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Xyius
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EDIT: Oops! I posted this in the wrong section! Meant to post in engineering section!
I am not sure if I am doing this problem correctly or not. It is very simple.
1. Homework Statement
A 2 port network consists of just a series inductor with ##Z=j 100 \Omega## in a ##50 \Omega## system. Determine the corresponding S-parameters.
S Parameter matrix equations
[tex]b_1 = S_{11}a_1 + S_{12}a_2[/tex]
[tex]b_2 = S_{21}a_1 + S_{22}a_2[/tex]
Where
[itex]a_1=\frac{V_1+Z_0I_1}{2\sqrt{Z_0}}[/itex],
[itex]a_2=\frac{V_2+Z_0I_2}{2\sqrt{Z_0}}[/itex],
[itex]b_1=\frac{V_1-Z_0I_1}{2\sqrt{Z_0}}[/itex],
[itex]b_2=\frac{V_2-Z_0I_2}{2\sqrt{Z_0}}[/itex].
First I let ##a_1=0## by terminating the line with an impedance equal to the characteristic impedance so ##V_1=-Z_0I_1##. Doing this allows me to use my first two equations above and solve for both ##S_{22}## and ##S_{12}##.
[tex]S_{22}=\frac{b_2}{a_2}=\frac{V_2-Z_0I_2}{V_2+Z_0I_2}=\frac{Z_{out}-Z_0}{Z_{out}+Z_0}[/tex]
When the circuit is terminated on the LHS with a resistor equal to the characteristic impedance, the output impedance is simply ##Z+Z_0##. Plugging everything in, this gives me.
[tex]S_{22}=\frac{1}{2}(1+j)[/tex]
This part wasn't a problem, getting ##S_{12}## is giving me confusion. The equation is, (we will use ##I_1=I_2##)
[tex]S_{12}=\frac{b_1}{a_2}=\frac{V_1-Z_0I_1}{V_2+Z_0I_2}=\frac{\frac{V_1}{I_2}-Z_0}{\frac{V_2}{I_2}+Z_0}.[/tex]
Because ##a_1=0## we know that ##V_1=-Z_0I_1=-Z_0I_2##, also ##V_2/I_2=Z_0##thus,
[tex]S_{12}=\frac{-2Z_0}{\frac{V_2}{I_2}+Z_0}=\frac{-2Z_0}{2Z_0}=-1.[/tex]
This is what I am confused about. Can the transmission coefficient be negative? Would that just mean that the transmitted energy simply has a ##\pi## phase shift? Am I doing this right?
The other two parameters can be obtained by similar means.
I am not sure if I am doing this problem correctly or not. It is very simple.
1. Homework Statement
A 2 port network consists of just a series inductor with ##Z=j 100 \Omega## in a ##50 \Omega## system. Determine the corresponding S-parameters.
Homework Equations
S Parameter matrix equations
[tex]b_1 = S_{11}a_1 + S_{12}a_2[/tex]
[tex]b_2 = S_{21}a_1 + S_{22}a_2[/tex]
Where
[itex]a_1=\frac{V_1+Z_0I_1}{2\sqrt{Z_0}}[/itex],
[itex]a_2=\frac{V_2+Z_0I_2}{2\sqrt{Z_0}}[/itex],
[itex]b_1=\frac{V_1-Z_0I_1}{2\sqrt{Z_0}}[/itex],
[itex]b_2=\frac{V_2-Z_0I_2}{2\sqrt{Z_0}}[/itex].
The Attempt at a Solution
First I let ##a_1=0## by terminating the line with an impedance equal to the characteristic impedance so ##V_1=-Z_0I_1##. Doing this allows me to use my first two equations above and solve for both ##S_{22}## and ##S_{12}##.
[tex]S_{22}=\frac{b_2}{a_2}=\frac{V_2-Z_0I_2}{V_2+Z_0I_2}=\frac{Z_{out}-Z_0}{Z_{out}+Z_0}[/tex]
When the circuit is terminated on the LHS with a resistor equal to the characteristic impedance, the output impedance is simply ##Z+Z_0##. Plugging everything in, this gives me.
[tex]S_{22}=\frac{1}{2}(1+j)[/tex]
This part wasn't a problem, getting ##S_{12}## is giving me confusion. The equation is, (we will use ##I_1=I_2##)
[tex]S_{12}=\frac{b_1}{a_2}=\frac{V_1-Z_0I_1}{V_2+Z_0I_2}=\frac{\frac{V_1}{I_2}-Z_0}{\frac{V_2}{I_2}+Z_0}.[/tex]
Because ##a_1=0## we know that ##V_1=-Z_0I_1=-Z_0I_2##, also ##V_2/I_2=Z_0##thus,
[tex]S_{12}=\frac{-2Z_0}{\frac{V_2}{I_2}+Z_0}=\frac{-2Z_0}{2Z_0}=-1.[/tex]
This is what I am confused about. Can the transmission coefficient be negative? Would that just mean that the transmitted energy simply has a ##\pi## phase shift? Am I doing this right?
The other two parameters can be obtained by similar means.