0
$\begingroup$

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 $$sE(s)=\frac{s}{s^2} \left[1-\frac{K(K_1s + K_2)}{Ts^2 + (K K_1 +1)s + K K_2}\right]$$ I multiplied so as to combine the two fractions: $$\frac{1}{s}\cdot\frac{Ts+ KK_1 + 1 + \frac{KK_2}{s}}{Ts+ KK_1 + 1 + \frac{KK_2}{s}}$$ And I'm left with $$\frac{Ts+ KK_1 + 1 + \frac{KK_2}{s} - KK_1s - KK_2}{Ts^2 + (K K_1 +1)s + K K_2}$$

Unfortunately, I can't figure how to simplify out the numerator.

The calculator states the answer is $$\frac{Ts+1}{Ts^2+(K K_1 +1)s+K K_2}$$ I can deal with that, as the limit will simply be $\frac{1}{KK_2}$. But what are the steps in between? I don't know how to get rid of the $KK_x$ terms in the numerator as I need to.

Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ You made a mistake when you added the fractions. I'm not sure why you divided by $s$ everywhere. Making a common denominator gives: $$\frac{1}{s}\left(\frac{(Ts^2+(KK_1+1)s+KK_2)-(K(K_1s+K_2))}{Ts^2+(KK_1+1)s+KK_2}\right).$$ $\endgroup$ – kccu Apr 28 '17 at 21:02
1
$\begingroup$

I believe your mistake is somewhere in how you combined the fractions initially. Here's what the simplification should look like: \begin{align*} sE(s)&=\frac{s}{s^2} \left[1-\frac{K(K_1s + K_2)}{Ts^2 + (K K_1 +1)s + K K_2}\right]\\ &=\frac{1}{s} - \frac{K(K_1s + K_2)}{s(Ts^2 + (K K_1 +1)s + K K_2)}\\ &=\frac{Ts^2 + (K K_1 +1)s + K K_2-K(K_1s + K_2)}{s(Ts^2 + (K K_1 +1)s + K K_2)}\\ &=\frac{Ts^{2}+s}{s(Ts^2 + (K K_1 +1)s + K K_2)}\\ &=\frac{Ts+1}{Ts^2 + (K K_1 +1)s + K K_2} \end{align*}

$\endgroup$
  • $\begingroup$ Dang, I can't believe I missed that. I've spent over an hour reworking this thing too. Thank you... and boy do I feel stupid. $\endgroup$ – Asinine Apr 28 '17 at 21:04
  • $\begingroup$ You're welcome, it happens to the best of us. $\endgroup$ – DMcMor Apr 28 '17 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.