U niversity of S outhern
C alifornia
School Of Engineering
Department Of Electrical
Engineering
EE 402:
Homework Assignment #06
(Due 11/19/2003)
Fall, 2003
Choma
Problem #25:
By inspection:
(a)
Z1Z2
+Zo Z2
Z1
Zo   2
Z1
2
+Zo  Z2
2
Z
Z

Zin = 1 +Z2 ||  1 +Zo  & Zin =Zo (25.1)
2
 2

(25.2)
Z1  Z1

 Z1Z2
+Zo Z2 (25.3)
 Zo   +Zo  Z2  
2  2
2



Z12
Z 
Zo  Z1Z2 
Zo  Z1Z2 1  1  (25.4) {ANS}
4
 4Z2 
2
(b) When a cascade structure is in need, one can independently design each block
of the whole network; hence, one can simply cascade them to each other to
meet given requirement because each block from input will see the
characteristic impedance.
(c)
Z1 =sL Z2 = 1
sC
Z1Z2 = L
C
&
Z1
Z2
=s 2 LC (25.5)
2
L  
L
2
 s LC 
 
Zo =
1+ 
Ro =
ωc =
(25.6)


C 
2
C
LC

 
 
Zo =R o
  s 2 
1+    (25.7){ANS}
  ωc  


In (25.7), one can scrutinize that if the network operates much below the cut off (
ωc )frequency of the characteristic impedance, the input impedance will always be
resistive and equal to load resistance which is Ro. This means our lumped circuit
tries to be like a match terminated transmission line.
(d) Since the load termination is not exactly as the characteristic impedance Zo,
we must derive input impedance when output is terminated with low frequency
characteristic impedance. Thus:
sL
sL 1  sL

+
Zin = + ||  +R o  (25.8)  Zin 
2
2 sC  2

sL
+R o
2
(25.9)
s 2 LC
1
+sR o C
2
 sL s 2 LC s3L2 C 
s3L2 C s 2 R o LC
sL+R o +
+
1+ R + 2 + 4R 
o
o 
4
2
Zin =
= Ro 
(25.10)
s 2 LC
s 2 LC


1+
+sR o C
 1+ 2 +sR o C 
2


2
3

 s   s 
 s  
1+2 
 +2   +2   
ω c   ωc 
L
2


 ωc   (25.11)
=R o C= LC=  Zin =R o 
2

Ro
ωc 
 s   s 


1+2 
+2
  


 ωc   ω c 
  ω 2
 ω   ω 3  
1-2   +j2 
 -  
ω
ω

c
c



  ωc   
Zin (jω)

=
 (25.12)
2
Ro
ω
 ω 


1-2   +j2 



 ωc 
 ωc 


Zin
1
-2jy3
ω
Ro
where y=
(25.14)
11 
(25.13)  ρ11 =
2
3
Zin
2(1-2y )+2j(2y-y )
ωc
1
Ro
One can straightforwardly use matlab or a similar program to plot the imaginary
and real parts of this reflection coefficient. Hence, the result will be as in figure
(P.25d).
(e) In the network of fig.3.17b, every element has c (transmission)-parameters.
Also, the overall c-parameters of the network will be tantamount to
multiplication of these c parameter matrices. Hence,
 s 2 LC
 s 2 LC  
sL
sL
1+
+
1+


2
2
1 sL   1 0 1 sL 
2
2 


(25.16)
2
2  (25.15) C 
C




s 2 LC
0 1  sC 1 0 1 
 sC

1+
2


Figure 25d: The reflection coefficient vs. normalized frequency
V
Vo
Ro
=
(25.17) o =
Vi c11R o +c12
Vi

Vo
=
Vi
1
L
2
and
=R o C= LC=
(25.18)
2
L 2 LC 3 CL
Ro
ωc
1+s +s
+s
Ro
2
4R o
1
2
 s   s 
 s 
1+2 
 +2   +2  
 ωc   ωc 
 ωc 
3
(25.19)
(f)
Similarly, by using c matrix and the c-parameters based circuit analysis, one can
arrive at the following results:
V
Vo
Ro
=
(25.20) o =
Vs
Vs c11R o +c12   c21R o +c22  R o
Vo
=
Vs
12
(25.21)
 L R o C  2 LC 3 CL2
1+s 
+
+s
 +s
2 
2
8R o
 2R o
1
2
 s   s   s 
1+2 
 +2   +  
 ωc   ωc   ωc 
3
(25.22)
(g) The delay is negative derivative of phase of the transfer function with respect
to ω. Thus, one can do the subsequent mathematical manipulation:
 2y-2y3 
d
2 
 1-2y 
3
Vo
dφ
dy
-1  2y-2y 
 D(y)=
=φ=-tan 
D(y)=2
2 
Vs
dy
 2y-2y3 
 1-2y 
1 
2 
 1-2y 
D(y)=
4 y4  2 y2  2
1-2y    2y-2y 
2
2
3
2
(h) We have two specs, namely, D(0) =20psec and Ro=50. Thus, there are two
equations that relate these two specifications. Hence,
D(0)=
2
2
L
 ωc  1011 rad sec  ωc 
 1011 & R o 
 50 
ωc
C
LC
C=.1pF&L=1nH
And the simulation result will be as in the succeeding lines:
Phase Response
Magnitude Response
The simulation results agree with the analysis done sections (e) and (g). Also, one will
view in the magnitude response there is peeking because the relation from Vi to Vo is not
exactly a Butterworth relation. Nevertheless, it gives a flat delay within bandwidth of
interest.
Netlist
The simulation of the third order Bessel filter
* Lossless
.opt post
Vi 1 0 AC 1
L1 1 2 .5n
C1 2 0 .4p
L2 2 3 .5n
Rl 3 0 50
** use the following function find the phase and delay of
specific node
.plot AC VP(3) VT(3)
.op
.ac dec 100 0 100g
.end
Delay Response
Problem #26:
When we replace the impedance elements of expression derived in
section (a) previous question, we will get the characteristic impedance of the system.
Hence,
(a)



 4+ 1-m 2 s 2 LC   L 
 Z1Z2 = 
  
4
1-m 2 


  C 

1
Z1 =smL Z2 =
+
sL  
smC
4m
 Z1
s 2 m 2 LC
=
 4Z
2
2
 2 4+ 1-m s LC


Based on the previous derivation:
Zo =R o
2

 s  

  
  s 2  
ω
2 1-m 2
2m

1+    1+  2  2   Zo (0)=R o ω1 
& ω2 
  ω1  
LC
LC

  1+  s  




 ω1  

From the previous equation it follows that:
  s   s 
Zo =R o 1+   +  
  ω1   ω2 

2
2

  Zo =R o


ωc 


s2
1+
2

ω1ω2 


 ω 2 ω 2
1
2

ω1ω2
ω  ω2
2
1
2
=







2
LC
for ωc >>ω  Zo (s)  R o
Thus, for the cut-off frequency there will not be any improvement; however, after the
cut-off frequency one will observe better and sharper attenuation owing to the
resonance of L1 and mC elements.
(b) If one makes the similar analysis done in the section (e) and (g) of the
previous question, He will find out that :
D(0)=m LC=20pSec R o 
L
 50 Let m=.5 Hence,
C
mL=1nH mC=.4pF L1 =.177nH . Using these values, the following netlist is
set up and simulated, and delay phase and magnitude response graphs are
obtained.
Netlist
Improved k-constant T network
.opt post
Vi in 0 AC 1
L1 in 2 .5n
C1 2 3 .4p
L3 3 0 .177n
L2 2 out .5n
Rl out 0 1e14 ** Used to find input reflection coefficient
** use the following function find the phase and delay of specific node
.net V(out) Vi Rout=50 Rin=50 ** Used to find input reflection coefficient
.plot AC S11(I) S11(R)
.plot AC VP(out) VT(out)
.op
.ac dec 100 0 100g
.end
Magnitude Response
Delay Response
Phase Response
Input Reflection Coefficient
(c) Using the same netlist, the preceding graph is obtained. The important point is
about this graph that it is has zero reflection within a bandwidth. This simply
implies that our input port see the characteristic impedance for some
frequencies.
Problem #27:
In order to find, s-parameters of the network we need to terminate both input and
output ports with characteristic impedances. However, before doing so, we can
simply unearth z-parameters of the network and then convert z-parameters to sparameters. Since this is already done in the previous homework, conversion can
simply be referred. Thus,
Io: Output current into the network.
Vo: Output voltage with respect to ground
Ii: Input current into the network
Vi: Input voltage with respect to ground.
Vi =z11Ii +z12 Io
Vo =z 21Ii +z 22 Io
Let set Io to zero!
Ii =V(g m +g h ) 
Vi 1  rg (g m +g h )


 Vi  V 1  rg (g m +g h )  z11 
V=Vi  Ii rg 
Ii
(g m +g h )


 sCg  g m 
Vo

 z 21 
Ii sCg (g m +g h )
sCg Vo  V  sCg  g m  
Vg m  sCg  V-Vo 
Let set Ii to zero!

sCg +g h 
Vo


 sCg Vo  V  sCg +g h  z 22 
Vg h  sCg  Vo  V  
Io sCg (g m +g h )

Io =V(g m +g h )
Io =V(g m +g h )
Vi =V
z12 
Vi
1

Io (g m +g h )
From the previous homework we have:
S11 
 z11  R 0  R 0 +z 22   z12 z 21
 z11 +R 0  R 0 +z 22   z12 z 21
S21 
2R 0 z 21
2R 0 z12
S12 
 z11 +R 0  R 0 +z22   z12z21
 z11 +R 0  R 0 +z22   z12z21
Problem #28:
S22 
 z 22  R 0  R 0 +z11   z12 z 21
 z11 +R 0  R 0 +z 22   z12 z 21
2
ρ +G T  1
2
4R SR L
 R +R 
4R V
G T = S o = S 6L
R L Vs
1+ω
2
2
 R L  RS 
6

 ω
R +R
2
ρ   S L 6
1+ω
 R L  RS  3 

s 
R S +R L 


ω=-js ρ 
2s3 +2s 2 +2s+ 4 3
2
3 
1+2s+2s +s  ZINnormalized =
2s 2 +2s+ 2 3

1 ρ

ZINnormalized 
1 ρ

Applying the long division:
Zα =50Ω
ZINnormalized = s+
1
3
1
s+
2 2s+2
Also we have Cα =
Lα =
1
=1.33pF
Zα ωα
Zα
=3.32nH
ωα
Thus, the network will be as in the following figure:
Netlist
Butterworth Network
.opt post
Vi 1 0 AC 1
Rs 1 2 50
L1 2 3 3.32n
C 3 0 1.99p
L2 3 4 6.63n
Rl 4 0 100
.op
.ac lin 100 0 100g
.end
Magnitude Response