CHAPTER 6
INTRODUCTION TO SYSTEM IDENTIFICATION
Broadly speaking, system identification is the art and science of using measurements obtained
from a system to characterize the system. The characterization of the system is usually in some
mathematical form. The limited cases considered here will use differential equations, in
particular, first and second order differential equations. When the form of the differential
equation is known the system identification problem is reduced to that of parameter
identification.
Present industrial practice presents several situations where system identification is used. An
important application is in industrial controls. Before a controller can be designed some things
must be known about the system which is to be controlled. Many systems do not lend themselves
to modeling and the most effective way to find out about the system is to make measurements
and apply the methods of system identification. The use of the methods covered in this course
and even more sophisticated methods such as finite element methods for modeling real
engineering systems, even simple ones, yield only approximate results and the models must be
adjusted using data obtained from the system. For most mechanical systems there are no
analytical methods for predicting system damping so that engineering judgment or system
identification methods must be used.
The measurements which are used for system identification can arise in one of several ways.
For large systems such as a building, ambient data is used. That is, natural excitations such as
wind, are used to excite the system. Even for uncontrolled random excitations such as this,
spectra that show the average distribution of response signal power as a function of frequency
can be used to identify system characteristics. These methods will not be discussed further here.
We will use several controlled inputs to give system responses which are easier to analyze. These
would include a step input (such as a sudden change in temperature of a thermo system), a snap
back (such as deflecting a spring-mass system and then suddenly releasing it), an impulse (such
as striking a spring-mass system with a sharp blow), or sinusoidal input. The selection of which
input to use is a function of your ability to generate the input and record and analyze the
response.
These notes will only cover 1st and 2nd order systems. Real engineering systems are rarely 1st
or 2nd order systems so the practical utility of these simple systems is questionable. Fortunately,
from an analysis point of view, even complex mechanical systems can be represented by several
connected first and second order systems. Consider as an example the measurement system
shown in Figure 1. The first component is an accelerometer which is a second order system, it is
connected to an amplifier which is a first order system, and a recording device which can be
6-2
modeled as a second order system. The total measurement system is thus fifth order but can be
modeled as three simpler systems connected in series.
acceleration
accelerometer
2nd Order
Figure 1:
amplifier
1st Order
recorder
2nd Order
deflection
A simple system for measuring acceleration.
Having identified models of the system sub-components, they can be combined to form a
model of the entire system. This is not always straightforward because systems interact when
they are connected. Interaction between sub-components is discussed in the next chapter. At the
end of the last chapter we showed how to combine descriptions of sub-system behavior to form a
description of the total system behavior, for the case when the interaction effects can be ignored.
In this case, often encountered in practice because measurement system sub-components are
usually designed to virtually eliminate these interaction effects, the derivation of the total system
model from the sub-component models is very straightforward.
So while the systems and methods that we look at in detail in this chapter are fairly simple, by
breaking down a complicated system into simpler sub-components, we can use these system
identification methods to identify characteristics of more complicated systems. Thus, these
simple methods are also often used by industry.
6-3
First Order Systems:
The differential equation is given by
y(t) y(t)
(1)
K x(t)
where y(t) is the system response
x(t) is the excitation
is the time constant, an indication of how fast the system responds
K is the static sensitivity.
Thermocouples, amplifiers, resistance temperature devices and RC circuits are examples of
systems whose behavior can be modeled with a first order differential equation such as equation
(1 ).
Step Response of a First Order System
Consider first the step response, that is, the response of the system subjected to a sudden
change in the input which is then held constant.
x(t) Cus (t)
where C is the magnitude of the step. us (t) is the unit step function defined below,
us (t) 0 for t 0, and us (t) 1 for t 0
Assume that the system is at rest at time t = 0, i.e., y(0) = y0 . As t gets large, y(t) yf , its
final value. By solving the differential equation, the response after the step was input is found to
be:
t
y(t)
yf 1 e
t
yoe
(2)
This can be written in the form:
t
y(t)
yf
yo yf
steady state
response
e
initial condition or
transient response
Note that if x(t) = 0 the response is that of a system with just the initial condition of
y(0) yo . Also, yf KC, i.e., the step size times the static sensitivity. The response is shown
in Figure 2. After τ seconds the response has moved 63.2% of the way to its final value. From
equation (2), we show that:
change in y(t)
final change
y(t) y o
yf y o
t
1 e
0.632 when t
6-4
Similarly when t 2 , the response has moved 86.5% of the way to its final value. So by
measuring when the response has moved 63.2% of the way to its final value, we can estimate ,
the time constant.
Figure 2:
Another way of determining
The Step Response of a 1st Order System.
is described below. Rewrite equation (2) in the form,
t
y(t) yf
e
y o yf
Take the natural log or log10 of both sides of this equation to yield,
1
ln
y o yf
y(t) yf
or
1
2.3026 log10
respectively. This is a linear equation in time, t, with slope
or with a slope
1
y o yf
y(t) yf
if you took the natural logarithm,
1
if you took the log10 .
2.3026
In
yf y o
yf y(t)
(3)
Slope = τ-1
NOTE: In X = 2.3026 log10 X
time - seconds
Figure 3: Calculating the Time Constant
6-5
In an experiment you can sample the response, extract the data after the step has been applied,
and fit a straight line to the rearranged (equation (3)) data plotted versus time. The inverse of the
estimated gradient will yield an estimate of the time constant, . The linearity of the rearranged
data is a measure of how well the system is represented by a 1st order differential equation. It is
best not to use data very close to the final value because the calculation: (yf y(t)) yields very
small values, and will be prone to error.
Frequency Response of a 1st Order System (response to a sinusoidal excitation)
A first order system is subjected to a sinusoidal excitation, A sin( t) . We will assume that
sufficient time has elapsed for the transient response to die out, the steady state response to a sine
wave excitation is,
KA
yss (t)
1 (
)2
sin ( t
)
(4)
Equation (4) defines the frequency response of the system, i.e., the amplitude and phase of the
output from the system as a function of frequency. Using complex notation, we can relate this to
the complex frequency response function of the system, T( j )
T( j )
K
1 j
(5)
The modulus or magnitude of the frequency response function is T( j ) and equals the ratio of
the amplitude of the sine wave coming out of the system to the amplitude of the sine wave going
into the system. So from equation (4), this yields
T( j )
K
1 (
)
(6)
2
which could have been derived by taking the magnitude of the right hand side of equation (5).
T( j ) is sometimes referred to as the dynamic gain.
The difference in phase between the sine waves going into and coming out of the system,
denoted by in equation (4), is the phase of T( j ) . Here, we will use the notation arg(T( j ))
to denote the phase of the complex function T( j ) .
arg(T( j ))
tan 1(
)
6-6
Plots of T( j ) and arg(T( j )) versus
(Bode plots) are shown in Figures 4(a) and (b). Note
that both the frequency ( ) axes are logarithmic, and the magnitude is plotted in decibels:
20 log10 T( j )
2
10 log10 T( j ) It is also traditional in controls applications to plot the phase
in degrees, though it is not incorrect to plot the phase in radians. In this example was set to .01
seconds and K was set to 10. Note that at higher frequencies, the slope of the magnitude plot rolls
off at a constant rate of 20 dB every time the frequency increases by a factor of 10. We say that
the roll-off is 20dB/decade.
Figure 4: 1st Order System Frequency Response: (a) Magnitude and (b) Phase
T( j ) can be derived directly from the differential equation shown in equation (1). Having this
frequency response function, we can calculate its magnitude and phase at the frequency of the
sine wave input to the system and then, can immediately write down the steady state response of
the system to this sine wave:
x(t)
Asin( t)
and
y(t)
A T( j ) sin ( t arg(T( j )))
(7)
To generate T( j ) from the differential equation we assume an input and an output of the form:
e j t and y(t)
x(t)
T( j ) e j t
(8)
respectively. We substitute these into the differential equation, equation (1), differentiating the
functions as necessary. This yields:
j
1 T( j ) e j t
K ej t
Comparing coefficients of e j t and rearranging yields equation (5). We will use this technique
again later in this chapter to derive the frequency response function of a second order system.
6-7
From the frequency response plots in Figure 4, we see that the output amplitude drops by only
3db over the frequency range 0 < ω < 1/τ. τ in this example was 0.01 and so 1/τ = 100 rad/s. This
is normally termed the bandwidth of the system, and c 1/ is referred to as the cut-off
frequency. Over this same frequency range the phase lag varies from 0 to 45 . When
1/
c , the phase lag of 45 indicates a shift of the output relative to the input in time.
The time shift is calculated by dividing the phase lag in radians by the frequency. Hence, here the
time shift would be /(4 c ) seconds. At this frequency the amplitude would be K/ 2 times the
input amplitude. However, there would be no influence on the overall shape of the signal, i.e., the
output would have the same sinusoidal form as the input.
Consider, however, the case where the input consisted of a sum of two sinusoids, one at
0.5/ and another at
3/ . Now the phase lag dependence on ω becomes important and
the output signal will suffer phase distortion. There will also be amplitude distortion. This
distortion is caused by the two sinusoidal components in the signal being treated differently by
the system: different gains in amplitude and different phase shifts. This is illustrated in Figure 5,
where a signal:
x(t)
5sin 50t 10sin 300t
is shown along with the response of the first order system to this signal. The system
characteristics are as above: K=10 and c 100rad/s . The response is:
y(t) 5 T( j50) sin 50t arg(T( j50))
y(t)
Figure 5:
50
sin 50t tan 1(.5)
1.25
10 T( j300) sin 300t arg(T( j300))
100
sin 300t tan 1(3)
10
The input to and response of a first order system
with c 100 rad/s and system gain = 10
6-8
These distortion effects will be even more prominent when the system is subjected to more
complex signals that contain many sinusoidal components. Each frequency component will
experience a different dynamic gain and phase lag, so the resulting output signal may be quite
unlike the input. This is of considerable significance in instrumentation applications, where we
would like the output of the measurement system to have the same shape as the input signal.
With an ideal measurement system, all frequency components in the input signal will have the
same gain and the same delay in time applied to them as they pass through the system; this will
preserve the shape of the signal. In reality, this is not achievable exactly, and we would specify
some error tolerance in gain and in phase (time delay) and design the measurement system to be
within these tolerances over the region of frequencies contained in the input signals we are trying
to measure.
Example
A transducer used in a measurement system can be modeled by a first order differential equation.
It has a time constant of 0.25 seconds and a static sensitivity of 5.0 Volts/input units.
a. Write down the differential equation of the system and also the frequency response
function.
b. What is the cut-off frequency of the transducer?
c. A very low frequency sine wave fluctuation is input into the transducer and the output
amplitude noted. Gradually the frequency is increased. At what frequency will the
amplitude be half what it was at very low frequencies?
d. What is the frequency range over which the amplitude distortion would be less than 1%?
What is the phase shift at the upper end of this frequency range?
Solution
a. K = 5 and τ = 0.25. The differential equation describing the system is therefore:
0.25
dy(t)
dt
y(t)
5 x(t)
where y(t) is the output of the transducer and x(t) the input. The frequency response
function is therefore,
K
1 j
5
1 j0.25
b. The cut-off frequency is the inverse of the time constant and hence equals 4 rad/s.
c. At low frequencies
0 , therefore the frequency response function is approximately K =
5. So we wish to find at what value of ω does the magnitude of the frequency response
function equal half of this, i.e., 2.5. Actually, we set up the equation for the magnitude
squared, which is:
(2.5)
K2
2
1 (
which implies,
52
)2
1 0.0625 2
6-9
1 0.0625 2
4,
and hence
48 rad/s
d. 1% amplitude distortion means that the magnitude of the frequency response function is
either 0.99 or 1.01 times the magnitude at zero frequency or DC. From Figure 4 we can
see that the magnitude of the frequency response function drops as frequency (ω)
increases. At DC (ω = 0) the magnitude of the frequency response function is equal to K =
5. So we wish to find the frequency at which the magnitude is 0.99 5 = 4.95, or the
magnitude squared is 24.5025. Therefore we need to solve:
24.5025
52
1 0.0625 2
This results in ω = 0.5700 rad/s. So the frequency range where amplitude distortion
would be less than 1% is 0 to 0.5700 rad/s. The phase shift at the upper frequency is:
phase
tan 1
tan 1(0.25 0.57)
0.1415 rads or 8.110
Using the Frequency Response for System Identification of a first order system
If we have a system or part of a measurement system and we have a way of generating plots of
the frequency response experimentally, we can use the plots to estimate the static sensitivity and
the time constant. The first thing to do, however, is to verify that the frequency response function
is that of a first order system. The plots should look like those in Figure 4, i.e., the magnitude (in
dB) will look flat at lower frequencies and then start to roll off at 20 dB/decade at higher
frequencies. The phase should start at 0° and at high frequencies flatten out to -90°. Once you
have confirmed that the system appears to be first order, you can then use it to estimate K and τ.
At DC (ω = 0 rad/s), the magnitude of the frequency response function is K or 20log10 K .
This gives you the static sensitivity. At the cut-off frequency,
c
1/ , the magnitude is
K/ 2 or 3dB down from 20log10 K . So we find at what frequency (in rad/s) the magnitude has
dropped 3dB from its level at very low frequencies, the inverse of this frequency gives us the
time constant, τ.
τ can also be estimated from the phase plot. When
c
1/ the phase is 45 .
Arg(T( j c ))
tan 1( c )
tan 1 1
45 . So identify at what frequency (in rad/s) the
phase shift is -45°, take the inverse of this frequency and that will give you an estimate of τ.
To generate the frequency response function experimentally follow the procedure outlined
below. The procedure will be very similar for higher order systems, the only difference being
how you might select the frequency range of interest.
6-10
Generate a very low frequency sine wave as input to your system. Note the frequency and
the amplitude of the response. Slowly increase the frequency and monitor the response of
the system on the scope. When the amplitude drops to approximately 1/100th to 1/1000th
of its initial value, note the frequency. This now defines the frequency span of your plots
of the frequency response function.
Select a sequence of frequencies that cover this range. Because we usually plot on
logarithmic frequency axes, the frequencies selected would normally be equally spaced on
a logarithmic scale. (E.g. 1 10 100 1000 10000 are equally spaced on a logarithmic scale.)
Set the input to one of the selected frequencies ( i ) and let the system response come to
steady state. The response should be a sine wave of the frequency you are putting into the
system. If it is not, the transient response may be still contributing, the system may be
nonlinear or the signal you are measuring may be corrupted with noise.
Note the amplitude of the input and the output: Ain and Aout , respectively. Now
A
calculate, M 20log10 out .
Ain
Measure the phase difference between the two signals. You can do this by using a
Lissajous plot on the scope (see Laboratory 1 of this course), or you can measure the
difference in zero crossing times between the two signals. The output will be delayed
relative to the input. Multiply the time difference by the frequency of the sine wave; the
phase, in radians, is minus this value. Convert to degrees.
P
i
(time difference) 180
Plot ( i , M) on the magnitude plot and ( i , P) on the phase plot.
Repeat the last four steps until all the selected frequencies have been input to the system.
It is possible to automate this procedure using a computer, if you have some way of controlling
the amplitude and frequency of the sine wave generator from the computer. You will use a
LABVIEW VI in your laboratory that does this. However, you should take care with three things
when using this VI or similar computer programs.
1. You must allow the system response to reach steady state after any changes are
made to the input signal, before any measurements are taken.
2. You must check that the system is behaving linearly.
3. You must check that the amplitude of the response is not too large to measure
with your computer and data acquisition system, or so small that quantization
noise is a problem.
6-11
Second Order Systems
The differential equation is given by:
y(t) 2
where
2
n y(t)
n y(t)
K 2n x(t)
(9)
y(t) is the system response (output)
x(t) is the excitation (input)
n is the undamped natural frequency, rad/sec
is the fraction of critical damping or damping ratio
K is the static sensitivity.
Many accelerometers, mass-spring-dampers, galvanometers, force transducers and LRC circuits
can be modeled as second order systems.
Snap Back Response
The snap back response is equivalent to an initial displacement with no forcing function. That
is, x(t) 0, y(0) 0 and y(0) yo. The solution is given by
y(t)
yo e
n t cos
2
1
n
t
yo e
n t cos
d
t
(10)
which is plotted in Figure 6.
Figure 6: Snap Back Response (
.08, and n
The period, Td , and the damped natural frequency,
Td
2
;
60 rad/s).
d , are given by:
d
n 1
2
(11)
d
n t . We can pick out this envelope
Note that the envelope of the response is given by y o e
function approximately by tracking the peaks of y(t). If one peak occurs at t i , a peak will also
appear n cycles later at ti n ti nTd . The peak values above the final steady state value, yf ,
at these two times are:
6-12
yi
yf
n ti
yo e
and
yi n
yf
n (t i nTd )
yo e
(12)
We include the subtraction of yf , the final steady state value, in order to give formulas that are
applicable to the snap back response, the impulse response and the step response. For the snap
back and impulse responses, yf is zero. From these expressions we can write down the formulas
for the logarithmic decrement method.
2
n Td
1
y yf
1n i
n
yi n yf
2
1
(13)
where δ is the logarithmic decrement that quantifies how much the response decays over time.
Therefore:
1
2
2
2
if
(14)
.1
Thus, the decay rate of the response is purely a function of the damping ratio.
Note that equations (13)-(14) can be used for the step response and the impulse response, but
be careful and remember to subtract any DC offset from your signal. Even if the signal should
not, theoretically, have an offset, data acquisition and instrumentation often introduces offsets
which should be subtracted.
Impulse Response
The impulse response is equivalent to an initial velocity with no forcing function. That is
x(t) = 0, y(0) = 0 and y(0) yo . The solution is given by
y(t)
yo
e
nt
sin ( n 1
2
(15)
t)
d
Step Response
The step response corresponds to x(t) C us (t), y(0) 0, y(0) 0 and yf y( ) K C ,
where K is the static sensitivity of the system. us (t) denotes the unit step function. The solution
is
y(t)
yf 1
e
nt
1
Figure 7 is a plot of the step response.
2
sin ( d t
)
where
tan
1
1
2
(16)
6-13
Figure 7:
Step Response of a 2nd Order System (ζ =.08, and ( n = 60 rad/s).
Note that the overshoot (OS) can be used to estimate the damping ratio.
1
y
1n f
yo
OS
(17)
This formula can be derived by finding out where the maxima of y(t) yf occur, i.e. by
differentiating:
y(t) yf
yf
nt
e
2
1
sin ( d t
(18)
)
with respect to t and setting the result to zero. Note that we have set y i to zero here to simplify
calculations. The resulting expression will include a tan term which results in multiple solutions
for t:
dt
where n is any integer.
n
Choosing n = 1, and substituting back into the expression for y(t) yf
overshoot in terms of the damping ratio.
OS
yf
e
1
However, sin( )
1
2
OS , will give the
nt
2
, see equation (16), and
sin( )
n
d/ 1
2
d if we assume that
1 2 is approximately equal to 1. Furthermore, recall that at the maximum
Substituting this into the equation above will give:
OS
yf e
Taking the natural logarithm of both sides and rearranging will give equation (17).
dt
.
6-14
Frequency Response of a 2nd Order System (Response to sinusoidal input)
In this case the excitation takes the form x(t) = B sin t and the initial conditions are such
that there is no transient response, or the transient response can be assumed to have decayed
away to zero. The steady state response is given by
y(t)
B T( j ) sin
t arg(T( j ))
where T( j ) is the frequency response function of the second order system,
(19)
denotes
magnitude and arg(.) denotes phase. To calculate T( j ) we follow the procedure we adopted
with the first order system. That is assume that the input and output of the system have the
following forms:
x(t)
ej t
T( j ) e j t
and y(t)
respectively. Substitute into equation (9), the differential equation describing the behavior,
differentiating y(t) as necessary. This yields:
2
2j
n
2
n
T( j ) e j t
K 2n e j t
(20)
Equating coefficients of e j t and rearranging yields the frequency response of the second order
system:
T( j )
K 2n
2
n
2
2j
(21)
n
In Figure 8(a) is shown the system frequency response magnitude which is also the amplitude of
the steady state response divided by the amplitude of the input (B), plotted against the frequency
of the input, ω. In Figure 8(b) is shown the phase response, i.e., a plot of versus ω. The peak
response occurs when:
r
1 2 2
n
(22)
This is referred to as the resonant natural frequency, and can be derived by finding the minimum
of the magnitude squared of the denominator of equation (21), i.e., differentiating it with respect
to ω, setting the result to zero and solving for ω. Note that when
is greater than
1
2 2 0.7071 , this expression becomes imaginary, indicating that there will be no peak
observed in the frequency response. Note that at frequencies well above the natural frequency,
the roll-off of the magnitude is 40 dB/decade, i.e., if the frequency increases by a factor on 10 the
magnitude drops by 40 dB.
6-15
Figure 8:
Frequency Response of a Second Order System
( .08, and n 60 rad/s)
Example
Some accelerometers can be modeled as second order systems. The one being considered here
has a natural frequency of 1000 Hz, and is used to measure accelerations of frequencies below
100 Hz. The damping ratio is 0.05 and the sensitivity (K) is 2V/g.
a. Write down the differential equation describing the accelerometer behavior, and write down
its frequency response function.
b. What is the steady state response of this accelerometer to an acceleration of the form:
x(t) = 0.1 sin(20t- /2 ) + 0.25cos(1200t) .
c. Would you expect to see a peak in the magnitude of the frequency response function? Why?
d. Since the upper frequency limit, specified by the manufacturer is 100 Hz, what is the
maximum percent amplitude distortion you will find when using the accelerometer in
accordance with the manufacturer's specifications? What is the maximum phase distortion?
Solution
a. From equation (9) with K = 2, n 2 1000 rad/s and
describing the accelerometer behavior is:
y 200 y 4 106 2 y 8 106 2 x(t)
0.05 , the differential equation
6-16
The frequency response function, from equation (21), is:
T( j )
8 106 2
4 106 2
2
j200
b. The steady state response is:
y(t)
0.1 T( j20) sin 20t
T( j20)
2
arg(T( j20))
8 106 2
4 106 2
400
0.25 T( j1200) cos 1200t arg(t( j1200))
and T( j1200)
j4000
8 106 2
4 106 2 1.44 106
j2.4 105
Calculating the magnitude and phase of each of these expressions gives:
T( j20)
and
T( j1200)
2.0000 V/g, arg(T( j20))
0.0003 rad
2.0753 V/g, arg(T( j1200))
0.0198 rad
Note that the units for the magnitude are the same as for the sensitivity, K. Substituting in
the expression for y(t) above gives the steady state response to this input, x(t), that consists
of two oscillating components. Note that the cosine term has a frequency that has a
frequency outside the manufacture's specified frequency range.
y(t)
0.2000 sin 20t
2
0.0003
c. Since the damping ratio 0.05 is smaller than 2
r
0.5188 cos(1200t 0.0198)
1
2 then
r has a real value, because
1
(1 2 2 ) 2 n . Therefore a peak will appear in the frequency response magnitude.
d. The question being asked here is how different is the amplification of components of
frequencies close to 100 Hz from the amplification of the DC component. You notice
from Figure 8 that when the damping ratio is sufficiently low, the magnitude increases as
frequency increases towards the resonant natural frequency. Simultaneously the phase is
decreasing from 0°.
When
0 rad/s, from equation (21) we can see that the frequency response function is
K = 2. At 100 Hz = 200 π rad/s, the frequency response function is:
T( j200 )
8 106 2
4 106 2
4 104 2
2
j4 104 2
0.99
j0.01
Therefore, the magnitude and the phase of the frequency response function at the upper
6-17
end of the manufacturer's range are:
T( j200 )
2.0201 and arg(T( j ))
0.01 rad
0.57
Therefore the maximum percentage amplitude distortion is:
(2.0201 2)
100
2
2%
The maximum phase distortion in the manufacturer's specified operating region also
occurs at 100 Hz, and is -0.57°.
The frequency response function can be generated experimentally by using the procedure
described for first order systems. The only problem, as with first order systems, may be your
ability to present and measure sinusoidal inputs to the system being investigated. When using the
frequency response behavior to estimate the characteristics of a system you think is second order,
you must first make certain that the system is indeed second order. As we have noted above,
when damping is high there will not be a peak in the magnitude of the frequency response
function, so at first glance the system may appear to look first order. Further investigation is
needed. The characteristics to look for are:
1. The magnitude should appear flat (on a dB scale) at lower frequencies. At higher
frequencies the magnitude should roll off at 40 dB/decade.
2. The phase should start close to 0° at low frequencies and at very high frequencies should be
close to -180°.
Now, having confirmed that the system is second order, we can proceed to estimate K,
.
n and
Methods of Calculating Natural Frequency and Damping by using e.g. a Frequency
Generator and an Oscilloscope or a Data Acquisition System.
Use an oscillator to generate a sine wave and put this through your system. Measure the input
to your system on the 1st channel of the oscilloscope and the system response on the 2nd
channel.
To find the natural frequency set the scope to the x vs. y setting and increase the frequency
until the phase delay is 90° between the 2 signals. When this occurs a circle appears on the scope
(see laboratory 1). The frequency of the sine wave will be n , the natural frequency of the
second order system.
Having calculated
n , there are several ways of calculating the damping ratio, ζ.
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AMPLIFICATION METHOD
T( j )
1
2
n
T( j )
0
• Set input to a low frequency sine wave (
magnitude of response (linear not in dB).
• Set the input frequency to
(linear not in dB).
0) , measure
n , measure magnitude of response
HALF POWER METHOD
2
2 n
1
2 n
• Plot out magnitude of the frequency response function versus
frequency around resonance, as shown in Figure 9(a). To do
this measure the amplitude of the input (X) and the amplitude
of the steady state response (Y), at each frequency. Plot Y/X
versus frequency.
• Evaluate the parameters shown in Figure 9(a) and substitute
into the formula. 1,2 are the frequencies at which the
magnitude has dropped by a factor of 1/ 2 from its peak
value at the resonant natural frequency. On a dB scale this
corresponds to a drop of 3 dB from the peak value.
SLOPE OF PHASE ANGLE METHOD
1
1
d
n d
n
• Plot out the phase, in radians, of the frequency response
function around resonance on a linear frequency axis
(radians/s), as shown in Figure 9(b). To do this you need to
calculate the phase delay between the output and input signals.
• Find the slope of the phase (d / d ) at
n and substitute
into formula.
NOTE: In practice the differences between n , d and r should not be used to estimate ζ,
because the errors in measuring n , d and r will result in large errors in the estimate of ζ.
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Figure 9:
Calculating the damping ratio: (a) half-power method and (b) slope
of the phase method. Note all axes are linear, frequency is in
radians/second and phase is in radians.