damping factor and dampted frequency
- Thread starter ham3388
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Why bother to calculate the damping factor since it is VERY low, maybe 2.
Most audio amplifiers today have a better damping factor of 200 to thousands so the output does not resonate.
Why bother to calculate the damping factor since it is VERY low, maybe 2.
Most audio amplifiers today have a better damping factor of 200 to thousands so the output does not resonate.
I used Microsoft Paint program to add some contrast to the image that was grey on grey.
Attachments
From the rather fuzzy graph I obtained 4 pairs of (time,value) data
t=0.26 ms V=0.44 V
t=0.76 ms V=0.35 V
t=0.52 ms V=0.12 V
t=1.02 ms V=0.18 V
From these data and some maths [not shown] it's possible to show the steady state final value is ~250mV. This is also fairly obvious from the graph trend to a steady-state final value of that order.
One may then find the decay time factor a=1/τ using
\(ln\( \frac{0.44-0.25}{0.35-0.25}\)=a\(0.76-0.26\)\times 10^{-3}\)
or
\(a=\frac{1}{\(0.76-0.26\)\times 10^{-3}} \[ ln\( \frac{0.44-0.25}{0.35-0.25}\) \]=2000 \times ln{\( \frac{0.19}{0.1} \)}\)
But
\(a=\zeta \omega_o\)
where
\(\omega_o=\sqr{{\omega_d}^2+a^2}\)
Using the above data the damped period is clearly Td= 0.5ms
Hence
\(\omega_d=\frac{2\pi}{T_d}=\frac{2\pi}{0.5\times10^{-3}}=12566 \ rad \ sec^{-1}\)
Which is the same as the OP's result.
This should provide sufficient alternative data to enable a cross check of the OP's working - which BTW looks to have some errors. For instance - a damping factor ζ of 0.86 would probably not produce the extended oscillatory result shown in the graph.
t=0.26 ms V=0.44 V
t=0.76 ms V=0.35 V
t=0.52 ms V=0.12 V
t=1.02 ms V=0.18 V
From these data and some maths [not shown] it's possible to show the steady state final value is ~250mV. This is also fairly obvious from the graph trend to a steady-state final value of that order.
One may then find the decay time factor a=1/τ using
\(ln\( \frac{0.44-0.25}{0.35-0.25}\)=a\(0.76-0.26\)\times 10^{-3}\)
or
\(a=\frac{1}{\(0.76-0.26\)\times 10^{-3}} \[ ln\( \frac{0.44-0.25}{0.35-0.25}\) \]=2000 \times ln{\( \frac{0.19}{0.1} \)}\)
But
\(a=\zeta \omega_o\)
where
\(\omega_o=\sqr{{\omega_d}^2+a^2}\)
Using the above data the damped period is clearly Td= 0.5ms
Hence
\(\omega_d=\frac{2\pi}{T_d}=\frac{2\pi}{0.5\times10^{-3}}=12566 \ rad \ sec^{-1}\)
Which is the same as the OP's result.
This should provide sufficient alternative data to enable a cross check of the OP's working - which BTW looks to have some errors. For instance - a damping factor ζ of 0.86 would probably not produce the extended oscillatory result shown in the graph.
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Perhaps it would be worth establishing a common basis for analysis - otherwise we might be at cross purposes.
I'm basing an analysis on the typical second order characteristic polynomial of the form
\(s^2+2as+{\omega_o^2}\)
where damping factor ζ fits according to the relationship
\(a=\zeta \omega_o\)
and the damped frequency is related to a and ωo by
\(\omega_d^2=\omega_o^2-a^2\)
Is this what your equations would be based upon?
I'm basing an analysis on the typical second order characteristic polynomial of the form
\(s^2+2as+{\omega_o^2}\)
where damping factor ζ fits according to the relationship
\(a=\zeta \omega_o\)
and the damped frequency is related to a and ωo by
\(\omega_d^2=\omega_o^2-a^2\)
Is this what your equations would be based upon?
It's such a poor image it's hard to extract the actual values. For my eye, the first peak is around 440mV and the steady state final value is about 250mV. So the difference would give Vpeak=190mV. But that probably wouldn't account for the disparity in our respective damping factor calculations.
The question doesn't really spell out the nature of the damped oscillatory response.
One might consider it to be typical of the voltage step [input step magnitude E volts] response of a series RLC circuit with the output across the capacitor.
Such a response is of the form
\(V_c(t)=E \[ 1-\frac{\omega_o}{\omega_d}e^{-at}\sin{\( \omega_dt+\phi \)} \]\)
The damping term a may be found from
\(a=-\frac{2}{T_d}\ln{\( \frac{V_{peak}-V_{\infty}}{V_{\infty}} \)}\)
Where Td is the damped frequency period and V∞ is the output steady-state value - these values [including Vpeak] must be determined from direct measurements performed on the output response plot.
From this one may deduce the natural frequency ωo as ..
\(\omega_o=\sqr{\omega_d^2 + a^2}\)
Where
\(\omega_d=\frac{2\pi}{T_d}\)
Finally the damping factor ζ can be found from ...
\(\zeta=\frac{a}{\omega_o}\)
One might consider it to be typical of the voltage step [input step magnitude E volts] response of a series RLC circuit with the output across the capacitor.
Such a response is of the form
\(V_c(t)=E \[ 1-\frac{\omega_o}{\omega_d}e^{-at}\sin{\( \omega_dt+\phi \)} \]\)
The damping term a may be found from
\(a=-\frac{2}{T_d}\ln{\( \frac{V_{peak}-V_{\infty}}{V_{\infty}} \)}\)
Where Td is the damped frequency period and V∞ is the output steady-state value - these values [including Vpeak] must be determined from direct measurements performed on the output response plot.
From this one may deduce the natural frequency ωo as ..
\(\omega_o=\sqr{\omega_d^2 + a^2}\)
Where
\(\omega_d=\frac{2\pi}{T_d}\)
Finally the damping factor ζ can be found from ...
\(\zeta=\frac{a}{\omega_o}\)
I see several things wrong. Some of it is simply "eye calibration."
1. The peak is 430 mV
2. The SS value is 250 mV
3. The delta peak = 430-250 = 180 mV
4. The ratio of delta peak to SS is 180/250 = 0.720
5. Your equation for damping ratio is missing a square.
6. Damping Ratio = { 1+ [ln(∏/0.720)]^2 }^(-0.5) = 0.104
7. Peak time appears to be 0.26 ms.
8. Damped frequency = ∏ / 0.000,26 = 12,080 rad/s
9. Natural frequency = (12,080) * sqrt(1-0.104^2) = 12,030 rad/s
You know the damping ratio can't be close to 0.86, because that is nearly critically damped and would have almost zero overshoot. Snce the damping ratio is really about 0.1, it means the damped and natural frequencies are nearly the same -- which they are within 0.5% of each other.
One final comment. You might notice I'm not using a lot of significant figures. In fact, I might be pushing it a little with 4 digits on the frequencies. Since your measurements are based on the human eye reading a plot, you can only guestimate the values. For exmple, if the peak is judged to be 430, how can I be certain it isn't 429 or even 433? There is an error to what we can read. Does it make sense to say the damping ratio is 0.103,999? It makes more sense to say it's 0.10 or 0.104 at best. I'm mentioning this because of your original frequency calculations, e.g. 12,566.
1. The peak is 430 mV
2. The SS value is 250 mV
3. The delta peak = 430-250 = 180 mV
4. The ratio of delta peak to SS is 180/250 = 0.720
5. Your equation for damping ratio is missing a square.
6. Damping Ratio = { 1+ [ln(∏/0.720)]^2 }^(-0.5) = 0.104
7. Peak time appears to be 0.26 ms.
8. Damped frequency = ∏ / 0.000,26 = 12,080 rad/s
9. Natural frequency = (12,080) * sqrt(1-0.104^2) = 12,030 rad/s
You know the damping ratio can't be close to 0.86, because that is nearly critically damped and would have almost zero overshoot. Snce the damping ratio is really about 0.1, it means the damped and natural frequencies are nearly the same -- which they are within 0.5% of each other.
One final comment. You might notice I'm not using a lot of significant figures. In fact, I might be pushing it a little with 4 digits on the frequencies. Since your measurements are based on the human eye reading a plot, you can only guestimate the values. For exmple, if the peak is judged to be 430, how can I be certain it isn't 429 or even 433? There is an error to what we can read. Does it make sense to say the damping ratio is 0.103,999? It makes more sense to say it's 0.10 or 0.104 at best. I'm mentioning this because of your original frequency calculations, e.g. 12,566.
It's a pain often trying to pin down these wretched equations ...
The actual equation should read ....
\(\LARGE{\zeta=\frac{1}{\sqr{1 + {\( \frac{\pi}{\ln{ \( \frac{V_{pk} - V_{SS}}{V_{SS}} \) }}\)}^2 }}}\)
Where Vpk is the output response peak value and Vss is the steady-state [final] output value
Yeah, that's right. Looks better in LaTex or whatever you call it. I had the equation correct before I transferred it from paper to keyboard. I admire your ability to be able to use it so well. I really struggle to write equations with that program.
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