Question 3 This is a series single stub tuner question. We shall follow the notation and general procedure as explained in the first part of the solution for question 2. Step 1: Find a normalized load impedance/admittance 𝑌𝐿 = 𝐺𝐿 + 𝑗𝜔𝐵𝐿 = 1

250 + 𝑗 2𝜋 × 500 × 106 3.82 × 10−12 = 0.004 + 𝑗0.012 ;

𝑦𝐿 = 𝑌𝐿𝑌0

= 0.004 + 𝑗0.0121/50 = 0.20 + 𝑗0.60 .

(Plotted as point A on the Smith Chart) Step 2: Transform into impedance (Move to Point B on the Smith Chart) Step 3: Move towards generator until 𝑟 = ℜ 𝑧 = 1 (black circle) Option 1: 𝑑1 = 0.5 − 0.338 + 0.194 = 0.356[𝜆] (from Point B to C) Option 2: 𝑑2 = 0.5 − 0.338 + 0.306 = 0.468[𝜆] (from Point B to D) Step 4: Determine the admittance required from the stub Option 1:

From point C we have 𝑧𝐵 ≅ 1 + 𝑗2.4 , so 𝑥𝑠 ≅ −2.4 . However, since the stub has a characteristic impedance 𝑍0′ = 35[Ω], which is than the main transmission line, we need to renormalize as follows before we can use the Smith chart to find out the desired length of the stub. 𝑥′𝑠 = 𝑥𝑠 𝑍0

𝑍0′ = −3.43 .

This can be achieved with an open circuited stub of length 𝑙1 = 0.272𝜆 − 0.25𝜆 = 0.022𝜆 (start at E, the open circuit point for impedance, and follow the red arrow). Of course one can also use a short circuited stub which is 0.25λ longer in length.

Option 2:

From point D we have 𝑧𝐵 ≅ 1 − 𝑗2.4 , so 𝑥𝑠 ≅ 2.4 ⇒ 𝑥′𝑠 = 2.4 𝑍0𝑍0′ = 3.43. This can be

achieved with a short circuited stub of length 𝑙2 = 0.204𝜆 (start at F, the short circuit point for impedance, and follow the purple arrow). Of course one can also use an open circuited stub which is 0.25λ longer in length.

Question 4 For this shunt double stub tuner question our subscript notations will follow that of (Cheng Fig. 5-41, p. 506)

Step 1: Find a normalized load impedance/admittance 𝑧𝐿 = 𝑍𝐿

𝑍0= 100+𝑗100

300 = 0.33 + 𝑗0.33 (Point A on the Smith Chart) Step 2: Transform into admittance (Move to Point B (𝑦𝐿 = 1.5 − 𝑗1.6) on the Smith Chart) Step 3: Include the effect of stub B Since we don’t know the length of stub B, we can only say that it contributes to the load susceptance (the imaginary part of the load admittance). Thus the susceptance looking into the parallel combination of stub B and the load can be anywhere on the red circle. Step 4: Move 3λ/8 towards generator Follow the green arrow to move from the red circle to the yellow circle. Step 5: Identify points for which 𝑔 = ℜ 𝑦 = 1 (ie. Intersections with black circle) They are points C (𝑦𝐵1 = 1 + 𝑗1.6) and D (𝑦𝐵2 = 1 + 𝑗0.43). Step 6: Find points corresponding to C and D at the location of the load We do this by rotating points C and D about the origin through a distance 3λ/8 towards the load, as indicated by the purple arrows. The resulting points and their admittances are E (𝑦𝐴1 = 1.5 −𝑗1.9) and F (𝑦𝐴2 = 1.5 − 𝑗0.25). Step 7: Find the lengths for Stubs A and B We use stub B to match points C and D to the origin, very similar to how we used the stub in the case of single stub tuning: 1 = 𝑦𝐵1 + 𝑦𝑠𝐵1 ⇒ 𝑦𝑠𝐵1 = 1 − 𝑦𝐵1 = −𝑗1.6 ; 𝑦𝑠𝐵2 = 1 − 𝑦𝐵2 = −𝑗0.43. We use stub A to go from point B to points E and F. Their admittances can be found by: 𝑦𝐴1 = 𝑦𝐿 + 𝑦𝑠𝐴1 ⇒ 𝑦𝑠𝐴1 = 𝑦𝐴1 − 𝑦𝐿 = −𝑗0.3 ; 𝑦𝑠𝐴2 = 𝑦𝐴2 − 𝑦𝐿 = 𝑗1.35.

Part a) We find the stub lengths by tracing an arc from the short circuit admittance point (G) to locations with the desired susceptances, as denoted in the Smith chart. The corresponding lengths are: 𝑙𝐴1 = 0.454 − 0.25 = 0.204[𝜆] ; 𝑙𝐵1 = 0.34 − 0.25 = 0.09[𝜆] ; 𝑙𝐴2 = 0.5 − 0.25 + 0.15 = 0.4[𝜆] ; 𝑙𝐵2 = 0.436 − 0.25 = 0.186[𝜆] . Part b) If we use open-circuited stubs instead of short-circuited stubs, the lengths will differ from those in part a) by λ/4: 𝑙𝐴1 = 0.204 + 0.25 = 0.454[𝜆] ; 𝑙𝐵1 = 0.09 + 0.25 = 0.34[𝜆] ; 𝑙𝐴2 = 0.4 − 0.25 = 0.15[𝜆] ; 𝑙𝐵2 = 0.186 + 0.25 = 0.436[𝜆] .

Problem 3 solution:

Problem 4 Solution:

Solution problem 7 :

Solution problem 8: