U niversity of S outhern
C alifornia
School Of Engineering
Department Of Electrical
Engineering
EE 402:
Homework Assignment #04
(Due 10/13/2003)
Fall, 2003
Choma
Problem #16:
(a)
In this part, we have three equations with three unknowns, namely, L, C,
Rc which are correlated. By solving these three equation, we will arrive an
unique answer as follows:
Zin (s ) =
sL + Rc + RL
1 + s LC + sC ( Rc + RL )
2
ω0 =
QL =
1
L
1
⇒ Zin ( jω0 ) =
+
C ( Rc + RL ) jω0C
LC
ω0 L
1
2
2
2
⇒ ( ω0 L ) = ( QL Rc ) ⇒ 2 2 = ( Q L Rc ) = 16 Rc 2
Rc
ω0 C
L
1
= 2 2 = RS ( Rc + RL ) = 50 ( 20 + Rc )
C ω0 C
50 ( 20 + Rc ) = 16Rc
(b)
2
è
ANSè
Netlist
Up-Conversion Matching Filter
.opt post
Vin 1 0 Ac 1
Rs 2 1 50
C 2 0 4.136p
L 2 3 6.12n
Rc 3 4 9.613
Rl 4 0 20
.ac dec 100 0 4g
.end
Rc = 9.62 O
C = 4.14 pF
L = 6.12 nH
Problem #17:
(a)
The same method as in the previous problem shall be applied. Therefore:
ω0 =
( Rc + RL )
LC
{( R
L=
c
RL
& Reff=
+ RL ) RL }
ω02C
⇒ L2
L
+ Rc = R S
CRL
{( R
=
c
+ RL ) RL }
2
ω0 4C 2
{( Rc + RL ) RL } = ( Q R )2 = 25R 2
ωL
2
2
QL = 0 ⇒ ( ω0 L ) = ( QL Rc ) ⇒
L c
c
Rc
ω02C 2
2
L {( Rc + RL ) RL }
=
= RL ( RS − Rc )
C
ω02C 2
( QL Rc )
2
= RL ( RS − Rc ) {( Rc + RL ) RL } = ( RS − Rc )( Rc + RL )
25 Rc 2 = ( 50 − Rc )( 200 + RL ) ⇒ R c = 16.94O L = 13.48nH C = 1.88pF ANS
(b)
Netlist
Down-Conversion Matching Filter
.opt post
Vin 1 0 Ac 1
Rs 2 1 50
C 4 0 1.88p
L 2 3 13.48n
Rc 3 4 16.94
Rl 4 0 200
.ac dec 100 0 4g
.end
Problem #18:
(a) In this problem, there are two equations but three unknowns. Hence, we
need to pick up a reasonable value for one of the unknown parameters. Thus:
 C 
R S =R L  1 + B 
 CA 
2
ω0 =
1
C C
where Ceff = A B
C A + CB
LCeff
Let L=
100
nH=2.533 nH è Ceff = 2.5 pF ç ANS
2
4π
2
 C 
C
75=25  1 + B  ⇒ B = 3 − 1 ⇒ CA =
CA
 CA 
3
Ceff
3 −1
CA =5.92 pF CB = 4.33pF ç ANS
Netlist
(b)
C-Tapped Matching Filter
.opt post
Vin 1 0 Ac 1
Rs 2 1 75
CB 3 0 4.33 p
CA 2 3 5.92 p
L 2 0 2.533n
Rl 3 0 25
.ac dec 100 1 4g
.end
Problem #19:
(a)
By inspection, input impedance can be expressed as in the following
equation.


RL
Zin ( s ) = (1 sC A ) //  sL +

1 + sRL CB 

Zin ( s ) =
(b)
(1 + s LC
(
) + sR (C
sL + RL 1 + s 2 LCB
2
A
L
B
)
+CA + s 2 LCB C A
)
ç ANS
When we set the resonance at the specified frequency, we shall arrive at
following results.
C 
Zin ( jω01 ) =
+ RL  B 
2
 CA

 CA 
( jω01 ) 

 ( CB +CA ) 
1
2
á ANS
This impedance is series connection of a resistor and capacitor.
(c)
Zin ( jω02 ) =
1 − CB
C A L CA
+
( jω02 ) CA RL ç ANS
This impedance is series combination of a resistor and, capacitor or inductor
because depending on values of CB and CA we may have either inductor or
capacitor.
(d)
CB = CA ⇒ Zin ( jω02 ) =
L=
(e)
1
ω0 C
2
⇒
L CA
RL ç ANS
L
1
1
= 2 2 = RL RS ⇒ C =
C ω0 C
ω0 RL RS
Based on the previous derivation, one can easily compute first the value of
capacitors, and then the value of inductor.
(i)
C=1.13 pF
L=5.63 nH
(ii)
C=1.5 pF
L=16.88 nH
(f)
Netlist
PI-(for Up-Conversion) Matching Filter
.opt post
Vin 1 0 Ac 1
Rs 2 1 100
CB 2 0 1.13p
CA 3 0 1.13p
L 2 3 5.63n
Rl 3 0 50
.ac dec 100 1 4g
.end
Netlist
PI-(for Down-Conversion) Matching Filter
.opt post
Vin 1 0 Ac 1
Rs 2 1 75
CB 2 0 1.5p
CA 3 0 1.5p
L 2 3 16.88n
Rl 3 0 150
.ac dec 100 1 4g
.end
Wave
Symbol
D0:A0:A
Result (lin)
60
55
50
45
40
35
30
25
20
15
10
5
0
0
up-conversion matching filter
3g
Problem 16
Zin @ resonant
frequency is
matched with Rs
2g
Frequency (lin) (HERTZ)
Current X=1.00e+09
Current Y=4.99e+01
Derivative=-9.70e-08
1g
4g
Wave
D0:A0:Z
Symbol
Result (lin)
220
210
200
190
180
170
160
150
140
130
120
110
100
90
80
70
60
50
0
Current X=1.00e+09
Current Y=4.90e+01
Derivative=-1.79e-08
up-conversion matching filter
2g
Frequency (lin) (HERTZ)
Problem 17
@resonant
frequency Zin is
matched with Rs
1g
3g
4g
Wave
D0:A0:Z
Symbol
Result (lin)
85
80
75
70
65
60
55
50
45
40
35
30
25
20
15
10
5
0
0
1g
c-tapped matching filter
Current X=2.00e+09
Current Y=7.54e+01
Derivative=-1.16e-07
2g
Frequency (lin) (HERTZ)
Problem18
@resonant
frequency Zin is
75 ohm that is
equal to Rs
3g
4g
Wave
Symbol
D0:A0:Rin
D0:A0:Xin
Result (lin)
120
100
80
60
40
20
0
-20
-40
-60
-80
-100
0
@2gHz Zin=Rl
Real PART
Imaginary PART
2g
Frequency (lin) (HERTZ)
Current X=2.00e+09
Current Y=-1.16e-01
Derivative=-7.95e-08
Current X=2.00e+09
Current Y=1.00e+02
Derivative=1.01e-07
pi-(for up-conversion) matching filter
PROBLEM19 F i
Rl=50 Rs=100
1g
3g
4g
Wave
Symbol
D0:A0:Rin
D0:A0:Xin
Result (lin)
200
180
160
140
120
100
80
60
40
20
0
-20
-40
-60
-80
-100
-120
-140
-160
-180
3g
PROBLEM 19 f ii
pi-(for down-conversion) matching filter
Observe @ resonant real part is 75 ohms
2g
Frequency (lin) (HERTZ)
Current X=1.02e+09
Current Y=-5.82e-02
Derivative=-1.64e-09
Current X=1.00e+09
Current Y=7.58e+01
Derivative=1.64e-07
1g
Observe @ resonant imaginary is zero
0
4g