Laplace transform of periodic function

Let f be a periodic function of period T >0, which take value zero on the negative real axis.
Suppose that

$$f_0(t)=f(t)(h(t-T)-h(t))$$

is integrable.
Then Laplace transform of f

$$\mathcal{L}(f(t))(s)=\int_{0}^{+\infty}f(t)e^{-st}dt=\frac{1}{1-e^{-sT}}\int_{0}^{T}f(t)e^{-st}dt=\frac{1}{1-e^{-sT}}\mathcal{L}(f_0(t))(s)$$

Indeed we have

$$\mathcal{L}(f(t))(s)=\int_{0}^{+\infty}f(t)e^{-st}dt=\sum_{n=0}^{+\infty}\int_{nT}^{(n+1)T}f(t)e^{-st}dt=\sum_{n=0}^{+\infty}\int_{0}^{T}f(\tau+nT)e^{-s(\tau+nT)}d\tau$$

=

$$=\sum_{n=0}^{+\infty}e^{-nT}\int_{0}^{T}f(\tau)e^{-s\tau}d\tau=\int_{0}^{T}f(\tau)e^{-s\tau}d\tau\sum_{n=0}^{+\infty}e^{-nsT}=\frac{1}{1-e^{-sT}}\int_{0}^{T}f(\tau)e^{-s\tau}d\tau$$
$$\mathcal{L}(|\sin t| h(t))(s)=\int_{0}^{+\infty}\sin te^{-st}dt=\frac{1}{1-e^{-s\pi}}\int_{0}^{\pi}\sin t e^{-st}dt=\frac{1+e^{-s\pi}}{1-e^{-s\pi}}\frac{1}{s^2+1}$$
Average: 4.5 (448 votes)

Back to top