3/25/2009
Example Signal Flow Graph Analysis
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Example: Analysis Using
Signal Flow Graphs
Below is a single-port device (with input at port 1a) constructed
with two two-port devices ( Sx and Sy ), a quarter wavelength
transmission line, and a load impedance.
j2
A=λ 4
Z0
Z0
Sx
port 1x
(input)
port 2x
Γ L = 0 .5
Sy
port 1y
port 2y
Where Z 0 = 50Ω .
The scattering matrices of the two-port devices are:
⎡0.35 0.5 ⎤
Sx = ⎢
0 ⎥⎦
⎣ 0.5
⎡ 0 0.8 ⎤
Sy = ⎢
⎥
⎣0.8 0.4 ⎦
Likewise, we know that the value of the voltage wave incident on
port 1 of device Sx is:
a1x
Jim Stiles
V01+x ( z1x = z1xP )
j2
j 2
=
=
V
5
50
Z0
The Univ. of Kansas
Dept. of EECS
3/25/2009
Example Signal Flow Graph Analysis
2/6
Now, let’s draw the complete signal flow graph of this circuit,
and then reduce the graph to determine:
a) The total current through load ΓL .
b) The power delivered to (i.e., absorbed by ) port 1x.
The signal flow graph describing this network is:
S21x
a1x
S11x
b2x
a1y
S21y
e
ΓL
S12y
−j βA
a2x
b2y
S22y
S11y
S22x
S12x
b1x
e −j βA
a2y
b1y
Inserting the numeric values of branches:
a1x = j
2
5
0.5
0.35
b1x
Jim Stiles
b2x
−j
a1y
0.8
0.4
0.0
0.0
0.5
−j
a2x
b2y
0.5
0. 8
b1y
The Univ. of Kansas
a2y
Dept. of EECS
3/25/2009
Example Signal Flow Graph Analysis
3/6
Removing the zero valued branches:
a1x = j
2
0.5
5
b2x
a1y
−j
0.8
0.4
0.35
0.5
b1x
−j
And now applying “splitting” rule 4:
2
5
0.5
b2x
−j
a2x
Followed by the “self-loop” rule 3:
a1x = j
2
5
0.8
b2y
( 0.4 ) 0.5 = 0.2
0.5
b1x
a2y
b1y
a1y
−j
0.35
0.5
0.8
a2x
a1x = j
b2y
0.5
b2x
−j
0.5
0.8
a2y
b1y
a1y
0. 8
= 1. 0
1 − 0.2
b2y
0.35
b1x
Jim Stiles
0.5
−j
a2x
0.5
0.8
b1y
The Univ. of Kansas
a2y
Dept. of EECS
3/25/2009
Example Signal Flow Graph Analysis
4/6
Now, let’s used this simplified signal flow graph to find the
solutions to our questions!
a) The total current through load ΓL .
The total current through the load is:
I L = −I ( z 2y = z 2yP )
V02+ y ( z 2y = z 2yP ) −V02− y ( z2y = z2yP )
=−
=−
=
Z0
a2y − b2y
Z0
b2y − a2y
50
Thus, we need to determine the value of nodes a2y and b2y. Using
the “series” rule 1 on our signal flow graph:
a1x = j
2
5
− j 0.5
b2y
0.35
b1x
0.5
− j 0.4
Note we’ve simply
ignored (i.e.,
neglected to plot)
the node for
which we have no
interest!
a2y
From this graph we can conclude:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
3/25/2009
Example Signal Flow Graph Analysis
5/6
⎛j 2⎞
⎟⎟ = 0.1 2
⎝ 5 ⎠
b2y = − j 0.5 a1x = − j 0.5 ⎜⎜
and:
(
)
a2y = 0.5 b2y = 0.5 0.1 2 = 0.05 2
Therefore:
IL =
b2y − a2y
=
50
( 0.1 − 0.05 )
2
50
=
0.05
= 10.0 mA
5
b) The power delivered to (i.e., absorbed by ) port 1x.
The power delivered to port 1x is:
Pabs = P + − P −
=
V1x+ ( z1x = z1xP )
2Z 0
2
=
a1x − b1x
2
−
V1x− ( z1x = z1xP )
2
2Z 0
2
2
Thus, we need determine the values of nodes a1x and b1x. Again
using the series rule 1 on our signal flow graph:
a1x = j
2
5
0.35
−0.1
b1x
Jim Stiles
Again we’ve
simply ignored
(i.e., neglected to
plot) the node for
which we have no
interest!
The Univ. of Kansas
Dept. of EECS
3/25/2009
Example Signal Flow Graph Analysis
6/6
And then using the “parallel” rule 2:
a1x = j
2
5
0.25 = 0.35 − 0.1
b1x
Therefore:
b1x = 0.25 a1x = 0.25 ( j
2
5
) = j 0.05
2
and:
Pabs =
Jim Stiles
j
2
2
5
− j 0.05 2
2
2
=
0.08 − 0.005
= 37.5 mW
2
The Univ. of Kansas
Dept. of EECS