Arectemieryf0

2022-07-20

Not seeing steps between factoring fractions
I'm looking to solve the limit of the following error function as s goes to 0, but I'm failing on factoring things out. My calculator (TI-89) gives me a nice form I can use, but I cannot manipulate things quite right. Can anyone point out what I'm doing wrong/missing?
The function starts as
$sE\left(s\right)=\frac{s}{{s}^{2}}\left[1-\frac{K\left({K}_{1}s+{K}_{2}\right)}{T{s}^{2}+\left(K{K}_{1}+1\right)s+K{K}_{2}}\right]$
I multiplied so as to combine the two fractions:
$\frac{1}{s}\cdot \frac{Ts+K{K}_{1}+1+\frac{K{K}_{2}}{s}}{Ts+K{K}_{1}+1+\frac{K{K}_{2}}{s}}$
And I'm left with
$\frac{Ts+K{K}_{1}+1+\frac{K{K}_{2}}{s}-K{K}_{1}s-K{K}_{2}}{T{s}^{2}+\left(K{K}_{1}+1\right)s+K{K}_{2}}$
Unfortunately, I can't figure how to simplify out the numerator.
The calculator states the answer is
$\frac{Ts+1}{T{s}^{2}+\left(K{K}_{1}+1\right)s+K{K}_{2}}$
I can deal with that, as the limit will simply be $\frac{1}{K{K}_{2}}$. But what are the steps in between? I don't know how to get rid of the $K{K}_{x}$ terms in the numerator as I need to.

lelapem

Expert

I believe your mistake is somewhere in how you combined the fractions initially. Here's what the simplification should look like:
$\begin{array}{rl}sE\left(s\right)& =\frac{s}{{s}^{2}}\left[1-\frac{K\left({K}_{1}s+{K}_{2}\right)}{T{s}^{2}+\left(K{K}_{1}+1\right)s+K{K}_{2}}\right]\\ & =\frac{1}{s}-\frac{K\left({K}_{1}s+{K}_{2}\right)}{s\left(T{s}^{2}+\left(K{K}_{1}+1\right)s+K{K}_{2}\right)}\\ & =\frac{T{s}^{2}+\left(K{K}_{1}+1\right)s+K{K}_{2}-K\left({K}_{1}s+{K}_{2}\right)}{s\left(T{s}^{2}+\left(K{K}_{1}+1\right)s+K{K}_{2}\right)}\\ & =\frac{T{s}^{2}+s}{s\left(T{s}^{2}+\left(K{K}_{1}+1\right)s+K{K}_{2}\right)}\\ & =\frac{Ts+1}{T{s}^{2}+\left(K{K}_{1}+1\right)s+K{K}_{2}}\end{array}$

Do you have a similar question?