Engineering System Investigation Process
Engineering
System
Investigation
Process
LRC Circuit
Electrical
System
START HERE
Physical
System
System
Measurement
Parameter
Identification
Physical
Model
Mathematical
Model
Measurement
Analysis
Mathematical
Analysis
Comparison:
Predicted vs.
Measured
YES
Design
Changes
Is The
Comparison
Adequate ?
NO
Foundations of Engineering
1
Physical System for Investigation
iin
iout
L
R
ein
C
eout
LRC Circuit
Electrical System
Foundations of Engineering
2
Physical Modeling
• Simplifying Assumptions
– Resistor and Capacitor are pure and ideal; Inductor is ideal
but has resistance
– Voltage source is ideal and supplies the intended voltage to
the circuit no matter how much current (and thus power)
this might require
– Measuring device is ideal and does not load the circuit by
drawing any current
iout = 0
iR
L
iL
R
ein
Foundations of Engineering
iC
C
eout
3
Model Parameter Identification
• Measure component values using the DMM by
connecting the components across the Current HI and
To DMM
Current LOW inputs
To DMM
Current HI
Current HI
R
• LRC Circuit
– R = Potentiometer (500 Ω)
– L = 22 mH (nominal)
– C = 0.1 μF (nominal)
Foundations of Engineering
C
To DMM
Current LOW
To DMM
Current LOW
To DMM
Current HI
L
To DMM
Current LOW
4
Mathematical Modeling of System
iR
iout = 0
L
iL
R
ein
iC
C
eout
Basic Component
Equations
(Constitutive Equations)
KVL
di L
dt
eR  iR R
eL  L
deC
iC  C
dt
ein  eR  eL  eC  0
di
ein  Ri  L  eout  0
KCL i R  i L  i C i out  0
dt
d  deout 
 deout 
2
ein  R  C
  L C
  eout  0
LCD
 RCD  1 eout  ein

dt  dt 
 dt 
eout
1
d 2 eout
deout

LC 2  RC
 eout  ein
2
e
LCD
 RCD  1
in
dt
dt
Foundations of Engineering
5
Another Approach: Impedance + Voltage Divider
e  iR
R
Impedance:
ein
i
1/CD
LD
di
e  L  LDi
dt
de
i  C  CDe
dt
e
R
i
e
 LD
i
e
1

i CD
1
CD
eout
eout
1


ein R  LD  1
LCD 2  RCD  1
CD
 LCD2  RCD  1 eout  ein
d 2 eout
de
LC 2  RC out  eout  ein
dt
dt
Foundations of Engineering
6
Mathematical Analysis and Prediction
• Analysis & Prediction
• LabVIEW Prediction
d 2 eout
deout
LC 2  RC
 eout  ein
dt
dt
eout
1

ein LCD2  RCD  1
L = 22 mH
Foundations of Engineering
C = 0.1 μF
Differential Equation
Transfer Function
R = 100, 200, 300, 400 Ω
7
MatLab Step and Frequency Response Plots
Unit Step Response
LRC Circuit
Frequency Response
LRC Circuit
Foundations of Engineering
8
LabVIEW
Control
Design
Toolkit
Analysis
&
Prediction
Transfer Function
1
LCD2  RCD  1
Step Response
&
Frequency
Response
Foundations of Engineering
9
2nd-Order
Dynamic
System Model
2
n =
a0
 undamped natural frequency
a2
a1

 damping ratio
2 a 2a 0
d q0
dq 0
a 2 2  a1
 a 0q 0  b0qi
dt
dt
b0
1 d 2 q 0 2 dq 0
KS 
 steady-state gain


q

K
q
a0
0
S i
2
2
n dt
n dt
qo
KS

1
2
qi
D

D 1
2
n
n
Step Response
of a
2nd-Order System
Foundations of Engineering
10
1.8
tr 
rise time
n
Unit Step Response
of a
2nd – Order
Dynamic System
Foundations of Engineering
4.6
ts 
settling time
n

Mp  e
1 2
 0    1
 

 1 

0.6


overshoot
 0    0.6 
11
Frequency Response
of a
2nd – Order
Dynamic System
Foundations of Engineering
-40 dB per decade slope
12
Some Observations
• When a physical system exhibits a natural
oscillatory behavior, a 1st-order model (or even a
cascade of several 1st-order models) cannot
provide the desired response. The simplest model
that does possess that possibility is the 2nd-order
dynamic system model.
• This system is very important in control design.
– System specifications are often given assuming that the
system is 2nd order.
– For higher-order systems, we can often approximate the
system with a 2nd-order transfer function.
Foundations of Engineering
13
• Damping ratio  clearly controls oscillation;  < 1 is
required for oscillatory behavior.
• The undamped case ( = 0) is not physically realizable
(total absence of energy loss effects) but gives us,
mathematically, a sustained oscillation at frequency n.
• Natural oscillations of damped systems are at the
damped natural frequency d, and not at n.
d  n 1   2
• In hardware design, an optimum value of  = 0.64 is
often used to give maximum response speed without
excessive oscillation.
• At  = n, the phase angle is exactly -90º.
Foundations of Engineering
14
• Undamped natural frequency n is the major factor
in response speed. For a given  response speed is
directly proportional to n.
• Thus, when 2nd-order components are used in
feedback system design, large values of n (small
time lags) are desirable since they allow the use of
larger control effort before stability limits are
encountered.
• For frequency response, a resonant peak occurs for
 < 0.707. The peak frequency is p and the peak
amplitude ratio depends only on .
p  n 1  2
Foundations of Engineering
2
peak amplitude ratio 
KS
2 1   2
15
• Bandwidth
– The bandwidth is the frequency where the amplitude ratio drops
by a factor of 0.707 = -3dB of its gain at zero or low-frequency.
– For a 1st -order system, the bandwidth is equal to 1/ .
– The larger (smaller) the bandwidth, the faster (slower) the step
response.
– Bandwidth is a direct measure of system susceptibility to noise,
as well as an indicator of the system speed of response.
– For a 2nd-order system:
BW  n 1  2 2  2  4 2  4 4
– As  varies from 0 to 1, BW varies from 1.55n to 0.64n. For
a value of  = 0.707, BW = n. For most design considerations,
we assume that the bandwidth of a 2nd-order system can be
approximated by n.
Foundations of Engineering
16
Summary
Transfer Function
eout
1

ein LCD 2  RCD  1
KS
=
1 2 2
D 
D 1
2
n
n
Frequency
Domain
Performance
1.8
tr 
n
4.6
ts 
n
Time Domain
Performance

Mp  e
1
2
Foundations of Engineering
Hardware
Parameters
Model
Parameters
1
R C
n 

KS  1
LC
2 L
p  n 1  2 2
peak amplitude ratio 
KS
2 1   2
d  n 1   2
BW  n 1  2 2  2  4 2  4 4
17
L
ein
R
i
C
eout
R = 200 Ω
C = 0.1 μF
L = 22 mH
Unit Step Response
1
n 
 21320 rad / sec
LC
 3393 Hz
R C
 0.213
2 L
KS  1

tr 
1.8
 0.084 m sec
n
ts 
4.6
 1.01 m sec
n

Mp  e
Foundations of Engineering
1 2
 0.504
18
Frequency Response
-40 dB per decade slope
p  n 1  2 2 =20330 rad/sec  3236 Hz
peak amplitude ratio 
KS
2 1  
2
 2.403  7.61 dB
d  n 1   2  20830 rad / sec  3315 Hz
BW  n 1  2 2  2  4 2  4 4  32057 rad / sec  5120 Hz
Foundations of Engineering
19
L
R
LRC Electrical System
K
B
fo
ein
i
C
eout
M
Mass-Spring-Damper
Mechanical System
+v
fi
ein  e L  e R  e C  0
fi  f K  f B  f M  0
di
ein  L  Ri  e out  0
dt
d  deout 
 de out 
ein  L  C
  R C
  e out  0
dt  dt 
 dt 
d 2 eout
de out
LC 2  RCdt
 eout  ein
dt
dt
eout
1

ein LCD 2  RCD  1
f i  Kx  Bv  M v  0
Foundations of Engineering
f 
f 
fi  fo  B  o   M  o   0
K
K
 
 
M
B
fo  fo  fo  fi
K
K
fo
1

fi M D2  B D  1
20
K
K
2nd-Order
Dynamics Systems
a0
n 
 undamped natural frequency
a2
a1
d 2q 0
dq 0

 damping ratio
a 2 2  a1
 a 0q 0  b0qi
2 a 2a 0
dt
dt
b0
1 d 2 q 0 2 dq 0
 steady-state gain

 q 0  K Sq i K S 
2
2
a0
n dt
n dt
Electrical System
Mechanical System
d eout
de out
LC 2  RC
 eout  ein
dt
dt
1
R C
n 

KS  1
LC
2 L
M d 2 f o B df o

 fo  fi
2
K dt
K dt
K
B 1
n 

KS  1
M
2 KM
2
Foundations of Engineering
21
Measurements Using LabVIEW
• LRC Circuit
– Step Response using LabVIEW
– Frequency Response using LabVIEW
Foundations of Engineering
22
Foundations of Engineering
23
Function
Generator
FUNC OUT
ein
-15 V
1
8
2
7
3
6
4
5
Time Response
ELVIS Connections
+ 15 V
eout
R
Buffer Op-Amp
L
Oscilloscope
Channel B+
Power Ground
Oscilloscope
Channel B-
C
Power Ground
Foundations of Engineering
24
Foundations of Engineering
25
Function
Generator
FUNC OUT
ein
-15 V
1
8
2
7
3
6
4
5
Frequency Response
ELVIS Connections
+ 15 V
eout
R
Buffer Op-Amp
Analog Input
Signal
ACH1+
L
Power Ground
Analog Input Signals
ACH0- and ACH1-
Analog Input
Signal
ACH0+
C
Power Ground
Foundations of Engineering
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Foundations of Engineering
27