Quarter Wave and Single Stub Matching

Discussion in 'Homework Help' started by xz4chx, Apr 23, 2014.

  1. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    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}

    -------------------------------------------------------------------------
    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 <br />
\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 <br />
Y = \frac{1}{Z(-d)} = .01073 - .00575j<br />
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<br />
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
     
    Last edited: Apr 24, 2014
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    You could do to make your Tex notations a bit clearer

    Why not for the first part

    Z_{ot}=\sqrt{75 \times 37.5} \text{\ ohms}

    Yes - the shorted stub is not necessary with the quarter wave transformer only needing to match 37.5Ω [resistive]. I suppose if the prof insists on a stub at the line split, you could use a superfluous shorted quarter wave section as the stub [Z=∞].

    I don't get similar solution(s) to yours for the alternative stub + line match. However I did it by simple trial and error on a Smith Chart.

    Of course impedances must change as you vary frequency. If you do the sums for a normalized frequency of unity then as the normalized frequency varies (more or less than unity) the original physical line lengths viewed as factors of normalized wavelength, necessarily change.
     
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  3. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    Were the values for part 2 off a large amount? If so i will probably redo my calculations as i might have plugged it into my calculator wrong. The  Z_{ot} makes sense, i just wanted to make sure that all i needed was to match the resistive element.

    The graphs still don't make sense to me. I am not quite understanding. I need to know what formulas to use but i also want to understand this.

    If  f_0 is a designed frequency. Do i just choose a number for frequency or is a certain value the right value for the graph?

    http://www.ece.rutgers.edu/~orfanidi/ewa/ch12.pdf
    On page 4 of this page, do i use these formulas for my calculations?

    I'm struggling to understand this as we didn't go over this in class. And struggled with the concept of the transmission line with the reflection coefficients, etc...

    Thank you for taking the time to help me.
     
  4. t_n_k

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    Mar 6, 2009
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    The correct section in the Rutgers chapter would be 12.8 - "Single Stub Matching". Example 12.8.1 provides a useful comparison for your project example analysis. The essential difference in your case is the lack of any imaginary term in the load impedance.

    As I said your notations aren't particularly clear. What do you get for the load reflection coefficient - i.e. the reflection coefficient at the line bifurcation with the twin 75 Ω loads? What are the line and stub lengths you obtain in terms of normalized wavelength? Or at least the ratio of line and stub lengths? Given there are two possible solutions the correct choice will impact the bandwidth. I believe the solution with shorter overall lengths would give the greater bandwidth - as defined on the basis of VSWR of 2.0.

    The fact that you haven't treated this in class puzzles me and the further issue of your not having a clear understanding of transmission lines makes life even more difficult for you.

    I'm reluctant to supply any actual numbers to you as this is a homework project.

    Are you familiar with the Smith Chart?
     
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  5. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    Yes, I am familiar with the concepts of Smith Charts. We are suppose to do all the calculations by hand. With smith charts, all we went over was matching the single stub matcher and quarter wave transformer. Besides that, we did not go into anything else. So you would use the smith chart and find the circle that represents the reflection coefficient. What does this circle represent though i don't know.
    After that you would match it to SWR = 1.0 and find the distance from those admittance values. Thats all we talked about.
    We didn't really discuss the calculations that take place by hand.

    Our teacher is hard to follow in his lectures and even watching youtube videos i'm still quite confused.

    I go the values for line and stub lengths in terms of wavelengths. How does this relate to it in terms of normalized wavelengths?
     
  6. t_n_k

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    Mar 6, 2009
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    If you are familiar / confident with the Smith Chart manipulations for design of a single stub match then you should be able to compare your values derived from mathematical formulas with a Smith Chart approach. If the values are inconsistent then this might suggest the likely error in the formula based method.

    You didn't answer my question about the load reflection coefficient value.

    The circle that represents a constant reflection coefficient (or VSWR) on the Smith Chart shows how the reflection coefficient phase angle at a point (x) on the feed-line varies as the feed-line section length interposed between that point (x) and the load varies. The reflection coefficient magnitude doesn't change - since it's constant :rolleyes:.

    I said normalized wavelength as the project pre-supposes a non-specific operating frequency or rather a normalized frequency. Perhaps my use of "normalized" in that context is superfluous.
     
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  7. xz4chx

    Thread Starter Member

    Sep 17, 2012
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     \Gamma_L = \frac {Z_L - Z_o}{Z_L + Z_o}

    This would give you  -\frac {1}{3}
     
  8. t_n_k

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    Mar 6, 2009
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    That's correct.

    So if you follow the example 12.8.1 in the Rutgers notes you can determine the possible βl and βd values. As a starting point it should now be clear to you that

    \| \Gamma_L \|= \frac{1}{3}

    \text{\theta_L=\pi \ [radians]}
     
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  9. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    Because I am using a short circuit in parallel i would use

     \beta l= \frac {1}{2}[\theta_L \frac{+}{ \ } acos(-|\Gamma_L|)]<br />
<br />
\beta d = arctan(-\frac {1}{2} tan(2\beta l - \theta_L))

     \beta = \frac{2\pi}{\lambda}

     \theta_L = \pi<br />
|\Gamma_L|=\frac{1}{3}<br />
<br />
\beta l = 2.52611 \ and \ .61548<br />
\beta d = .95533 \  and \  -.95532

     l = .40204\lambda \ and \ .09796\lambda<br />
d = .15204\lambda \ and \ (-.15204\lambda + .5\lambda)= .34796\lambda
     
    Last edited: Apr 25, 2014
  10. t_n_k

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    Check your βd calculations. I believe they're incorrect.
     
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  11. xz4chx

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    Sep 17, 2012
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    Sorry about that. I edited the post.

    Since you want to use the shortest length, you would use
     l =.09796\lambda \ and \ d=.34796\lambda
     
    Last edited: Apr 25, 2014
  12. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    So now the SWR value would be 1 on the main feeding line because of the matching stub

     SWR = \frac{1 + \Gamma_L} {1-\Gamma_L}

    You would have a new reflection coefficient because you added in the single stub matcher.
     
    Last edited: Apr 25, 2014
  13. t_n_k

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    Try again. Still appear to be incorrect values.

    I have βd as ±0.9553

    Your d of 0.11599 \text{\lambda} would make the SWR about 5 rather than unity.
     
    Last edited: Apr 25, 2014
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  14. xz4chx

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    Sep 17, 2012
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    I didn't multiple my value by 2 for the formula... I feel really smart.
     
  15. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    I would like to thank you for taking all this time to help me. This is starting to make more sense and i really appreciate this.
     
  16. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    Okay, so everything from there i understand. Now the main feeding line has a VSWR of 1 because the reflection coefficient is 0. How does this change with respect to frequency when i thought

       VSWR = \frac {1+|\Gamma_{newL}|}{{1-|\Gamma_{newL}|}

    And the magnitude of this never changes i thought
     
  17. t_n_k

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    Glad to have been of assistance.

    I'm going out shortly, but will respond to your other post on the effect of frequency changes on VSWR - when I get the opportunity. In the mean time, perhaps other AAC members may have some thoughts as well.
     
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  18. t_n_k

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    I think the question you are asking is that once the reflection coefficient has been zeroed by the tuned stub or the quarter wave transformer, then why would changing the frequency alter the reflection coefficient?

    Probably the simplest thing to look at is the quarter wave transformer. Without the quarter wave transformer made from a Tx line section with characteristic impedance \text{Z_{o\small{T}}=\sqrt{75 \times 37.5}=53.033 \ ohms} the reflection coefficient would be -1/3. Once the aforementioned section is interposed between the 75Ω feed-line and the "load", the reflection coefficient observed in the 75Ω feed-line would be zero [VSWR=1]. The name of the transformer itself gives a rather broad clue - "Quarter Wave Transformer". What does that mean? At the frequency of operation the section length is exactly one quarter of the equivalent wavelength. At any frequency other than the design frequency, the length won't be exactly one quarter of a wavelength.

    Take a specific example. At 300 MHz operating frequency [\text{\lambda=1.0 meter}] a quarter wave transformer will be close to 1/4 meter or 250 mm long - assuming a line velocity factor of 1.0. Suppose we keep the same transformer but change the frequency to 240 MHz. At 240 MHz [\text{\lambda=1.25 meter}]this section is still physically 250 mm long but no longer a quarter wavelength long - rather it is only 0.20 wavelengths long. So at 240MHz it won't serve as a true quarter wave transformer and the matching achieved at 300MHz no longer applies, with the reflection coefficient at the 75Ω feed-line / quarter wave transformer interface now being non-zero.
     
    Last edited: Apr 25, 2014
  19. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    First, I thought because everything was in terms of wavelength so a design frequency was not selected in the process. So how is the plot of SWR designed. Is it considering the magnitude and phase for the reflection coefficient? And what is the relationship of the reflection coefficient angle and the values calculated?
     
  20. t_n_k

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    If you refer back the the Rutgers chapter on Impedance Matching you will see in example 12.8.2 the relevant formulas for relating reflection coefficient and frequency. That example shows plots of reflection coefficient magnitude vs normalized frequency for the various types of stub matches.

    It should only be a few computational steps for you to (at least) plot the graph for reflection coefficient vs normalized frequency for your project example. Then it should be another reasonable step to plotting VSWR vs normalized frequency since you know the relationship between VSWR and reflection coefficient.
     
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