π
<-

## REVARES2

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Catégorie :Category: mViewer GX Creator App HP-Prime
Auteur Author: rafaellefrio
Type : Application
Page(s) : 6
Taille Size: 17.73 Ko KB
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### Description

Chapter 16, Solution 96.

If Vo is the voltage across R, applying KCL at the non-reference node gives
Vo Vo 1 1
Is sC Vo sC V
R sL R sL o

Is sRL Is
Vo
1 1 sL R s 2 RLC
sC
R sL

Vo sL I s
Io 2
R s RLC sL R

Io sL s RC
H (s ) 2 2
Is s RLC sL R s s RC 1 LC

The roots
-1 1 1
s1, 2 2
2RC (2RC) LC
both lie in the left half plane since R, L, and C are positive quantities.

Thus, the circuit is stable.
Chapter 16, Solution 97.

3 1
(a) H1 (s) , H 2 (s )
s 1 s 4

3
H (s ) H 1 (s ) H 2 (s )
(s 1)(s 4)

A B
h(t ) L-1 H (s) L-1
s 1 s 4

A 1, B -1

h (t ) (e -t e -4t ) u(t )

(b) Since the poles of H(s) all lie in the left half s-plane, the system is stable.
Chapter 16, Solution 98.

Let Vo1 be the voltage at the output of the first op amp.

Vo1 1 sC 1 Vo 1
,
Vs R sRC Vo1 sRC

Vo 1
H (s )
Vs s R 2C 2
2

t
h (t )
R C2 2

lim h ( t ) , i.e. the output is unbounded.
t

Hence, the circuit is unstable.
Chapter 16, Solution 99.

1
1 sL sL
sL || sC
sC 1 1 s 2 LC
sL
sC

sL
V2 1 s 2 LC sL
V1 sL 2
s RLC sL R
R 2
1 s LC
1
V2 s
RC
V1 1 1
s2 s
RC LC

Comparing this with the given transfer function,
1 1
2 , 6
RC LC

1
If R 1 k , C 500 F
2R
1
L 333.3 H
6C
The circuit is transformed in the s-domain as shown below.

1/sC 2

1/sC 1 R2

Vi
Vo
R1 +

1
R1
1 sC1 R1
Let Z1 R 1 //
sC1 1 1 sR1C1
R1
sC1

1
R2
1 sC2 R2
Z 2 R 2 //
sC2 R 1 1 sR2C2
2
sC2
This is an inverting amplifier.

R2 1 1
s s
Vo Z2 1 sR2C2 R2 R1C1 R1C1 C1 R1C1
H (s)
Vi Z1 R1 R1 R2C2 s 1 C2 s 1
1 sR1C1 R2C2 R2C2
Comparing this with
( s 1000)
H (s )
2( s 4000)
we obtain:
C1
1/ 2 C2 2C1 20 F
C2
1 1 1
1000 R1 100
R1C1 1000C1 10 x10 x10 6
3

1 1 1
4000 R2 12.8
R2C2 4000C2 4 x10 x 20 x10 6
3
Chapter 16, Solution 101.

We apply KCL at the noninverting terminal at the op amp.
(Vs 0) Y3 (0 Vo )(Y1 Y2 )
Y3 Vs - (Y1 Y2 )Vo
Vo - Y3
Vs Y1 Y2

Let Y1 sC1 , Y2 1 R 1 , Y3 sC 2
Vo - sC 2 - sC 2 C1
Vs sC1 1 R 1 s 1 R 1C1

Comparing this with the given transfer function,
C2 1
1, 10
C1 R 1 C1

If R 1 1k ,
1
C1 C2 100 F
10 4

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95 octets bytes appslist.txt

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