- #1
amcavoy
- 665
- 0
I need to show that for f(t)=f(t+T) on [0,infty), that the Laplace Transform is:
[tex]\mathcal{L}\left[f(t)\right]=\frac{\int_0^Te^{-st}f(t)\,dt}{1-e^{-sT}}.[/tex]
The first thing I did was to write the transform as:
[tex]\mathcal{L}\left[f(t)\right]=\sum_{n=0}^{\infty}\int_{nT}^{\left(n+1\right)T}e^{-st}f(t)\,dt.[/tex]
Am I on the right track here? It looks like the formula given to me (that I need to show) is an infinite geometric series multiplied by the integral in the numerator. However, I am unable to get what I have into something of that form. Any ideas?
Thank you.
[tex]\mathcal{L}\left[f(t)\right]=\frac{\int_0^Te^{-st}f(t)\,dt}{1-e^{-sT}}.[/tex]
The first thing I did was to write the transform as:
[tex]\mathcal{L}\left[f(t)\right]=\sum_{n=0}^{\infty}\int_{nT}^{\left(n+1\right)T}e^{-st}f(t)\,dt.[/tex]
Am I on the right track here? It looks like the formula given to me (that I need to show) is an infinite geometric series multiplied by the integral in the numerator. However, I am unable to get what I have into something of that form. Any ideas?
Thank you.