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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

PL510
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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.

Dirceu Rodrigues Jr
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  • This is right. The point of finding the % error in this example, is to determine the loss of accuracy that is introduced due to the change in road grade. The author then uses this figure to illustrate the benefit of adding closed loop gain. The best way to make this comparison is to find the % error in both cases (open loop & closed loop) and compare the results. – Blair Fonville Sep 12 '18 at 04:05