# Laplace transform of averaging operator

I am studying dc-dc converter now. I got a problem with Laplace transform of the averaging operator as in the image below.

Can anyone help me derive the Laplace transform result $G_{av}(s)$ as in the image?

Here's the outline of the argument, feel free to fill in the details.

The averaging operator is like a convolution with a "square" pulse of height $1/T_s$ supported on the interval $[-T_s/2, T_s/2]$.

You can express the square pulse as a sum of two heaviside step functions.

Finally, recall the Laplace transform of a step function $\mathcal{L} \{H(t-T_s/2)\}(s) = \frac{e^{-sT_s/2}}{s}.$

• Thanks. That is an interesting approach. I am stuck at express square pulse as sum of two heaviside step functions. Because there is a negative part in the pulse, I express it as H(-t) -H(-t-T/2) + H(t) - H(t-T/2). Let ignore 1/T for now. Is that right? Dec 6, 2016 at 21:45
• You want to go "up" at $-T_s/2$, so that means you need $H(t+T_s/2)$. Then you need to go "down" at $T_s/2$, which means you need to subtract out $H(t-T_s/2)$. Dec 6, 2016 at 22:03
• Oh, yeah. Look simple! I made it complicated. I will study the convolution part now. Dec 6, 2016 at 22:36

$$int_{t-T_s/2}^{t+T_s/2}=int_{-\infty}^{t+T_s/2}-int_{-\infty}^{t-T_s/2}$$

Take the Lapalce transform of each term

$$exp^{+sT_s/2}/s-exp^{-sT_s/2}$$ since the Lapalce transform of $$int_{-\infty}^{t}$$ is $$1/s$$ and the time shifting operation $$x(t+-T_s/2)$$----> $$exp^{+-T_s/2}$$