The terminating impedance ZL is 1100+ j100 Ω and the character impedance Z0 of the line and stub is 50Ω. The first stub is placed at 0.40λ away from the load. The spacing between the two stubs 3λ/8. Determine the length of the short circuited stubs when the match is achieved. -

Mumbai University > Electronics and telecommunication > Sem 7 > Microwave and Radar Engineering

Marks: 10 Marks

Year: Dec 2015

Given

$Z_L = 1100+ j100 Ω \\ Z_0 = 50Ω$

Step 1:

Step 2:

Normalized $Z_L$

$\overline{z_L}=\dfrac{Z_L}{Z_0} =\dfrac{100+j100}{50}=2+2j$

Step 3:

Draw VSWR 0 and get π

$\overline{yL}=0.25-j0.25$ (From smith chart)

Step 4:

To find $\overline{yd1}$

Move on VSWR 0 in clockwise direction from $\overline{y_L}$ by distance of 0.4λ.

$\overline{yd1}=0.55-j1.1$

Step 5:

To draw rotated unity 0.

Distance between two stub is $3λ/8.$

i.e $\dfrac{3 \times 720}{8}=270^0$

Rotate actual unity 0 by $270^0$ toward load i.e anticlockwise.

Step 6:

To find $\overline{y11}$

Move on r = 0.55 in clockwise direction till it intersect rotated unity to get $\overline{y11}$

$\overline{y11} = 0.55 – j0.1$

Step 7:

To find $\overline{y_{S1}}$

WKT, $\overline{y11} =\overline{yd1}+\overline{ys1} \\ 0.55 – j0.1 = 0.55 – j1.1 + \overline{ys1} \\ \overline{ys1}= +1.0j$

Step 8:

To find ‘l1’

Plot $\overline{ys1}$ and move in anticlockwise direction till short ckt point on smith chart.

l1=0.375λ

Step 9:

To find $\overline{yd2}$

Rotate $\overline{y11}$ by a length of $3λ/8 ( \ \ \ or \ \ \ 270^0)$ in clockwise direction.

$\overline{yd2}=1-j0.6$

Step 10:

To find $\overline{ys2}$.

WKT, $\overline{y22}= \overline{yd2}+ \overline{ys2} \\ 1 + j0 = 1-j0.6 + \overline{ys2}$

$∴ \overline{ys2}= +j0.6$

Step 11:

To find ‘l2’

Plot $\overline{ys2}$ on smith chart and move in anticlockwise till short ckt point on smith Chart.

l2 = 0.336λ

Step 12:

Final diagram