Query regarding steady state error in a velocity control system

1,789
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Main Question or Discussion Point

Hi

I was reading BC Kuo's Automatic Control Systems where I came across a solved problem (page 369 of 7th edition) regarding velocity control. I have a problem understanding how the steady state error has been computed. The original problem and its solution as given in the book are quoted below.

Let the feed-forward transfer function be

[tex]G(s) = \frac{1}{s^2(s+12)}[/tex]

and the feedback transfer function be

[tex]H(s) = K_{t}s[/tex]

where [itex]K_{t}[/tex] is the tachometer constant.

Let [itex]K_{t} = 10[/itex] V/rad-sec. This means that for a unit step of 1 V, the desired velocity in the steady state is 1/10 or 0.1 rad/sec, since when this is achieved, the output voltage of the tachometer would be 1 V and the steady state error would be zero. The closed loop transfer function of the system is

[tex]M(s) = \frac{Y(s)}{R(s)} = \frac{G(s)}{1+G(s)H(s)} = \frac{1}{s(s^2+12s+10)}[/tex]

For a unit step input [itex]R(s) = 1/s[/itex]. The output time response is

[tex]y(t) = 0.1t - 0.12 - 0.000796e^{-11.1t} + 0.1208e^{-0.901t}[/tex]

for [itex]t \geq 0[/itex]

Since the exponential terms of [itex]y(t)[/itex] all diminish as [itex]t \rightarrow \infty[/itex], the steady state part of [itex]y(t)[/itex] is [itex]0.1t-12[/itex]. Thus the steady state error of the system is

[tex]e_{ss} = \lim_{t\rightarrow \infty}\left[0.1t - y(t)\right] = 0.12[/tex]
I am not clear about how the steady state error has been computed here. I understand that the dominating terms as [itex]t\rightarrow \infty[/itex] are the linear term and the constant term, but how does the limit of the (0.1t -y(t)) term represent steady state error? How is the reference signal equal to 0.1t?

Thanks in advance.
 

Answers and Replies

1,789
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Anyone?
 
1,789
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I haven't been able to figure this out yet, so I would be really grateful if someone could look into it for me and tell me what the mistake in my understanding/interpretation is?
 
1,789
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Got it...thanks to varunag :-)

It should be 10V/rad/sec. The output is the angle measure.
 
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For the later occurrences, steady state means change of the function goes to zero when t goes to infinity right? That means if I take the derivative of the function it should go to zero since there is a steady state(existence must be checked of course!) then derivative means multiplying with s right?

[tex]\lim_{t\to \infty} f(t)\star u(t) = \lim_{s\to 0}sf(s)u(s)\[/tex]

If the limit exists it will give you the SS error.
 

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