Download Project 1 - City Tech OpenLab

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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:
𝐙𝐋−𝐙𝐨
𝟓𝟎−𝐣𝟐𝟓−𝟓𝟎
a) 𝚪𝐋 = 𝐙𝐋+𝐙𝐨 = 𝟓𝟎−𝐣𝟐𝟓+𝟓𝟎= 0.242˂-76°
b) 𝑽𝑺𝑾𝑹 =
𝟏+ 𝜞𝑳
𝟏− 𝜞𝑳
𝟐𝝅
c) Since 𝜝𝒍 = (
𝐙𝐢𝐧 =
𝝀
=
𝟏+.𝟐𝟒𝟐
𝟏−.𝟐𝟒𝟐
𝝀
= 1.64
𝝅
) ( 𝟖) = 𝟒
𝐙𝐨(𝐙𝐋+𝐣𝐙𝐨𝐭𝐚𝐧(𝛃𝐥))
𝐙𝐨+𝐣𝐙𝐋𝐭𝐚𝐧(𝛃𝐥)
=
𝛑
𝟒
𝛑
[𝟓𝟎+𝐣(𝟓𝟎−𝐣𝟐𝟓) 𝐭𝐚𝐧(𝟒)]
𝟓𝟎[𝟓𝟎−𝐣𝟐𝟓+𝐣𝟓𝟎 𝐭𝐚𝐧( )]
=Zin=30.8-j3.9 Ω
Now with Smith Chart
a) 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
[(. 𝟏 + 𝒋𝟔. 𝟐𝟖) ∗ (𝟐𝟎𝟎 ∗ 𝟏𝟎𝟔 ) ∗ (𝟏 ∗ 𝟏𝟎−𝟔 )]
𝑹´𝒋𝝎𝑳´
𝒛𝟎 = √
=√
[(𝟏𝟎 ∗ 𝟏𝟎𝟔 + 𝒋𝟔. 𝟐𝟖) ∗ (𝟐𝟎𝟎 ∗ 𝟏𝟎𝟔 ) ∗ (𝟏𝟎 ∗ 𝟏𝟎−𝟗 )]
𝑮´𝒋𝝎𝑪´
. 𝟏 + 𝒋𝟏𝟐𝟓𝟔
= √ −𝟓
= 𝟏𝟎𝜴
𝟏𝟎 + 𝒋𝟏𝟐. 𝟓𝟔
𝚪𝐋 =
𝐙𝐋 − 𝐙𝐨 𝟐𝟓 + 𝐣𝟏. 𝟐𝟓𝟕 − 𝟏𝟎
𝟏𝟓. 𝒋𝟏. 𝟐𝟓𝟕
=
=
= 𝟎. 𝟕𝟑 < 𝟐. 𝟕𝟒°
𝐙𝐋 + 𝐙𝐨
𝟐𝟓 + 𝐣𝟏. 𝟐𝟓 + 𝟏𝟎
𝟑𝟓 + 𝒋𝟏. 𝟐𝟓𝟕
𝝀=
𝒁𝒊𝒏
𝒄 𝟑 ∗ 𝟏𝟎𝟖
=
= 𝟏. 𝟓𝒎
𝒇 𝟐 ∗ 𝟏𝟎𝟖
𝒍
𝟏
=
= 𝟎. 𝟔𝟔𝟕 𝒐𝒓 𝒍 = 𝟎. 𝟔𝟔𝟕𝝀
𝝀 𝟏. 𝟓𝒎
𝟐𝞹
𝞫𝒆 = ( ) ∗ 𝟎. 𝟔𝟔𝟕𝞴 = 𝟏. 𝟑𝟑𝟒𝞹
𝞴
= (𝟏𝟎) ∗ [(𝒁𝒍 + 𝒁𝟎 𝐭𝐚𝐧 𝒚𝒍)/(𝒁𝟎 + 𝒁𝒍 𝐭𝐚𝐧 𝒚𝒍)]
= (𝟏𝟎) ∗ [(𝟐𝟓 + 𝒋𝟏. 𝟐𝟓𝟕
+ 𝟏𝟎𝐭𝐚𝐧(𝟐𝟒𝟎. 𝟏𝟐)/(𝟏𝟎 + (𝒋𝟏. 𝟐𝟓𝟕 + 𝟐𝟓) 𝐭𝐚𝐧 𝟐𝟒𝟎. 𝟏𝟐)]
(𝟒𝟐. 𝟒𝟎 + 𝒋𝟏. 𝟐𝟓𝟕)
𝒁𝒊𝒏 =
𝟕. 𝟖𝟏 + 𝒋𝟒𝟑. 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 cm
Length =2.86 λ =.36 λ
Now Zin normalized is zin=.923 –j.675
Draw the gamma circle, the initial coordinate
correspond to zin is .143 λ. Use wavelength
toward the load.
(.143+.36) λ =.503 λ=.03 λ
From 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 TLine, 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.5
We 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=1.4
The actual reactance is XC =1.4* 100=240 Ω
C=1/2WXc
a)
𝒓 = 𝒋𝒙 = 𝟏 = 𝒋𝟏. 𝟒 𝒘𝒊𝒕𝒉 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒐𝒓 𝒂𝒅𝒅𝒆𝒅 𝟏 + 𝒋𝟏. 𝟒 − 𝒋. 𝟏𝟒
𝑿𝒄 = 𝟏. 𝟒 ∗ 𝟏𝟎𝟎 = 𝟏𝟒𝟎𝜴
−𝑱
𝟏
=
= 𝟏𝟖. 𝟗𝟔µ𝑭
𝝎𝑪𝑿𝟎 𝟐𝞹 ∗ 𝟔𝟎 ∗ 𝟏𝟒𝟎
𝒅 =. 𝟓𝟎𝟎𝞴+. 𝟏𝟕𝟑𝞴−. 𝟒𝟏𝟗𝞴−. 𝟐𝟓𝟒𝞴 =. 𝟐𝟓𝟒𝞴
𝑪=
b)
𝒓 = ±𝒋𝒙 = 𝟏 − 𝒋𝟏. 𝟒 𝒘𝒊𝒕𝒉 𝒊𝒏𝒅𝒖𝒄𝒕𝒐𝒓 𝒂𝒅𝒅𝒆𝒅 𝟏 − 𝒋𝟏. 𝟒 + 𝒋. 𝟏𝟒
𝑿𝒍 = 𝟏. 𝟒 ∗ 𝟏𝟎𝟎 = 𝟏𝟒𝟎𝜴
𝑿𝒍
𝟏
𝑳=
=
=. 𝟑𝟕𝟏𝑯
𝟐𝞹𝒇 𝟔. 𝟐𝟖 ∗ 𝟔𝟎
𝒅 =. 𝟓𝟎𝟎𝞴+. 𝟑𝟐𝟕𝞴−. 𝟒𝟏𝟗𝞴 =. 𝟒𝟎𝟖𝞴
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 = √𝒁𝟎 𝒁𝑳 = 𝟏𝟕𝟎 ∗ 𝟓𝟎 = 𝟗𝟐. 𝟐𝜴
𝟏𝟕𝟎
𝑵𝒐𝒓𝒎𝒂𝒍𝒂𝒊𝒛𝒆𝒅 𝒕𝒉𝒆 𝒍𝒐𝒂𝒅 𝒁𝒍 =
= 𝟑. 𝟒
𝟓𝟎
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.
𝚪𝐋 =
𝐙𝐋 − 𝐙𝐨 𝟏𝟓𝟎 − 𝟓𝟎
=
= 𝟏/𝟐
𝐙𝐋 + 𝐙𝐨 𝟏𝟓𝟎 + 𝟓𝟎
𝐙𝐬−𝐙𝐨
𝟑𝟎−𝟓𝟎
𝚪𝐬 = 𝐙𝐬+𝐙𝐨 = 𝟑𝟎+𝟓𝟎 = −𝟏/𝟒
𝐙𝐨
𝟓𝟎
Vo+= Vs𝐙𝐬+𝐙𝐨= 𝟏𝟎 𝟑𝟎+𝟓𝟎=6.25 v
T=l/up=.5 ns
At t=.25 ns, Vmiddle= (Vo+) 𝚪𝐥=6.25*1/2=3.125v