![Step responses for a second order system defined by the transfer function: H ( s ) = ω n 2 s 2 + 2 ζ ω n s + ω n 2 {\displaystyle H(s)={\frac {\omega _{n}^{2}}{s^{2}+2\zeta \omega _{n}s+\omega _{n}^{2}}}} where ζ {\displaystyle \zeta } is the damping ratio and ω n {\displaystyle \omega _{n}} is the undamped natural frequency. The equations were obtained from here, plotted using maxima and edited in a text editor to insert the Greek alphabets in the plot. The equations are: h u n d e r ( t ) = 1 − 1 β ( ζ ) e − ζ ω n t sin ( β ( ζ ) ω n t + θ ( ζ ) ) , β ( ζ ) = 1 − ζ 2 , θ ( ζ ) = tan − 1 ( β ( ζ ) ζ ) {\displaystyle h_{under}(t)=1-{\frac {1}{\beta (\zeta )}}e^{-\zeta \omega _{n}t}\sin \left(\beta (\zeta )\omega _{n}t+\theta (\zeta )\right),\beta (\zeta )={\sqrt {1-\zeta ^{2}}},\theta (\zeta )=\tan ^{-1}\left({\frac {\beta (\zeta )}{\zeta }}\right)} h u n ( t ) = 1 − cos ( ω n t ) {\displaystyle h_{un}(t)=1-\cos \left(\omega _{n}t\right)} h c r i t ( t ) = 1 − e − ω n t ( 1 + ω n t ) {\displaystyle h_{crit}(t)=1-e^{-\omega _{n}t}\left(1+\omega _{n}t\right)} h o v e r ( t ) = 1 + ω n 2 ζ 2 − 1 ( e − s 1 t s 1 − e − s 2 t s 2 ) , s 1 = ( ζ + ζ 2 − 1 ) ω n , s 2 = ( ζ − ζ 2 − 1 ) ω n {\displaystyle h_{over}(t)=1+{\frac {\omega _{n}}{2{\sqrt {\zeta ^{2}-1}}}}\left({\frac {e^{-s_{1}t}}{s_{1}}}-{\frac {e^{-s_{2}t}}{s_{2}}}\right),s_{1}=\left(\zeta +{\sqrt {\zeta ^{2}-1}}\right)\omega _{n},s_{2}=\left(\zeta -{\sqrt {\zeta ^{2}-1}}\right)\omega _{n}}](http://upload.wikimedia.org/wikipedia/commons/thumb/4/4f/Second_order_transfer_function.svg/631px-Second_order_transfer_function.svg.png)
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Date/Time | Thumbnail | Dimensions | User | Comment | |
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current | 15:57, 7 June 2011 | ![]() | 631 × 356 (80 KB) | Krishnavedala | figure correction after correcting the equations |
19:25, 26 May 2011 | ![]() | 800 × 400 (36 KB) | Krishnavedala | erroneous under-damped curve corrected. | |
18:28, 26 May 2011 | ![]() | 800 × 400 (36 KB) | Krishnavedala |
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