Smith Chart Introduction
Smith Chart Introduction
Smith Chart Introduction
I(0)
+
Zg V (0)
V (0) ZL =
F
- I(0)
l 0
normalized load ZL
zL = rL + jxL
impedance: Z0
load reflection V (0) ZL Z0 zL 1
L (d = 0) = + = =
coefficient: V (0) ZL + Z0 zL + 1
TL terminated by arbitrary load
I(d)
Zg +
V (d) ZL
F
-
l 0
|1
+
voltage (
d)|
minimum
(d) voltage
Vmin 1 |L |
maximum
Vmax 1 + |L |
Separate real and imaginary parts of to find special cases:
(r + jx) 1
(d) = r + ji =
(r + jx) + 1
r2 1 + x2 2x
= + j
(r + 1)2 + x2 (r + 1)2 + x2
r1
(d) = + j0
r+1 r=
(open ckt)
r Re{}
0 -1
1/3 -1/2
r=1
r=0 (matched)
1 0 (short ckt)
3 1/2
1
Separate real and imaginary parts of to find special cases:
(r + jx) 1
(d) = r + ji =
(r + jx) + 1
r2 1 + x2 2x
= + j
(r + 1)2 + x2 (r + 1)2 + x2
x2 1 2x
|(d)| = | 2 +j 2 |=1 x=1
x +1 x +1
x=
x ! Re{}
0
1 /2 x=0
x = x = 1
0
-1 /2
Separate real and imaginary parts of z to find general contours:
1 + (r + ji ) 1 2r 2i + j2i
z = r + jx = =
1 (r + ji ) (1 r )2 + 2i
contours of constant r? equate real parts of above expression
r 2 2 1 2
(r ) + i = ( )
1+r 1+r
1 + (r + ji ) 1 2r 2i + j2i
z = r + jx = =
1 (r + ji ) (1 r )2 + 2i
contours of constant x? equate imag. parts of above expression
2 1 2 1 2
(r 1) + (i ) = ( )
x x
By drawing the
SWR circle on the
chart, we can
quickly relate (d)
with the unique
line impedance
z(d) = r + jx at that
point
x values
r values
-x values
Distance ( )
towards
generator
Distance ( )
towards
load
The standing wave ratio
is read off of the chart
by noting the r value
where a constant
circle intersects the
r axis
1) SWR = Zmax/Z0
= zmax
= rmax
2) SWR = Z0/Zmin
= 1/zmin
1/SWR SWR
= 1/rmin
EXAMPLE:
If the effective reflection
coefficient on a
piece of 50 line
is =0.4+j0.2, what
is the corresponding
line impedance
26.6
at that point ? i = 0.2
. 4 47
1) Find on the | =0
| r = 0.4
Smith Chart
EXAMPLE:
If the effective reflection
coefficient on a
piece of 50 line x = 1.0
is =0.4+j0.2, what
is the corresponding
line impedance
at that point ?
1) Find on the
Smith Chart
2) Read r and x
r =2.0
off of chart
3) Use Z0 to re-
normalize
Z = 50 (2.0 + j1.0)
= 100.0 + j 50.0
16
EXAMPLE:
A load with ZL=50-j25 is
attached to a lossless,
100 T-L. Find Z(d) at
d = 0.4
1) Normalize ZL
zL = 0.5-j0.25
2) Find zL on the
Smith Chart
z = 0.5-j0.25
3) Rotate along
constant by
0.4
4) Read off new
0.05
values of z
4) Use Z0 to re- z = 0.95-j0.77
normalize
1 1 (d) 1 L ej2d
y(d) = = =
z(d) 1 + (d) 1 + L ej2d
)
= z(d ) since: ej2(d 4 =e j2d
ej2( 4)
4
= ej2d e
= ej2d
Given Z = 95+j20 on
a 50 line, find Y
1) Find z
z = 1.9+j0.4
2) Draw circle
3) Draw line through
origin
4) Find intersection
with circle
5) Read off y
y = 0.5-j0.1
6) Renormalize y
Y = y/Z0
= 10-j2 mS
Ycalc = 10.1-j2.12 mS