# Query regarding steady state error in a velocity control system

## 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

$$G(s) = \frac{1}{s^2(s+12)}$$

and the feedback transfer function be

$$H(s) = K_{t}s$$

where $K_{t}[/tex] is the tachometer constant. Let [itex]K_{t} = 10$ 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

$$M(s) = \frac{Y(s)}{R(s)} = \frac{G(s)}{1+G(s)H(s)} = \frac{1}{s(s^2+12s+10)}$$

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

$$y(t) = 0.1t - 0.12 - 0.000796e^{-11.1t} + 0.1208e^{-0.901t}$$

for $t \geq 0$

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

$$e_{ss} = \lim_{t\rightarrow \infty}\left[0.1t - y(t)\right] = 0.12$$
I am not clear about how the steady state error has been computed here. I understand that the dominating terms as $t\rightarrow \infty$ 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?

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Anyone?

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?

Got it...thanks to varunag :-)

It should be 10V/rad/sec. The output is the angle measure.

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?

$$\lim_{t\to \infty} f(t)\star u(t) = \lim_{s\to 0}sf(s)u(s)\$$

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