lab2ee361group8

DaNang university of technology ***************

EE 361 LAB 2

Group 8

Member Pham Thi Trang Ong Thi Hoang Anh Duong Viet Le Ngoc Tan

*************DaNang 2012************

Question 1: For such a transmission line, what are the inductance and capacitance per meter? Answer: The inductance per meter is 𝑍

L’ = 𝑢 0 = 2 𝑝

3

50𝛺 ×3×10 8 𝑚 /𝑠

= 2.5 × 10−7 (𝐻/𝑚)

The capacitance per meter is: C’ = 𝑍

1

=

0𝑢𝑝

1 2 3

50𝛺× ×3×10 8 𝑚 /𝑠

= 10−10 (𝐹/𝑚) = 100 (𝑝𝐹/𝑚)

Question 2: For a different coaxial cable, μ = μ0 and ε = 3ε0. What is b/a if Z0 = 50 Ω? Answer: We have: 𝐿′ = 𝐶′

𝑍0 =

𝜇 𝑏 2𝜋 𝑙𝑛 𝑎 = 1 𝑙𝑛 𝑏 2𝜋𝜀 2𝜋 𝑎 𝑏 𝑙𝑛 𝑎

𝜇 𝜀

From that, we can infer the ratio of b and a by the below equation: 𝑏 𝜀 3 × 8.854 × 10−12 (𝐹/𝑚) = 𝑒𝑥𝑝 𝑍0 × 2𝜋 × = 𝑒𝑥𝑝 50𝛺 × 2𝜋 × ≈ 4.235 𝑎 𝜇 1.257 × 10−6 (𝐻/𝑚)

Question 3: If b = 3 mm in question 2.2, what is a? Answer: From Question 2, we have: 𝑎=

𝑏 3𝑚𝑚 = ≈ 0.7084𝑚𝑚 4.235 4.235

Question 4: At 200 MHz, and with 𝜇𝑝 = 2/3 c, what is the wavelength in the transmission line? Answer: The wavelength in the transmission line is: 2 × 3 × 108 𝑚/𝑠 𝜇𝑝 3 𝜆= = = 1(𝑚) 𝑓 200 × 106 𝐻𝑧 Question 5: What is the time delay associated with λ/16? Answer: ZG 0V

50

T1 Input 0V

ZL Load 0V

100

TD = {delay } Z0 = 50 V1 1Vac 0Vdc

0V

0V

0V

PARAMETERS: 0V

0

delay = 5ns

0

0

0

The time delay associated with λ/16 is: 𝜆 𝐿 𝐿 1 1 16 𝑇𝑑 = = = = = = 3.125 × 10−10 𝑠 = 0.3125𝑛𝑠 6 𝜇𝑝 𝜆𝑓 𝜆𝑓 16𝑓 16 × 200 × 10 𝐻𝑧 Using SPICE, we have the following table: No. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

delay (ns) 0.3125 0.625 0.9375 1.25 1.5625 1.875 2.1875 2.5 2.8125 3.125 3.4375 3.75 4.0625 4.375 4.6875 5

length(m) 0.0625 0.125 0.1875 0.25 0.3125 0.375 0.4375 0.5 0.5625 0.625 0.6875 0.75 0.8125 0.875 0.9375 1

Input V(mV) 628.864 526.996 399.889 333.373 399.975 527.159 628.964 666.664 628.979 527.061 399.898 333.333 399.906 527.061 628.987 666.651

Input (mA) 8.0013 10.542 12.58 13.333 12.579 10.538 7.9988 6.6668 7.9985 10.541 12.58 13.333 12.58 10.541 7.9983 6.6672

From the table, using Excel, we get:

Input (mA) 14 12 10 8 6

Input (mA)

4 2 0 0

0.5

1

1.5

Load (mV) 666.667 666.667 666.667 666.667 666.667 666.667 666.667 666.667 666.667 666.667 666.667 666.667 666.667 666.667 666.667 666.667

Load(mA) 6.6667 6.6667 6.6667 6.6667 6.6667 6.6667 6.6667 6.6667 6.6667 6.6667 6.6667 6.6667 6.6667 6.6667 6.6667 6.6667

Input V(mV) 700 600 500 400 300

Input V(mV)

200 100 0 0

0.5

1

1.5

Load(mA) 8 7 6 5 4 Load(mA)

3 2 1 0 0

0.2

0.4

0.6

0.8

1

1.2

Load (mV) 800 700 600 500 400 Load (mV)

300 200 100 0 0

0.5

1

1.5

Question 6: Use PSPICE, Excel, or Matlab to plot the magnitude of the voltage at Input as a function of length. From the Voltage Values on the plot and the relationship: 𝑉𝑆𝑊𝑅 =

𝑉𝑚𝑎𝑥 𝑉𝑚𝑖𝑛

,

determine the VSWR, and from the VSWR calculate |𝛤|. Answer:

From the simulation of Vinput, we have: Vmax = 666.173mV Vmin = 335.080mV So: VSWR =

Vmax Vmin

666.173

= 335.080 =1.99

And we also get: 1+|Γ|

VSWR = 1−|F| = 1.99

=> |Γ| = 0.331

Question 7: Use PSPICE, Excel, or Matlab to plot the magnitude of the current at Input as a function of length. From the Current Values on the plot, determine the VSWR, and from the VSWR calculate |Γ|. Do the voltage and current yield the same VSWR and |Γ|? Answer:

From the simulation of Iinput, we have: Imax =13.333mA Imin = 6.684mA So: VSWR =

Imax Imin

=

13.333 6.684

=1.99

And we also get: 1+|Γ|

VSWR = 1−|F| = 1.99

=> |Γ| = 0.331

Comment: The result of VSWR and |Γ| in Question 6 and Question 7 are the same.

Question 8: Plot the magnitude of the impedance at Input as a function of length using the data you collected with PSPICE. Plot the Real and Imaginary Parts of the Impedance using PSPICE and also plot impedance using a Smith Chart. Answer: The magnitude of the impedance at Input:

The Real part of the impedance at Input:

The Imaginary part of the impedance of the Input:

The Smith chart:

From the Smith chart, the normalized load impedance is zL =ZL/ Z0 = ZL/50 =2 + j0 => the load impedance is ZL = 100+j0 Question 9: Using the scales at the bottom of the Smith Chart, find the VSWR and |Γ|. Do they agree with your previous answers? Answer: From the below Smith Chart, VSWR= 1.99 and |Γ| = 0.333. Therefore, we totally agree with our previous answers.

Question 10: Compute Γ and VSWR directly using equations (2.6) and (2.7) below. Do these agree with your measurements from question 6, 7 & 8? Answer: We have: 𝑍 −𝑍

Γ = 𝑍𝐿 +𝑍0 = 𝐿

0

100𝛺−50𝛺 100𝛺+50𝛺

=

1 3

And we also get: 1+ 𝛤 1 + 1/3 = =2 1− 𝛤 1 − 1/3 Comment: The results of Γ and VSWR in Question 10 and Question 6, 7 and 8 are the same. 𝑉𝑆𝑊𝑅 =

Question 11: Plot the voltage magnitude at Load as a function of length. How does the voltage change with length? From this, how do you think the power delivered to the load will change with length? Answer:

-

From the simulation, we see that the load voltage is nearly unchanged with length. Since the value of the load and load voltage are unchanged, the power delivered to the load (100𝛀) is nearly unchanged with length (by the formula P = V2/R).

Question 12: If you have 1 meter of the coaxial cable described in question 4, at what frequency does it have length λ/2? At what frequency does it have length 2.5λ? (Note that we are NOT changing the physical length of the line, only it’s “electrical length” as defined above.) Answer: From Question, we have f = 200MHz. The coaxial cable has length λ/2 at 1 200𝑀𝐻𝑧 𝑓1 = 𝑓 = = 100𝑀𝐻𝑧 2 2 The coaxial cable has length 2.5λ at 𝑓1 = 2.5𝑓 = 2.5 × 200𝑀𝐻𝑧 = 500𝑀𝐻𝑧

Question 13: Plot the magnitude of the voltage at Input for the different “lengths” (remember that you are really just adjusting the frequency) properly relabeling the horizontal axis. (You can do this by hand or by using text boxes in Pspice.) Does this agree with your plot in question 6? What is the VSWR? Answer:

This plot is totally agree with our plot in Question 6. We have: 𝑍 −𝑍

Γ = 𝑍𝐿 +𝑍0 = 𝐿

0

100𝛺−50𝛺 100𝛺+50𝛺

≈ 0.333

From the simulation, we have: Vmax = 666.667mV Vmin = 333.333mV So the VSWR is: 𝑉𝑆𝑊𝑅 =

𝑉𝑚𝑎𝑥 666.667𝑚𝑉 = ≈2 𝑉𝑚𝑖𝑛 333.333𝑚𝑉

Question 14: Plot the magnitude of the voltage at Input, and compare to the previous case of 100 Ω. From the plot, what is the VSWR? On a Smith Chart, what similarity is there between the 100 Ω and 25 Ω cases? Answer: The schematic of the circuit:

ZG 0V

50

T1

ZL

Input 0V

Load 0V

25

TD = {delay } Z0 = 50 V1 1Vac 0Vdc

0V

0V

0V

PARAMETERS: 0V

0

delay = 5ns

0

0

0

The simulation of magnitude of voltage at input:

The plot of Question 14 has the same shape as Question 13 but two plots are out of phase. We have: Γ=

𝑍𝐿 −𝑍0 𝑍𝐿 +𝑍0

=

25𝛺−50𝛺 25𝛺+50𝛺

≈ −0.333

From the simulation, we have: Vmax = 666.667mV Vmin = 333.333mV So the VSWR is: 𝑉𝑚𝑎𝑥 666.667𝑚𝑉 = ≈2 𝑉𝑚𝑖𝑛 333.333𝑚𝑉 Thus, VSWR is the same for ZL = 100𝛀 and ZL = 25𝛀. Comment for the similarity of Smith chart: On the Smith Chart, the similarity between the 100 Ω and 25 Ω cases is that their impedance, VSWR and Γ don’t have imaginary part. 𝑉𝑆𝑊𝑅 =

Question 15: Plot the magnitude of the voltage at Input. From the plot, find the VSWR. From equations (2.6) and (2.7) calculate the VSWR. Do these two results agree? Answer: The schematic of the circuit: ZG 0V

50

T1

ZL

Input 0V

Load 0V

0.001

TD = {delay } Z0 = 50 V1 1Vac 0Vdc

0V

0V

0V

PARAMETERS: delay = 5ns

0V

0

0

0

0

The simulation of the magnitude of voltage at input:

-

From the simulation, we have: Vmax = 1V Vmin = 0V So the VSWR is: 𝑉𝑆𝑊𝑅 =

-

𝑉𝑚𝑎𝑥 1𝑉 = =∞ 𝑉𝑚𝑖𝑛 0𝑉

By calculation: From Eq. (2.6) and (2.7), we get: 𝑍 −𝑍

Γ = 𝑍𝐿 +𝑍0 = 𝐿

And we also get:

0

0.01𝛺−50𝛺 0.01𝛺+50𝛺

≈ −1

1+ 𝛤 1+1 = =∞ 1− 𝛤 1−1 Therefore, both results from simulation and calculation are the same. 𝑉𝑆𝑊𝑅 =

Question 16: Plot the magnitude of the voltage at Input. Find the VSWR. Also, calculate the VSWR. Do these two results agree? Answer: The schematic of the circuit: ZG 0V

50

T1 Input 0V

ZL Load 0V

1M

TD = {delay } Z0 = 50 V1 1Vac 0Vdc

0V

0V

PARAMETERS: delay = 5ns

0V

0

0

0

The simulation of the magnitude of the voltage at input:

-

From the simulation, we have: Vmax = 1V Vmin = 0V So the VSWR is: 𝑉𝑆𝑊𝑅 =

-

0V

By calculation: From Eq. (2.6) and (2.7), we get:

𝑉𝑚𝑎𝑥 1𝑉 = =∞ 𝑉𝑚𝑖𝑛 0𝑉

0

𝑍 −𝑍

Γ = 𝑍𝐿 +𝑍0 = 𝐿

0

10 6 𝛺−50𝛺 10 6 𝛺+50𝛺

≈ 1

And we also get: 1+ 𝛤 1+1 = =∞ 1− 𝛤 1−1 Therefore, both results from simulation and calculation are the same. 𝑉𝑆𝑊𝑅 =

Question 17: How are the plots from Question 15 and Question 16 similar? How are these two impedances related on the Smith Chart? Answer: - From the figure of question 15&16, the similar is that they have the same amplitute but different in phase and the phase difference is λ\4 - Two impedances are on unit circle but one in the left side(Γ =-1) and the other is in the right side(Γ =1)

Question 18: Using the equations above and below (and possibly equations from the book), calculate V+, V-, Vin and VL. How do the calculated Vin and VL compare to your Pspice results? Answer: We have: 𝐿 1.65𝑚 𝑇𝑑= = = 8.25 × 10−9 𝑠 = 8.25𝑛𝑠 2 𝑢𝑝 8 3 × 3 × 10 𝑚/𝑠 ZG 0V

T1

ZL

Input 0V

50

Load 0V

100

TD = {delay } Z0 = 50 V1 1Vac 0Vdc

0V

0V

0V

PARAMETERS: delay = 8.25ns

0V

0

0

0

0

a/ By calculating: Firstly, we have: Γ=

𝑍𝐿 −𝑍0 𝑍𝐿 +𝑍0

=

100𝛺−50𝛺 100𝛺+50𝛺

=

1 3

From Γ, we can calculate Z in: 𝑍𝑖𝑛 = 𝑍0

in which:

𝛽𝑙 =

2𝜋 𝜆

𝑙=

2𝜋𝑓 𝜇𝑝

2𝜋×200𝑀𝐻𝑧

𝑙=2 3

×3×10 8 𝑚/𝑠

1 + 𝛤𝐿 exp⁡ (2𝑗𝛽𝑙) 1 − 𝛤𝐿 exp⁡ (2𝑗𝛽𝑙) × 1.65𝑚 = 3.3𝜋

So: 𝑍𝑖𝑛 ≈ 0.6749 + 0.4814𝑗 Next, we find V in using equation:

𝑍𝑖𝑛 ≈ 0.013 + 0.0094𝑗 𝑍𝑖𝑛 + 𝑍𝑔 From calculated results above, we can calculate V +: 𝑍𝑖𝑛 1 𝑉 + = 𝑉𝑔 ≈ −0.006 − 0.016𝑗 𝑍𝑖𝑛 + 𝑍𝑔 exp −𝑗𝛽𝑙 + 𝛤𝐿 exp⁡ (𝑗𝛽𝑙) Thus, we have the value of Vload: VL = 𝑉0+ 1 − Γ ≈ −0.006 − 0.00214𝑗 b/ By simulation: The simulation of Vin: 𝑉𝑖𝑛 = 𝑉𝑔

The simulation of VL:

Comment: The calculated Vin and VL are not the same with our Pspice results since some errors occurs in calculating and rounding

Question 19: Calculate Pav (average power delivered to the load), Piav and Prav (average power in the incident and reflected waves, respectively). (Equations for Pav, Piav and Prav are listed in Chapter 2 of the text.) Show that conservation of power been satisfied. Answer: We have: 𝑉0+ 2 𝑖 𝑃𝑎𝑣 = 2𝑍0 𝑖 From the result in Question 18, we get: 𝑃𝑎𝑣 ≈ 2.956 × 10−6 𝑊 We also have: 𝑟 𝑖 𝑃𝑎𝑣 = −|Γ|2 𝑃𝑎𝑣 ≈ −3.284 × 10−7 𝑊 And 𝑖 𝑟 𝑃𝑎𝑣 = 𝑃𝑎𝑣 + 𝑃𝑎𝑣 ≈ −2.628 × 10−6 𝑊

Question 20: Use SPICE to find the magnitude of V+ and V- at Input. Use these values to compute Piav and Prav. Do these match your answers in Questions 18 and 19? Answer: The simulation of V+:

The simulation of V-:

Comment: From the simulation from SPICE, we see clearly that the calculated value of question 18 and 19 are not matched with what we receive from simulation since some errors occurs in calculating and rounding.