elec351 lecture notes set 20trueman/elec351/elec351_2018_20.pdf• transmission lines in the time...

ELEC351 Lecture Notes Set 20

1

Exam Questions from Previous Years, with Solutions.

This file has questions selected from Final Exams from previous years.

Best is to solve the problem yourself and then compare your solution with these lecture notes.

Make-up Tutorial Tuesday December 4, 2018

The make-up tutorial will be at the same time and in the same room as the usual Monday tutorial.

Note that there will be tutorials on Monday December 3 and on Tuesday December 4.

On-line course evaluation: November 13 to December 3, 2018.The final exam in ELEC 351 is December 13, 2018 from 2:00 to 5:00.

Best Practice to Prepare for the Final Exam

• Engineering is ‘learn by doing’ so the best practice to study for a final exam is to do a lot of problems.

• Study materials:

• Solve all the examples that were solved in the lecture notes.

• Homework assignments with solutions: excellent preparation for the final exam!

• Workshop problems: re-solve the workshop problems

• Mid-term exam: Don’t assume that problems like those on the mid-term will NOT appear on the final! In fact the final exam may have problems that are very like those you found on the mid-term exam.

Topics for Exam QuestionsExam questions are drawn from the major topic areas in the course. These include (but are not necessarily limited to) the following:

• Transmission lines in the time domain: step response, pulse response, time domain reflectometry.

• Transmission lines in the frequency domain: input impedance, solve a TL circuit, standing wave maxima and minima, impedance matching quarter wave transformer and series line, single stub matching, double stub matching.

• Plane waves: lossy media, intrinsic impedance and propagation constant, penetration depth, attenuation with distance, normal incidence, oblique incidence.

• Waveguides: TE and TM modes, cutoff frequency, dominant mode, phase constant, wave impedance.

• Antennas: Hertzian dipole, half wave dipole, directional patterns, radiated power, power density, received power, Friis transmission equation.

∆= 𝐿𝐿 − 𝑧𝑧 is the distance from the observer at 𝑧𝑧 to the load at 𝐿𝐿.Make the incident and the reflected waves out of phase for a minimum:

𝑍𝑍𝐴𝐴𝐴𝐴 = 𝑅𝑅𝑐𝑐𝑍𝑍𝐿𝐿 + 𝑗𝑗𝑅𝑅𝑐𝑐 tan𝛽𝛽𝐿𝐿𝑅𝑅𝑐𝑐 + 𝑗𝑗𝑍𝑍𝐿𝐿 tan𝛽𝛽𝐿𝐿

= 300300 + 𝑗𝑗300 tan𝛽𝛽𝐿𝐿300 + 𝑗𝑗300 tan𝛽𝛽𝐿𝐿

= 300 Ω

𝑍𝑍𝐿𝐿 = 300 Ω

Solve a T.L. Circuit with a 300-ohm load:

𝑉𝑉 𝑧𝑧 = 𝑉𝑉+𝑒𝑒−𝑗𝑗𝛽𝛽𝛽𝛽 + 𝑉𝑉−𝑒𝑒𝑗𝑗𝛽𝛽𝛽𝛽 where

Procedure for solving a transmission-line circuit:1.The input impedance.2.The voltage at the input V(0).3.The travelling-wave amplitude 𝑉𝑉+.4.The reflected travelling-wave amplitude 𝑉𝑉−.5.The voltage at the load V(L).6.The power delivered to the load.

𝑉𝑉𝑆𝑆=10 vamplitude

𝑉𝑉(0)

𝑅𝑅𝑐𝑐 = 50

𝑍𝑍𝐿𝐿 = 300

𝑅𝑅𝑆𝑆 = 52 Ω

𝑉𝑉(𝐿𝐿)

𝐿𝐿=1.12 cmLj

LeVV β2−+− Γ=

Procedure for solving a transmission-line circuit: (notes set 11)1.The input impedance. Done2.The voltage at the input V(0). Next step3.The travelling-wave amplitude 𝑉𝑉+.4.The reflected travelling-wave amplitude 𝑉𝑉−.5.The voltage at the load V(L).6.The power delivered to the load.

Procedure for solving a transmission-line circuit: (notes set 11)1.The input impedance. Done2.The voltage at the input V(0). Done3.The travelling-wave amplitude 𝑉𝑉+. Next step𝑉𝑉 𝑧𝑧 = 𝑉𝑉+𝑒𝑒−𝑗𝑗𝛽𝛽𝛽𝛽 + 𝑉𝑉−𝑒𝑒𝑗𝑗𝛽𝛽𝛽𝛽 where𝑉𝑉 𝑧𝑧 = 𝑉𝑉+𝑒𝑒−𝑗𝑗𝛽𝛽𝛽𝛽 + 𝑉𝑉+Γ𝐿𝐿𝑒𝑒−𝑗𝑗𝑗𝛽𝛽𝐿𝐿𝑒𝑒𝑗𝑗𝛽𝛽𝛽𝛽𝑉𝑉 0 = 𝑉𝑉+(1 + Γ𝐿𝐿𝑒𝑒−𝑗𝑗𝑗𝛽𝛽𝐿𝐿)

LjLeVV β2−+− Γ=

𝑉𝑉 𝑧𝑧 = 𝑉𝑉+𝑒𝑒−𝑗𝑗𝛽𝛽𝛽𝛽 + 𝑉𝑉−𝑒𝑒𝑗𝑗𝛽𝛽𝛽𝛽

At 𝑧𝑧 = 𝐿𝐿:

None of these!

Impedance Matching with a Line Stretcher

Continued…

Start at the load value of 𝑦𝑦𝐿𝐿 = 0.292 + 𝑗𝑗𝑗.331 and rotate on a constant Γ to the 𝑔𝑔 = 1 circle at 𝑦𝑦1 =1+j1.45

Stub: start at the short circuit at 0.25𝜆𝜆and rotate around the outside of the Smith Chart to − j1.45 at 0.346𝜆𝜆.

Draw the “rotated g=1” circle. Find the load admittance at 0.6-j1.65. Follow the constant g=0.6 circle until it intersects the “rotated g=1 circle” at 0.6-j0.33. The stub admittance needed is 𝑦𝑦𝑠𝑠= 0.6 − 𝑗𝑗𝑗.33− 0.6 − 𝑗𝑗𝑗.65 = 𝑗𝑗𝑗.32

For the stub start at the short circuit at 0.25𝜆𝜆 and rotate around the outer circle to j1.32 at 0.146𝜆𝜆The stub length is (0.5+0.146)𝜆𝜆-0.25𝜆𝜆=0.396𝜆𝜆.Rotate 0.6-j0.33 around the chart to the g=1 circle at 1+j0.71. The stub admittance needed is -j0.71. Start at 0.25𝜆𝜆 find -0.71 at 0.402𝜆𝜆.

The cross shows the location of the load admittance 0.3+j0.4 (normalized).Follow a constant g=0.3 circle around to put the admittance of the load-plus-stub onto the rotated g=1 circle, see next slide.

The cross shows the location of the admittance of the load in parallel with the stub, on the “rotated g=1 circle” at 0.3+j0.287. We need the stub to change the load of 0.3+j0.4 mS into 0.3+j0.287. The stub admittance is ys=0.3+j0.287-(0.3+j0.4)=-j0.113

Go around the outside of the Smith Chart from the short circuit at LOAD at 0.25 wavelengths to make the stub admittance -j0.113 at INPUT at 0.4815 wavelengths. The line length is 0.4815-0.2500 = 0.2315 wavelengths or 0.736 cm.

Go around the Smith Chart on a constant Γ circle from the “rotated g=1 circle” at 0.3+j0.287 to the g=1 circle at 1+j1.37 normalized admittance. The input stub must have admittance -j1.37 (normalized).

The input stub must have admittance -j1.37 (normalized). Go around the Smith Chart from the short circuit at 0.25 wavelengths to input admittance –j1.37 at 0.3515 wavelengths. The stub length is 0.3515-0.2500=0.1015 wavelengths or 0.322 cm.

Plane Waves

5.5What is the amplitude of the electric field at the observer?

𝑓𝑓 =2.65 GHz𝜆𝜆1 = 𝑐𝑐

𝑓𝑓= 𝑗.9979𝑥𝑥108

𝑗.65𝑥𝑥109= 0.1131 m

𝛽𝛽1 = 360°𝜆𝜆1

= 3600.1131

= 3,182 degrees/meter

𝐸𝐸1 𝑧𝑧 = 𝐸𝐸𝑖𝑖𝑒𝑒−𝑗𝑗𝛽𝛽1𝛽𝛽 + 𝐸𝐸𝑟𝑟𝑒𝑒𝑗𝑗𝛽𝛽1𝛽𝛽 𝐸𝐸𝑗 = 𝐸𝐸𝑡𝑡𝑒𝑒−𝑗𝑗𝛽𝛽2𝛽𝛽

where 𝐸𝐸𝑖𝑖 = 10 V/m.

5.1 𝜂𝜂𝑗 = 𝜂𝜂0𝜖𝜖𝑟𝑟1

= 376.739

= 125.58 ohms

5.2 Γ = 𝜂𝜂2−𝜂𝜂1𝜂𝜂2+𝜂𝜂1

= 1𝑗5.58−376.731𝑗5.58+376.73

= −0.5000

5.3 𝐸𝐸1 𝑧𝑧 = 𝐸𝐸𝑖𝑖𝑒𝑒−𝑗𝑗𝛽𝛽1𝛽𝛽 + 𝐸𝐸𝑟𝑟𝑒𝑒𝑗𝑗𝛽𝛽1𝛽𝛽The incident field is 𝐸𝐸𝑖𝑖𝑒𝑒−𝑗𝑗𝛽𝛽1𝛽𝛽 and the phase of the incident field is −𝛽𝛽1𝑧𝑧 = −3,𝑗82𝑥𝑥 −0.022 = +70.0 degrees

5.4 𝐸𝐸1 𝑧𝑧 = 𝐸𝐸𝑖𝑖𝑒𝑒−𝑗𝑗𝛽𝛽1𝛽𝛽 + 𝐸𝐸𝑟𝑟𝑒𝑒𝑗𝑗𝛽𝛽1𝛽𝛽The reflected field is 𝐸𝐸𝑟𝑟𝑒𝑒𝑗𝑗𝛽𝛽1𝛽𝛽 where 𝐸𝐸𝑟𝑟 = Γ𝐸𝐸𝑖𝑖 = −0.5𝑗𝑗𝑗𝑥𝑥𝑗𝑗 = −5 = 5𝑒𝑒𝑗𝑗𝑗𝑗 = 5𝑒𝑒𝑗𝑗180And 𝛽𝛽1𝑧𝑧 = 3,𝑗82𝑥𝑥 −0.022 = −70.0 degreesSo the reflected field is 𝐸𝐸𝑟𝑟𝑒𝑒𝑗𝑗𝛽𝛽1𝛽𝛽 = 5𝑒𝑒𝑗𝑗180𝑒𝑒𝑗𝑗(−70) = 5𝑒𝑒𝑗𝑗110And the phase of the reflected field is 110 degrees

𝑇𝑇 =2𝜂𝜂𝑗

𝜂𝜂1 + 𝜂𝜂𝑗=

2𝑥𝑥𝑗25.58376.73 + 125.58

= 0.5000

𝐸𝐸𝑡𝑡 = 𝑇𝑇𝐸𝐸𝑖𝑖 = 0.5𝑗𝑗𝑗𝑥𝑥𝑗𝑗 = 5 V/m𝑆𝑆𝑎𝑎𝑎𝑎 = 𝐸𝐸𝑡𝑡2

𝑗𝜂𝜂2= 52

𝑗𝑥𝑥1𝑗5.58= 0.09954 = 99.54 mW

5.5 𝐸𝐸1 𝑧𝑧 = 𝐸𝐸𝑖𝑖𝑒𝑒−𝑗𝑗𝛽𝛽1𝛽𝛽 + 𝐸𝐸𝑟𝑟𝑒𝑒𝑗𝑗𝛽𝛽1𝛽𝛽where −𝛽𝛽1𝑧𝑧 = +70.0 degrees and 𝐸𝐸𝑟𝑟 = −5 hence𝐸𝐸1 𝑧𝑧 = 10𝑒𝑒𝑗𝑗70 − 5𝑒𝑒𝑗𝑗 −70 = 1.637 + 𝑗𝑗𝑗𝑗.13 = 14.23∠83.2°

5.5What is the amplitude of the electric field at the observer?

Plane Waves

From question 5.1:

Reuse the numbers:

𝑓𝑓𝑐𝑐𝑗0 = 𝑐𝑐𝑗

𝑗𝑎𝑎

𝑗= 𝑐𝑐

𝑗𝑗𝑎𝑎

= 𝑐𝑐𝑎𝑎

= 0.30.00𝑗54

= 118.11 GHz

𝑓𝑓𝑐𝑐30 = 𝑐𝑐𝑗

3𝑎𝑎

𝑗= 𝑐𝑐

𝑗3𝑎𝑎

= 177.2 GHzSo three modes propagate.

Antennas

(amplitude)

We have 314 mV/m amplitude but the question asks for RMS so the answer is “none of these”.

E is amplitude!

Antennas

Good Luck!