Assignment 4 Solutions - Advanced Microwave Measurements | ECE 451, Assignments of Electrical and Electronics Engineering

Material Type: Assignment; Class: Adv Microwave Measurements; Subject: Electrical and Computer Engr; University: University of Illinois - Urbana-Champaign; Term: Fall 2008;

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Pre 2010

Uploaded on 03/10/2009

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ECE 451 Homework 4 Solutions
Problem 1
The voltage and current on a lossless transmission line are
() zz
Vz Ve Ve
γ
γ
+− −+
=+
()
() zz
o
Iz Y Ve Ve
γγ
+− −+
=−
Let port 1 be located at z=-l with voltage and current
1ll
VVe Ve
γ
γ
++ −−
=+
and
()
1ll
o
IYVe Ve
γγ
++ −−
=−
Let port 2 be located at z=0 with voltage and current
2
VVV
+−
=+
and
()
2o
I
yV V
+−
=−
where the current is reversed to take into account that positive current enters the DUT for
an admittance matrix. Solving for V+ and V- yields
22
2
o
VzI
V+
=
and
22
2
o
VzI
V+
=
Eliminating V+ and V- from the equations for port 1 and solving for the currents yields
22 22
122
ll
oo
VZI VZI
Ve e
γ
γ
+
−+
⎛⎞⎛⎞
=+
⎜⎟⎜⎟
⎝⎠⎝⎠
22 22
122
ll
oo
oo
VZI VZI
I
YeYe
γ
γ
+
−+
⎛⎞⎛⎞
=−
⎜⎟⎜⎟
⎝⎠⎝⎠
so that
12 2
cosh( ) sinh( )
o
VV l ZI l
γ
=−
pf2

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ECE 451 Homework 4 Solutions

Problem 1

The voltage and current on a lossless transmission line are

z z V z V e V e

  • − γ − + γ = +

z z I z Yo V e V e

  • − γ − +γ = −

Let port 1 be located at z=-l with voltage and current

1

l l V V e V e

    • γ − − γ = +

and

1 (^ )

l l I Yo V e V e

    • γ − −γ = −

Let port 2 be located at z=0 with voltage and current

V 2 V V

  • − = +

and

I 2 yo ( V V )

  • − = −

where the current is reversed to take into account that positive current enters the DUT for

an admittance matrix. Solving for V

+ and V

- yields

2 2

2

V z I o V

and

2 2

2

V z I o V

Eliminating V

+ and V

- from the equations for port 1 and solving for the currents yields

2 2 2 2 1 2 2

V Z Io l V Z Io l V e e

⎛ −^ ⎞ (^) + γ ⎛ + ⎞ − γ = (^) ⎜ ⎟ +⎜ ⎟ ⎝ ⎠ ⎝ ⎠

2 2 2 2 1 2 2

o l o l o o

V Z I V Z I

I Y e Y e

⎛ −^ ⎞ (^) + γ ⎛ + ⎞ − γ = (^) ⎜ ⎟ − ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

so that

V 1 = V 2 cosh( γ l ) − Z Io 2 sinh( γ l )

I 1 = Y Vo 2 sinh( γ l ) − I 2 cosh( γ l )

from which

2 11 2

cosh( ) cosh( )

sinh( ) sinh( )

o

o

I l Y l Y Z I l l

and

2 21 2 sinh( ) sinh( )

o

o

I Y

Y

Z I γ l γ l

Using symmetry and reciprocity,

cosh( ) 1

sinh( ) 1 cosh( )

o Y l

l l

Y

Problem 2

j j S j j

⎡ +^ − ⎤

⎣ −^ + ⎦

Convert to Z using

( )( )

- Z = Zo I + S I - S

Invert Z to get Y

j j

j j

⎡ −^ + ⎤

⎣ +^ − ⎦

Y

Since for a lossless line,

cos( ) 1

sin( ) 1 cos( )

o jY l

l l

Y

We obtain

j

Y

So, Yo sin( β l ) = 0.0143216and cos( β l ) = 0.070737leading to β l =1.5 and Yo = 0.

or Zo = 69.65Ω. Since β = ω ε eff / c , ε eff = 4