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

  • 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


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.


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.

  • 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

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