Impulse response convolution and normalization2 when I take inverse Laplace transform of a system transfer function \

Lets say LPF whose TF is 

$$\frac{Y(s)}{X(s)} \triangleq H(s) = \frac{W}{s+W} $$

the inverse Laplace/impulse response is 

$$h(t) = We^{-Wt}u(t) $$

where $u(t)$ is the Heaviside unit step function:

$$ u(t) \triangleq \begin{cases}
1 \qquad &amp; t \ge 0 \\
0 \qquad &amp; t &lt; 0 \\
\end{cases} $$

Now to see the system response to a square wave $x(t)$ with $|x(t)| = 1$, I need to convolve 

$$ y(t) = h(t) \star x(t) $$

Now if you look at $h(t)$, the maximum amplitude of $h(t)$ is 

$$\max{|h(t)|} = W $$

Then $y(t)$ is amplified by $W$ times. 

So what is happening? how do I normalize this?


Should I normalize this by $\max{|h(t)|}$ or
Should I normalize this with 


$$W_{z_1} W_{z_2} \cdots W_{z_n}/(W_{p_1} W_{p_2} \cdots W_{p_n})$$

(product of zeroes)/(product of poles) of transfer function?

Why is this even happening?
I think you're likely forgetting how the anti-derivative of $h(t)$ affects the gain of the convolution operation.  Recall that somewhere in your convolution integral, you'll be taking an integral of the form $\int We^{W\tau} d\tau$.  The Chain Rule requires the  $W$ in the exponent must appear as a $1/W$ factor after integrating.  This factor cancels the $W$ multiplier in $h(t)$ giving unity gain at DC.