Question: List the performance specifications of Time Response Analysis and derive any four of them.
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Mumbai University > Electronics Engineering > Sem 4 > Linear Control Systems

Topic : Time Response Analysis

Difficulty : High

Marks : 10M

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modified 6 months ago by gravatar for tanya.tanyabarnwal tanya.tanyabarnwal0 written 7 months ago by gravatar for Abhishek Tiwari Abhishek Tiwari50
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Specifications of Time response analysis :-

  1. Delay Time $T_d$

  2. Rise Time $T_r$

  3. Peak Time $T_p$

  4. Peak overshoot $M_p$

  5. Settling Time $T_s$

Derivations ::

I) Rise time $(t_{r})$:

The unit step response of second order system for under-damped case is given by

$$c(t)=1-\cfrac{e^{- \varepsilon_{1} \omega_{n} t_{r}}}{\sqrt{1-\varepsilon_{1}^{2}}} \sin{(\omega_{d} t + \theta)}$$

At $t=t_{r}, c(t)=c(t_{r})=1$.

$$\therefore c(t_{r})=1-\cfrac{e^{- \varepsilon_{1} \omega_{n} t_{r}}}{\sqrt{1-\varepsilon_{1}^{2}}} \sin{(\omega_{d} t_{r} + \theta)}=1$$ $$\therefore \cfrac{e^{- \varepsilon_{1} \omega_{n} t_{r}}}{\sqrt{1-\varepsilon_{1}^{2}}} \sin{(\omega_{d} t_{r} + \theta)}=0$$

Since, $$-e^{- \varepsilon_{1} \omega_{n} t_{r}} \neq 0$$

then the term, $$\sin{(\omega_{d} t_{r} + \theta)}=0$$

When, $\phi =0, \pi, 2\pi, 3\pi,....$

$\therefore \sin{\phi}=0$

$\therefore \omega_{d} t_{r} + \theta=\pi$

$\therefore \omega_{d} t_{r} =\pi - \theta$

$\therefore t_{r}=\cfrac{\pi - \theta}{\omega_{d}}$

$\therefore$Rise Time$=t_{r}=\cfrac{\pi - \theta}{\omega_{d}}$

Here, $\theta=\tan^{-1}{(\cfrac{\sqrt{1-\varepsilon_{1}^{2}}}{\varepsilon})}$

$\omega_{d}=\omega_{n} \sqrt{1-\varepsilon_{1}^{2}}$

II) Peak time $(t_{p})$:

To find the expression for peak time, $t_{p}$, differentiate between c(t) with respect to t and equal to 0, i.e., $\cfrac{d}{dt} c(t) |_{t=t_{p}}=0$

The unit step response of under damped second order system is given by

$c(t)=1-\cfrac{e^{- \varepsilon_{1} \omega_{n} t_{r}}}{\sqrt{1-\varepsilon_{1}^{2}}} \sin{(\omega_{d} t + \theta)}$

Differentiating c(t) with respect to $t$, $\cfrac{d}{dt}c(t)=-\cfrac{e^{- \varepsilon_{1} \omega_{n} t_{r}}}{\sqrt{1-\varepsilon_{1}^{2}}} (-\varepsilon_{1} \omega_{n}) \sin{(\omega_{d} t + \theta)} + (-\cfrac{e^{- \varepsilon_{1} \omega_{n} t_{r}}}{\sqrt{1-\varepsilon_{1}^{2}}}) \cos{(\omega_{d} t + \theta)} \omega_{d}$ Put $\omega_{d}=\omega_{n} \sqrt{1-\varepsilon_{1}^{2}}$ $\therefore \cfrac{d}{dt}c(t)=\cfrac{e^{- \varepsilon_{1} \omega_{n} t_{r}}}{\sqrt{1-\varepsilon_{1}^{2}}} (\varepsilon_{1} \omega_{n}) \sin{(\omega_{d} t + \theta)} - \cfrac{\omega_{n} \sqrt{1-\varepsilon_{1}^{2}}}{\sqrt{1-\varepsilon_{1}^{2}}} e^{- \varepsilon_{1} \omega_{n} t_{r}} \cos{(\omega_{d} t + \theta)}$ =$\cfrac{\omega_{n} e^{- \varepsilon_{1} \omega_{n} t_{r}}}{\sqrt{1-\varepsilon_{1}^{2}}} [\varepsilon_{1} \sin{(\omega_{d} t + \theta)}-\sqrt{1-\varepsilon_{1}^{2}} \cos{(\omega_{d} t + \theta)} ]$ =$\cfrac{\omega_{n} e^{- \varepsilon_{1} \omega_{n} t_{r}}}{\sqrt{1-\varepsilon_{1}^{2}}} [\cos{\theta} \sin{(\omega_{d} t + \theta)}-\sin{\theta} \cos{(\omega_{d} t + \theta)} ]$ =$\cfrac{\omega_{n} e^{- \varepsilon_{1} \omega_{n} t_{r}}}{\sqrt{1-\varepsilon_{1}^{2}}} [\sin{(\omega_{d} t + \theta)} \cos{\theta} - \cos{(\omega_{d} t + \theta)} \sin{\theta} ]$ =$\cfrac{\omega_{n} e^{- \varepsilon_{1} \omega_{n} t_{r}}}{\sqrt{1-\varepsilon_{1}^{2}}} [\sin{((\omega_{d} t + \theta)-\theta)}] $ =$\cfrac{\omega_{n} e^{- \varepsilon_{1} \omega_{n} t_{r}}}{\sqrt{1-\varepsilon_{1}^{2}}} \sin{(\omega_{d} t)}$ At $t=t_{p}$, $\cfrac{d}{dt}c(t)=0$ $\therefore \cfrac{\omega_{n} e^{- \varepsilon_{1} \omega_{n} t_{r}}}{\sqrt{1-\varepsilon_{1}^{2}}} \sin{(\omega_{d} t_{p})}=0$ Since, $e^{-\varepsilon_{1} \omega_{n} t_{p}} \neq0$ the term $\sin{(\omega_{d} t_{p})}=0$ When $\phi=0, \pi, 2\pi, 3\pi,....$ $\sin{\phi}=0$ $\therefore \omega_{d} t_{p} = \pi$ $\therefore t_{p}=\cfrac{\pi}{\omega_{d}}$ $\therefore$ Peak time$=t_{p}=\cfrac{\pi}{\omega_{d}}$ $\therefore \omega_{d}=\omega_{n} \sqrt{1-\varepsilon_{1}^{2}}$ III) **Settling time $(t_{s})$** The response of second order system has two components. They are: 1) Decaying exponential component $ \cfrac{e^{- \varepsilon_{1} \omega_{n} t}}{\sqrt{1-\varepsilon_{1}^{2}}}$ 2) Sinusoidal component, $\sin{(\omega_{d} t + \theta)}$ In this, the decaying exponential term dampens or reduces the oscillations produced by sinusoidal component. Hence, the settling time is decided by the exponential component. The settling time can be found out by equating exponential component to percentage tolerance errors. For 2% tolerance error band at $t=t_{s}, \cfrac{e^{- \varepsilon_{1} \omega_{n} t_{s}}}{\sqrt{1-\varepsilon_{1}^{2}}}=0.02$ For least value of $\varepsilon_{1}$, $e^{- \varepsilon_{1} \omega_{n} t_{s}}=0.02$ Taking natural log on both sides, $ln[e^{- \varepsilon_{1} \omega_{n} t_{s}}]=ln[0.02]$ $- \varepsilon_{1} \omega_{n} t_{s}=ln[0.02]$ $- \varepsilon_{1} \omega_{n} t_{s}=-4$ $t_{s}=\cfrac{-4}{- \varepsilon_{1} \omega_{n} }$ $t_{s}=\cfrac{4}{ \varepsilon_{1} \omega_{n} }$ IV) **Peak Overshoot $(M_{p})$:** The unit step response of a second order system is given by: $c(t)=1-\cfrac{e^{- \varepsilon_{1} \omega_{n} t}}{\sqrt{1-\varepsilon_{1}^{2}}} \sin{(\omega_{d} t + \theta)}$----(1) At $t=\infty$ in equation (1), $c(\infty)=c(t)=1-\cfrac{e^{- \varepsilon_{1} \omega_{n} (\infty)}}{\sqrt{1-\varepsilon_{1}^{2}}} \sin{(\omega_{d} t + \theta)}$ $c(\infty)=c(t)=1-0=1$ At $t=t_{p}$ in equation (1), $c(t)=c(t_{p})=1-\cfrac{e^{- \varepsilon_{1} \omega_{n} t_{p}}}{\sqrt{1-\varepsilon_{1}^{2}}} \sin{(\omega_{d} t_{p} + \theta)}$ But $t_{p}=\cfrac{\pi}{\omega_{d}}$ $c(t)=c(t_{p})=1-\cfrac{e^{- \varepsilon_{1} \omega_{n} \cdot \cfrac{\pi}{\omega_{d}}}}{\sqrt{1-\varepsilon_{1}^{2}}} \sin{(\omega_{d} \cdot \cfrac{\pi}{\omega_{d}} + \theta)}$ $c(t)=c(t_{p})=1-\cfrac{e^{- \varepsilon_{1} \omega_{n} \cdot \cfrac{\pi}{\omega_{d}}}}{\sqrt{1-\varepsilon_{1}^{2}}} \sin{(\pi+ \theta)}$ But we know, $\sin{(\pi + \theta)} =-\sin{\theta}$ $\omega_{d}=\omega_{n} \sqrt{1-\varepsilon_{1}^{2}}$ $c(t)=c(t_{p})=1-\cfrac{e^{- \varepsilon_{1} \omega_{n} \cdot \cfrac{\pi}{\omega_{n} \sqrt{1-\varepsilon_{1}^{2}}}}}{\sqrt{1-\varepsilon_{1}^{2}}} (-\sin{\theta})$ $c(t)=c(t_{p})=1+\cfrac{e^{- \cfrac{ \varepsilon_{1} \pi}{ \sqrt{1-\varepsilon_{1}^{2}}}}}{\sqrt{1-\varepsilon_{1}^{2}}} \sin{\theta}$ On constructing a right angled triangle, we get $sin{\theta}=\sqrt{1-\varepsilon_{1}^{2}}$ ![enter image description here][1] $c(t)=c(t_{p})=1+\cfrac{e^{- \cfrac{ \varepsilon_{1} \pi}{ \sqrt{1-\varepsilon_{1}^{2}}}}}{\sqrt{1-\varepsilon_{1}^{2}}} \cdot\sqrt{1-\varepsilon_{1}^{2}}$ $c(t)=c(t_{p})=1+e^{- \cfrac{ \varepsilon_{1} \pi}{ \sqrt{1-\varepsilon_{1}^{2}}}}$ $\therefore$Percentage peak overshoot$=M_{p}=\cfrac{c(t_{p}-c(\infty))}{c(\infty)} \times 100=\cfrac{1+e^{- \cfrac{ \varepsilon_{1} \pi}{ \sqrt{1-\varepsilon_{1}^{2}}}}-1}{1} \times 100$

$\therefore \%M_{p}=e^{- \cfrac{ \varepsilon_{1} \pi}{ \sqrt{1-\varepsilon_{1}^{2}}}}\times 100$

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modified 6 months ago by gravatar for Abhishek Tiwari Abhishek Tiwari50 written 6 months ago by gravatar for tanya.tanyabarnwal tanya.tanyabarnwal0
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