The impedance matching network shown in the figure is to match a loss...
Given
Lossless line characteristic impedance Z0 = 50Ω
Open circuit stub characteristic impedance ( Z1)oc = 75Ω
Short circuit stub characteristic impedance ( Z1)sc = 75Ω
Method 1 : The given arrangement of transmission line is shown below,
Input Impedance of Λ\2 long line-1,
Thus total impedance at terminal QR is,
ZQR = ZL || (Zin)Line-1= ZL || ∞= ZL
Now arrangement of Transmission Line becomes as,
Thus Input Impedance of line 2 is,
Input Impedance of Line 3 is,
Thus total impedance at terminal PS is,
ZPS = (Zin)Line-2 || (Zin)Line-3
Now Transmission Line arrangement becomes as-
Hence ZPS work as load for main Transmission line.
For matching of main Transmission line with load ( ZPS) ,
Hence, the correct option is (A)
Method 2 :
From given arrangement, it is clear that,
So, input impedance of both line-1 and line-3 are ∞ (i.e. open circuit) so it does not make any effect on main transmission line, so given transmission line configuration becomes as,
Thus Input Impedance at terminal PS is,
So ( Zin)PS work as load for main transmission line so arrangement of transmission line becomes as,
Hence ( Zin)PS work as load for main Transmission line so the condition, for matching the main Transmission line with load ( Zin)PS is
Hence, the correct option is (A)