Inverse Laplace transform of double pole
Here's the transfer function:

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

(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.

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

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

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

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:



Here's the output:



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