0
\$\begingroup\$

I've been attempting to prove the taught result that: $$T_p=\frac{\pi}{\omega_n\sqrt{1-\zeta^2}}\textrm{.}$$

From the time-domain form of an underdamped response, we have $$c(t)=K_1e^{-\sigma_dt}\cos(\omega_dt-\phi)\textrm{.}$$ In an attempt to find maxima, the first derivative has been found as $$\frac{dc(t)}{dt}=K_1\left(-\sigma_de^{-\sigma_dt}\cos(\omega_dt-\phi)-\omega_de^{-\sigma_dt}\sin(\omega_dt-\phi)\right)\textrm{.}$$ From here, assuming \$K_1\textrm{, }e^{-\sigma_dt}\ne0\$, the maxima are found at $$\omega_d\sin(\omega_dt-\phi)+\sigma_d\cos(\omega_dt-\phi)=0\textrm{.}$$ Re-arranging for \$t\$, we find $$t=\frac{-\tan^{-1}\left(\frac{\sigma_d}{\omega_d}\right)+\phi +n\pi}{\omega_d}\textrm{.}$$ The denominator clearly aligns with the taught result, as the damped frequency is related as such to the natural frequency. However, I do not see how the numerator aligns to \$\pi\$. Where does the \$\tan\$ term go? I'm aware that we're hunting for the first peak, so it makes sense for \$n=0/1\$? Any help would be appreciated.

\$\endgroup\$
  • \$\begingroup\$ google.com/… \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 16 at 23:46
  • 2
    \$\begingroup\$ @TonyStewartSunnyskyguyEE75 That's a completely inappropriate comment. Considering this is by no means a post which shows no research effort, I'd have thought a person of your experience would have thought better than to do that. There are one or two terms which I need help resolving, and I have given all steps I have taken so far. \$\endgroup\$ – Benjamin Crawford Ctrl-Alt-Tut May 16 at 23:52
  • \$\begingroup\$ Doesn't the peak time corresponds to one-half cycle of the frequency of a damped oscillation? Perhaps you mean settling time which has different threshold criteria. I did not understand your assumptions as valid. The error signal is i.stack.imgur.com/N1mDy.gif \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 17 at 0:00
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 Yes, that is true, and I'm attempting to prove that mathematically, incorporating exponential decay. Which assumption do you not agree is valid? Amplitude K1 would not be zero, else there would be no signal to analyse in the first place. Exponential term would not reach zero until infinite time... The rest is rigid calculus. And no, I'm definitely talking about peak time, hence finding the maxima... \$\endgroup\$ – Benjamin Crawford Ctrl-Alt-Tut May 17 at 0:06
  • \$\begingroup\$ Sorry, I would just look up how its done .dademuch.com/2017/11/14/underdamped-second-order-system maybe here \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 17 at 0:09
0
\$\begingroup\$

The inverse tangent is cancelled by the phase shift angle. To see it you should present EVERY term as expressions of the natural frequency and the damping factor. The basic 2nd order low-pass system doesn't have other parameters.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

I was assuming that the \$\phi\$ phase constant was arbitrary, but it is the result of a contraction from: $$\cos\omega_n\sqrt{1-\zeta^2}t+\frac{\zeta}{\sqrt{1-\zeta^2}}\sin\omega_n\sqrt{1-\zeta^2}t\textrm{.}$$ This can be represented as $$\frac{1}{\sqrt{1-\zeta^2}}\cos\left(\omega_n\sqrt{1-\zeta^2}-\phi\right),$$ with \$\phi=\tan^{-1}{\frac{\zeta}{\sqrt{1-\zeta^2}}}\$.

It is also fairly trivial to find that $$\frac{\sigma_d}{\omega_d}=\frac{\zeta\omega_n}{\omega_n\sqrt{1-\zeta^2}}=\frac{\zeta}{\sqrt{1-\zeta^2}},$$ so the \$\tan^{-1}\$ term and the \$\phi\$ term cancel in the original equation, leaving $$\frac{n\pi}{\omega_n\sqrt{1-\zeta^2}}.$$ For \$n=0\$, this is the lowest point before the response begins, so for \$n=1\$, this gives the first peak value, as expected. Additional thanks to other answers which gave guidance.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ I was especially wanting to call your attention to the unitless portion of the general solution:$$\propto \left\{1-e^{{\zeta \frac{t}{\tau}}} \cdot \left( \cos \left[ \sqrt{1-\zeta^2} \cdot \frac{t}{\tau} \right] + \frac{\zeta}{\sqrt{1-\zeta^2}}\cdot\sin\left[\sqrt{1-\zeta^2} \cdot \frac{t}{\tau} \right] \right) \right\}$$Glad to see you have what you need. \$\endgroup\$ – jonk May 17 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.