# Numericals

**NUMERICALS**

**Solution:**

Insertion loss= -20log (P_{2}/P_{1})

0.5= -20log |S_{21}|

Or S_{21}= 10^{0.5/20}= 10^{-0.025}

Isolation=30dB= -20log(P_{1}/P_{2})

30= -20log |S_{12}|

S_{12}= 10^{-30/20}= 10^{-1.5}

**For Isolator s _{11}=S_{22}=0**

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

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[ᴩ_{1}b_{1}]^{2}

Power reflected from port 2= 1/2[ᴩ_{2}b_{2}]^{2}

Power delivered to load Z_{1 }at port 1= 60 ohm; is

P_{1}= power Incident-Reflected power

P_{1}= 1/2[b_{1}]^{2}- 1/2[ᴩ_{1}b_{1}]^{2}

P_{1}= 1/2[b_{1}]^{2 }[1- |ᴩ_{1}|^{2}]

Power delivered to load Z_{2 }at port 2= 75 ohm; is

P_{2}= power Incident-Reflected power

P_{2}= 1/2[b_{2}]^{2}- 1/2[ᴩ_{2}b_{2}]^{2}

P_{2}= 1/2[b_{2}]^{2 }[1- Іᴩ_{2}І^{2}]

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

|ᴩ_{1}|= |60-50|/|60+50|= 10/110= 1/11 & | P_{1}|^{2}=1/121

|ᴩ_{2}|=|75-50|/|75+50|= 25/125= 1/5 & | P_{1}|^{2}=1/25

Therefore,

P_{1}= 1/2[b_{1}]^{2 }[1- |ᴩ_{1}|^{2}]

= 10*10^{-3}[1-1/121]

P1= 9.92mW

& P2=1/2[b_{2}]^{2 }[1- Іᴩ_{2}І^{2}]

= 10*10^{-3}[1-1/25]

P2= 9.6mW

**Solution:**

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

Here,

a_{1 =}ᴩ_{1}b_{1}

a_{2 =}ᴩ_{2}b_{2}

|a_{3}|^{2}_{ =}1W

a_{4=}ᴩ_{4}b_{4}

Where a_{1}, a_{2}, a_{3}, a_{4} be the normalized input voltage & b_{1}, b_{2}, b_{3}, b_{4} are output voltage.

a_{1}=0.5b1

a_{2}=0.6b2

|a_{3}|^{2}_{ =}1W

a_{4}=0.8b4

Therefore,

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

Insertion loss=

Isolation= -20 log(P1/P2)=

Reflection coefficient= (S-1/S+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

Practically Isolation=30-40 dB

Insertion loss <1 dB

VSWR < 1.5 can be achieved.

**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

_{}

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

Then,

S_{13} = S_{21} = S_{32} = 1

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

[ s ] = 0 S_{12} S_{13}

S_{12} 0 0

S_{13} S_{23 }0

For a lossless junction, the [ s ] is unitary

S . S*= I

S_{12 . }S*_{12 }+ S_{13. }S*_{13 = }1 - (1)

S_{12 . }S*_{12 }+ S_{23. }S*_{23 = }1 - (2)

S_{13 . }S*_{13 }+ S_{23. }S*_{23 = }1 - (3)

S_{13 . }S*_{23 }= S_{12. }S*_{23 }= S_{12 . }S*_{13 = }1_{ }-_{ }(4)

If S_{12} ≠ 0, equation (4) gives S_{13}= 0 = S23

But this does not satisfy equation (3)

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

**Solution**: C = 10 dB

P_{i} = 250 mW

C = 10 log10(P_{i}/P_{f})

10/10= log10(P_{i}/P_{f})

Taking antilog

10^{1} = P_{i}/P_{f}

P_{f} = P_{i}/10 = 250/10 =~~ ~~25 mW.

O/p power in auxillary arm = P_{f} = 25 mW -----forwarded or coupled power.

Received power= P_{r} = P_{i} - P_{f} – P_{b} ------ (Pb=0)

= 250mW – 25mW

P_{r} = 225 mW