(New page: <math>\int_{-\infty}^{\infty} h(\tau)\,\mathrm{d}\tau \,=\, \int_{0}^{\infty} h(\tau)\,\mathrm{d}\tau \;\;\;\;\; \because h(t)=0 \;\;\; \forall t<0</math> <math>\Rightarrow \int_{0}^{\inf...)
 
Line 1: Line 1:
 
<math>\int_{-\infty}^{\infty} h(\tau)\,\mathrm{d}\tau \,=\, \int_{0}^{\infty} h(\tau)\,\mathrm{d}\tau \;\;\;\;\; \because h(t)=0 \;\;\; \forall t<0</math>
 
<math>\int_{-\infty}^{\infty} h(\tau)\,\mathrm{d}\tau \,=\, \int_{0}^{\infty} h(\tau)\,\mathrm{d}\tau \;\;\;\;\; \because h(t)=0 \;\;\; \forall t<0</math>
  
<math>\Rightarrow \int_{0}^{\infty} \mathrm{e}^{-\tau}\,\mathrm{d}\tau</math>
+
<math>\Rightarrow \int_{0}^{\infty} \mathrm{e}^{-\tau}\,\mathrm{d}\tau \,=\, \left.-\mathrm{e}^{-\tau}\right|_{0}^{\infty}</math>
  
 
<math>\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</math>
 
<math>\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</math>
  
<math>\Rightarrow \int_{-\infty}^{t} \mathrm{e}^{t-\tau}\,\mathrm{d}\tau</math>
+
<math>\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}</math>

Revision as of 18:07, 10 February 2013

$ \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