February 5, 2006
P.P.19.11
V1 − 0
R1
0 − V2
I1 =
R2
I1 =
Also,
⎯
⎯→ V1 = I 1 R 1
⎯
⎯→ V2 = - I 1 R 2
Comparing these with
V1 = z 11 I 1 + z 12 I 2
V2 = z 21 I1 + z 22 I 2
shows that
z 11 = R 1 ,
z 21 = - R 2 ,
z 12 = z 21 = 0
Hence,
⎡ R1
[z ] = ⎢
⎣ - R2
0⎤
0 ⎥⎦
Since Δ z = z 11 z 22 − z 12 z 21 = 0 , [z ]-1 does not exist . Consequently, [y ] does not exist .
P.P.19.12
This is a series connection of two two-ports.
For N a ,
For N b ,
z 12 a = z 21a = 20 ,
z 12 b = z 21b = 50 ,
Thus,
[z ] = [z a ] + [z b ]
z 11a = 20 − j15 ,
z 11b = 50 + j40 ,
z 22 a = 20 + j10
z 22 b = 50 − j20
⎡ 20 − j15
20 ⎤ ⎡50 + j40
50 ⎤
[z ] = ⎢
+⎢
⎥
20 + j10 ⎦ ⎣ 50
50 − j20 ⎥⎦
⎣ 20
⎡ 70 + j25
70 ⎤
[z ] = ⎢
70 − j10 ⎥⎦
⎣ 70
V2
z 12 Z L
=
Vs (z 11 + Z s )(z 22 + Z L ) − z 12 z 21
V2
(70)(40)
=
Vs (70 + j25 + 5)(70 − j10 + 40) − 4900
V2
2800
=
Vs 8250 − j750 + j2750 + 250 − 4900
V2
2800
=
= 0.6799∠ - 29.05°
Vs 3600 + j2000
P.P.19.13
We convert the upper T network N a to a Π network, as shown below.
25 S
-j5 S
j5 S
y 1 y 2 + y 2 y 3 + y 3 y 1 (-j5)(j5) + (j5)(1) + (1)(-j5)
=
= -j5
j5
y2
yb = 5 ,
y c = 25
ya =
For N a ,
y 12 a = -25 = y 21a ,
y 11a = 25 − j5 ,
y 22 a = 25 + j5
⎡ 25 − j5
- 25 ⎤
[y a ] = ⎢
25 + j5⎥⎦
⎣ - 25
For N b ,
y 12 b = j10 = y 21b ,
y 11b = 2 − j10 = y 22 b
⎡ 2 − j10
j10 ⎤
[y b ] = ⎢
2 − j10 ⎥⎦
⎣ j10
Since N a and N b are in parallel, [y ] = [y a ] + [y b ]
⎡ 27 − j15 - 25 + j10 ⎤
[y ] = ⎢
⎥S
⎣ - 25 + j10 27 − j5 ⎦
P.P.19.14
Convert the left Π network to a T network.
(20)(50)
(20)(30)
R1 =
= 6, R2 =
= 10 ,
100
20 + 30 + 50
R3 =
(30)(50)
= 15
100
Putting this network into the given network produces the network shown below. This
may be regarded as a cascaded connection of T two-port networks.
6Ω
15 Ω
60 Ω
40 Ω
10 Ω
20 Ω
Na
Nb
For N a ,
Aa = 1+
Ca =
6
= 1 .6 ,
10
1
= 0 .1 ,
10
⎛6⎞
B a = 15 + ⎜ ⎟ ( 25) = 30
⎝ 10 ⎠
15
Da = 1 +
= 2 .5
10
⎡1.6 30 ⎤
[Ta ] = ⎢
⎥
⎣0.1 2.5⎦
For N b ,
Ab = 1+
Cb =
40
= 3,
20
1
= 0.05 ,
20
⎛ 40 ⎞
B b = 60 + ⎜ ⎟ (80) = 220
⎝ 20 ⎠
60
Db = 1 +
=4
20
⎡ 3
220 ⎤
[Tb ] = ⎢
⎥
⎣ 0.05 4 ⎦
Hence,
220⎤
⎡1.6 30 ⎤ ⎡ 3
[T] = [Ta ][Tb ] = ⎢
⎥
⎢
⎥
⎣0.1 2.5⎦ ⎣0.05 4 ⎦
We can now use MATLAB to obtain T.
>> Ta=[1.6,30;0.1,2.5]
Ta =
1.6000 30.0000
0.1000 2.5000
>> Tb=[3,220;0.05,4]
Tb =
3.0000 220.0000
0.0500 4.0000
>> T=Ta*Tb
T=
6.3000 472.0000
0.4250 32.0000
472 Ω ⎤
⎡ 6.3
[T] = ⎢
32 ⎥⎦
⎣ 0.425 S
P.P.19.15
To obtain h11 and h 21 , simulate the schematic in Fig. (a) using PSpice.
(a)
Insert a 1-A dc current source to account for I 1 = 1 A . Also, include pseudocomponents
VIEWPOINT and IPROBE to display V1 and I 2 respectively. When the circuit is saved
and run, the values of V1 and I 2 are displayed on the pseudocomponents as shown in
Fig. (a). Thus,
I
V
h 21 = 2 = -0.6190
h11 = 1 = 4.238 Ω ,
1
1
To obtain h12 and h 22 , insert a 1-V dc voltage source at the output port to account for
V2 = 1 V . The pseudocomponents VIEWPOINT and IPROBE are included to display
V1 and I 2 respectively. After simulation, the schematic displays the results as shown in
Fig. (b).
V
I
h12 = 1 = -0.7143 ,
h 22 = 2 = -0.1429 S
1
1
(b)
Thus,
⎡ 4.238 Ω - 0.7143 ⎤
[h ] = ⎢
⎥
⎣ - 0.6190 - 0.1429 S ⎦
P.P.19.16
Insert a 1-A ac current source at the output terminals to account for
I 1 = 1 A . Include two VPRINT1 pseudocomponents to output V1 and V2 . For each
VPRINT1, set the attributes to AC = yes, PHASE = yes, and MAG = yes. In the AC
Sweep and Noise Analysis dialog box, set Total pt : 1, Start Freq : 60, and End Freq : 60.
The schematic is shown in Fig. (a).
(a)
Once the schematic is saved and run, the output results include :
FREQ
6.000E+01
VM($N_0002)
3.987E+00
VP($N_0002)
1.755E+02
FREQ
6.000E+01
VM($N_0003)
1.752E-02
VP($N_0003)
-2.651E+00
From this table,
z 11 =
V1
= 3.987 ∠175.5° ,
1
z 21 = 0.0175∠ - 2.65°
Similarly, insert a 1-A ac source at the output port with the two pseudocomponents in
place as in Fig. (a). The result is the schematic in Fig. (b).
(b)
When the schematic is saved and run, the output results include :
FREQ
6.000E+01
VM($N_0002)
1.000E-30
VP($N_0002)
0.000E+00
FREQ
6.000E+01
VM($N_0003)
2.651E-01
VP($N_0003)
9.190E+01
From this table,
z 12 =
V1
≅0
1
z 22 = 0.265∠91.9°
Thus,
⎤
⎡ 3.987 ∠175.5°
0
[z ] = ⎢
⎥Ω
⎣ 0.0175 ∠ - 2.65° 0.265 ∠91.9° ⎦
P.P.19.17
In this case, R s = 150 kΩ , R L = 3.75 kΩ .
h ie h oe − h re h fe = (6 × 10 3 )(8 × 10 -6 ) − (1.5 × 10 -4 )(200) = 18 × 10 -3
The gain for the transistor is given as,
- (200)(3750)
Av =
= Vo /Vb = –123.61
6000 + (18 × 10 - 3 )(3.75 × 10 3 )
To calculate the gain of the circuit we need to use,
–Vs + 150kIb + Vb = 0 or 0.002 = 150k(0.002/156k) – Vc/123.61
Vc = –9.506 mV which leads to the gain = –9.506/2 = –4.753
200
= 194.17
1 + (8 × 10 )(3.75 × 10 3 )
Ai =
-6
Z in = 150,000 + 6000 − (1.5 × 10 -4 )(194.17) ≅ 156 kΩ
150 × 10 3 + 6 × 10 3
(150 × 10 3 )(8 × 10 -6 ) − (1.5 × 10 -4 )(200)
156
=
kΩ = 128.08 kΩ
1.248 − 0.03
Z out =
P.P.19.18
Let D(s) = (s 3 + 4s) + (s 2 + 2)
Dividing both numerator and denominator by s 3 + 4s gives
2
3
s + 4s
H (s) =
s2 + 2
1+ 3
s + 4s
i.e.
y 21 =
-2
3
s + 4s
y 22 =
s2 + 2
s 3 + 4s
As a third order function, we can realize H (s) by the LC network shown in Fig. (a).
L3
L1
+
+
C2
V1
V2
−
1Ω
−
(a)
ZA =
1
s 3 + 4s
2s
= 2
=s+ 2
= s L3 + Z B
y 22
s +2
s +2
L3 = 1 H
ZB =
2s
s +2
2
L3
L1
ZB
C2
(b)
YB =
y22 = 1 / ZA
1
s2 + 2
1
1
=
= 0.5s + = s C 2 +
ZB
2s
s
YC
C 2 = 0.5 F
YC =
1
1
=
s L1 s
⎯
⎯→
L1 = 1 H
Hence,
L1 = 1 H ,
C 2 = 0 .5 F ,
L3 = 1 H