Electronic – speed control loop percent error

feedback

I'm having trouble understanding the percent error from the following closed loop feedback equations. \$w =1\$ is the road grade change and \$r=65\$ is the speed control input, and given closed loop output speed \$Y_{cl}\$ and closed loop error \$e_{cl}\$:
$$ \ Y_{cl} = \frac{100}{101}r-\frac{5}{101}w\
$$
$$ \ e_{cl} = \frac{r}{101}+\frac{5}{101}w\
$$
My question is why is the correct percent error equation as follows:
$$ \%error = 100\frac{\frac{100}{101}r – (\frac{100}{101}r – \frac{5}{101}w)}{\frac{100}{101}r}
\
$$
and not: $$ \ 100\frac{e_{cl}}{r}\
$$

thank you

Best Answer

This is an example that appears in the first chapter of the excellent book Feedback Control of Dynamic Systems, from Prof. Gene F. Franklin. The percent error on output is defined as:

$$ \%error = 100 \times\frac{(Y_{cl}\space when\space w=0)-(Y_{cl}\space when\space w\neq0)}{(Y_{cl}\space when\space w=0)}$$

Note that \$(Y_{cl}\space when\space w=0) = \frac{100}{101}\times r\$

Do the math.

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