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 deflection accelerometer amplifier recorder nd 2nd Order 1st Order 2 Order
Figure 1: 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.
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First Order Systems:
The differential equation is given by
y(t) y(t) Kx(t) (1) 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,
u(t)ss 0for t 0,andu(t) 1for 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: tt y(t) yfo 1 e y e (2)
This can be written in the form: t y(t) y e f yof y steady state initial condition or response 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:
changeiny(t) y(t) y t o 1 e 0.632 when t final change yfo y 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: The Step Response of a 1st Order System. Another way of determining is described below. Rewrite equation (2) in the form, t y(t) y e f yyof
Take the natural log or log10 of both sides of this equation to yield,
1 yo y f 1 y o y f ln or 2.3026 log10 (3) y(t) yff y(t) y
1 respectively. This is a linear equation in time, t, with slope if you took the natural logarithm, 1 or with a slope if you took the log . 2.3026 10
yy In fo Slope = τ-1 yf y(t)
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, Asin( 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) sin( t ) (4) 1 ( )2
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 )
K T(j ) (5) 1j
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
K T( j ) (6) 1 ( )2 which could have been derived by taking the magnitude of the right hand side of equation (5). 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 .
arg(T( j )) tan1 ( )
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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: 2 20 log10 T(j ) 10 log 10 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:
x(t) ej t and y(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 ) ej t K e j t
Comparing coefficients of ejt 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 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 /(4c ) 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) 5sin50t 10sin300t 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)) 10 T( j300) sin 300t arg(T( j300))
50 100 y(t) sin 50t tan11 (.5) sin 300t tan (3) 1.25 10
Figure 5: 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: dy(t) 0.25 y(t) 5 x(t) dt where y(t) is the output of the transducer and x(t) the input. The frequency response function is therefore, K5
1 j 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: K522 (2.5)2 1 ( )22 1 0.0625 which implies, 6-9
1 0.06252 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:
52 24.5025 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 tan11 tan (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 . 11 Arg(T( jcc )) tan ( ) tan 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.
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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 A out , respectively. Now Aout calculate, M 20log10 . 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.
(time difference) 180 P i
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: 22 y(t) 2n y(t) n y(t) K n x(t) (9) where 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
nntt2 y(t)yeo cos n 1 t ye o cos d t (10) which is plotted in Figure 6.
Figure 6: Snap Back Response ( .08, andn 60 rad/s).
The period, Td , and the damped natural frequency, d , are given by:
2 2 Td ; d n 1 (11) d
nt Note that the envelope of the response is given by yeo . We can pick out this envelope function approximately by tracking the peaks of y(t). If one peak occurs at ti , a peak will also appear n cycles later at ti n t i nT d . The peak values above the final steady state value, yf , at these two times are: 6-12
nit n(t i nT d ) yi y f y o e and y i n y f y o 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, is zero. From these expressions we can write down the formulas for the logarithmic decrement method.
2 1 yif y ndT 1n (13) 1 2 n yi n y f where δ is the logarithmic decrement that quantifies how much the response decays over time. Therefore:
1 2 if .1 (14) 22 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
yo nt 2 y(t) e sin (n 1 t) (15) d
Step Response
The step response corresponds to x(t) C usf (t),y(0) 0,y(0) 0 and y y( ) K C, where K is the static sensitivity of the system. us (t) denotes the unit step function. The solution is
t 2 e n 1 1 y(t) yfd 1 sin ( t ) where tan (16) 1 2
Figure 7 is a plot of the step response.
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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 y 1n fo (17) OS
This formula can be derived by finding out where the maxima of y(t) yf occur, i.e. by differentiating: e nt y(t)yf y f sin(t d ) (18) 1 2 with respect to t and setting the result to zero. Note that we have set yi to zero here to simplify calculations. The resulting expression will include a tan term which results in multiple solutions for t:
d tn where n is any integer.
Choosing n = 1, and substituting back into the expression for y(t) yf OS , will give the overshoot in terms of the damping ratio.
e nt OS yf sin( ) 1 2
2 2 However, sin( ) 1 , see equation (16), and n d/1 d if we assume that 2 1 is approximately equal to 1. Furthermore, recall that at the maximum dt . Substituting this into the equation above will give:
OS yf e Taking the natural logarithm of both sides and rearranging will give equation (17).
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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) BT(j )sin t arg(T(j )) (19) where T( j ) is the frequency response function of the second order system, denotes magnitude and arg(.) denotes phase. To calculate 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 and y(t) T( j ) e j t respectively. Substitute into equation (9), the differential equation describing the behavior, differentiating y(t) as necessary. This yields: 2 2 j t 2 j t 2jn n T( j ) e K n e (20)
Equating coefficients of ejtand rearranging yields the frequency response of the second order system:
K 2 T( j ) n (21) 22 nn2j
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:
2 rn12 (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 22 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.
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Figure 8: Frequency Response of a Second Order System ( .08, andn 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 0.05, the differential equation describing the accelerometer behavior is: y 200y 4106 2 y 810 6 2 x(t) 6-16
The frequency response function, from equation (21), is:
8 1062 T( j ) 4 106 2 2 j200 b. The steady state response is:
y(t) 0.1 T( j20) sin 20t arg(T( j20)) 0.25 T( j1200) cos 1200t arg(t( j1200)) 2
8 106 2 8 10 6 2 T( j20) and T( j1200) 4 106 2 400 j4000 4 10 6 2 1.44 10 6 j2.4 10 5
Calculating the magnitude and phase of each of these expressions gives: T( j20) 2.0000 V/g, arg(T( j20)) 0.0003 rad and T(j1200) 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 0.0003 0.5188 cos(1200t 0.0198) 2
1 2 c. Since the damping ratio 0.05 is smaller than 2 then r has a real value, because 1 2 2 rn(1 2 ) . 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:
8 1062 2 T( j200 ) 4 106 2 4 10 4 2 j4 10 4 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 n T( j )0 2 • Set input to a low frequency sine wave ( 0) , measure magnitude of response (linear not in dB).
• Set the input frequency to n , measure magnitude of response (linear not in dB).
HALF POWER METHOD
21 22nn • 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 1d
d n 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 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.