1
\$\begingroup\$

Given the simple transfer function of a double pole:

$$ H(s)=\frac{1}{(1+as)^2}=\frac{1}{1+s2a + s^2 a^2} = \frac{1}{1+s k_1 + s^2 k_2} $$

Its inverse Laplace transform is (e.g. [1]):

$$ h(t) = -\frac{\cdots}{\sqrt{k_1^2 - 4 k_2}} $$

The expression in the root becomes zero and hence h(t) undefined. The system is perfectly stable and legit.

I thought maybe there are multiple expressions going to undefined values such that I need to invoke de l'Hopital. However this is not the case.

What am I missing?

[1] http://www.wolframalpha.com/input/?i=inverse+laplace+transform&rawformassumption=%7B%22F%22,+%22InverseLaplaceTransformCalculator%22,+%22transformfunction%22%7D+-%3E%221%2F(1+%2B+s*k1+%2B+s%5E2*k2)%22&rawformassumption=%7B%22F%22,+%22InverseLaplaceTransformCalculator%22,+%22variable1%22%7D+-%3E%22s%22&rawformassumption=%7B%22F%22,+%22InverseLaplaceTransformCalculator%22,+%22variable2%22%7D+-%3E%22t%22&rawformassumption=%7B%22C%22,+%22inverse+laplace+transform%22%7D+-%3E+%7B%22Calculator%22%7D

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Can't you assign \$b=\frac{1}{a}\$ and then solve for \$\mathscr{L}^{-1}\frac{b^2}{\left(s+b\right)^2},\:a\ne 0\$? \$\endgroup\$
    – jonk
    Jun 1, 2018 at 4:45
  • 1
    \$\begingroup\$ Great point! I did that now and it fixed it. I am still curious though why the full-blown solution fails \$\endgroup\$
    – divB
    Jun 1, 2018 at 7:53

2 Answers 2

2
\$\begingroup\$

Here's the transfer function:

$$\begin{align*} H_s&=\frac{1}{\left(1+a\:s\right)^2} \end{align*}$$

(NOTE: It my help others to know that this transfer function can be generated, if \$a=\tau=R\: C\$, by an RC low-pass, followed by an ideal buffer to remove loading effects, followed by an identical RC low-pass. This is always critically damped.)

The solution to the inverse Laplace can be pursued in several different ways. But convolution leads to a little less writing.

$$\begin{align*} \mathscr{L}^{-1}\left\{H_s\right\}&=\mathscr{L}^{-1}\left\{\frac{1}{\left(1+a\:s\right)^2}\right\}\\\\ &=\mathscr{L}^{-1}\left\{\frac{1}{a^2\left(s+\frac{1}{a}\right)^2}\right\},\text{ set }b=\frac{1}{a}\\\\ &=\mathscr{L}^{-1}\left\{\frac{b^2}{\left(s+b\right)^2}\right\}\\\\ &=\mathscr{L}^{-1}\left\{\frac{b}{s+b}\cdot\frac{b}{s+b}\right\}\\\\&&\text{set }F_s&=\frac{b}{s+b}\\\\&&\therefore f_t&=b\:e^{\:-b\:t}\\\\ &=\mathscr{L}^{-1}\left\{F_s\: F_s\right\}\\\\ &=\left(f * f\right)_t\\\\ &=\int_0^t\:b\:e^{\:-b\:\left(t-v\right)}\:b\:e^{\:-b\:v}\:\text{d} v\\\\ &= b^2\int_0^t\:e^{\:-b\:\left(t-v\right)}\:e^{\:-b\:v}\:\text{d} v\\\\ &= b^2\int_0^t\:e^{\:-b\:t}\:\text{d} v\\\\ &= b^2\:e^{\:-b\:t}\int_0^t\:\text{d} v\\\\ &= b^2\:t\:e^{\:-b\:t}\\\\ &=\frac{1}{a}\:\frac{t}{a}\:e^\frac{-t}{a} \end{align*}$$

Or, using \$\tau=a\$,

$$\begin{align*} \mathscr{L}^{-1}\left\{\frac{1}{\left(1+\tau\:s\right)^2}\right\}&= \frac{1}{\tau}\:\frac{t}{\tau}\:e^\frac{-t}{\tau} \end{align*}$$

The above breaks into two parts, one with units (\$\frac{1}{\tau}\$) and one that is unitless (\$\frac{t}{\tau}\:e^\frac{-t}{\tau}\$.) If you Spice out the circuit I described in the above note (RC, buffer, RC) and hit it with a Dirac impulse (a nice tall spike for a very short time relative to the RC time constant), then you will see exactly this output using a .TRAN.

The peak should occur when the derivative is 0, or when \$t=\tau\$. So this means the peak of the curve that Spice shows should be \$\frac{1}{e\:\tau}\$. And obviously, this peak should occur at \$t=\tau\$ on the time plot. (The area in the Dirac impulse will equal the area under the time plotted curve.)


Let's do it. Here's the Spice schematic:

enter image description here

Here's the output:

enter image description here

Feel free to also plot the equation itself and see how well they match.

\$\endgroup\$
0
\$\begingroup\$

Remember that there are three conditions for a 2nd order system: overdamped, critically damped, and underdamped, which means that the denominator has three possible results:

  1. overdamped -- means the 4*k2 < k12 => the impulse response has a sinh() term, thus of the form exp(-t)*sinh(t);

  2. critically damped -- 4*k2 = k12 => the impulse response has only an exp(-t) term, while the denominator also simplifies to 2*k2;

  3. underdamped -- 4*k2 > k12 => the impulse response has is of the form exp(-t)*sin(t);

I could write the direct expressions, but that would be robbing you of the fun.

\$\endgroup\$
2
  • 2
    \$\begingroup\$ Well, I thought about that but the system I described is critically damped by definition! $$(1+as)^2 = (1+2as+a^2 s^2) = 1+s/(Q \omega_0) + s^2/\omega_0^2$$ with $$\omega_0=1/a$$ and $$Q=1/2$$ (!). The result I get is the most generic form so it should not fail for a narrower case (Q=1/2). \$\endgroup\$
    – divB
    Jun 1, 2018 at 7:52
  • \$\begingroup\$ @divB As I said, I would have robbed you of the fun of it, that's why I kept it in a general case, where you were supposed to figure out which case you're in -- you did (congratulations), the second, where the denominator is not zero, but the condition makes it transform into 2*k2. jonk gives a (very good) complete answer, but for cases like these, I usually try to avoid that. \$\endgroup\$ Jun 1, 2018 at 9:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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