Viewing blog entries in category: microwave engineering pozar 05 exercise 09

• How to solve Microwave Engineering Pozar chapter 05 exercise 09 with MATLAB

Question

Literature source [POZAR] available here:
https://www.amazon.co.uk/Microwave-Engineering-David-M-Pozar/dp/0470631554

There's also a solutions manual available here:
https://www.scribd.com/doc/176505749/Microwave-engineering-pozar-4th-Ed-solutions-manual

MATLAB script source pozar_05_exercise_09.m and all necessary support functions included in .zip here attached file pozar_05_exercise_09.zip
In the attached zip there is a detailed explanation in Adobe format detailing all lines of the solution.

An analytic solution with |s11| surface is show and the frequency response to valide the calculated lengths is obtained with different cross-sections of such surface.

The sought Open Ciruit shunt stub lengths are the shortest D1 D2 that null |s11|

The obtention of the |s11| surface nulls, the exact location where D1 D2 cancel reflections is found reversing the surface and catching peaks.
This is because when attemting too small D1 D2 steps, with conventional computer platforms MATLAB may reach default limits or borrow excessive
time.

The Laplacian obtained with del2 is here conveniently used to reduce the amount of peaks returned by command findpeaks.

1st solution, stubs with lengths below 0.5:

D11=0.086

D12=0.375

2nd solution
D21=0.199

D22=0.375

Some of the support functions used in previous questions have been ugraded to solve this exercise with the Smith chart.

1st stub, next to load, 1st length. The red is ZL and the green marker is YL=1/ZL.
the 1st stub has to bring YL to the magenta circle, to the intersection blue marker to the left hand side of YL.

1st stub, 2nd length, reaching the 2nd intersection now with opposite sign reactance.

2nd stub, 1st length, 3rd and 4th intersections again are the blue markers.

2nd stub, 2nd length

The resulting stub lengths match those obtained analytically.

Verifying results with frequency response:

It turns out that for a given pair D1 D2 the frequency resonse is already a particular cross section of surface |s11|.
Choosing 2 cross sections including those pairs D1 D2 reaching |s11| nulls,

And the extracted contours are

and