Magic T Numericals
Lossless Reciprocal and Nonreciprocal Two Port Network
Question 1: Consider a lossless two port network. (a) If the network is reciprocal, show that \(|S_{21}|^2 = 1 - |S_{11}|^2\). (b) If the network is nonreciprocal, show that it is impossible to have unidirectional transmission, where \(S_{12} = 0\) and \(S_{21} \neq 0\).
(a) Reciprocal Lossless Two Port Network
For a two port network, the general scattering matrix is written as
\[ [S]= \begin{bmatrix} S_{11} & S_{12}\\ S_{21} & S_{22} \end{bmatrix} \]
For a reciprocal network, the transmission from Port 1 to Port 2 is equal to the transmission from Port 2 to Port 1. Therefore,
\[ S_{12}=S_{21} \]
Hence, the scattering matrix becomes
\[ [S]= \begin{bmatrix} S_{11} & S_{21}\\ S_{21} & S_{22} \end{bmatrix} \]
Since the network is lossless, its scattering matrix must be unitary. Therefore,
\[ [S][S]^\dagger=[I] \]
For the first row or, equivalently, the first column of the unitary scattering matrix, the lossless condition gives
\[ |S_{11}|^2+|S_{21}|^2=1 \]
Rearranging this equation gives
\[ |S_{21}|^2=1-|S_{11}|^2 \]
Therefore,
\[ \boxed{|S_{21}|^2=1-|S_{11}|^2} \]
This result has a direct physical meaning. When a signal is incident at Port 1, the input power is divided between the power reflected back from Port 1 and the power transmitted to Port 2. Since the network is lossless, there is no power dissipated inside the network. Therefore, the reflected and transmitted powers must add to the total incident power.
(b) Nonreciprocal Lossless Two Port Network
For a nonreciprocal network, the forward and reverse transmission coefficients are not required to be equal. Therefore, in general,
\[ S_{12}\neq S_{21} \]
Suppose that unidirectional transmission is required. This means that transmission is allowed from Port 1 to Port 2, while transmission from Port 2 to Port 1 is zero. Thus,
\[ S_{12}=0 \]
and
\[ S_{21}\neq0 \]
For a lossless network, the scattering matrix is unitary. Therefore, the rows and columns of the scattering matrix must satisfy the corresponding power and orthogonality conditions.
Considering the first column of the scattering matrix gives
\[ |S_{11}|^2+|S_{21}|^2=1 \]
Since \(S_{21}\neq0\), the above equation shows that the first column contains a nonzero transmission component.
Now consider the second column. The lossless condition gives
\[ |S_{12}|^2+|S_{22}|^2=1 \]
Since the assumed unidirectional transmission condition is
\[ S_{12}=0 \]
we obtain
\[ |S_{22}|^2=1 \]
Therefore,
\[ |S_{22}|=1 \]
The columns of a unitary scattering matrix must also be orthogonal. Hence,
\[ S_{11}S_{12}^{*}+S_{21}S_{22}^{*}=0 \]
Using \(S_{12}=0\), this becomes
\[ S_{21}S_{22}^{*}=0 \]
But \(S_{22}\) has unit magnitude because \(|S_{22}|=1\). Therefore, \(S_{22}\neq0\), which requires
\[ S_{21}=0 \]
This contradicts the original assumption that
\[ S_{21}\neq0 \]
Hence, a lossless two port network cannot have ideal unidirectional transmission with \(S_{12}=0\) and \(S_{21}\neq0\).
Therefore,
\[ \boxed{S_{12}=0\quad\Rightarrow\quad S_{21}=0} \]
Thus, it is impossible for a lossless two port network to provide ideal unidirectional transmission in which one transmission coefficient is zero while the transmission coefficient in the opposite direction is nonzero.
Final Results
- For a reciprocal lossless two port network, \[ \boxed{|S_{21}|^2=1-|S_{11}|^2} \]
- For a lossless two port network, ideal unidirectional transmission with \(S_{12}=0\) and \(S_{21}\neq0\) is impossible.
- The unitary property of the scattering matrix is the key condition used in both results.
Power Distribution in a Lossless H Plane T Junction
Question 2: A 20 mW signal is fed into one of the collinear (port 1) of a lossless H-plane T-junction. Calculate the power delivered through each port when other ports are terminated in matched load.
Solution
Since the signal is applied at Port 1 and the other ports are terminated in matched loads, there are no incident waves entering the junction from Ports 2 and 3. Therefore,
\[ a_2=a_3=0 \]
For a lossless H-plane T-junction, the magnitude of the reflection coefficient at the excited collinear port is
\[ |S_{11}|=\frac{1}{2} \]
The incident power supplied to Port 1 is
\[ P_{\mathrm{in}}=20\,\text{mW} \]
For normalized incident and reflected waves, the power reflected at Port 1 is determined by \(|S_{11}|^2\). Therefore, the effective power entering the junction is
\[ P_1=P_{\mathrm{in}}\left(1-|S_{11}|^2\right) \]
Substituting the given values,
\[ P_1=20\left(1-\left(\frac{1}{2}\right)^2\right) \]
\[ P_1=20\left(1-\frac{1}{4}\right) \]
\[ P_1=20\left(\frac{3}{4}\right) \]
Therefore,
\[ \boxed{P_1=15\,\text{mW}} \]
Thus, 15 mW of the supplied power enters the junction, while the remaining 5 mW is reflected back toward the source.
Power Transmitted to Port 3
For the lossless H-plane T-junction, the power entering the junction is divided between the two output arms according to the corresponding scattering coefficients. The power transmitted to Port 3 is
\[ P_3=P_{\mathrm{in}}|S_{31}|^2 \]
For the given H-plane T-junction,
\[ |S_{31}|^2=\frac{1}{2} \]
Therefore,
\[ P_3=20\times\frac{1}{2} \]
\[ \boxed{P_3=10\,\text{mW}} \]
Power Transmitted to Port 2
Similarly, the power transmitted to Port 2 is
\[ P_2=P_{\mathrm{in}}|S_{21}|^2 \]
For the given H-plane T-junction,
\[ |S_{21}|^2=\frac{1}{4} \]
Therefore,
\[ P_2=20\times\frac{1}{4} \]
\[ \boxed{P_2=5\,\text{mW}} \]
Power Conservation Check
The power entering the junction is 15 mW. The transmitted powers through Ports 2 and 3 are 5 mW and 10 mW respectively. Therefore,
\[ P_1=P_2+P_3 \]
Substituting the calculated values,
\[ 15=5+10\,\text{mW} \]
Hence,
\[ \boxed{P_1=P_2+P_3} \]
The remaining 5 mW from the original 20 mW input is reflected at Port 1. Thus, the total input power is also verified as
\[ 20\,\text{mW}=5\,\text{mW reflected}+5\,\text{mW transmitted to Port 2}+10\,\text{mW transmitted to Port 3} \]
Therefore, the 20 mW input power is completely accounted for, confirming the power distribution in the lossless H-plane T-junction.
Power Delivered to Unequal Loads in an H Plane T Junction
Question 3: In an H-plane T-junction, compute power delivered to the loads of 40 ohms and 60 ohms connected to arms 1 and 2 when a 10 mW power is delivered to the matched port 3.
Solution
The H-plane T-junction divides the power entering the matched Port 3 equally between Arms 1 and 2. Therefore, when 10 mW power is delivered to Port 3, the power travelling toward each load is
\[ P_{\mathrm{in},1}=P_{\mathrm{in},2} =\frac{10}{2} =5\,\text{mW} \]
Thus, 5 mW is incident toward the 40 ohm load connected to Arm 1 and 5 mW is incident toward the 60 ohm load connected to Arm 2.
Power Incident on the 40 Ohm Load
The characteristic impedance of the transmission lines is taken as
\[ Z_0=50\,\Omega \]
The load connected to Arm 1 is
\[ Z_1=40\,\Omega \]
The reflection coefficient at this load is
\[ \Gamma_1= \frac{Z_1-Z_0}{Z_1+Z_0} \]
Substituting the values,
\[ \Gamma_1= \frac{40-50}{40+50} \]
\[ \Gamma_1= -\frac{10}{90} =-\frac{1}{9} \]
Therefore,
\[ |\Gamma_1|=\frac{1}{9} \]
and
\[ |\Gamma_1|^2= \frac{1}{81} \approx0.01234 \]
The incident power toward the 40 ohm load is 5 mW. The power delivered to the load is therefore
\[ P_1=P_{\mathrm{in},1}\left(1-|\Gamma_1|^2\right) \]
Hence,
\[ P_1=5\left(1-0.01234\right) \]
\[ P_1=5(0.98766) \]
Therefore,
\[ \boxed{P_1=4.9383\,\text{mW}} \]
Power Incident on the 60 Ohm Load
The load connected to Arm 2 is
\[ Z_2=60\,\Omega \]
The reflection coefficient at this load is
\[ \Gamma_2= \frac{Z_2-Z_0}{Z_2+Z_0} \]
Substituting the values,
\[ \Gamma_2= \frac{60-50}{60+50} \]
\[ \Gamma_2= \frac{10}{110} =\frac{1}{11} \]
Therefore,
\[ |\Gamma_2|=\frac{1}{11} \]
and
\[ |\Gamma_2|^2= \frac{1}{121} \approx8.264\times10^{-3} \]
The incident power toward the 60 ohm load is 5 mW. Hence, the power delivered to the load is
\[ P_2=P_{\mathrm{in},2}\left(1-|\Gamma_2|^2\right) \]
Substituting the value of the reflection coefficient,
\[ P_2=5\left(1-8.264\times10^{-3}\right) \]
\[ P_2=5(0.991736) \]
Therefore,
\[ \boxed{P_2\approx4.9587\,\text{mW}} \]
Reflected Power from Each Load
The reflected power from the 40 ohm load is
\[ P_{R1}=P_{\mathrm{in},1}|\Gamma_1|^2 \]
\[ P_{R1}=5(0.01234) \]
\[ \boxed{P_{R1}\approx0.0617\,\text{mW}} \]
Similarly, the reflected power from the 60 ohm load is
\[ P_{R2}=P_{\mathrm{in},2}|\Gamma_2|^2 \]
\[ P_{R2}=5(8.264\times10^{-3}) \]
\[ \boxed{P_{R2}\approx0.0413\,\text{mW}} \]
Final Answer
- Power incident toward the 40 ohm load = \(5\,\text{mW}\)
- Power delivered to the 40 ohm load = \(\boxed{4.9383\,\text{mW}}\)
- Power incident toward the 60 ohm load = \(5\,\text{mW}\)
- Power delivered to the 60 ohm load = \(\boxed{4.9587\,\text{mW}}\)
- Power reflected from the 40 ohm load = \(\boxed{0.0617\,\text{mW}}\)
- Power reflected from the 60 ohm load = \(\boxed{0.0413\,\text{mW}}\)
Although the H-plane T-junction divides the input power equally between the two arms, the actual power delivered to the loads is slightly different because the 40 ohm and 60 ohm loads have different reflection coefficients with respect to the 50 ohm characteristic impedance.
Power Calculation for a Terminated Magic T
Question 4: A magic-T is terminated at collinear ports 1 and 2 and difference port 4 by impedances of reflection coefficients Γ1 = 0.5, Γ2 = 0.6 and Γ4 = 0.8, respectively. If 1W power is fed at the sum port 3, calculate the power reflected at Port 3 and power transmitted to the other three ports.
Solution
For a matched Magic T having collinear ports 1 and 2, sum port 3, and difference port 4, the scattering matrix is
\[ [S]=\frac{1}{\sqrt{2}} \begin{bmatrix} 0 & 0 & 1 & 1\\ 0 & 0 & 1 & -1\\ 1 & 1 & 0 & 0\\ 1 & -1 & 0 & 0 \end{bmatrix} \]
Let \(a_1,a_2,a_3,a_4\) represent the normalized incident wave amplitudes at Ports 1, 2, 3, and 4, respectively, and let \(b_1,b_2,b_3,b_4\) represent the corresponding outgoing wave amplitudes.
The reflection coefficients of the terminations are given by
\[ \Gamma_1=0.5,\qquad \Gamma_2=0.6,\qquad \Gamma_4=0.8 \]
Since Ports 1, 2, and 4 are terminated, their incident waves are related to their outgoing waves by
\[ a_1=\Gamma_1b_1=0.5b_1 \]
\[ a_2=\Gamma_2b_2=0.6b_2 \]
\[ a_4=\Gamma_4b_4=0.8b_4 \]
Power is supplied at the sum Port 3, so \(a_3\) is the incident wave at Port 3. The incident power is given by
\[ P_i=\frac{1}{2}|a_3|^2 \]
Since the applied input power is 1 W,
\[ 1=\frac{1}{2}|a_3|^2 \]
Therefore,
\[ |a_3|^2=2 \]
Taking the incident voltage amplitude as positive real,
\[ a_3=\sqrt{2}\,\text{V} \]
Scattering Equations
From the scattering matrix, the outgoing waves are related to the incident waves by
\[ \begin{bmatrix} b_1\\ b_2\\ b_3\\ b_4 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 0 & 0 & 1 & 1\\ 0 & 0 & 1 & -1\\ 1 & 1 & 0 & 0\\ 1 & -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} a_1\\ a_2\\ a_3\\ a_4 \end{bmatrix} \]
Thus, the four wave equations are
\[ b_1=\frac{1}{\sqrt{2}}(a_3+a_4) \]
\[ b_2=\frac{1}{\sqrt{2}}(a_3-a_4) \]
\[ b_3=\frac{1}{\sqrt{2}}(a_1+a_2) \]
\[ b_4=\frac{1}{\sqrt{2}}(a_1-a_2) \]
Substituting the termination conditions gives
\[ b_1=\frac{1}{\sqrt{2}}(a_3+0.8b_4) \]
\[ b_2=\frac{1}{\sqrt{2}}(a_3-0.8b_4) \]
\[ b_3=\frac{1}{\sqrt{2}}(0.5b_1+0.6b_2) \]
\[ b_4=\frac{1}{\sqrt{2}}(0.5b_1-0.6b_2) \]
Solving for the Output Waves
Using \(a_3=\sqrt{2}\), the first two equations become
\[ b_1=1+0.4b_4 \]
\[ b_2=1-0.4b_4 \]
Substitute these expressions into the equation for \(b_4\):
\[ b_4= \frac{1}{\sqrt{2}} \left[ 0.5(1+0.4b_4) - 0.6(1-0.4b_4) \right] \]
Expanding,
\[ b_4= \frac{1}{\sqrt{2}} \left[ 0.5+0.2b_4-0.6+0.24b_4 \right] \]
Therefore,
\[ b_4= \frac{1}{\sqrt{2}} \left[ -0.1+0.44b_4 \right] \]
Solving for \(b_4\),
\[ b_4\left(1-\frac{0.44}{\sqrt{2}}\right) = -\frac{0.1}{\sqrt{2}} \]
Hence,
\[ \boxed{b_4\approx-0.126} \]
Now substitute this value into the expressions for \(b_1\) and \(b_2\).
\[ b_1=1+0.4(-0.126) \]
\[ \boxed{b_1\approx0.950} \]
Similarly,
\[ b_2=1-0.4(-0.126) \]
\[ \boxed{b_2\approx1.050} \]
Using these values in the equation for \(b_3\),
\[ b_3= \frac{1}{\sqrt{2}} (0.5b_1+0.6b_2) \]
Therefore,
\[ b_3\approx \frac{1}{\sqrt{2}} \left[ 0.5(0.950)+0.6(1.050) \right] \]
\[ \boxed{b_3\approx0.988} \]
Power Reflected at Port 3
The power reflected back toward the source at Port 3 is
\[ P_{R3}=\frac{1}{2}|b_3|^2 \]
Using \(b_3\approx0.988\),
\[ P_{R3} = \frac{1}{2}(0.988)^2 \]
Therefore,
\[ \boxed{P_{R3}\approx0.488\,\text{W}} \]
Power Transmitted to Port 1
The power travelling toward Port 1 is
\[ P_1=\frac{1}{2}|b_1|^2 \]
Using \(b_1\approx0.950\),
\[ P_1=\frac{1}{2}(0.950)^2 \]
Therefore,
\[ \boxed{P_1\approx0.451\,\text{W}} \]
Power Transmitted to Port 2
The power travelling toward Port 2 is
\[ P_2=\frac{1}{2}|b_2|^2 \]
Using \(b_2\approx1.050\),
\[ P_2=\frac{1}{2}(1.050)^2 \]
Therefore,
\[ \boxed{P_2\approx0.551\,\text{W}} \]
Power Transmitted to Port 4
The power travelling toward Port 4 is
\[ P_4=\frac{1}{2}|b_4|^2 \]
Using \(b_4\approx-0.126\),
\[ P_4=\frac{1}{2}(0.126)^2 \]
Therefore,
\[ \boxed{P_4\approx0.0079\,\text{W}} \]
Results
- Power reflected at Port 3 is approximately \(\boxed{0.488\,\text{W}}\).
- Power transmitted toward Port 1 is approximately \(\boxed{0.451\,\text{W}}\).
- Power transmitted toward Port 2 is approximately \(\boxed{0.551\,\text{W}}\).
- Power transmitted toward Port 4 is approximately \(\boxed{0.0079\,\text{W}}\).
The powers transmitted toward Ports 1, 2, and 4 are not equal because the terminations at these ports have different reflection coefficients. The reflected waves return to the Magic T and interact with one another, which changes both the transmitted waves and the power reflected back toward the source.