1. The problem statement, all variables and given/known data
Design a quarter wave transformer and single stub matcher that will match the design frequency.
For the quarter wave design and single stub matcher design, generate a plot of SWR on main feeding line vs. the normalized frequency \(\frac{f}{f_0}\) where \(f_0 \) is design frequency.
Calculate the percent bandwidths of each system \(\frac{f_2 - f_1}{f_0}\)
where \(f_1\) and \(f_2\) are the lower and upper frequencies for which SWR = 2.0.
2. Relevant equations
The equations can be seen on the 481Lecture8 for single stub matcher and quarter wave transformer i assume all i need is the\( Z_{ot} = \sqrt{R_L*Z_o}\) . The picture (Untitled) shows the equations of quarter wave.
3. The attempt at a solution
I am having a little trouble with this project. Using the two pictures seen in on page 3 (MyProject.pdf), we are suppose to build a quarter wave transformer and single stub matching. Since there is no reactive element, is the only thing needed for the quarter wave transformer is the value for \(Z_{ot} \)?
There is no length needed because there is no reactive element on the load right? Because the reflection coefficient angle is 0.
So \(Z_{ot} \) just equals \(\sqrt{75*37.5}\)because \(Z_L\) is just \( \frac{1}{75} + \frac{1}{75}\)
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For the single stub matcher first using the formula on the lecture notes (equation 5.9) and i get a value of t = +/- .707107
After that i use equation 5.10 I would get
\(\frac{d}{\lambda} = (\frac{1}{2\pi})tan^{-1}(.707107) and
\frac{d}{\lambda} = (\frac{1}{2\pi})(\pi + tan^{-1}(-.707107) \)
So \( d = .96679\lambda \) and\( d = 3.96801\lambda \)
(Equation 5.7)
After \( Z(-d) = 75*\frac{37.5 + j75*.707107}{75+j37.5*.707107} = 72.42639 + 32.78680j
Y = \frac{1}{Z(-d)} = .01073 - .00575j
Y* = .01073 + .00575j \)(for the short circuit stub)
(Also I remember from Smith charts that you can add as many multiples or .5λ but i don't know if that applies here)
\(l_{sc}= (\frac{1}{(2\pi})\tan^{-1}(\frac{.01073}{-.00575}) = -1.69467\lambda + 2\lambda = .30533\lambda
l_{sc}= (\frac{1}{(2\pi})\tan^{-1}(\frac{.01073}{.00575}) = 1.69467\lambda -1.5\lambda = .19467\lambda \)
I also don't understand the concept of the plot of SWR vs normalized frequency either. When you find your single-stub matcher and quarter wave transformer, i thought you didn't take into account the frequency as all values are in terms of wavelengths. So how does the SWR change in terms of normalized frequency.
Thank you for all the time and help
Zach
Design a quarter wave transformer and single stub matcher that will match the design frequency.
For the quarter wave design and single stub matcher design, generate a plot of SWR on main feeding line vs. the normalized frequency \(\frac{f}{f_0}\) where \(f_0 \) is design frequency.
Calculate the percent bandwidths of each system \(\frac{f_2 - f_1}{f_0}\)
where \(f_1\) and \(f_2\) are the lower and upper frequencies for which SWR = 2.0.
2. Relevant equations
The equations can be seen on the 481Lecture8 for single stub matcher and quarter wave transformer i assume all i need is the\( Z_{ot} = \sqrt{R_L*Z_o}\) . The picture (Untitled) shows the equations of quarter wave.
3. The attempt at a solution
I am having a little trouble with this project. Using the two pictures seen in on page 3 (MyProject.pdf), we are suppose to build a quarter wave transformer and single stub matching. Since there is no reactive element, is the only thing needed for the quarter wave transformer is the value for \(Z_{ot} \)?
There is no length needed because there is no reactive element on the load right? Because the reflection coefficient angle is 0.
So \(Z_{ot} \) just equals \(\sqrt{75*37.5}\)because \(Z_L\) is just \( \frac{1}{75} + \frac{1}{75}\)
-------------------------------------------------------------------------
For the single stub matcher first using the formula on the lecture notes (equation 5.9) and i get a value of t = +/- .707107
After that i use equation 5.10 I would get
\(\frac{d}{\lambda} = (\frac{1}{2\pi})tan^{-1}(.707107) and
\frac{d}{\lambda} = (\frac{1}{2\pi})(\pi + tan^{-1}(-.707107) \)
So \( d = .96679\lambda \) and\( d = 3.96801\lambda \)
(Equation 5.7)
After \( Z(-d) = 75*\frac{37.5 + j75*.707107}{75+j37.5*.707107} = 72.42639 + 32.78680j
Y = \frac{1}{Z(-d)} = .01073 - .00575j
Y* = .01073 + .00575j \)(for the short circuit stub)
(Also I remember from Smith charts that you can add as many multiples or .5λ but i don't know if that applies here)
\(l_{sc}= (\frac{1}{(2\pi})\tan^{-1}(\frac{.01073}{-.00575}) = -1.69467\lambda + 2\lambda = .30533\lambda
l_{sc}= (\frac{1}{(2\pi})\tan^{-1}(\frac{.01073}{.00575}) = 1.69467\lambda -1.5\lambda = .19467\lambda \)
I also don't understand the concept of the plot of SWR vs normalized frequency either. When you find your single-stub matcher and quarter wave transformer, i thought you didn't take into account the frequency as all values are in terms of wavelengths. So how does the SWR change in terms of normalized frequency.
Thank you for all the time and help
Zach
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