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The peak time of a second order system with a step input can be easily calculated as shown below. Is it possible to do the same when input is a rectangular pulse?

To explain, A second order system can be represented as: $$\frac{C(s)}{R(s)} = \frac{\omega^{2}_n}{s (s^2 + 2\zeta\omega_n s + \omega^2_n)}$$ The system is underdamped, so the time domain equation is: $$c(t) = 1- \frac{e^{-\zeta\omega_n t}}{\sqrt{1-\zeta^2}} \sin(\omega_n\sqrt{ (1-\zeta^2)} \cdot t + \phi)$$ At peak response, $$\frac{dc(t)}{dt} = 0$$ $$\frac{e^{-\zeta \omega_n t}}{\sqrt{1-\zeta^2}} \big[- \omega_n \sqrt{1-\zeta^2}\cdot \cos(\omega_n \sqrt{1-\zeta^2}t + \phi)+ \zeta \omega_n \sin (\omega_n \sqrt{1-\zeta^2}t + \phi)\big] = 0 $$ $$\tan[\omega_n \sqrt{1-\zeta^2}t + \phi] = \frac{\sqrt{1-\zeta^2}}{\zeta} = \tan\phi $$ $$[\omega_n \sqrt{1-\zeta^2}t + \phi] = n \pi$$ for n = 1,2,3,4... $$t_{peak} = \frac{\pi}{\omega_n \sqrt{1-\zeta^2}}$$

I would like to derive the same way but with rectangular pulse input, $$\frac{C(s)}{R(s)} = \bigg[\frac{e^{-as} - e^{-bs}}{s}\bigg] \frac{\omega^{2}_n}{(s^2 + 2\zeta\omega_n s + \omega^2_n)} $$ where a and b are the start and end time of the pulse. In the rectagular pulse case, it seems difficult to separate time variable from others with the exponential term staying on the left hand side. Is there a method or reference I can use?

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  • $\begingroup$ Please type up your question using MathJax, rather than expecting people to click on a link. $\endgroup$
    – user1729
    Nov 28 '19 at 16:44
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    $\begingroup$ Thank you! I have changed the question accordingly. I am new to this site. $\endgroup$
    – Jay
    Dec 5 '19 at 14:34
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    $\begingroup$ Is it possible to reopen the post? I have tagged more relevant groups now and made the question more explanatory. $\endgroup$
    – Jay
    Dec 6 '19 at 8:44
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Those exponential terms just introduce time delay. So, the response of the second system is $$ c(t)|_{t=t-a} - c(t)|_{t=t-b} $$ where $c(t)$ is defined above. If the on and off times of the rectangular input is long enough (meaning bigger than $t_{\operatorname{peak}}$) then peak time doesn't change.

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