I'm learning about convolution operations right now, and I have to find the discrete time convolution between x[n] = 2^nδ[n − 1] and h[n] = 0.4^nu[n]. As I think I understand, the convolution between 2^nδ[n] and 0.4^nu[n] is the same as 0.4^nu[n]*2^nu[n] because the impulse function is equal to 1, but for n-1, would the result be the same except for replacing n with n-1?
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Is $x[n] = 2^n \delta[n-1]$ just a weird way of writing $x[n] = 2 \delta[n-1]$? Assuming $\delta[n-1]$ equals $1$ for $n=1$ and $0$ otherwise. – Nov 15 '18 at 03:37
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Oh yeah, I think that makes sense. If that's the case, I think I know how to do the convolution. – Tom Nov 15 '18 at 03:42