User:M1ss1ontomars2k4/sandbox
From the definition of convolution, we know:
![{\displaystyle {\begin{aligned}(x*h)(n)&=\sum _{m=-\infty }^{\infty }{x[m]h[n-m]}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3894a9c70dd2340657feceec38d7ef547f14617f)
where is the shifted delta, . Let us substitute this in to obtain:
![{\displaystyle h[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89981bbbb05ffd469eeadb828c18359965985e46)
![{\displaystyle \delta [n-n_{0}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bdb0265027e056f16fce87ab282b57cb03c4f8c)
![{\displaystyle \sum _{m=-\infty }^{\infty }{x[m]\delta [(n-m)-n_{0}]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88626e95c89aacde3e136e3d9c04420d579cf8c6)
Let . Then we have:
![{\displaystyle {\begin{aligned}(x*h)(n)&=\sum _{m=-\infty }^{\infty }{x[m]\delta [(n-m)-n_{0}]}\\&=\sum _{m=-\infty }^{\infty }{x[(n-n_{0})-k]\delta [k]}\\&=\sum _{k=-\infty }^{\infty }{x[(n-n_{0})-k]\delta [k]}\\&=(x*\delta )(n-n_{0})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f5a5bbee7885daa81a5a23ebd57495006f6c00f)
i.e., the result of where is the shifted delta is simply the convolution of and , shifted by the same amount.
