- #1
maverick280857
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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.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Let 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<br /> <br /> M(s) = \frac{Y(s)}{R(s)} = \frac{G(s)}{1+G(s)H(s)} = \frac{1}{s(s^2+12s+10)}<br /> <br /> For a unit step input R(s) = 1/s. The output time response is<br /> <br /> y(t) = 0.1t - 0.12 - 0.000796e^{-11.1t} + 0.1208e^{-0.901t}<br /> <br /> for t \geq 0<br /> <br /> 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<br /> <br /> e_{ss} = \lim_{t\rightarrow \infty}\left[0.1t - y(t)\right] = 0.12 </div> </div> </blockquote><br /> 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?<br /> <br /> Thanks in advance.
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.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Let 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<br /> <br /> M(s) = \frac{Y(s)}{R(s)} = \frac{G(s)}{1+G(s)H(s)} = \frac{1}{s(s^2+12s+10)}<br /> <br /> For a unit step input R(s) = 1/s. The output time response is<br /> <br /> y(t) = 0.1t - 0.12 - 0.000796e^{-11.1t} + 0.1208e^{-0.901t}<br /> <br /> for t \geq 0<br /> <br /> 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<br /> <br /> e_{ss} = \lim_{t\rightarrow \infty}\left[0.1t - y(t)\right] = 0.12 </div> </div> </blockquote><br /> 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?<br /> <br /> Thanks in advance.