Guidelines: Please type all the solutions! For the Smith Chart use Tools/ Typewriter to inset symbols or text. Email and UPLOAD your project 1 on your ePortfolio. In addition to this poject you also need to upload Project 2 and the reflections on the seminars/workshops/field trips, and email them to me.

Team 2: Yevgeniy Morozov, Rupan Hossain, Esnan Ambeau

1. A source with 50 Ω source impedance drives a 50 Ω T-Line that is 1/8 of a wavelength long, terminated in a load ZL=50-j25 Ω. Calculate and use the Smith Chart to find:a)ΓL; b)VSWR; c)Zin seen by the source.

Answer:

a) Γ L=ZL−ZoZL+Zo

=50− j 25−5050− j 25+50 = 0.242˂-76°

b) VSWR=1+ΓL1−ΓL

=1+.2421− .242= 1.64

c) Since Βl=( 2πλ )( λ8 )=π4

Zin=Zo (ZL+ jZotan (βl ) )Zo+ jZLtan (βl )

=50 [50− j 25+ j50 tan( π4 )][50+ j (50− j25 ) tan( π4 )]

=¿Zin=30.8-j3.9 Ω

Now with Smith Charta) Step 1: We first locate the normalized load impedance zL

zL= 50/50 –j25/50 =1-j.5

Step 2: We drew a line from the center through ZL toward the angle of reflection coefficient. The angle is -74°.

OP1 is the distance from the center to the normalized zL OP2 is the distance from the Center toward the angle coefficient of reflection.

ΓL=OP2/OP1=2.45, angle =-74°b) For VSWR we drew the gamma circle and take the real value

where it intersects the real axis, therefore VSWR=1.66

c) For the Zin, ZL is at .353λ , so .353λ + .125λ = . 478λ toward the

generator from the load; at dis distance the Zin normalized is:

Zin= .62 –j.1 so the actual input impedance is Zin =(.62 –j.1) 50Ω

Zin= 31 – j5 Ω

2. A 1m long T-line has the following distributed parameters: R’=0.10 Ω/m, L’=1.0 µH/m, G’=10.0 µS/m, and C=10nF/m. If the line is terminated in a 25 Ω resistor in series with a 1.0nH inductor, calculate at 200MHz, ΓL and Zin.

Answer:

R = 0.1 Ω / m * 1 m = 0.1 Ω

L = 1.0 μH / m * 1 m = 1.0 μH

G = 10.0 μS / m * 1 m = 10.0 μS

C = 1.0 nF / m * 1 m = 1.0 nF

ZL = 25 + jwC = 25 + j * 2 * 3.1428 * 200 * 10 6 * (10-9) = 25 + j1.257 Ω

z0=√ R´ jωL´G´ jωC ´=√ [ ( .1+ j 6.28 )∗(200∗106 )∗(1∗10−6 ) ][ (10∗106+ j6.28 )∗(200∗106 )∗(10∗10−9 ) ]

=√ .1+ j 125610−5+ j12.56

=10Ω

Γ L=ZL−ZoZL+Zo

=25+ j1.257−1025+ j1.25+10

= 15. j 1.25735+ j 1.257

=0.73<2.74 °

λ= cf=3∗10

8

2∗108=1.5m

lλ= 11.5m

=0.667∨l=0.667 λ

β e=( 2πλ )∗0.667 λ=1.334 πZ¿=(10)∗¿

Z¿=(42.40+ j1.257 )7.81+ j 43.51

3. The input impedance for a 30cm length of lossless 100 Ω impedance T-line operating at 2.0 GHz is Zin=92.3-j67.5 Ω. The propagation velocity is 0.7c. Determine the load impedance.

Answer:

Firat find V, V=.7c=.7*3*(10^-8)=.21 *(10^-8) m/s

Λ λ=v/f= .21*10^8/2*10^9 =10.5 cmLength =2.86 λ =.36 λNow Zin normalized is zin=.923 –j.675Draw the gamma circle, the initial coordinate correspond to zin is .143 λ. Use wavelength toward the load.(.143+.36) λ =.503 λ=.03 λFrom that distance, we drew a line toward the gamma circle and the normalized load impedance zL =.5 –j.15

The actual load impedance is ZL = (.5 –j.15) *100=50 –j15 Ω

4. A matching network, using a reactive element in series with a length d of T-Line, is to be used to match a 35 – j50 load to a 100 T-Line. Find the through line length d and the value of the reactive element if (a) a series capacitor is used, and (b) a series inductor is used.

Answer: Locate ZL

ZL= 35/100 – j50/100 = .35 –j.5We drew the gamma circle and use the intersection of the gamma circle and the 1 circle to finde the imaginary point that intersection. The jx=1.4The actual reactance is XC =1.4* 100=240 ΩC=1/2WXca)r= jx=1= j1.4withcapacitor added1+ j1.4− j .14

X c=1.4∗100=140Ω

C= −JωC X0

= 12 π∗60∗140

=18.96 µF

d= .500λ+.173λ−.419 λ−.254 λ=.254 λb) r=± jx=1− j 1.4with inductor added 1− j1.4+ j .14

X l=1.4∗100=140Ω

L=X l

2π f= 16.28∗60

=.371H

d= .500λ+.327 λ− .419 λ=.408 λ

5. You would like to match a 170 load to a 50 T-Line. (a) Determine the characteristic impedance required for a quarter-wave transformer. (b)

What through-line length and stub length are required for a shorted shunt stub matching network?

Answer: Zq=√Z0Z L=170∗50=92.2Ω

Normalaized theload Z l=17050

=3.4

We drew the gammma circle then we locate gamma ϒl. We move clockwise to an interaction with Y=1+j5, the distance is d=.170𝞴 and the length l=.354𝞴-.250𝞴=0.104𝞴

6. Consider the circuit in the figure with the following values: Vs = 10 V, Zs = 30 , Zo = 50 , up = 0.666c, ZL = 150 , l = 10 cm for a 10 V pulse of duration 0.4 ns. Plot, out to 2 ns, (a) the voltage at the source end, (b) the voltage at the middle, and (c) the voltage at the load end of the T-Line.

Γ L=ZL−ZoZL+Zo

=150−50150+50

=1/2

Γ s=Zs−ZoZs+Zo

=30−5030+50

=−1/4

Vo+= Vs ZoZs+Zo= 10

5030+50=6.25 v

T=l/up=.5 ns

At t=.25 ns, Vmiddle= (Vo+) Γ l=6.25*1/2=3.125v