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Example Consider two lossless transmission lines with different real characteristic impedances Zo1 = R₁ and Z02= R₂ as shown in Fig. 2.11. We wish to place a capacitor in parallel with the second line to change L Z01= R1 and Figure 2.11: Lumped element matching example. the real part of the overall impedance, and then put an inductor in series to cancel the imaginary part of the impedance. The result should be such that the impedance seen by the first line equals the real value R₁, and therefore the reflected wave on the first line is eliminated. It is straightforward to show that this requires: R₁ = jwL+. We multiply both sides by 1+ jwCR2, and force the real and imaginary parts of the equation to hold separately. The result is: = R₁ R₂ L 1-w²LC = с R₂ 1+jwCR₂ R₁ (R₂ - R₁) W R₁ R2 We can see from the second of these that this circuit can work only if R₁ < R₂ (in the other case, we have to change the positions of the inductor and capacitor). Solving the first equation for L and substituting into the second to find C, we obtain: C = Z02=R₂ 1 R₂ - R₁ R₁ S (2.20)

Example Consider two lossless transmission lines with different real characteristic impedances Zo1 = R₁ and Z02= R₂ as shown in Fig. 2.11. We wish to place a capacitor in parallel with the second line to change L Z01= R1 and Figure 2.11: Lumped element matching example. the real part of the overall impedance, and then put an inductor in series to cancel the imaginary part of the impedance. The result should be such that the impedance seen by the first line equals the real value R₁, and therefore the reflected wave on the first line is eliminated. It is straightforward to show that this requires: R₁ = jwL+. We multiply both sides by 1+ jwCR2, and force the real and imaginary parts of the equation to hold separately. The result is: = R₁ R₂ L 1-w²LC = с R₂ 1+jwCR₂ R₁ (R₂ - R₁) W R₁ R2 We can see from the second of these that this circuit can work only if R₁ < R₂ (in the other case, we have to change the positions of the inductor and capacitor). Solving the first equation for L and substituting into the second to find C, we obtain: C = Z02=R₂ 1 R₂ - R₁ R₁ S (2.20)

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Tansmission Lines: How did my textbook get this answer for R1? I am confused by their example. Can someone explain"

Example Consider two lossless transmission lines with different real characteristic impedances Zo₁ = R₁ and Z02= R₂ as shown in Fig. 2.11. We wish to place a capacitor in parallel with the second line to change L Z01= R₁ and Figure 2.11: Lumped element matching example. the real part of the overall impedance, and then put an inductor in series to cancel the imaginary part of the impedance. The result should be such that the impedance seen by the first line equals the real value R₁, and therefore the reflected wave on the first line is eliminated. It is straightforward to show that this requires: R₁ = jwL+ We multiply both sides by 1+ jwCR2, and force the real and imaginary parts of the equation to hold separately. The result is: L L C R1 (R₂ - R₁) لها с R₂ 1+ jwCR₂ R₁ R₂ 1-w²LC We can see from the second of these that this circuit can work only if R₁ < R₂ (in the other case, we have to change the positions of the inductor and capacitor). Solving the first equation for L and substituting into the second to find C, we obtain: R₁ R₂ C = Z02=R₂ 1 WR₂ R₂ - R₁ R₁ (2.20)
Transcribed Image Text:Example Consider two lossless transmission lines with different real characteristic impedances Zo₁ = R₁ and Z02= R₂ as shown in Fig. 2.11. We wish to place a capacitor in parallel with the second line to change L Z01= R₁ and Figure 2.11: Lumped element matching example. the real part of the overall impedance, and then put an inductor in series to cancel the imaginary part of the impedance. The result should be such that the impedance seen by the first line equals the real value R₁, and therefore the reflected wave on the first line is eliminated. It is straightforward to show that this requires: R₁ = jwL+ We multiply both sides by 1+ jwCR2, and force the real and imaginary parts of the equation to hold separately. The result is: L L C R1 (R₂ - R₁) لها с R₂ 1+ jwCR₂ R₁ R₂ 1-w²LC We can see from the second of these that this circuit can work only if R₁ < R₂ (in the other case, we have to change the positions of the inductor and capacitor). Solving the first equation for L and substituting into the second to find C, we obtain: R₁ R₂ C = Z02=R₂ 1 WR₂ R₂ - R₁ R₁ (2.20)
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