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Definition: Let be $u$ and $v$ two sequences the convolution of these sequences is defined than $$h(m)=u(m)* v(m) = \sum_{s=-\infty}^{\infty}u(m-s)v(s).$$

Question: Show that $\sum_{m=-\infty}^{\infty}h(m)=\sum_{m=-\infty}^{\infty}u(m)\sum_{m=-\infty}^{\infty}v(m)$.

I want know if is posible solve this using convolution theorem for sequences (*) I trying ...

$$H(w)=U(w)V(w) = \sum_{m=-\infty}^{\infty}u(m)\exp(-j m w)\sum_{m=-\infty}^{\infty}v(m)\exp(-j m w),$$

but I don't what else to do.

*Theorem: The Fourier transform of $h(m)=u(m)* v(m)$ is $H(w)=U(w)V(w).$

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Why the minus one? – juaninf Nov 17 '12 at 23:09
One should write $(u*v)(m)$ rather than $u(m)*v(m)$. For example, suppose $m=3$, and $u(3)=12$, and $v(3)=17$. Then $u(m)*v(m)$ would be $12*17$, and that's nonsense. – Michael Hardy Nov 18 '12 at 1:44
up vote 1 down vote accepted

Hint: You've stated the convolution theorem as it applies to the discrete time Fourier transform (DTFT) of the convolution of two sequences $u(m), v(m)$: $$H(\omega)=U(\omega)V(\omega),$$ where $U, V, H$ are respectively the DTFT of $u, v$ and $u\ast v$.

Now recall the definition of the discrete Fourier transform: $$U(\omega)=\sum_{m=-\infty}^\infty u(m)e^{-i\omega m}.$$ What value does this give when $\omega=0$?

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