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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?

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  • \$\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
    Commented Jul 24, 2016 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
    Commented Jul 24, 2016 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
    Commented Jul 24, 2016 at 15:58

1 Answer 1

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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.

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  • \$\begingroup\$ thanks, I'll check out that well known limit then... it's been a while \$\endgroup\$
    – frankk
    Commented Jul 24, 2016 at 16:03
  • \$\begingroup\$ alright, taylor series expansion of the square rooted expression yields that limit... am good now, thanks. \$\endgroup\$
    – frankk
    Commented Jul 24, 2016 at 16:19

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