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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.

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?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]

Thanks in advance.