(New page: Given: <math>y(t)=x(t)*h(t)=\int_{k=-\infty}^{\infty}x(\tau)h(t-\tau)d\tau</math> #<math>x(t)*(h_1(t)+h_2(t))=\int_{k=-\infty}^{\infty}x(\tau)*(h_1(t-\tau)+h_2(t-\tau)d\tau</math> #<math>...) |
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#<math>x(t)*(h_1(t)+h_2(t))=\int_{k=-\infty}^{\infty}x(\tau)*(h_1(t-\tau)+h_2(t-\tau)d\tau</math> | #<math>x(t)*(h_1(t)+h_2(t))=\int_{k=-\infty}^{\infty}x(\tau)*(h_1(t-\tau)+h_2(t-\tau)d\tau</math> | ||
#<math>x(t)*(h_1(t)+h_2(t))=\int_{k=-\infty}^{\infty}(x(\tau)*h_1(t-\tau)+x(\tau)*h_2(t-\tau))d\tau</math> | #<math>x(t)*(h_1(t)+h_2(t))=\int_{k=-\infty}^{\infty}(x(\tau)*h_1(t-\tau)+x(\tau)*h_2(t-\tau))d\tau</math> | ||
| − | #<math>x(t)*(h_1(t)+h_2(t))=\int_{k=-\infty}^{\infty}x(\tau)*h_1(t-\tau)+\int_{k=-\infty}^{\infty}x(\tau)*h_2(t-\tau)d\tau</math> | + | #<math>x(t)*(h_1(t)+h_2(t))=\int_{k=-\infty}^{\infty}x(\tau)*h_1(t-\tau)d\tau+\int_{k=-\infty}^{\infty}x(\tau)*h_2(t-\tau)d\tau</math> |
#<math>x(t)*(h_1(t)+h_2(t))=x(t)*h_1(t)+x(t)*h_2(t)</math> | #<math>x(t)*(h_1(t)+h_2(t))=x(t)*h_1(t)+x(t)*h_2(t)</math> | ||
Latest revision as of 15:42, 24 June 2008
Given: $ y(t)=x(t)*h(t)=\int_{k=-\infty}^{\infty}x(\tau)h(t-\tau)d\tau $
- $ x(t)*(h_1(t)+h_2(t))=\int_{k=-\infty}^{\infty}x(\tau)*(h_1(t-\tau)+h_2(t-\tau)d\tau $
- $ x(t)*(h_1(t)+h_2(t))=\int_{k=-\infty}^{\infty}(x(\tau)*h_1(t-\tau)+x(\tau)*h_2(t-\tau))d\tau $
- $ x(t)*(h_1(t)+h_2(t))=\int_{k=-\infty}^{\infty}x(\tau)*h_1(t-\tau)d\tau+\int_{k=-\infty}^{\infty}x(\tau)*h_2(t-\tau)d\tau $
- $ x(t)*(h_1(t)+h_2(t))=x(t)*h_1(t)+x(t)*h_2(t) $
