#### Transcript Lecture 3. Dynamic Response Second Order Systems

```AOSC 634
Air Sampling and Analysis
Lecture 3
Measurement Theory
Performance Characteristics of Instruments
Dynamic Performance of Sensor Systems
Response of a second order system to
A step change
A ramp change
Copyright Brock et al. 1984; Dickerson 2015
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Dynamic Response
Sensor output in response to changing input.
Dynamic Characteristics of
Second Order Systems
d X
dX
2
2
+ 2Vw n
+ wn X = wn XI
2
dt
dt
2
EQ I
• Where wn is the undamped natural frequency, a
constant (s-1).
• z is the damping ratio, a unitless constant.
• We must solve an initial value problem.
2
Solving the differential equation
We will use the technique of variation of parameters to find
complementary solutions.
We must assume a time dependence of the form ert and substitute this
into Eq I.
The characteristic equation is:
r + 2Vwnr + w = 0
2
2
n
Each root gives rise to a solution; there are four.
3
Four roots of the characteristic equation
1. z = 0
2. 0 < z < 1
3. z = 1
4. z > 1
Xc(t) = C sin(wnt + q)
Xc(t) = C exp (−wnzn t) sin(wmt + q)
Where wm = wn(1 – z2)½ {= wn within 5% for z < 0.3}
Xc(t) = exp (−wn t) (At + B)
Where A and B are constants.
XC (t) = e-Vwnt éë Ae+t/t m + Be-t/t m ùû
4
4. z > 1
XC (t) = e
-Vwnt
éë Ae+t/t m + Be-t/t m ùû
Where tm = 1/wm
And the characteristic time is 1/wm
In dimensionless time wnt = t”
Critically damped systems are an ideal; in the real world only
overdamped and underdamped systems exist. We will focus
on underdamped systems such as the dew pointer (or a car
first order”.
5
Time Response of second order systems.
input and output are zero. X(0) = XI(0) = X0
Their first derivatives are likewise zero at time
zero.
dX dX I
dt
=
dt
=0
We will proceed as with the first order system
assuming a step change. Using the dimensionless form.
1. z = 0, no damping. X’(t”) = 1 – cos (t”)
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Time Response of second order systems.
2. 0 < z < 1.0, underdamped.
X ' (t" ) = 1 
e t "
1


 cos 1  
1   
where q = cos 1   
2 1/ 2
2 1/ 2

2 1/ 2
t"q

3. z = 0, critically damped.
X’(t”) = 1 – e-t”(t” + 1)
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Time Response of second order systems.
4. z > 1.0, overdamped.
1 n 1 / 2t "
X ' (t" ) = 1  (n  1) e

n
n 1
e
n 1 / 2t "
1  (1    2 )1/ 2
where v =
1  (1    2 )1/ 2
The damping number is n.
n +1
With z =
see Figure 2-11 of Brock et al.
2 v
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Output 
Response of a second order system to
an a step increase of input.
Undamped
Underdamped
Dimensionless time t” = wnt
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An example with doubly normalized time
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Notes on Figure 2-11.
• For all z > 0 the final state is XI(t”) for t” > 0
– The slope is real, continuous, and near zero where t”
<< 1.0. Contrast with first order.
• For z = 0, undamped systems, there is free
oscillation at wn.
• For 0 < z < 1, there is damping at a frequency of:
wn (1- V 2 )1/2 = wm
The modified (damped) natural freq in Hz.
• For z << 1, there is large overshoot and a long
time lag. L
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Notes on Figure 2-11, continued.
• For z << 1, there is large overshoot and a long time lag. L
– The amplitude of the oscillations decreases exponentially with a
time constant of z-1.
The extrema can be found:
Where the sub e represents extrema.
The extrema come on time at pt”.
t"e =
np +1
1- V 2
Where n is a positive integer.
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Practical application
• From the amplitude of the first extreme
(assume here a maximum) we can calculate
the damping ratio z:
X 'max
- Vp
X 'e (t") =
= 1+ e
XI
-1 æ X 'max ö
V = ln ç
-1÷
p è XI
ø
13
• From the time (in units of t” of the first
extreme (assume here a maximum) we can
calculate the natural undamped frequency wn:
tmax = p
t " º w nt
wn =
p
tmax
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• Note, the closer to z is to unity, and the smaller wn, the faster
X’(t”) approaches XI.
• Example using Figure 2-11. Try this yourself with a mm ruler.
Let’s check the curve with z = 0.10 for the first maximum.
Looking at a paper copy,
X’(t”)max = 60 mm
X’(t)final = XI = 35 mm
æ 60 ö
V = ln ç -1÷ = 0.107
p è 35 ø
-1
Close to the 0.100 value in the book.
If the max amplitude is twice the input then
(2/1 – 1) is 1 and z =0.
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• Example using Figure 2-11, continued.
Let’s look for the natural frequency, wn. Let the
time of the first max be 30 s, an arbitrary value.
p
wn =
» 0.10s-1 or 0.1 Hz.
30s
To get within e-1 of the final value requires:
t” = 1/z
= 10 = wnt = 0.1t and t = 100 s!
In general, the time to e-1 is (wnz)-1 for z < 0.3.
For z > 0.3, use wm.
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Summary
• Although less intuitive than first order
systems, second order systems lend
themselves to analysis of performance
characteristics.
• A step change is in some ways a worst case
scenario for overshoot. Any second order