# 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:

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

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?

1. Should I normalize this by $$\max{|h(t)|}$$ or
2. 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.