Revision as of 18:07, 10 February 2013 by Green26 (Talk | contribs)

$ \int_{-\infty}^{\infty} h(\tau)\,\mathrm{d}\tau \,=\, \int_{0}^{\infty} h(\tau)\,\mathrm{d}\tau \;\;\;\;\; \because h(t)=0 \;\;\; \forall t<0 $

$ \Rightarrow \int_{0}^{\infty} \mathrm{e}^{-\tau}\,\mathrm{d}\tau \,=\, \left.-\mathrm{e}^{-\tau}\right|_{0}^{\infty} $

$ \int_{-\infty}^{\infty} h(t-\tau)\,\mathrm{d}\tau \,=\, \int_{-\infty}^{t} h(t-\tau)\,\mathrm{d}\tau \;\;\;\;\; \because h(t)=0 \;\;\; \forall t<0 $

$ \Rightarrow \int_{-\infty}^{t} \mathrm{e}^{-(t-\tau)}\,\mathrm{d}\tau \,=\, \int_{-\infty}^{t} \mathrm{e}^{\tau-t}\,\mathrm{d}\tau \,=\, \left.-\mathrm{e}^{\tau-t}\right|_{-\infty}^{t} $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn