Convolution Theorem: Difference between revisions

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Convolution Theorem is as follows
Convolution Theorem is as follows
*<math>\mathcal{F}^{-1}\left[ X(f)H(f)\right] = x(t)\times h(t) = \int_{-\infty}^{\infty}x(\lambda)h(t-\lambda)\,d\lambda</math>
*<math>\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}x(\lambda)h(t-\lambda)\,d\lambda\right)e^{-j2\pi ft}\,dt</math>
*<math>\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}X(f'')e^{j2\pi f \lambda}\,df\int_{-\infty}^{\infty}H(f')e^{j2 \pi f'(t-\lambda)}\,df'\right) e^{-j2 \pi f t}\,dt\,d\lambda </math>
*<math>\int_{-\infty}^{\infty}X(f'')\int_{-\infty}^{\infty}H(f')\int_{-\infty}^{\infty} e^{j2\pi (f'-f)t} \, dt \int_{-\infty}^{\infty} e^{j2\pi (f''-f')}\,d \lambda \, df' \, {df''}</math>
*<math>\int_{-\infty}^{\infty}X(f'')\int_{-\infty}^{\infty}H(f')\delta(f-f') \delta(f''-f') \, df' \, {df''}</math>
*<math>\int_{-\infty}^{\infty}X(f')H(f')\delta(f-f')\, {df'}</math>
*<math>X(f)H(f)\,</math>


*<math> x(t) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x(\lambda)e^{-j2\pi f \lambda}\,d\lambda \,e^{j2\pi f t}\,df</math>
*<math> x(t) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x(\lambda)e^{-j2\pi f \lambda}\,d\lambda \,e^{j2\pi f t}\,df</math>

Latest revision as of 01:55, 13 October 2006

Convolution Theorem is as follows

  • 1[X(f)H(f)]=x(t)×h(t)=x(λ)h(tλ)dλ
  • (x(λ)h(tλ)dλ)ej2πftdt
  • (X(f)ej2πfλdfH(f)ej2πf(tλ)df)ej2πftdtdλ
  • X(f)H(f)ej2π(ff)tdtej2π(ff)dλdfdf
  • X(f)H(f)δ(ff)δ(ff)dfdf
  • X(f)H(f)δ(ff)df
  • X(f)H(f)


  • x(t)=x(λ)ej2πfλdλej2πftdf

Switching the order of integration

  • x(t)=x(λ)(ej2πf(tλ)df)dλ

Taking note of the fact that the inner integral simplifies to ej2πf(tλ)tdf=δ(tλ)=δ(λt)

  • x¯=i(x¯ai^)ai^=i(jxjaij)ai^

Work in progress
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