Convolution Theorem

Convolution Theorem is as follows

• ${\mathcal {F}}^{-1}\left[X(f)H(f)\right]=x(t)\times h(t)=\int _{-\infty }^{\infty }x(\lambda )h(t-\lambda )\,d\lambda$
• $\int _{-\infty }^{\infty }\left(\int _{-\infty }^{\infty }x(\lambda )h(t-\lambda )\,d\lambda \right)e^{-j2\pi ft}\,dt$
• $\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\left(\int _{-\infty }^{\infty }X(f'')e^{j2\pi f\lambda }\,df\int _{-\infty }^{\infty }H(f')e^{j2\pi f'(t-\lambda )}\,df'\right)e^{-j2\pi ft}\,dt\,d\lambda$
• $\int _{-\infty }^{\infty }X(f'')\int _{-\infty }^{\infty }H(f')\int _{-\infty }^{\infty }e^{j2\pi (f'-f)t}\,dt\int _{-\infty }^{\infty }e^{j2\pi (f''-f')}\,d\lambda \,df'\,{df''}$
• $\int _{-\infty }^{\infty }X(f'')\int _{-\infty }^{\infty }H(f')\delta (f-f')\delta (f''-f')\,df'\,{df''}$
• $\int _{-\infty }^{\infty }X(f')H(f')\delta (f-f')\,{df'}$
• $X(f)H(f)\,$

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

Switching the order of integration

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

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

• ${\bar {x}}=\sum _{i}({\bar {x}}-{\hat {a_{i}}}){\hat {a_{i}}}=\sum _{i}\left(\sum _{j}x_{j}a_{ij}\right){\hat {a_{i}}}$

Work in progress
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