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  1. #1
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    Single Stub matching

    Hello
    As you see this problem with single stub matching

    I matched zl=.5+1.5i to a normalized line (Z0=50)(I did it with smith chart and analytically) but as you see in the second picture as I calculate Zin it doesn't get me the same result (.5+1.5j).
    what is my wrong?

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  2. #2
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    Re: Single Stub matching

    Are you going in the same direction around the smith chart?



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  3. #3
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    Re: Single Stub matching

    Hello
    Yes I did



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  4. #4
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    Re: Single Stub matching

    The impedance of the shorted stub reduces to z_in = j*tan(2*pi*1.68) = j2.13.

    The input impedance of the load as viewed from the stub is z_in = ((0.5+j1.5) + j(50Ω)*tan(2*pi*0.28))/((50Ω) + j(0.5+j1.5)*tan(2*pi*0.28)) = 0.21 - j4.49.

    The two in parallel don't give 1... am I missing something?



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    Re: Single Stub matching

    I am a little confused. Why do you keep handwriting "Yin = Zin"? Yin is of course 1/Zin

    and in computing the impedance where the stub joins the main line, the ADMITTANCES add, not the impedances. You want normalized Yin total to equal 1 at that point.


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  6. #6
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    Re: Single Stub matching

    Hello dear biff44
    "Yin=Zin" it means that if we find normalized Zin=1 then Yin=1/Zin=1.
    Thanks for your response



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