λ/4 Transformer Analysis for Complex Load Matching

Assignment 4 λ/4 Transformer for Complex Loads University of Ottawa ELG 3106 November 22, 2023

Introduction In this assignment, A λ/4 transformer (see Figure 1) is used to match a line of lossless characteristic impedance Z 01 , with complex load Z L , to a feedline of lossless characteristic impedance, Z 01 . To do that, we need to find the transformer lossless characteristic impedance Z 02 and the load line length d with the help of Smith Chart and simulation app Module 2.7 provided in the textbook website. Theory The input impedance at the source end of a line of length l is given by equation 1 below: Z ¿ = Z 0 { Z L + j Z 0 tan ( βl ) Z 0 + j Z L tan ( βl ) } ( 1 ) In this assignment, we are considering the circuit shown in Figure 1, where Z 01 = Z in , therefore we can use equation 2 to equation the value of Z 02 that is derived by Z in = Z 0 2 /Z L so that we are allowed to find d max and d min to renormalize them to the original transmission line. Z 01 Z ( d ) = Z 02 2 ( 2 ) Figure 1. An in-series λ/4 transformer inserted at either d max or d min

Calculation and Simulation Results Part 1 Given Z 01 = 50Ω, Z L = (100 - j200) Ω, and Z in = Z 01 using Smith Chart, z 01 = (100 – j200)/50 = 2 – j4 Ω, the point is being plotted as point A in Figure 1 below: Figure 2. Smith Chart for d min From the WTG scale, since the closest voltage minimum is at 0λ (point B), d min = z 01 – 0λ, we can see that d min is around 0.218λ, and z(d 1 ) ≈ 0.096. Therefore, by equation 2, Z(d 1 ) = 4.8Ω and Z 02 = [Z in Z(d 1 )] 1/2 = [50(4.8)] 1/2 = 15.49Ω, then we renormalize the line to the original transmission line by plotting the following: z(d 1 ) = 0.096*Z 01 /Z 02 = 0.096 * 50/15.49 = 0.31 as point C, next we draw a SWR circle as shown in

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Figure 2 and plot point D to help us to renormalize with Z 01 . Therefore: z in = 3.22 *15.49/50 = 0.998 ≈ 1, renormalized with Z 01 , and Γ = 0.

Part 2 Next, we repeat the same step drawing the Smith Chart with z 01 = 2 – j4 Ω and plot it as point A, since the closest voltage minimum is at 0.25 λ (point B), so d max = 0.218λ + 0.25λ = 0.468λ, and z(d 2 ) ≈ 10.4, therefore Z(d 2 ) = 10.4*50 = 520Ω. Next, using equation 2, we found that Z 02 = [50*520] 1/2 = 161.25Ω, then we renormalize the line to the original transmission line by finding z(d 2 ) with respect to the original transmission line: z(d 2 ) = 10.4*50/161.25 = 3.225, and plot it as point C. Finally, after it is renormalized, using point D, we found that z in = 0.31 * 161.25/50 = 0.9998 ≈ 1. Figure 3. Smith Chart for finding d max

Figure 4. Simulation results of the first solution using Module 2.7 simulation Figure 5. Simulation results of the second solution using Module 2.7 simulation

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Figure 6. Summary of the simulation results using Module 2.7 simulation Calculation Results Simulation Results d min 0.218λ 0.2183λ Z(d 1 ) 4.8Ω 4.8059Ω Z 02 15.49Ω 15.5014Ω z(d 1 ) 0.31 0.31 z in (after renormalized) 1 1 d max 0.468λ 0.4683λ Z(d 2 ) 520Ω 520.1941Ω Z 02 161.25Ω 161.2752Ω z(d 2 ) 3.225 3.2255 z in (after renormalized) 1 1 Table 1. Comparison between Computation Results and Simulation Results

Discussion In this assignment, we aimed to compute Z(d max ) and Z(d min ) using the d max and d min values found. We can do that by first finding the distance between the value of z in and minimum/ maximum voltage, marking the point, drawing the SWR circle, finding the z value and thus the Z(d 1 ) or Z(d 2 ) value, then we use equation 2 to find out the value of Z 02 , and then renormalize the value of the line impedance z(d 1 ) or z(d 2 ) and z in with Z 01 to the original transmission line, in the other words, matching the transformer impedance value to the feedline, and we can confirm that by looking at the reflective coefficient at the end, if the result is 0, then it is renormalized. Looking at our results, we can say that we successfully compute the result considering the simulation results are almost the same as the results that we calculated.

Conclusion In this assignment, we investigated the λ/4 transformer and its design approach. The function of the transformer is impedance transformation, and its primary purpose is to match the impedance of the load to that of the transmission line, by doing that, we can suppress signal reflection and achieve maximum power transfer, this can be achieved by configuring the section of line to equal to one-fourth of the wavelength at the operating frequency, by doing that, according to equation 1, tan(βl) = tan(π/2) = ∞ and in order for dual impedance to match, we need to have the condition: Z in = Z 01 , and the section of line of λ/4 long impedance: Z 02 equals to sqrt(Z 01 Z L ), if Z 01 and Z L are both real, then Z 02 is real, then we can use a lossless line to perform the matching. If Z L is complex, it can be made real by adding a section of line to it in order to perform the matching.

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Assignment 4

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