Instrumentation: Circuit Analysis

This document provides an overview of circuit analysis in the field of instrumentation. It covers topics such as Laplace transform, nodal analysis, RL circuits, and Salley key band pass filter. The document includes solved examples and explanations for better understanding. Students can find study material and assignments related to electrical circuit analysis at Desklib.

INSTRUMENTATION 1
INSTRUMENTATION
By Name
Course
Instructor
Institution
Location
Date

INSTRUMENTATION 2
QUESTIONONE
Draw Laplace transform circuit of the given circuit,
S is the complex frequency;
As for t ¿ 0: 124 (t) Short circuit
And assume circuit work in steady state So ;
I(0) = I (0) =i(o) =0omp
For t¿ 0 ; 124 (t) = 12 Volt
And from ;

INSTRUMENTATION 3
Apply nodal analysis at node 1;
V 1
12 + V 1
2 + V 1−12
6 = 0
V1 [ V 1
12 + 1
2 + 1
6 ] =2
V1= 2.666 volt
I(∞) = 2.666
2 amp
I(∞ ¿ =1.333 amps
As we know that inductor current i(t) for t¿ 0 for RL circuit with source given by .
I(t) = (∞ ¿ + ( i(0)- (∞ ¿ e
−t
τ
τ is time constant = L
Req

INSTRUMENTATION 4
Req is calculated across inductor by deactivating voltage source by short circuit as shown below;
Therefore τ =3
2 11

INSTRUMENTATION 5
So: i(t) = 1.333+(0-1.333) e
−t
3
2
Ui(t) = 1.333((1- e
−2 t
3 ) amp for t¿ 0
As ; V3Ω = L d
dt i(t)
= 3* d
dt ( 1.333*((1-1.333) e
−2 t
3
= 3*1.333(-(- 2
3) e
−2 t
3
V3 Ω=2.666 e
−2 t
3 for t ¿ 0
As i3 Ω= V 3 Ω
3 amp
= 2.666
3 e
−2 t
3
i3 Ω = 0.006e
−2 t
3

INSTRUMENTATION 6
Apply KCL at node a
i2= i(t) +i3 Ω
= 1.333(1-e
−2 t
3 ) + 0.006 e
−2 t
3
So
V= 2* i2 Ω [ 1.333(1- e
−2 t
3 ) +0.006 e
−2 t
3 ]
V(t) = 2.666(1-e
−2 t
3 )+1.772 e
−3 t
2 for t¿ 0
Question 2
V0= 10+5e-50t
Vo(0) = 10+5 =15
At t=0
Vc= Ce - t
R 1 ×1 0−6 +A ( comparing this with V0(t)
R1= 1
50× 10−6 = 20kΩ

INSTRUMENTATION 7
Vx= 1 k
1k + R 2 × Vo(t)
Vx=Vc(o+) = 1 k
1k + R 2 ×Vo(0+)
For t¿ 0 during steady state
1k + R 2
1 k =5
1k +R2 =5
R2=4kΩ
R2=4 Ω
For t¿ 0

INSTRUMENTATION 8
2=R1×1×10-6 dvc
dv +Vc
dvc
dv + 1
R 1 ×1 0−6
= 2
R 1 ×1 0−6
QUESTION 3
The circuit in S domain using Laplace

INSTRUMENTATION 9
Io(S) = Is(S)×(R// 1
CS )/ (R// 1
CS ) +LS
Io
Is = H(S)
R × 1
CS
R+ 1
CS
R × 1
CS + LS
= R
S2 RLC + SL+ R
=
1
LC
S2 RLC + S
RC + 1
LC
Characteristic equation S2 RLC + S
RC + 1
LC
Standard form is S2 + ω
Q S +ω 02 =0
S0, ω= 1
√ LC
ω
Q = 1
RC
Given ωo=100 Rads/ sec, Q=1 1
LC =1002 and 1
RC =100

INSTRUMENTATION 10
LC= 1
1002 = 10-4
RC=0.01
Choose C= 1μF
R= 0.01
C = 10k Ω
Given Is(t)= 2u(t)= 1 = s(S) = 2
S
H(s)= 1002
S2+ 100 S+10 02
= 2 ×104
S (S2 +100 S +10 04 )
= A
S + BS+C
S2+ 100 S+10 04
1002 = S2 +100 S +10 04 + (BS+C) S
Equating S2 coefficient
A+B
B=- A
100A+C =0 A=1
B=- A
C=- 100A = -100

INSTRUMENTATION 11
Io(S)= 1
S - S +100
S2+100 S+104
1
S - S+50+50
( S +50 )2 +¿ ¿
Inverse laplace transform
Io(t) = e−50 t cos (86.6 ¿+ 0.5774e−50 t sin(86.6 ¿ U(t)
QUESTION 4

INSTRUMENTATION 12
OVs
R 1 + 0−U
R 3 + O−Uy
R 2 =0
Uy( R 4+ R 5
R 4 R 5 ) = U 0
R5
Uy= U 0 R 4
R 4+ R 5
U 0 R 4
RR 4+5 × 1
R 2 = - U 0
R3 - U 5
R1
U0 ( 1
R 3 + 1
R 3 . 1
R 4+R 5 ) = −Us
R 1
U0 ( R 1
R 3 + R 1
R 3 . R 4
R 4+R 5 ) = −Us
R 1
So gain= Uo
Us =
−1
( R 1
R 3 + R 1
R 3 . R 4
R 4+R 5 )
For Uo= -20 Us
Consider ( R 1
R 3 = R 1
R 2 ) = K ( Say )
R4=R5
Gain =-20 =
−1
K +k . 1
2

INSTRUMENTATION 13
30 k = 1
K= 1
30
Therefore, R1
R 3 = R 1
R 2 = 1
30
And R 3
R1 = R 2
R 1 = 30
1
R1=10kΩ, R2=R3=30kΩ
R4=R5 = 10kΩ
QUESTION FIVE

INSTRUMENTATION 14
For an Ideal OPAMP
Vp=Vn and Ip =In =0
Kcl at node Vn
Vn
10 k + Vn−V
R + In =0
V= Vn ×[ 1+ R
10 k ]
V=Vn× 10 k
R +10 k
Vp=vn = Vn× 10 k
R +10 k

INSTRUMENTATION 15
Kcl node Vp
Vp−Vi
1 k + Vp
I /CS + Ip =0
( 10-3+SC) VP = 10-3 VI
(10-3 +SC) = 10 k
R +10 k × SL+SL
5 k V0 = 10-3 VI
Vo
VI = 5(R+10 k )
10 k ( SL+5 k)¿ ¿
Vo
VI = 5(R+10 k )
( 10 k ) LC ¿ ¿
15 ×1 06
( S+2000)( S+5000),
5 k
L =2000
L=2.5
10−3
C ,
C= 0.2 μF
5(R+10 k )
( 10 ) LC = 15×106
R=5 kΩ

INSTRUMENTATION 16
H(S) = 15 ×106
(S+200)( S+5000)
H(S) = 15 × 106
(S2 +7000 S +107 )
S2 +7000 S +107, and the standard form is S2 + ω
Q S +ω 02
=0
ω o2= 107
ω= 316.278 rad/sec
ω
2 π = f0 = 503.29
Quality factor Q= ω
7000 = 0.4517
Gain K= 15×106
QUESTION SIX

INSTRUMENTATION 17

INSTRUMENTATION 18

INSTRUMENTATION 19

INSTRUMENTATION 20
BODE PLOTE

INSTRUMENTATION 21
QUESTION 7

INSTRUMENTATION 22
Salley Key band Pass filter is

INSTRUMENTATION 23

INSTRUMENTATION 24

INSTRUMENTATION 25
The Salley key Band pass filter is as below;
C

INSTRUMENTATION 26
Bibliography
CHITRA, M., 2018. ELECTRICAL CIRCUIT ANALYSIS. 2nd ed. Hull: PHI Learning Pvt. Ltd.
Ergul, O., 2017. Introduction to Electrical Circuit Analysis. 1st ed. Chicago: John Wiley & Sons.