Determining Input Impedance of given lumped circuit elements with TL segments. A lossless TL, with a characteristic load impedance of ZL=100+j75Ω is shown below. Calculate the input impedance if the L-C circuit shown is inserted at a point 0.12λ away from the load end. Use Smith Chart and show the impedances or admittances at Load, A, B, C points with clear explanations for each, leading to a calculation of the input impedance Zin.

From the beginning

Determining Input Impedance of given lumped circuit elements connected arbitrarily. On a Smith Chart, obtain and show all the impedance values at points a, b, c, d and Zin of the circuit below at the frequency operation of f=2GHz. d c b a

a b za=31.25/50=0.625 ya=1/0.625=1.6 BcL=ωcL=24 mS bcL=50BcL =1.2 yb=1.6+j1.2 za=0.625 ya=1.6 zb=0.4-j0.3 BcL=ωcL=24 mS bcL=50BcL =1.2 yb=1.6+j1.2 Zb=0.4-j0.3 Normalized by 50

d c b a xL1=ωL1/50=1.1 zc=0.4+j0.8 yc=0.5-j1.0 bc=50ωc=1.5 yd=0.5+j0.5 bc=50ωc=1.5 yd=0.5+j0.5 zd=1-j1 yin=zin=1.0 zd=1.0-j1.0 xL2=ωL2/50=1 zin=yin=1 yc=0.5-j1.0 Normalized by 50

Find the input impedance in terms of magnitude and phase of the following network at an operating frequency of 950 MHz. Z0=50

0.1 λ zR=75/50=1.5 yR=1/1.5=0.667 z=0.4402+j1.6292 xL=ωL/50=23.88/50 xL =0.4775 bL=-2.094 yL=0.667-j2.094 zL=0.1381+j0.4336 zL=0.1381+j0.4336 z= 0.5013+j0.1580 zR=1.5 yR=0.667 Move from toward generator 0.1λ y= 1.8146 - j0.5720 z=0.4402+j1.6292 y= 0.1546 - j0.5720 yL=0.667-j2.094 zR=30/50=0.6 yR=1/0.6=1.66 y= 1.8146-j0.5720 z= 0.5013+j0.1580 y= 0.1546 - j0.5720

Continued from Go 0.25λ toward generator to find: z=1.8146-j0.57191 0.25 λ Go 0.25λ toward generator to find: z=1.8146-j0.57191 y=0.5013+j0.1580 Note: a 0.25λ TL is the same as changing from impedance to admittance or vice versa A 0.1 λ shorted TL has an input impedance of: z= j0.72654 y= -j1.3764 y=0.5013 - j1.2184 z=0.2888 + j0.7019 y= 0.5013+j0.1580 z= 0.5013+j0.1580 z= 1.8146-j0.57191 As in case: z= 0.5013 - j1.2184 zin = 0.5013 - j2.8937 y= 0.5013 - j1.2184 z= 0.5013 - j1.2184 xc=-1/50ωc=-1.6753 zin=0.5013 - j2.8937 50Zin=25.065-j144.69

Single Stub matching network design with fixed characteristic impedances For a load impedance of ZL=(60-j45)Ω, design two single-stub matching networks that transform the load to a Zin=(75+j90)Ω input impedance. Assume that both stub and transmission line shown below have a chracteristic impedance of Z0=75Ω.

Try to move the load admittance to SWR circle of input impdance by adding the admittance of the stub. The two possibilities for the susceptance value of the stub are:

The two other possibilities are made by using short circuit stubs With open circuit Reactances are: The two lengths are: lSA,o=0.317-0.25 =0.067λ lSB,o=0.25+0.087 =0.337λ The two other possibilities are made by using short circuit stubs

Design of a double-stub matching network It is assumed that in the double-stub matching network seen below, the lengths of the TLs are l1=λ/8 and l3=l2=3λ/8. Find the lengths of the short circuited stubs that match the load impedance ZL=(50+j50)Ω to a 50Ω input impedance. The chracteristic line impedance for all components is Z0=50Ω

MAKE SURE THAT YD IS NOT INSIDE THE FORBIDDEN REGION The difference between YD and YC is: Forbidden Region Therefore the length of first stub is lS1 = 0.074λ. The difference between YB and YA is: Therefore the length of second stub is lS2 = 0.051λ.