Convolution Theorem

Convolution Theorem is as follows

• ${{ℱ}}^{-1}[X(f)H(f)]=x(t)\times h(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )h(t-\lambda )d\lambda$
• $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )h(t-\lambda )d\lambda )e^{-j2\pi ft}dt$
• $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f^{″})e^{j2\pi f\lambda }df\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}H(f^{\prime })e^{j2\pi f^{\prime }(t-\lambda )}d{f}^{\prime })e^{-j2\pi ft}dtd\lambda$
• $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f^{″})\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}H(f^{\prime })\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi (f^{\prime }-f)t}dt\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi (f^{″}-f^{\prime })}d\lambda d{f}^{\prime }d{f}^{″}$
• $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f^{″})\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}H(f^{\prime })\delta (f-f^{\prime })\delta (f^{″}-f^{\prime })d{f}^{\prime }d{f}^{″}$
• $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f^{\prime })H(f^{\prime })\delta (f-f^{\prime })d{f}^{\prime }$
• $X(f)H(f)$

• $x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )e^{-j2\pi f\lambda }d\lambda e^{j2\pi ft}df$

Switching the order of integration

• $x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )\left(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi f(t-\lambda )}df\right)d\lambda$

Taking note of the fact that the inner integral simplifies to $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi f(t-\lambda )t}df=\delta (t-\lambda )=\delta (\lambda -t)$

• $\overline{x}=\sum _i(\overline{x}-\widehat{a_i})\widehat{a_i}=\sum _i\left(\sum _j{x}_j{a}_{ij}\right)\widehat{a_i}$

Work in progress
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