NUMERICALS

Example 1: An isolator has an insertion loss of 0.5dB & an isolation of 30dB. Determine the scattering matrix of the isolator if the isolated ports are perfectly matched to the junction.

Solution:

Insertion loss= -20log (P2/P1)

0.5= -20log |S21|

Or S21= 100.5/20= 10-0.025

Isolation=30dB= -20log(P1/P2)

30= -20log |S12|

S12= 10-30/20= 10-1.5

For Isolator s11=S22=0

[S]=\begin{bmatrix} 0 &10^{-1.5} \\ 10^{^{-0.025}} & 0 \end{bmatrix}

Example 2:-In an H-plane Tee junction, 20mW power is applied to port 3 that is perfectly matched to the junction. Calculate the power delivered to the load 60 ohm; & 75 ohm; connected to port 1 & port 2.

Ans:- The [S] for H-plane Tee with port 3 matched to the junction is,

[S]=\begin{bmatrix} \frac{1}{2}& \frac{-1}{2} &\frac{1}{\sqrt{2}} \\ \frac{-1}{2} & \frac{1}{2} &\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} & 0 \end{bmatrix}

Power at port 3 get equally divided into port 1& 2 that is 20mW power at port 3 delivers 10mW at ports 1 & 2.

Power reflected from port 1= 1/2[ᴩ1b1]2

Power reflected from port 2= 1/2[ᴩ2b2]2

Power delivered to load Zat port 1= 60 ohm; is

P1= power Incident-Reflected power

P1= 1/2[b1]2- 1/2[ᴩ1b1]2

P1= 1/2[b1][1- |ᴩ1|2]

Power delivered to load Zat port 2= 75 ohm; is

P2= power Incident-Reflected power

P2= 1/2[b2]2- 1/2[ᴩ2b2]2

P2= 1/2[b2][1- Іᴩ2І2]

Now taking the characteristic impedance of the line as 50 ohm;

|ᴩ1|= |60-50|/|60+50|= 10/110= 1/11 & | P1|2=1/121

|ᴩ2|=|75-50|/|75+50|= 25/125= 1/5 & | P1|2=1/25

Therefore,

P1= 1/2[b1][1- |ᴩ1|2]

= 10*10-3[1-1/121]

P1= 9.92mW

& P2=1/2[b2][1- Іᴩ2І2]

= 10*10-3[1-1/25]

P2= 9.6mW

Example 3:-The co-linear ports 1 & 2 of a magic Tee are terminated by impedance of reflection coefficient ᴩ1=0.5 & ᴩ2=0.6. The difference port 4 is terminated by an impedance with reflection coefficient of 0.8. If 1W power is fed at sum port 3, calculate the power reflected at port 3 & the power divisions at the other port.

Solution:
The input and output relations for a Magic Tee is given by

\begin{bmatrix} b1\\ b2\\ b3\\ b4 \end{bmatrix}=\begin{bmatrix} 0 &0 &\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \\ 0& 0&\frac{1}{\sqrt{2}} &\frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} &0 & 0\\ \frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}&0 & 0 \end{bmatrix}.\begin{bmatrix} a1\\ a2\\a3 \\ a4\end{bmatrix}

Here,

a1 =1b1

a2 =2b2

|a3|2 =1W

a4=4b4

Where a1, a2, a3, a4 be the normalized input voltage & b1, b2, b3, b4 are output voltage.

a1=0.5b1

a2=0.6b2

|a3|2 =1W

a4=0.8b4

Therefore,

\begin{bmatrix} b1\\ b2\\ b3\\ b4 \end{bmatrix}=\begin{bmatrix} 0 &0 &\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \\ 0& 0&\frac{1}{\sqrt{2}} &\frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} &0 & 0\\ \frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}&0 & 0 \end{bmatrix}.\begin{bmatrix} 0.5b1\\ 0.6b2\\1b3 \\ 0.8b4\end{bmatrix}

Similarly, for b1, b2, b3, b4, we get, using cramer's rule

b1=0.6566V

b2=0.7576V

b3=0.6536V

b4=0.0893V

P1=|b1|2=0.4309W

P2=|b2|2=0.5738W

P3=|b3|2=0.3065W

P4=|b4|=0.0079W

Example 4:Determine the [S] of a 3-port circulator given insertion loss of 0.5 dB, isolation of 20dB &VSWR=?

S=\begin{bmatrix} S_{11}& S_{12} & S_{13} \\ S_{21}& S_{22} & S_{23} \\ S_{31}& S_{32} & S_{33} \end{bmatrix}

Insertion loss= I=-20\log_{10}(\frac{P2}{P1})

Isolation= -20 log(P1/P2)=Isolation=-20\log_{10}(\frac{P1}{P2})

Reflection coefficient= (S-1/S+1)

NOTE—
Insertion loss--- I/p applied at port 1 &o/p taken at port2
Isolation---If i/p applied at port2 there will be no o/p at port 1

Given—

Insertion loss= 0.5 dB = -20 log[S21]

Or [S21] = 10(-0.5/20) = [S21 ] = 10-0.025

Similarly, [S21] = [S32] = [S13] = 10-0.025

Isolation = 20dB = -20log[S12]

[S12] = 10(-20/20) = 10-1

[S12] = 0.1

Similarly,[S23] = [S31] = 0.1

Reflection coefficient= (S-1)/(S+1) = (2-1)/(2+1)=1/3

Reflection coefficient=0.33333

Hence

S=\begin{bmatrix} 0.33& 0.1 & 10^{-0.025} \\ 10^{-0.025}& 0.33 & 0.1 \\ 0.1 &10^{-0.025} & 0.33 \end{bmatrix}

Practically Isolation=30-40 dB

Insertion loss <1 dB

VSWR < 1.5 can be achieved.

Example 5:Obtain the scattering matrix for a 3-port circulator & also prove that it is impossible to construct a perfectly matched losslss, reciprocal 3-port junction.

Solution:
A 3-port circulator can be obtained by a 180° H-plane wavdeguide with a central ferrite part. A axial magnetic field is applied along the axis of this part for proper operation of the circulator.

For a perfectly matched lossless non-reciprocal 3-port circulator, the [s] is given by

S=\begin{bmatrix} 0 & 0 & S_{13}\\ S_{21}&0 &0 \\ 0 & S_{32}& 0 \end{bmatrix}

The phase angles of S13, S21 & S32 can be made zero if the terminal planes are properly chosen.

Then,

S13 = S21 = S32 = 1

S=\begin{bmatrix} 0 & S_{12}& S_{13}\\ S_{12}&0 &0 \\ S_{13} & S_{23}& 0 \end{bmatrix}

For a perfectly matched 3-port junction, the [ s ] is:

[ s ] = 0 S12 S13

S12 0 0

S13 S23 0
 

For a lossless junction, the [ s ] is unitary

S . S*= I

S12 . S*12 + S13. S*13 = 1 - (1)

S12 . S*12 + S23. S*23 = 1 - (2)

S13 . S*13 + S23. S*23 = 1 - (3)

S13 . S*23 = S12. S*23 = S12 . S*13 = 1 - (4)

If S12 ≠ 0, equation (4) gives S13= 0 = S23

But this does not satisfy equation (3)

Hence a reciprocal lossless 3-port ciculator can’t be perfectly matched.

Example: If the incident power of 10dB directional coupler is 250mW. Calculate:
1] The o/p power in the main arm.
2] The o/p power in the auxillary arm.

Solution: C = 10 dB

Pi = 250 mW

C = 10 log10(Pi/Pf)

10/10= log10(Pi/Pf)

Taking antilog

101 = Pi/Pf

790653079.pngPf = Pi/10 = 250/10 = 25 mW.

O/p power in auxillary arm = Pf = 25 mW -----forwarded or coupled power.

Received power= Pr = Pi - Pf – Pb ------ (Pb=0)

829744851.png829744851.png= 250mW – 25mW

Pr = 225 mW

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Created by Sujit Wagh on 2017/07/27 19:24