Dynamic Response - Ch 11
Chapter 11
Dynamic Behavior of
Measurement Systems
21.1
System Response
• you have studied 1st order systems before
(Diff. Eq., ME 372, ECE 320)
– 1 energy ____________ element
– 1 element which _______________________
energy
• a _________ - ___________ combination
• a “bulb in tube” thermometer
• a resistor - capacitor combination
Units - Resistance
R
Units - Capacitance
V
 ohms 
I
► From
ECE 320, for a simple resistor
V = IR
► V is the voltage drop (in volts)
volts)
► I is the current through the resistor (in
amps)
amps)
► R is the resistance (in ohms)
Units
volt amp  sec
RC  ohm * farad 
*
 sec
• if R has units of ohms
• and C has units of farads
• then  has units of seconds
 = RC
C  farad 
► From
ECE 320, for a simple capacitor
dV
C
I
dt
►C
or
V 
1
C
t
0 I dt  V( 0 )
is the capacitance (in farads)
First Order System
Resistor _______________ energy
Capacitor ___________ energy
Vin
Vout
dVout
1
1

Vout 
Vin
RC
RC
dt
Dynamic Response - Ch 11
System Response
► Equation
System Response
for a “generic” 1st order system:
► Solution
by
where
y = dependent variable (velocity, voltage, …)
for a step (constant) input is given
where
►
 = time constant
►
f(t) = forcing function (input)
►
1st Order System Step Response Characteristics
y is the limiting or final (steady(steady-state) value
y0 is the initial value at t=0
 is the time constant (units of seconds)
1st Order System Step Response Characteristics
= 80 units
► y0 = 20 units
► at t =  (one time constant),
► Response
will be _________% closer to the
final value after 1 time constant, 
► y
100
80
y  20  0.632  80  20  _______
60
40
20
Estimate the time
constant from the graph
0
0
1
2
3
4
5
Time, t (seconds)
100
Output, P
Output, P
21.2
80
60
40
20
0
0
1
2
3
Time, t (seconds)
4
5
Dynamic Response - Ch 11
1st Order System Step Response Characteristics
will be 63.2% closer to the final
value after 1 time constant, 
Output, P
► Response
1st Order System Step Response Characteristics
► y
= -60 units
► y0 = 40 units
► at
40
30
20
10
0
-10
-20
-30
-40
-50
-60
21.3
t =  (one time constant),
y  40  0.632  (60)  40  _________
0
0.2
0.4
0.6
0.8
1
Estimate the time
constant from the graph
1.2
Time, t
1st Order System Step Response Characteristics
line tangent to the response at t = 0 will
intersect the final value at t = 
Output, P
►A
40
30
20
10
0
-10
-20
-30
-40
-50
-60
0
0.2
0.4
0.6
0.8
1
Output, P
Time, t
40
30
20
10
0
-10
-20
-30
-40
-50
-60
0
0.2
0.4
0.6
Time, t
0.8
1
1.2
1.2
Dynamic Response - Ch 11
RC Time Constant
Theoretical Time Constant
4
Vin
3
+
Ei
Vout
2
1
Volts
21.4
R
+
Eo
i
C
-
-
a) What is the theoretical
time constant, RC, if
R = 47 k in parallel with
22 k, C = 0.01 F?
0
b) Sketch the response of the system (Eo) if
Ei is a step input from -3 volts to +9 volts
(y0=-3V, y∞=+9V), be sure to label both
axes of your sketch!
Input
Output
-1
-2
-3
0
0.02
0.04
0.06
0.08
0.1
Time, sec
1st Order System - Harmonic Response
► at
What if you give a 1st order system a
sinusoidal (harmonic) input?
“low” frequencies,
dy 1
 y yi sin t
dt 
Output, P
where
yi = amplitude of input
 = frequency of input
2.5
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-2.5
0
0.2
0.4
0.6
0.8
1
Time, t
5
Input and Output, volts
4
3.5
Is there any
Input
uncertainty Output
in the
time constant, ?
2
1.5
1
0.5
0
0.00
0.02
0.04
0.06
0.08
Time, sec
0.10
0.12
0.14
2.5
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-2.5
0
0.2
0.4
0.6
0.8
1
1.2
Time, t
y ( )  y   y0  y  e  /  4.70  1.25  4.70 e 1  3.43
4.5
3
1.2
“high” frequencies
Uncertainty in Time Measurements
Determine Time Constant from
Experimental Data
2.5
► at
Output, P
1st Order System - Harmonic Response
0.16
Time
sec
0.024
0.025
Input
volts
1.25
4.70
Output
volts
1.25
1.25
0.026
:
0.056
0.057
4.70
:
4.70
4.70
1.36
:
3.39
3.42
0.058
4.70
3.47
Dynamic Response - Ch 11
Spring - Mass - Damper
2nd Order System Response
► Typical
2nd order systems
f(t)
+v
 2 energy storing elements
►must
21.5
+x (measured from static
equilibrium position)
M
store ________________________ of energy!
 1 element which dissipates (or removes) energy
________ - ______ - damper combination
► a resistor - ________ - capacitor (RLC)
combination
x  v
►a
2nd Order System
1
 Kx  Bx  f (t )
M
 
 x  
 
 
 x  
 

x  

y  2
where:

1
 x   f(t )
M


1
 x   f(t )
M

Mechanical Oscillation
 1N  s 2 
 
X kg 
 kg  m 
y  F (t )
K
 natural frequency
M
Mechanical Oscillation
,m
Units of mass: kg
y 
B
 damping ratio
2 MK

n 
,k
Units of spring stiffness: N/m
N
 
k
m
  2 
m  N s 


 m 
1
 Kx  Bv  f (t ) 
M
“Generic” 2nd order system:
Rearrange the equation above,

x  

v 
Generic 2nd Order System
Substitute the definition for v into the second equation,
v  x 
B
K
,m
Units of weight: lb
W  mg  m 
W
g
,k
Units of spring stiffness: lb/in

k
m
Frequency
in
 lb 
 
k  in  1


m lb  s 2
k
k


m W g
 lb 
 
in
  
lb 
in 

s2 

1
s2
Dynamic Response - Ch 11
2nd Order Systems
► Three
2.0
possible solutions:
21.6
2nd Order System Response
Theoretical response - continuous
1.5
1.0
Output, y
 2 distinct, real roots
 2 repeated, real roots
 2 complex conjugate roots
0.5
0.0
-0.5
-1.0
-1.5
-2.0
0.0
0.2
0.4
0.6
0.8
1.0
Time, sec
2.0
2nd Order System Response
Measured response -___________________
1.5
Output, y
1.0
0.5
0.0
-0.5
-1.0
-1.5
-2.0
0.0
0.2
0.4
0.6
0.8
1.0
Time, sec
Time
sec
¦
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.2
¦
0.79
0.8
0.81
0.82
0.83
0.84
0.85
0.86
¦
Time
sec
¦
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.2
¦
0.79
0.8
0.81
0.82
0.83
0.84
0.85
0.86
¦
Data from SignalExpress
Output
volts
¦
0.514
1.110
1.542
1.752
1.715
1.440
0.968
0.368
¦
0.034
0.434
0.768
0.989
1.069
0.999
0.791
0.475
¦
“0” peak between 0.15 sec & 0.17 sec
Data from SignalExpress
Output
volts
¦
0.514
1.110
1.542
1.752
1.715
1.440
0.968
0.368
¦
0.034
0.434
0.768
0.989
1.069
0.999
0.791
0.475
¦
“0” peak at T0~0.16 sec
“4” peak at T4~0.83 sec
Uncertainty Analysis
n   d   
u
 T4  T0 0 - 4cycles- 1


T0
T4  T0 2

“4” peak between 0.82 sec & 0.84 sec
2
 - 4cycles

T4



4cycles  uT
 uT


T0  T4  T0 2  4cycles 
0
2
  uT0    uT4 
  

 

 T0    T4  
4 cycles
T4  T0
0
 T T 
0 
 4
 uT

T4  T4  T0
4
Dynamic Response - Ch 11
2nd Order System Response –
Log Decrement
2.0
Estimate of Damping Ratio, , from
“Log Decrement”
Log decrement, , is the natural log (ln) of the ratio of
two successive “peak-to-peak” amplitudes :
1.5
1.0
 An 
  log decrement
 A n 1 
0.5
  ln
0.0
-0.5
Damping ratio, , can be estimated from:
-1.0
 
-1.5
-2.0
0.0
0.2
0.4
0.6
0.8

4 2   2
1.0
Time, sec
Time
sec
¦
0.32
0.33
0.34
¦
0.40
0.41
0.42
¦
0.49
0.50
0.51
¦
0.57
0.58
0.59
In-Class Exercise
Data from SignalExpress
Output
volts
¦
1.444
1.559
1.455
¦
-1.275
-1.451
-1.421
¦
1.328
1.365
1.212
¦
-1.195
-1.291
-1.205
Work in groups of two students to find the natural frequency
and the damping ratio from this 2nd order system response:
3.0
2.0
An = 1.559-(-1.451) = 3.010
1.0
Output, y
Output, y
21.7
0.0
-1.0
An+1 = 1.365-(-1.291) = 2.656
-2.0
-3.0
0.00
0.25
0.50
0.75
1.00
Time, sec
1.25
1.50
1.75
2.00