Proof: y(t)=x(t)*(h_1(t)*h_2(t))=(x(t)*h_1(t))*h_2(t)

Given:

  1. $ y(t)=x(t)*h(t)=\int_{k=-\infty}^{\infty}x(\tau)h(t-\tau)d\tau $
  2. $ y(t)=x(t)*h(t)=h(t)*x(t) $ commutative property of convolution for continuous time

Steps:

  1. $ x(t)*(h_1(t)*h_2(t))=x(t)*(h_2(t)*h_1(t)) $ commutative property of convolution for continuous time
  2. $ x(t)*(h_1(t)*h_2(t))=x(t)*\int_{-\infty}^{\infty}h_2(\tau)h_1(t-\tau)d\tau $
  3. $ x(t)*(h_1(t)*h_2(t))=\int_{-\infty}^{\infty}x(\mu)\int_{-\infty}^{\infty}h_2(\tau)h_1(t-\tau-\mu)d\tau d\mu $
  4. $ x(t)*(h_1(t)*h_2(t))=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x(\mu)h_2(\tau)h_1(t-\tau-\mu)d\tau d\mu $
  5. $ x(t)*(h_1(t)*h_2(t))=\int_{-\infty}^{\infty}h_2(\tau)\int_{-\infty}^{\infty}x(\mu)h_1(t-\tau-\mu)d\mu d\tau $
  6. $ x(t)*(h_1(t)*h_2(t))=h_2(t)*\int_{-\infty}^{\infty}x(\mu)h_1(t-\mu)d\mu $
  7. $ x(t)*(h_1(t)*h_2(t))=h_2(t)*(x(t)*h_1(t)) $
  8. $ x(t)*(h_1(t)*h_2(t))=(x(t)*h_1(t))*h_2(t) $ commutative property of convolution for continuous time

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009