W.C.Chew
ECE 350 Lecture Notes
8. Examples on Using the Smith Chart
(a) Find the voltages at A on the transmission line.
Z0 = 50 W, v = 1.5 × 108 ms–1
20 Ω
Vs =
10 sin ωt
volts
Zs
A
B
ZL (30+j25) Ω
z = –l = –1 m
25 MHz
z=0
The voltage source sets up a forward going and a backward going wave
on the transmission lines. Hence,
V (z) = V0 e;jz + v V0 ejz :
(1)
The corresponding current is
(2)
I (z) = ZV0 e;jz ; v ZV0 ejz :
0
0
In impedance at position Z is
;jz + v ejz
1 + ;(z) Z (z) = VI ((zz)) = Z0 ee;jz ;
=
Z
(3)
0
v ejz
1 ; ;(z)
where
;(z) = v e2jz v = ZZL +; ZZ0 :
(4)
l
0
We can use the Smith Chart to nd Z (;l). To use the Smith Chart, we have
to normalize all the impedances with respect to the characteristic impedance
of the line. Hence,
ZnL = ZZL = 30 +50j 25 = 0:6 + j 0:5:
(5)
0
We can locate ZnL on the Smith Chart which is the complex ; plane. ;(0)
or v can also be deduced from the Smith Chart. Since ;(z) is given by (4),
at z = ;l, we have
;(;l) = v e;2jl :
(6)
1
At f = 25MHz, and with v = 1:5 108 ms;1 , = v=f = 6m. Then
l = 2 l = 3 l. Therefore,
(7)
;(;l) = v e;j 23 l :
2
At z = ;l = ;1m ;(;1) = v e;j 3 . From the Smith Chart, we can read
Zn(;1) = 2:15 ; j 0:3 or Z(;1) = 107:5 ; j15 :
(8)
So, an equivalent circuit for the point A is:
20 Ω
A
Zs
Vs = 10 sin ωt
Z(–1) (107.5–j15) Ω
In phasor representation, VS = 10e;j 2 = ;j 10. Hence,
;j 7:9
107
:
5
;
j
15
108
:
54
e
Z
(
;
1)
VA = VS Z + Z (;1) = ;j 10 127:5 ; j 15 = 128:38e;j6:7 e;j90 10V
S
= 8:5e;j91:2 V:
(9)
Since
VA = V (;1) = V0 ej 1 + ;(;1)]
(10)
we can nd Vo from the above. Once Vo is found, we can nd VB from
VB = V (0) = Vo 1 + v ]:
(11)
j1
j0.5
j2
ρ
V
j0.2
zL
o
120
0
0.2
0.5
1
2
Γ(-l)
Z(-l)
−j0.2
−j0.5
−j2
−j1
2
(b) Find ZL from VSWR and dmin using a Smith Chart
The voltage on the transmission line is
V (z) = Vo(e;jz + v e+jz ) = Voe;jz 1 + ;(z)]:
(12)
If V (z) = jV (z)jej(z) , the real time voltage can be written as
V (z t) = <ejV (z)jej(z)ej!t ] = jV (z)j cos !t + (t)]:
(13)
Hence the amplitude of the real time voltage is proportional to jV (z)j which
is the voltage standing wave pattern.
|V(z)|
Vmax
Rnmax = 2.5
= VSWR
Γ(z) for
voltage min.
ρv
z = –dmin
Vmin
5λ/16
–dmin
0
load
z
2.5
Re Γ
toward
load
For example, we may be given that the VSWR = 2:5 on the line, Zo =
75, and dmin = 5=16, in order to nd ZL.
First, we note that jV (z)j / j1 + ;(z)j where ;(z) = v e2jz . Note that
Vmin occurs when ;(z) is purely negative. When z varies, ;(z) traces out a
constant circle on the Smith Chart, since j;(z)j = jv j is independent of z.
Since the j;(z)j circle must intersect the real ; axis at Rn = 2:5 since the
VSWR= 2:5, we can deduce that magnitude of j;(z)j = jv j. Since z = ;dmin
point corresponds to ;(z) as shown above, and the load is 5=16 from the
dmin point, we can gure out v 's location on the Smith Chart. We can read
o ZnL = 1:4 + j 1:1 on the Smith Chart. Hence ZL = (105 ; j 82:5).
3
j1
j0.5
j2
ρ
j0.2
V
Z=-dmin
0
0.2
0.5
1
R n=2.5=VSWR
2
5λ
16
−j0.2
−j0.5
−j2
−j1
4