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cjs94
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Homework Statement
From the waveform shown below, estimate
a) the damping ratio ζ (you may compare response with a standard chart);
b) the forced or damped frequency of oscillation; and
c) the natural or undamped frequency of oscillation.
Homework Equations
Since the waveform is under damped, I'm attempting to use the logarithmic decrement method, described here: http://en.wikipedia.org/wiki/Logarithmic_decrement
[tex]\sigma = \frac{1}{n}\ln\frac{x(t)}{x(t + nT)}[/tex]
[tex]\zeta = \frac{1}{\sqrt{1 + \left(\frac{2\pi}{\sigma}\right)^2}}[/tex]
[tex]f_d = \frac{1}{T}[/tex]
[tex]f_n = \frac{f_d}{\sqrt{1 - \zeta^2}}[/tex]
The Attempt at a Solution
I have estimated the first two peaks from the graph as:
[tex]p_1 = 0.438\text{ V} \text{ at } 0.27\text{ ms}[/tex]
[tex]p_2 = 0.350\text{ V} \text{ at } 0.77\text{ ms}[/tex]
Using the above equations:
[tex]\begin{align}
\sigma &= \ln\left(\frac{p_1}{p_2}\right)\\
&= \ln\left(\frac{0.438}{0.350}\right)\\
&= 0.224\\
\text{and}\\
\zeta &= \frac{1}{\sqrt{1 + \left(\frac{2\pi}{0.224}\right)^2}}\\
&= 0.0356\\
f_d &= \frac{1}{0.77 \times 10^{-3} - 0.27 \times 10^{-3}}\\
&= 2\text{ kHz}\\
f_n &= \frac{2000}{\sqrt{1 - 0.0356^2}}\\
&= 2001\text{ Hz}
\end{align}
[/tex]
The problem is that I'm not sure I believe the results. I'm trying to verify the results by putting them back into the second order characteristic equation:
[tex]\begin{align}
\text{C.E.} &= s^2 + 2\zeta{\omega}_{n}s + {\omega}_{n}^2\\
&= s^2 + (2 \times 0.0356 \times 2\pi \times f_n)s + (2\pi \times f_n)^2\\
&= s^2 + 895s + 158071624
\end{align}
[/tex]
then simulating that with a Laplace block in PSpice. However, the simulated waveform doesn't match the one above. The frequency is correct, but the damping ratio is too low -- playing about with the numbers, I find I need to increase the damping ratio to approximately ##2.8\zeta## to get the waveform looking correct.
I don't know if there is a problem in my method and the results are wrong, or if my simulation is in error (or possibly both!). Can someone please help?
Thanks,
Chris
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