Smith chart
Smith chart is a graphical aid for solving transmission line
problems very easily which would be tedious to solve by using analytical
methods.
1.
Smith chart consist of resistance and
reactance circles.
r=Rl\Zo
x=Xl\Zo
where Zl=Rl+jXl
hence in smith chart. This process is called “normalization”.
and the ratio (Zl\Zo) is called “per unit
load impedence” or “normalized impedence”.
2.
Plotting
of impedence: Any complex impedence can be shown by a single point on the
Smith chart . this point is
intersection of “r=Rl\Zo” and “x=jXl\Zo”.
EXAMPLE : Let Zl = 120+j160 Ω and Z0=400 Ω
Normalising we
get,
r=0.3
x=0.4
zl=0.3+j0.4Ω shown on Smith chart.
BLACK MAGIC DESIGN
3. VSWR Determination:
With “O” as centre and OA as radius a circle is drawn which cuts right side of the horizontal axis
at “r” value of 3.9 (shown as point B)
This gives value of VSWR = ρ = 3.9
4. Determination of Γ in magnitude and
direction:
The line oa in figure is produced to meet the
outer circle at point C
The angle at point C= Ɵ=133.2° is the phase
angle of reflection coefficient.
To find magnitude of Γ , the linear scale at bottom of chart
is refered which is marked
“voltage reflection coefficient”. With O as centre and
radius exactly wqual to “OA”,
An arc is cut on left side.
The value of | Γ
| is read as 0.6
Therefore Γ=0.6∟133.2°
5. Location of
Vmax and Vmin :
The constant ρ circle intersects the central horizontal axis at
Points D and B.
Vmax = Rmax\Zo =rmax = 3.91
Vmin = Rmin\Zo = rmin = 0.26
6. Open and short circuit line:
At point F on the horizontal axis r=∞ ,
x=∞
this represents open circuit termination of line.
At point E in the horizontal axis r=0 ,
x=0 this represents short circuit
termination of line.
7. Movement along periphery :
Movement in clockwise direction from point
E results in no of wavelengths towards generator
Movement in anti-clockwise direction from
point E results in no of wavelengths towards load.
One full rotation on pheriphery corresponds to λ\2.
8. Conversion of impedence to admittance:
To find admittance of an impedence at point
a, the point a is rotated through constant S-circle
By an amount λ\4
which is equivalent to 180°.
The point yl is diametrically opposite to “A” or (zl) has a
value,
yl = 1.2-j1.6.
yl = 1.2-j1.6.
