
First consider the case where the generator or source driving the
line is matched to the line characteristic impedance Z 0 as shown
below in Figure 1.15.
Figure 1.15: Lossless line terminated with impedance Z 0
• For an arbitrary load impedance, Z L , the boundary conditions
will require both forward (+z) and backward (-z) propagating
waves to exist
• At z = 0 , the voltage and current equations with P = jQ are
+ – jQ ) 0 *
V)0* = V e
– jQ ) 0 *
+V e
= VL
(1.98)
and
1 + – jQ ) 0 * 1 – jQ ) 0 *
I ) 0 * = ----- V e
– ----- V e
= IL
Z0
Z0
(1.99)
• By convention, the voltage reflection coefficient at the load is
defined as
–
V< = ----+
V
ECE 5250/4250 Microwave Circuits
(1.100)
1–37
Review of Transmission Line Theory
• From the voltage and current equations we can now solve for
< in terms of Z L and Z 0
+
+ <V
) 0 * = Z = --------------------------------V ) 1 + < * - = Z 1---------------------L
01 – <
+
I)0*
V ) 1 – < * 3 Z0
(1.101)
ZL – Z0
< = -----------------ZL + Z0
(1.102)
or
• The voltage transmitted to the load due to the incident voltage wave can be defined in terms of the voltage transmission
coefficient T
+
V L = TV = ) 1 + < *V
+
(1.103)
so
TV1+<
(1.104)
• At a discontinuity in a transmission line system, such as load
Z L terminating the line at z = 0 , the ratio of power incident
to the power reflected is a quantity of interest
• A common scalar network analyzer measurement is return
loss (RL) which is 10log10 of this power ratio
• The return loss is related to < in the following way
P incident /
.
RL = 10log 10 -------------------, P reflected-
(1.105)
where
1–38
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
+2
P incident
1
+ + *
1V
= --- Re C V ) I * D = --- -----------2
2 Z0
–2
1
– – *
1V
P reflected = --- Re C V ) I * D = --- ----------2
2 Z0
(1.106)
• Now
+2
P incident
V
1-------------------- = ------------ = -------2
P reflected
–2
<
V
(1.107)
RL = – 20log 10 < dB
(1.108)
so
• If Z L = Z 0 then < = 0 and the magnitude of the voltage
+
along the line is just that of the incident wave which is V
• In general < $ 0 so
+ – jQz
V)z* = V e
+ jQz
(1.109)
j2Qz
(1.110)
+ <V e
and
V) z* = V
by writing < = We
jX
+
1 + <e
(polar form) then we can write
V) z* = V
+
1 + We
j ) X + 2Qz *
(1.111)
which inspired the vector diagram of V ) z * shown below
ECE 5250/4250 Microwave Circuits
1–39
Review of Transmission Line Theory
Figure 1.16: Vector diagram showing V ) z *
• With the aid of the vector diagram shown above, it is clear
that V ) z * takes on maximum and minimum values of
V ) z * max = V
+
V ) z * min = V
+
1 + W % X + 2Qz = 2nY
1 – W % X + 2Qz = 2nY + Y
(1.112)
where n is an integer
• The variation in V ) z * is sinusoidal with the distance
between maxima and between minima each being
d = Y 3 Q= Z 3 2
• Recall that Q = 2Y 3 Z where Z is the wavelength of TEM
waves in the medium
• The incident and reflected voltage waves interfere to produce
the voltage standing-wave pattern shown below in Figure
1.17
1–40
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
Figure 1.17: Voltage wave along the line for various real loads.
• The ratio of V ) z * max to V ) z * min is defined as the voltage
standing-wave ratio (VSWR), or simply SWR
+ W- = --------------1+ <
VSWR = 1----------1–W
1– <
(1.113)
• Comparison of termination characterization parameters
<
RL
VSWR
0.0
+G
1.0
0.1
+20 dB
1.2222
0.3162
+10 dB
1.9520
0.5012
+6 dB
3.0095
0.8913
+1 dB
17.3910
0.9441
+0.5 dB 34.7532
ECE 5250/4250 Microwave Circuits
1–41
Review of Transmission Line Theory
The analysis of a terminated transmission line up to this point has
assumed that the source impedance and line characteristic
impedance are equal. This simplifies the analysis of voltage and
current along the line since under this assumption the wave
reflected at the load is completely absorbed when it arrives at the
source.
• To consider the mismatched source case, the impedance seen
looking into an arbitrarily terminated transmission line is
helpful
Figure 1.18: Circuit for finding the input impedance of a line terminated in ZL.
Figure 1.25: Circuit for finding the input impedance of a line terminated in ZL.
• To find Z in , simply form the ratio V ) z * 3 I ) z * with z = – l
Z in
) z *= V
---------I)z*
+ jQl
z = –l
– – j Ql
V e +V e = Z 0 -------------------------------------+ jQl
– – j Ql
V e –V e
(1.114)
Recall that the load reflection coefficient is given by
–
ZL – Z0
V
< L = ------ = -----------------+
ZL + Z0
V
1–42
(1.115)
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
so
Z L – Z 0 –jQl
e + ------------------ e
ZL + Z0
= Z 0 -------------------------------------------– jQl
Z
–
Z
jQl
L
0
e – ------------------ e
ZL + Z0
jQl
Z in
(1.116)
Finally after rearranging using Euler’s identity for tan( ) Z in
reduces to
Z in
Z L + jZ 0 tan Ql
= Z 0 ---------------------------------Z 0 + jZ L tan Ql
(1.117)
• Two special cases of interest that will be considered later
result when the operating frequency is such that l = Z 3 2 or
l = Z34
• The input impedances for these cases are
Z in ) l = Z 3 2 * = Z L
2
Z
Z in ) l = Z 3 4 * = -----0ZL
(1.118)
• The input impedance of open and short circuit terminated
transmission lines will also be of interest
Z in
Z in
ZL ' 0
ZL ' G
ECE 5250/4250 Microwave Circuits
= jZ 0 tan Ql
= – jZ 0 cot Ql
(1.119)
1–43
Review of Transmission Line Theory
• The reflection coefficient at any z = – l can also be obtained
by noting that
V
+
V
+ jQl
z = –l
= V e
(1.120)
– – j Ql
–
z = –l
= V e
so
– – j Ql
e - = < e – j2Ql
<)l* = V
---------------L
+ jQl
V e
(1.121)
• The above result will appear later during the discussion of
scattering parameters.
Terminated Lossless Line with Arbitrary Source Impedance
The circuit of interest is shown below in Figure 1.26
Figure 1.19: Terminated line with arbitrary source impedance.
• One approach to this problem is to consider all the reflections
and re-reflections and then sum the infinite series
• Since the infinite series is geometric and < 5 1 , it is summable in closed form
1–44
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
• A second approach is to use the boundary conditions at the
source and load terminations to obtain the steady-state solution directly
• Using (1.117) for Z in we can write
Z in
+ jQl
– – jQl
V ) z = l * = V g ------------------- = V e + V e
Z in + Z g
–
(1.122)
+
• Now since V = V < L
Z in
+ jQl
– jQl
V g ------------------- = V @ e + < L e A
Z in + Z g
(1.123)
– jQl
Z in
e
V = V g ------------------- ----------------------------Z in + Z g 1 + < e –j2Ql
(1.124)
so
+
L
• Now substitute
– j2Ql
Z in
1 + <L – e
= Z 0 ---------------------------------– j2Ql
1 – <L e
(1.125)
into (1.124) to obtain
– jQl
Z0 e
+
-------------------------------------------------------------------------------------------V = Vg
– j2Ql
– j2Ql
@ Z0 ) 1 + <L e
* + Zg ) 1 – <L e
*A
– jQl
Z0
e
= V g ------------------ ----------------------------------–
j2Ql
Z0 + Zg 1 – < < e
(1.126)
g L
ECE 5250/4250 Microwave Circuits
1–45
Review of Transmission Line Theory
where < g is the reflection coefficient looking towards the
generator
• Finally, we can write for any – l 5 z 5 0
j2Qz
Z0
1 + <L e
- e – jQ ) z + l *
V ) z * = V g ------------------ ----------------------------------Z 0 + Z g 1 – < < e –j2Ql
(1.127)
g L
An equation for I ) z * may be obtained in a similar manner
Source to Load Power Transfer Considerations
• The power delivered to the load via the lossless line is just
the input power which is given by
*
1--*
1--- =
V
) –l * :
P = Re C V ) – l *I ) – l * D = > V ) – l * ---------------- 9
2
2?
Z in 7
1
2
= --- V g
2
= 1 :
Z in
-9
Re
------------------> -----*
Z in + Z g
? Z in 7
2
(1.128)
• Let Z in = R in + jX in and Z g = R g + jX g , then
R in
2
1
P = --- V g -----------------------------------------------------------2
2
2
)R + R * + )X + X *
in
g
in
(1.129)
g
– As a special case suppose that Z L = Z 0
– Clearly < L = 0 , Z in = Z 0 , and
Z0
2
1
-------------------------------------P = Vg
2
2
2
)Z + R * + X
0
1–46
g
(1.130)
g
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
– As another special case suppose that Z in = Z g
– Note that the condition Z in = Z g can be satisfied through
adjustment of Ql and Z 0
Rg
2
1--P = V g -------------------------2
2
2
4)R + X *
g
(1.131)
g
– How do we obtain maximum power transfer?
– For Z g fixed we know from circuit theory that for maximum power transfer we choose Z in = Z *g
– The resulting power delivered is as expected
1
2 1
P = --- V g --------2
4R g
(1.132)
– Note that if Z g is real then (1.130) and (1.131) yield the
maximum power transfer result
Terminated Lossy Line
• The analysis of a terminated lossy line is very similar to the
lossless case except now jQ must be replaced by O = P + jQ
• It will be assumed that P is small enough to imply that Z 0
can still be considered real
Figure 1.20: Terminated lossy line.
ECE 5250/4250 Microwave Circuits
1–47
Review of Transmission Line Theory
• To begin with we can immediately write
+
V)z* = V @e
– Oz
Oz
+ <L e A
(1.133)
+
V – Oz
Oz
I ) z * = ------ @ e – < L e A
Z0
• Additionally < L and Z in respectively become
< ) l * = <L e
– j2Ol
= @ <L e
– j2Ql
Ae
– 2Pl
(1.134)
and
Z in
Z L + Z 0 tanh Ol
= Z 0 ---------------------------------Z 0 + Z L tanh Ol
(1.135)
• The power delivered to the line input is
1
P in = --- Re C V ) – l *I * ) – l * D
2
2
=
8
>
8
?
+
V = 2Pl
– j2Ql
– j2Ql * :
2 – 2Pl
----------=
e – <L e
+ <L e
– ) <L e
* 9
2Z 0 >?
7
+2
2 2Pl
<)l* e
2 2Pl
V
= ------------ @ 1 – < ) l * Ae
2Z 0
(1.136)
• The power actually delivered to the load is
+2
1
V
2
P L = --- Re C V ) 0 *I * ) 0 * D = ------------ @ 1 – < L A
2Z 0
2
1–48
(1.137)
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
• The power lost in the line is
+2
V
2Pl
– 2Pl
2
P loss = P in – P L = ------------ @ ) e – 1 * + < L ) 1 – e
* A (1.138)
2Z 0
Line Attenuation Calculation
The perturbation method of loss analysis assumes that the fields
associated with the lossy line are very similar to the fields of the
lossless line.
• The power flow along the line is given by
P ) z * = P0 e
– 2Pz
(1.139)
where P 0 is the power input at z = 0 , and P is the attenuation parameter of interest
• The power loss per unit length is
(P- = 2PP e –2Pz = 2PP ) z *
P l = –--------0
(z
(1.140)
• Rearranging we obtain
Pl ) z *
Pl ) z = 0 *
P = ------------- = ---------------------2P ) z *
2P 0
(1.141)
• Note that this analysis assumes that P 0 is known and P l can
be found from the fields of the lossless line
ECE 5250/4250 Microwave Circuits
1–49
Review of Transmission Line Theory
Dispersion
• In the first order approximation to O given by (1.96) we
found that
Q = M
RG
LC – ---------------------2
2M LC
(1.142)
• If Q is not of the form Q = aM , then the phase velocity
+ p = M 3 Q $ constant
• If + p is a function of frequency then the frequency components associated with a broadband signal will arrive at the
load end of the line at different times
• This phenomenon is known as dispersion.
• In linear system theory we would say the system has nonlinear phase
2
• Fortunately since RG is typically much less than 2M LC ,
we see that for low loss lines Q is very nearly linear in M
Example: Consider the transmission line circuit shown below in
Figure 1.21.
Figure 1.21: Match terminated low loss line.
1–50
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
• We can immediately write that
1
– Pl
v L ) M * = --- v g ) M *e e
2
– jM
RG
LC – --------------------2
2M LC
(1.143)
• The transfer function denoted H ) M * = H ) 2Yf * is thus given
by
vL ) M *
1 – Pl
H ) M * = -------------= --- e e
2
vg ) M *
– jM
RG
LC – --------------------2
2M LC
(1.144)
• The magnitude and phase are
1
H ) M * = --- e –Ml
2
RGl NH ) M * = – Ml LC + -----------------2M LC
(1.145)
• For RG $ 0 we see that the phase function is composed of
linear and nonlinear components
Distortionless Line
• It is possible for a lossy line to have a linear phase factor if
the line parameters satisfy
R
--- = G
---L
C
(1.146)
• To demonstrate this insert (1.146) into (1.96)
ECE 5250/4250 Microwave Circuits
1–51
Review of Transmission Line Theory
2
R
R – 2j ------O = jM LC 1 + -----------2 2
ML
M L
R
= jM LC 1 – j ------ML
= R C
---- + jM LC = P + jQ
L
132
(1.147)
• Clearly, Q is now a linear function of frequency and the
attenuation, P , is nearly constant with frequency
• Note that the line resistance is usually a weak function of frequency
Example: RG-58/U Coax
• To better illustrate the impact of dispersion, consider the special case of RG-58/U coaxial cable
• In the Pozar text (and Collin) the transmission parameters of
coax are shown to be
Coax
b
a
B
L = ------ ln b--2Y a
2Y!" C = -------------ln b 3 a
R s . 1 1/
----R = - --- + --2Y , a b-
G = 2YM!""
----------------ln b 3 a
where R s is the surface resistivity of the conductor, and is
given by
Rs =
1–52
MB
------- ,
2T
(1.148)
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
where here B = B 0 B r = B 0 with B 0 = 4Y R 10
–8
• For RG-58/U the dielectric diameter is 2b = 0.116 in and the
dielectric material is polyethylene having ! r = 2.25 @ 10
GHz and !"" = 0.004! 0 ! r
• Since the characteristic impedance is nominally 50 ohms we
can determine the inner radius, a , by setting
60 b
L- = -------Z 0 S --ln --- = 50 ,
C
!r a
thus 2a = 0.033 in
• The transmission line parameters are given by:
!r Z0
B- b
----L =
ln --- = --------------- = 250 nH/m
2Y a
c
!
2Y!"
C = --------------- = --------r = 100 pf/m
cZ 0
ln b 3 a
(1.149)
R s . 1 1/
–4
R = ------ --- + --- = 1.273 R 10 R f ohms/m
2Y , a b– 13
G = 2YM!""
----------------- = MC!""
--------------- = 2.513 R 10 R f s/m
ln b 3 a
!"
• To obtain the frequency response of a one meter section of
properly terminated RG-58/U cable, we simply insert the
above transmission line parameters into
ECE 5250/4250 Microwave Circuits
1–53
Review of Transmission Line Theory
vL ) f *
1 –Ol
H ) f * = ------------ = --- e
2
vg ) f *
1
RG + j2Yf ) RC + LG *
= --- exp – j2Yf LC 1 + ----------------------------------------------------- l
2 2
2
– 4Y f LC
(1.150)
making sure to explicitly include the frequency dependence
on R and G , i.e., R = R ) f * and G = G ) f *
• The equivalent impulse response of the system can be
obtained by inverse Fourier transformation, i.e.,
–1
h)t* = F CH)f*D =
G
[–G
H ) f *e
j2Yft
df
(1.151)
• The response of the system to an arbitrary pulse waveform,
v g ) t * = p ) t * , can be obtained in the transform domain, or by
direct convolution
–1
v L ) t * = p ) t * * h ) t * = F C P ) f *H ) f * D
(1.152)
where P ) f * = F C p ) t * D
• The following results were obtained using Mathematica
– Magnitude and phase frequency response for l = 2 m
– Impulse response computed using a 1024 point inverse
Fourier transform (IFFT) for l = 2% 10% and 50 m
– Simulated pulse response to a 500 Mb/s return-to-zero
(RZ) data pattern of the form ...000101001000... (each
pulse is 2ns long)
1–54
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
+
vL ) t * = y ) y *
Z0
p)t*
Z0
l
p)t*
2
2 ns
6
10
16
t (ns)
• Mathematica modeling
ECE 5250/4250 Microwave Circuits
1–55
Review of Transmission Line Theory
• Frequency response
0
Gain in dB less
the 1/2 loss factor
of 6 dB
Gain (dB)
!2
!4
!6
!8
!10
107
1–56
108
109
1010
1011
f (Hz)
ECE 5250/4250 Microwave Circuits
Sinusoidal Signals on Transmission Lines
• The phase response and pulse response
0.035
Phase (deg)
0.030
0.025
Linear phase
term removed
0.020
0.015
0.010
0.005
108
109
1010
1011
f (Hz)
Real
0.6
The model makes it
impossible to eliminate
the imaginary part
0.4
0.2
10
20
30
40
50
t (ns)
!0.2
!0.4
Imaginary?
– The input is a real signal, but the frequency response is not
conjugate symmetric, so the impulse response is complex
– This makes the pulse response contain real and imaginary
parts as well
ECE 5250/4250 Microwave Circuits
1–57
Review of Transmission Line Theory
• ADS schematic when using the lossy coax model
Input
t
R
VtPulse R1
SRC1
R=50 Ohm
Delay=6 nsec
Width=2 nsec
t
VtPulse
SRC2
Delay=10 nsec
Width=2 nsec
Output
COAX_MDS
TL1
A=16.5 mil
Ri=58 mil
Ro=60 mil
L=2 meter
T=2.0 mil
Cond1=5.8E+7
Cond2=5.8E+7
Mur=1.0
Er=2.25
TanD=0.004
R
R2
R=50 Ohm
TRANSIENT
Tran
Tran1
StopTime=50.0 nsec
MaxTimeStep=0.1 nsec
t
VtPulse
SRC3
Delay=16 nsec
Width=2 nsec
• Input and output waveforms with ADS are more realistic, but
a warning regarding a complex impulse response is posted
1.0
Input
Output
0.8
Output
Input
0.6
0.4
0.2
-0.0
-0.2
0
5
10
15
20
25
30
35
40
time, nsec
1–58
ECE 5250/4250 Microwave Circuits
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