1
\$\begingroup\$

I'm working my way through "Foundations of Analog and Digital Circuits (Agarwal)" and am, for the first time, stumped.

Specifically, they analyze an over-damped, undriven series RLC circuit in section 12.2.2. To illustrate the circuit's behavior when R grows very large, the limits of the roots s1 and s2 as R tends to infinity are computed.

Like so:

Limits of s1 and s1 as R tends to infinity

I'm fine with the result for s1. The one for s2 has me baffled and particularly so in light of the result for s1. For 12.70, I see the expression in parentheses evaluating to '1 - 1' when R grows very large (just as it evaluated to '1 + 1' in the case of 12.69).

What am I missing?

\$\endgroup\$
  • \$\begingroup\$ Assume that nobody has a copy of the book and post the actual circuit and maths behind the formulas you are currently showing. \$\endgroup\$ – Andy aka Jul 24 '16 at 15:15
  • \$\begingroup\$ @frankk: The expressions under the square root symbols are identical for both \$s_1\$ and \$s_2\$, so how can one become 1+1 and the other 1-1? I don't understand your "baffling". \$\endgroup\$ – Curd Jul 24 '16 at 15:34
  • \$\begingroup\$ @Curd Whoops, my bad. I was thinking of the expression in parentheses but for some reason referred to the one under the square root. \$\endgroup\$ – frankk Jul 24 '16 at 15:58
0
\$\begingroup\$

When alpha grows the expression under square root do not tend to 1-1, it tends to 1-0 or better to

$$ \sqrt{1-\left(\frac{\omega_0}{\alpha}\right)^2}\rightarrow1-\frac{1}{2}\left(\frac{\omega_0}{\alpha}\right)^2 $$

it's a well known limit.

$$ a\rightarrow 0\quad\quad\Rightarrow\quad\quad\sqrt{1\pm a}\rightarrow1\pm\frac{a}{2}$$

In case 12.69 we disregard the (small) fraction against 2.

In case 12.70 the (small) fraction is the only thing remaining.

\$\endgroup\$
  • \$\begingroup\$ thanks, I'll check out that well known limit then... it's been a while \$\endgroup\$ – frankk Jul 24 '16 at 16:03
  • \$\begingroup\$ alright, taylor series expansion of the square rooted expression yields that limit... am good now, thanks. \$\endgroup\$ – frankk Jul 24 '16 at 16:19

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.