AMME 2700 Instrumentation Assignment Solution - Circuits Analysis

This document presents a comprehensive solution to an instrumentation assignment, addressing several key concepts in electrical circuit analysis. The solution begins with the application of Laplace transforms and nodal analysis to determine circuit behavior, followed by calculations involving inductor currents and voltage drops. The assignment then delves into circuit analysis in the S-domain, including characteristic equations, standard forms, and the application of inverse Laplace transforms. Further, the solution explores operational amplifier (op-amp) circuits, calculating component values and analyzing circuit behavior. It covers the analysis of op-amp circuits, including gain calculations and component determination, and culminates with Bode plots and the design of a Salley Key bandpass filter, providing a complete and detailed solution to the assignment questions.

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