Underdamped response

  • Thread starter magnifik
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  • #1
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An underdamped waveform has the general form

Ke-[tex]\sigma[/tex]tsin([tex]\omega[/tex]t - [tex]\varphi[/tex])

From the graph, determine K, [tex]\sigma[/tex], [tex]\omega[/tex], and [tex]\varphi[/tex]
2w4mn9s.jpg


i know f = 1/t
t = .004 s
f = 250 Hz
[tex]\omega[/tex] = 2[tex]\pi[/tex]f
[tex]\omega[/tex] = 1570
[tex]\varphi[/tex] is clearly 0

How do i determine K and [tex]\sigma[/tex]??
 

Answers and Replies

  • #2
33
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Interesting problem. Nothing will ever change the frequency, right? Right.

Take the value after going [tex]2\pi[/tex] and plug in the time that took. Since the [tex]\sin(\omega-\phi[/tex] will be the same, we can write

[tex]\rm{Value}\, \rm{beginning}\,=\rm{Value}\, \rm{end}\, 2\pi \exp[-\sigma t][/tex]

Finding [tex]\kappa[/tex] from the remaining variables should be easy.
 
  • #3
360
0
Interesting problem. Nothing will ever change the frequency, right? Right.

Take the value after going [tex]2\pi[/tex] and plug in the time that took. Since the [tex]\sin(\omega-\phi[/tex] will be the same, we can write

[tex]\rm{Value}\, \rm{beginning}\,=\rm{Value}\, \rm{end}\, 2\pi \exp[-\sigma t][/tex]

Finding [tex]\kappa[/tex] from the remaining variables should be easy.


but isn't the value at the beginning 0?
 
  • #4
1,097
3
An underdamped waveform has the general form

Ke-[tex]\sigma[/tex]tsin([tex]\omega[/tex]t - [tex]\varphi[/tex])

From the graph, determine K, [tex]\sigma[/tex], [tex]\omega[/tex], and [tex]\varphi[/tex]
2w4mn9s.jpg


i know f = 1/t
t = .004 s
f = 250 Hz
[tex]\omega[/tex] = 2[tex]\pi[/tex]f
[tex]\omega[/tex] = 1570
[tex]\varphi[/tex] is clearly 0

How do i determine K and [tex]\sigma[/tex]??


Is there any additional information given in the question? Does it mention what type of circuit the graph is from, or any other parameters?
 
  • #5
360
0
no, there is nothing else mentioned. it just gives you the general form and the graph of the waveform. the y-axis is current and the x-axis is time in milliseconds.
 
  • #6
33
0
but isn't the value at the beginning 0?

Right. The zero values don't help much. Pick a pair of peaked values.
 
  • #7
764
71
I estimated the peak at t = 5 ms is roughly 65 % of the peak at t = 1ms. (.65 is about 0.55/0.85)

0.65 e-[tex]^{\sigma0.001}[/tex] = e-[tex]^{-\sigma0.005}[/tex]

With sigma determined,set the expression for one of the peaks equal to the estimated value at that peak, and get K.
 
Last edited:
  • #8
360
0
I estimated the peak at t = 5 ms is roughly 65 % of the peak at t = 1ms. (.65 is about 0.55/0.85)

0.65 e-[tex]^{\sigma0.001}[/tex] = e-[tex]^{-\sigma0.005}[/tex]

With sigma determined,set the expression for one of the peaks equal to the estimated value at that peak, and get K.

using your method i got [tex]\sigma[/tex] = 107.695729
however, the correct answer is [tex]\sigma[/tex] = 140
 
  • #9
764
71
It was based on an estimation, and I haven't checked your math. I suppose you could get a better estimation using the distance in pixels (maybe rotate the plot level first), or by putting a ruler down on the page, and recalculate based on those measurements. In practical situations, there may be cursors on a digital equipment that can read out exact values. It seems to me like if they want exact value solutions, they should provide you with the exact values for two data points.
 
Last edited:

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