Chapter 9.

Transmission Lines

Objectives
• Transmission lines are used to transmit electric
energy and signals from one point to another,
specifically from a source to a load.
• In this chapter, we investigate wave phenomena
in transmission lines.
– To understand how to treat transmission lines as
circuit elements possessing complex impedances
that are functions of line length and frequency.
– To understand wave propagation on lines, including
cases in which losses may occur.
– To learn methods of combining different
transmission lines to accomplish a desired objective.
– To understand transient phenomena on lines.
 Physical description of
transmission line propagation

• Consider a uniform transmission line

– Consisting of two parallel perfect conductors
– TEM wave: voltage between conductors and
current along the line  transverse
components of E and H

The transmission line equations

• Consider a differential length Δz of a transmission

line that is described by the following four
parameters.
– R: resistance per unit length (both conductors) (Ω/m)
– L: inductance per unit length (both conductors) (H/m)
– G: conductance per unit length (S/m)
– C: capacitance per unit length (F/m)
Using Kirchoff's voltage law:
∂i ( z , t )
v( z , t ) − RΔz ⋅ i ( z , t ) − LΔz − v( z + Δz , t ) = 0
∂t
Using Kirchoff's current law:
∂v( z + Δz , t )
i ( z , t ) − GΔz ⋅ v( z + Δz , t ) − C Δz − i ( z + Δz , t ) = 0
∂t

∂v( z , t ) ∂i ( z , t ) 
= − Ri ( z , t ) − L
∂z ∂t 
 "Telegrapher" equations
∂i ( z , t ) ∂v( z , t ) 
= −Gv( z , t ) − C
∂z ∂t 

++++++++++++++++++
∂ 2V ∂I ∂2 I ∂2 I ∂V ∂ 2V
= − R − L and = − G − C
∂z 2 ∂z ∂t ∂z ∂z 2 ∂z ∂t ∂z
 ∂ 2V ∂ 2V ∂V
 2 = LC 2
+ ( LG + RC ) + RGV
∂z ∂t ∂t
 2 2
← general wave equations for the transmission line
 ∂ I = LC ∂ I + ( LG + RC ) ∂I + RGI
 ∂z 2 ∂t 2 ∂t

Lossless propagation
∂ 2V ∂ 2V
= LC
∂z 2 ∂t 2
 z  z
→ V ( z , t ) = f1  t −  + f 2  t +  = V + + V − , where v, the wave velocity, is constant.
 ν  ν
z z
t − : forward z propagation, t + : backward z travel
ν ν
∂f1 ∂f1 ∂ (t − z / ν ) 1
= = − f1'
∂z ∂(t − z / ν ) ∂z ν
∂f1 ∂f1 ∂ (t − z / ν )
= = f1'
∂t ∂(t − z / ν ) ∂t
∂ 2 f1 1 " ∂ 2 f1
= f1 and = f1"
∂z 2 ν 2 ∂t 2

1 "
2 1
f = LCf1"
ν
1
ν=
LC
 ∂V ∂I
 ∂z = − L ∂t
 ← telegraphist's equations under lossless condition (R=G=0)
 ∂I = −C ∂V
 ∂z ∂z

∂I 1 ∂V 1
=− = ( f1' − f 2' )
∂t L ∂z Lv
1   z  z 
I ( z, t ) =  f1  t −  − f 2  t +   = I + + I −
Lv   v   v 
L
Z 0 = Lv = ← characteristic impedance
C
V + = Z0 I +
V − = −Z0 I −

Lossless propagation of sinusoidal

voltages
 ( z , t ) = V0 cos ω (t ± z / ν p )+φ  = V0 cos [ωt ± β z +φ ]
f ( z , t ) = V0 cos(ωt − β z ) ← forward z propagation
b ( z , t ) = V0 cos(ωt + β z ) ← backward z propagation
ω
β≡ ← phase constant
νp
f ( z , 0) = b ( z , 0) = V0 cos(β z ) ← t = 0
2π νp
λ= =
β f
β z = 2mπ
ωt − β z = ω (t − z / ν p ) = 2mπ
 Complex analysis of sinusoidal
waves
e ± jx = cos( x ) ± j sin ( x )
1 1
cos( x ) = Re[e ± jx ] = (e jx + e − jx ) = e jx + c.c.
2 2
1 jx − jx 1 jx
sin ( x ) = ± Im [e ± jx ] = (e − e ) = e + c.c.
2j 2j
1
 ( z , t ) = V0 cos [ωt ± β z +φ ] = ( V0 e jφ ) e ± j β z e jωt + c.c.
2 
V0
± jβ z jωt
Vc ( z , t ) = V0 e e ← complex instantaneous voltage
Vs ( z ) = V0 e ± j β z ← phasor voltage under sinusoidal steady-state conditions
1
 ( z , t ) = V0 cos [ωt ± β z +φ ] = Re[Vc ( z , t )] = Vc + c.c.
2
1
 ( z , t ) = V0 cos [ωt ± β z +φ ] = Re[Vs ( z )e jωt ] = Vs ( z )e jωt + c.c.
2

Example
Two voltage waves having equal frequencies and amplitudes propagate in opposite
directions on a lossless transmission line. Determine the total voltage as a function
of time and position.

Solution .
VsT ( z ) = V0 e − j β z + V0 e + j β z = 2V0 cos( β z )
 ( z , t ) = Re[2V0 cos( β z )e jωt ] = 2V0 cos( β z )cos(ωt )
← standing wave
 Transmission line equations and
their solutions in phasor form
∂ 2 ∂ 2 ∂
2
= LC 2 + ( LG + RC ) + RG←general wave equation for the transmission line
∂z ∂t ∂t
∂ 2Vs
= −ω 2 LCVs + jω ( LG + RC )Vs + RGVs
∂z 2
d 2Vs
= ( R + jω L )(G + jωC )Vs = γ 2Vs γ : propagation constant (complex)
dz 2   
Z Y

γ = ( R + jω L )(G + jωC ) = ZY = α + j β
Vs ( z ) = V0+ e −γ z + V0− e + γ z
I s ( z ) = I 0+ e −γ z + I 0− eγ z
1 1
 ( z , t ) = I 0 cos(ωt ± β z +ξ ) = ( I 0 e jξ ) e ± j β z e jωt + c.c. = I s ( z )e jωt + c.c.
2
 2
I0

∂  ∂  ∂Vs
= −  R + L  = −( R + jω L ) I s = − ZI s
∂z  ∂t  ∂z
∂  ∂  ∂I s
= −  G + C  = −(G + jωC )Vs = −YVs
∂z  ∂t  ∂z
−γ V0+ e −γ z + γ V0− eγ z = − Z ( I 0+ e −γ z + I 0− eγ z )
V0+ V− Z Z Z
Z0 = = − 0− = = =
I0+
I0 γ ZY Y
R + jω L
Z0 = = Z 0 e jθ
G + jωC

Example
A lossless transmission line is 80 cm long and operates at a frequency of 600 MHz.
The line parameters are L = 0.25μ H/m and C = 100 pF/m. Find the characteristic
impedance, the phase constant, and the phase velocity.

Solution .
L 0.25 × 10−6
Z0 = = = 50Ω
C 100 × 10−12
γ = α + j β = ( R + jω L )(G + jωC ) = jω LC
β = ω LC = 2π (600 ×106 ) (0.25 ×10−6 )(100 ×10−12 ) = 18.85 rad/m
ω 2π (600 ×106 )
νp = = = 2 ×108 m/s
β 18.85
Lossless and low-loss propagation
Vs ( z ) = V0+ e −α z e − jβ z + V0− eα z e j β z
 ( z , t ) = V0+ e −α z cos(ωt − β z ) + V0− eα z cos(ωt + β z ), α : attenuation coefficient
Lossless in propagation (R=G=0) → α = 0 → γ = j β = jω LC
++++++++
In the low-loss approximation, R<<ω L and G<<ωC ,
 R  
1/ 2
G  
1/ 2

γ = α + j β = [( R + jω L )(G + jωC )] = jω LC 1 +

1/ 2
 1 +  
 jω L   jωC  

2
x x
1+ x ≈ 1+− ( x << 1)
2 8
 R R2   G G 2 
γ ≈ jω LC 1 + + 2 2  1 + + 2 2 
 j 2ω L 8ω L   j 2ωC 8ω C  
 1  R G  1  R 2 2 RG G 2  
γ = α + j β ≈ jω LC 1 + + +  − + 
 j 2ω  L C  8ω 2  L2 LC C 2  

1 C L  1 G R  
2

α ≈  R +G  and β ≈ ω LC 1 +  −  
2 L C  8  ωC ω L  

Z0 =
R + jω L
= 
(
jω L 1 + )≈
L  1 + j 2ω L + 8ω 2 L2 
R R R2
jω L

( )
G + jωC G
jωC (1 + )
C  1 + j 2GωC + G22 2  ( )
jωC  8ω C 
L  1  1  R G  G2  j  G R  
2

Z0 ≈ 1 + 2   +  − 2  + −  ← 1 / (1 + x ) ≈ 1 − x, where x << 1.
C  2ω  4  L C  C  2ω  C L  

Example
Suppose in a certain transmission line G = 0, but R is finite-valued and satisfies the
low-loss requirement, R << ω L. Use Eq. (56) to write the approximate magnitude
and phase of Z 0 .

Solution .
L  R  jθ
Z 0 (G = 0 ) ≈ 1 − j 2ω L  = Z 0 e
C  
where Z 0 ≈ L / C , and θ =tan −1 ( − R / 2ω L ).
 Power transmission and loss
characterization
Instantaneous power:  ( z , t ) =  ( z , t )  ( z , t ) = V0 I 0 e −2α z cos(ωt − β z )cos(ωt − β z + θ )
1 T
T 0
Time-averaged power:   = V0 I 0 e −2α z cos(ωt − β z )cos(ωt − β z + θ )dt ← T=2π /ω

1 T1
V0 I 0 [ cos( 2ωt − 2 β z + θ ) + cos(θ )] dt
T 0 2
  =
2
1 1 V0 −2α z
  = V0 I 0 e −2α z cosθ = e cosθ [W]
2 2 Z0
V0 −α z − j β z
Vs ( z ) = V0 e −α z e − j β z and I s ( z ) = I 0 e −α z e − j β z = e e where Z 0 = Z 0 e jθ
Z0
1
  = Re{Vs I*s}
2
2
1  −α z − j β z V0*  1 V V
*
 1 V0 −2α z
  = Re V0 e e − jθ
e−α z e + jβ z  = Re  0 0 e −2α z e jθ  = e cosθ
2  Z0 e  2  Z 0  2 Z 0
 ( z ) =  (0) e −2α z
 ( z )
= e−2α z = 10− kα z
 (0)
e −2 = 10− k  k = log10 (e 2 ) = 0.869
  (0)    (0)   V ( 0)  −α z
Power Loss(dB) = 10log10   = 8.69α z , Power Loss(dB) = 10log10   = 20log10  0  where V0 ( z ) = V0 (0) e .
  ( z )    ( z )   V0 ( z ) 

Example
A 20 m length of transmission line is known to produce a 2.0 dB drop in power from
end to end, (a ) What fraction of the input power reaches the output? (b) What fraction
of the input power reaches the midpoint of the line? (c ) What exponential attenuation
coefficient, α , does this represent?

Solution .
 ( 20)
(a) = 10−0.2 = 0.63
 (0)
(b ) 2 dB in 20 m → 0.1 dB/m. For 10 m span, the loss is 1.0 dB. 10−0.1 = 0.79
  (0) 
(c ) Power Loss(dB) = 10log10   = 8.69α z
  ( z ) 
2.0 dB
α= = 0.012 [Np/m]
(8.69 dB/Np )( 20 m )
 Wave reflection at discontinuities
Vi ( z ) = V0i e −α z e− j β z
Vr ( z ) = V0 r e +α z e + j β z
VL = V0i + V0 r
1 V 1
I L = I 0i + I 0 r =
Z0
[V0i − V0 r ] = ZL = Z [V0i + V0 r ]
L L

V0 r Z L − Z 0
Γ≡ = = Γ e jφr ← reflection coefficient
V0i Z L + Z 0
VL = V0i + ΓV0i
VL 2Z L
τ≡ = 1+ Γ = = τ e jφt ← transmission coefficient
V0i Z0 + Z L
2
1 V0V0* −2α L jθ  1 V0 −2α L
i  = Re  e e = e cosθ ← incident power
2  Z 0  2 Z 0
2 2
1  (ΓV0 )(Γ*V0* ) −2α L jθ  1 Γ V0 −2α L
r  = Re  e e = e cosθ ← reflected power
2  Z0  2 Z 0
r  2
= ΓΓ* = Γ
i 
t  2
= 1− Γ
i 
Z 02 − Z 01
Γ=
Z 02 + Z 01

Example
A 50 Ω lossless transmission line is terminated by a load impedance, Z L = 50 −
j 75 Ω. If the incident power is 100 mW, find the power dissipated by the load.

Solution .
Z − Z 0 50 − j 75 − 50
Γ= L = = 0.36 − j 0.48 = 0.60e − j .93
Z L + Z 0 50 − k 75 + 50
2
t  = (1 − Γ )i  = [1 − (0.60)2 ](100) = 64 mW
 Example
Two lossy lines are to be joined end-to-end. The first line is 10 m long and has a loss
rating of 0.20 dB/m. The second line is 15 m long and has a loss rating of 0.10 dB/m.
The reflection coefficient at the junction (line 1 to line 2) is Γ = 0.30. The input
power (to line 1) is 100 mW. (a ) Determine the total loss of the combination in dB.
(b) Determine the power transmitted to the output end of line 2.

Solution .
 1   1 
( a ) L j (dB) = 10log10   = 10log10  = 0.41 dB
 1− Γ 
2
 1 − 0.09 
 
Lt (dB) = (0.20)(10) + 0.41 + (0.10)(15) = 3.91 dB
(b ) Pout = 100 ×10−0.391 = 41 mW

Voltage standing wave ratio

(VSWR)
VsT ( z ) = V0 e − j β z + ΓV0 e j β z ← load at z =0
Z L − Z0
Γ= = Γ e jφ
Z L + Z0
VsT ( z ) = V0 (e − j β z + Γ e j ( β z +φ ) ) = V0 (e − j β z − Γ e − j β z + Γ e − j β z + Γ e j ( β z +φ ) ) = V0 ( e− j β z − Γ e− j β z ) + V0 Γ (e − j β z + e j ( β z +φ ) )
VsT ( z ) = V0 (1 − Γ )e − j β z + V0 Γ e jφ / 2 (e − j β z e − jφ / 2 + e j β z e jφ / 2 )
VsT ( z ) = V0 (1 − Γ )e − j β z + 2V0 Γ e jφ / 2 cos( β z + φ / 2)
 ( z , t ) = Re[VsT ( z )e jωt ] = V0 (1 − Γ )cos(ωt − β z ) + 2 Γ V0 cos( β z + φ / 2)cos(ωt + φ / 2) (1)
 
traveling wave standing wave

VsT ( z ) = V0 e − j β z (1 + Γ e j ( 2 β z +φ ) ) VsT ( z ) max = V0 (1 + Γ ) VsT ( z ) min = V0 (1 − Γ )

1
zmin = − (φ + ( 2m + 1)π ) ( m = 0, 1, 2, ) (2) ← 2β zmin + φ = −( 2m + 1)π , 2π / β = λ
2β
VsT ( zmin ) = V0 (1 − Γ )
 ( zmin , t ) = ±V0 (1 − Γ )sin (ωt + φ / 2) ← substituting (2) into (1)
1
zmax = − (φ + 2mπ ) ( m = 0, 1, 2, ) ← 2β zmax + φ = −2mπ
2β
VsT ( zmax ) = V0 (1 + Γ )
 ( zmax , t ) = ±V0 (1 + Γ )cos(ωt + φ / 2)
VsT ( zmax ) 1 + Γ
s≡ = ← VSWR
VsT ( zmin ) 1 − Γ
21
Γ=0

22
Γ=1
23
Γ=-1

Γ=0.5
24
 Example
Slotted line measurements yield a VSWR of 5, a 15 cm spacing between successive
voltage maxima, and the first maximum at a distance of 7.5 cm in front of the load.
Determine the load impedance, assuming a 50 Ω impedance for the slotted line.

Solution .
15 cm spacing between maxima → λ /2, f = c / λ = 1 GHz
First maxima at 7.5 cm ← at a distance of λ /4 from the load
→ voltage minimum occurs at the load
s −1 5 −1 2
Γ = = =
s +1 5 +1 3
2 Z − Z0
Γ=− = L
3 Z L + Z0
1 50
Z L = Z0 = = 10 Ω
5 5
 10.11 Transmission lines of finite
length

VsT ( z ) = V0+ e − j β z + V0− e j β z

I sT ( z ) = I 0+ e − j β z + I 0− e j β z
VsT ( z ) V0+ e − j β z + V0− e j β z
Zw ( z) ≡ = ← wave impedance
I sT ( z ) I 0+ e − j β z + I 0− e jβ z
 e − j β z + Γe j β z 
Z w ( z ) = Z 0  − jβ z jβ z 
← V0− = ΓV0+ , I 0+ = V0+ / Z 0 , and I 0− = −V0− / Z 0
 e − Γ e 
 Z cos( β z ) − jZ 0sin ( β z ) 
Z w ( z ) = Z0  L  ← Euler identity, Γ = ( Z L − Z 0 ) / ( Z L + Z 0 )
 Z 0 cos( β z ) − jZ L sin ( β z ) 
 Z cos( β l ) + jZ 0sin ( β l )   Z L + jZ 0 tan( β l ) 
Z in = Z 0  L  = Z0   ← at z = -l
 Z 0 cos( β l ) + jZ L sin ( β l )   Z 0 + jZ L tan( β l ) 

2π mλ
βl = = mπ ( m = 0, 1, 2, ) ← half-wavelength line, or an integer multiple thereof
λ 2
 Z in (l = mλ / 2) = Z L
2π λ π
βl = ( 2m + 1) = ( 2m + 1) ( m = 0, 1, 2, ) ← odd multiple of a quarter wavelength
λ 4 2
Z 02
 Z in (l = λ / 4) =
ZL
++++++++++++++++++++++

Z01 Z02, λ/4 Z03

ZL=Z03
Zin

Z 03 + jZ 02 tan β 2l
Z in = Z 02
Z 02 + jZ 03 tan β 2l
Z 022
Z in (line 2) =
Z 03
No reflection if Z in = Z 01 .
Z 02 = Z 01Z 03
 Some transmission line examples

2π
λ = v / f = 2.5 m, β l = ⋅ 2 = (0.8π )2 = 1.6π = 288o = 0.8λ
λ ( = 2.5 m )
Vin = 30 cos ( 2π 108 t ) V
VL = 30 cos ( 2π 108 t − 1.6π ) V ← delayed in phase by 1.6π rad
Vin
I in = = 0.1 cos ( 2π 108 t ) A
300
I L = 0.1 cos ( 2π 108 t − 1.6π ) A
1
Pin = PL = × 30 × 0.1 = 1.5 W
2
−−−−−−−−−−−−−−−−−−−−−−−−−−−
150 − 300 1
Γ= = − ⇐ connecting a second receiver: load impedance of 150 Ω
150 + 300 3
1 + 13
s= =2
1 − 13

Z L + jZ 0 tan β l 150 + j 300 tan 288

Z in = Z 0 = 300 = 510∠ − 23.8 = 466 − j 206 Ω
Z 0 + jZ L tan β l 300 + j150 tan 288 

Vs , in 60
I s , in = = = 0.0756∠15.0 A
Z in 300 + 466 − j 206
1 2 1
Pin = I s , in Rin = × (0.0756)2 × 466 = 1.333 W
2 2
2
1 Vs , L
1.333 / 2 = 0.667 = : for one receiver
2 300
Vs , L = 20 V
1
zmax = − (φ + 2mπ ) ( m = 0, 1, 2, ) ← Γ = Γ e jφ
2β
zmax = −0.625 and − 1.875 m ← β = 0.8π and φ = π (Γ = −1 / 3)
zmin = 0 and − 1.25 m
Vs , in = I s , in Z in = (0.0756∠15.0 )(510∠ − 23.8 ) = 38.5∠ − 8.8
(
VsT = e − j β z + Γe j β z V0+ )
(
Vs , in = e j β l + Γe − j β l )V 0
+
← z = −l
Vs , in 38.5∠ − 8.8
V0+ = = = 30.0∠72.0 V
e j β l + Γe − j β l e j1.6π − 13 e − j1.6π
Vs , L = (1 + Γ )V0+ = 20∠72 = 20∠ − 288 ← z = 0
 Example
In order to provide a slightly more complicated example, let us now place a purely
capacitive impedance of − j300 Ω in parallel with the two 300 Ω receivers. We are
to find the input impedance and the power delivered th each receiver.

Solution .
150( − j 300) − j 300
ZL = = = 120 − j 60 Ω
150 − j 300 1 − j 2
120 − j 60 − 300 −180 − j 60
Γ= = = 0.447∠ − 153.4
120 − j 60 + 300 420 − j 60
1 + 0.447
s= = 2.62
1 − 0.447
(120 - j 60) + j 300 tan 288
Z in = 300 
= 755 − j138.5 Ω ← electrical length of line: 288
300 + j (120 - j 60) tan 288
VTh 60
I s , in = = = 0.0564∠7.47 A
ZTh + Z in 300 + 755 − j138.5
1
Pin = (0.0564)2 (755) = 1.200 W, each receiver gets only 0.6 W
2

Example
As a final example, let us terminate our line with a purely capacitive impedance, Z L =
− j 300 Ω. We seek the reflection coefficient, the VSWR, and the power delivered to
the load.

Solution .
− j 300 − 300
Γ= = − j1 = 1∠ − 90
− j 300 + 300
1 + − j1
s= =∞
1 − − j1
− j 300 + j 300 tan 288
Z in = 300 = j 589
300 + j ( − j 300) tan 288
Input Impedance of a Short-Circuited TL

l
Z L + jZ 0 tan β l
Z in (l ) = Z 0
Z 0 + jZ L tan β l
Z in → Z0 s.c Z L =0

Z in (l ) = jZ 0 tan β l = jX in

 2π 
 βl = ⋅l 
X in  λ 
 1 π
Z0  l = λ → βl = 
 4 2
 l = 1 λ → βl = π 
 
 2 

Input Impedance of an Open-Circuited TL

l
Z L + jZ 0 tan β l
Z in (l ) = Z 0
Z 0 + jZ L tan β l
Z in → Z0 o.c Z L =∞

Z in (l ) = − jZ 0 cot β l = jX in

 2π 
 βl = ⋅l 
 λ 
X in
 1 π
Z0  l = λ → βl = 
 4 2
1
 l = λ → βl = π 
 
 2 
 The Smith Chart

Smith Chart
• A graphical aid for
solving transmission
line problems
• Microwave and RF
engineers can
develop intuition
about transmission
line and impedance
matching problems
by learning to think
in terms of the
Smith chart.

36
 Derivation of the Smith chart
• A polar plot of the voltage reflection coefficient, Γ
• The real utility of the Smith chart lies in the fact that it can be used to convert from
reflection coefficient to normalized impedances (or admittances), and vice versa.
 mapping normalized impedances on a complex plane for the reflection coefficient

If a lossless line of characteristic impedance Z 0 is terminated with a load impedance Z L ,

zL − 1
Γ= = Γ e jθ , where z L = Z L / Z 0 .
zL + 1
1 + Γ e jθ
zL =
1 − Γ e jθ

Let Γ = Γ r + jΓi , and z L = rL + jxL .

(1 + Γ r ) + jΓi
rL + jxL =
(1 − Γ r ) − jΓi

1 − Γ 2r − Γi2 2Γi
rL = , xL =
(1 − Γ r ) 2 + Γi2 (1 − Γ r ) 2 + Γi2

2 2 2 2
 rL   1   1   1 
( Γ r − 1)
2 2
 Γr −  + Γi =   , +  Γi −  =   .
 1 + rL   1 + rL   x L   xL 
37

Constant Resistance Circles

2 2
 rL  2  1 
 Γr −  + Γi =  
 1 + rL   1 + rL 

r = 0  center: (0, 0), radius: 1

r = 0.5  center: (1/3, 0), radius: 2/3
r = 1  center: (1/2, 0), radius: 1/2
r = 2  center: (2/3, 0), radius: 1/3
r = ∞  center: (1, 0), radius: 0
Constant Reactance Circles
2 2
 1   1 
( Γ r − 1) +  Γi −  =  
2

 xL   xL 

x = 0  center:(1, ∞), radius: ∞

x = 0.5  center:(1, 2), radius: 2
x = −0.5  center:(1, -2), radius: 2
x = 1  center:(1, 1), radius: 1
x = 1  center:(1, -1), radius: 1
x = 2  center:(1, 1/2), radius: 1/2
x = −2  center:(1, -1/2), radius: 1/2
x = ∞  center:(1, 0), radius: 0
 Admittance Smith chart
The input impedance of a load z L connected to a λ / 4 line:
zin = 1/ z L

λ / 4 transformation → rotating the chart by 180

A given impedance (or admittance) point ↔ the corresponding admittance (or impedance) point.
 
across the center of the chart

A positive imaginary part → an inductive reactance or a capacitance susceptance

A negative imaginary part → an capacitive reactance or a inductive susceptance

Z-Y
Smith chart
 Example
The use of the transmission line chart is best shown by example. Let us again consider
a load impedance, Z L = 25 + j 50 Ω, terminating a 50 Ω line. The line length is
60 cm and the operating frequency is such that the wavelength on the line is 2 m. We
desire the input impedance.

Solution .
z L = 0.5 + j1 → A
Read Γ =0.62∠82
Read 0.135 on the wtg
(wavelength toward generator) scale
l / λ = 0.6 / 2 = 0.3
0.135 + 0.3 = 0.435
Read zin = 0.28 − j 0.40 → Z in = 14 − j 20

50 Ω slotted line → VSWR 2.5, minimum at 47 cm at a certain load

frequency 400 MHz → wavelength 75 cm ← λ0 = c / f
short circuit to the load → minimum at 26 cm
26 − 37.5 (half wavelength) = -11.5 cm ← load is located at -11.5 cm
The minimum is 47-(-11.5)=58.5 cm from the load or 21 cm from the load (subtracting one-half wavelength)
The voltage maximum is 21-(37.5/2)=2.25 cm from the load or 2.25/75=0.030 wavelength from the load
 Microstrip Matching Networks
Advantages:
They are easily fabricated using printed circuit techniques. Placement of devices are easily.

45

Match ZL to the 50 Ω line by placing a short-circuited stub (50 Ω) of length d1 a distance d from the load.
Determine d1 and d.

The resultant input impedance must be 1+j0

The input admittance of the length d containing the load must be 1+jbin
for the addition of the input admittance of the jbstub
to produce a total admittance of 1+j0.

zL = 2.1 + j 0.8
yL = 1 / ( 2.1 + j 0.8 ) ← adding one-quarter wavelength on the Smith chart
yL = 0.41 − j 0.16
1 + j 0.95 at wtg=0.16 ((0.5-0.47)+0.16=0.19) → ystub = − j 0.95 at wtg=0.379 (0.379-0.25=0.129 or 9.67 cm)

 1 - j 0.95 at wtg=0.34 → ystub = j 0.95 at wtg=
 Microstrip Matching Networks

z1 = 1 + j 2 zL = 1 + j 2

wtg1 =
y2 = 1 − jb2
wtg 2 =
d1 = ( 0.5 − wtg1 ) + wtg 2 = 0.397

47

wtg1′ = 0.25
y2′ = jb2
wtg 2′ =
d 2 = wtg 2′ − wtg1′ = 0.156
 EXAMPLE A 50[Ω]-lossless line is terminated in ZL = 17.5−j55[Ω]. Find the
location and length of a short-circuited stub for the impedance matching.

Solution
 Point A is the normalized load impedance
zL = (17.5−j55)/50 = 0.35−j1.1

Move to B for the normalized load admittance

yL = 0.26+j0.83 with wtg=0.113λ

Move to C and meet r =1 circle, where

x =2.2 with wtg=0.191λ

wtg from B to C  l = 0.078λ

 A short circuit has ZL = 0 and YL = ∞ (point D )
Move to E and meet b = −2.2 circle
wtg from D to E  d = 0.068λ

Introduction to Engineering
Chapter 9 49
Electromagnetics

Single-stub shunt tuning

• Design two
single-stub
shunt
tuning
networks
for ZL=20-
j15 Ω, Ζ0=
50 Ω

50
z1 = ( 20 − j15 ) / 50 = 0.4 − j 0.3
wtg1 =
y2 = 1 − j1.05
wtg 2 =
d1 = ( 0.5 − wtg1 ) + wtg 2 = 0.141

wtg1′ = 0.25
y2′ = j1.05
wtg 2′ =
d 2 = wtg 2′ − wtg1′ = 0.129
 Single-stub shunt tuning using short-circuited stub

• ZL=60-j80 Ω
• Z0= 50 Ω
• 2 GHz
• Load is
matched at 2
GHz with a
resistor and a
capacitor in
series.
• Plot |Γ| from 1
to 2 GHz.
53

The Normalized Impedance and Admittance Smith Chart

• ZY Smith chart : superimposing Z Smith chart and Y Smith

chart rotated by 180°
• Impedance and admittance values are in the same position.
54
 Impedance Matching Networks

55

z1 = (10 + j 25 ) / 50 = 0.2 + j 0.5

z2 = 0.2 − j 0.4
xC = x2 − x1 = −0.9
1
X C = xC ⋅ 50 = −45 = −
ωC
( )
C = 1/ 2π ⋅109 ⋅ 45 = 3.54 pF

y2 = ( 0.02 + j 0.04 ) ⋅ 50 = 1 + j 2
y3 = 1
bL = b3 − b2 = −2
1
BL = bL / 50 = −0.04 = −
ωL
( )
L = 1/ 2π ⋅109 ⋅ 0.04 = 3.98 nH