Let $x[n]$ be real signal and $y[n]=\exp(j3\pi n)$ be a complex signal
Would the convolution between those two signals be $$x[n] * \Re(y[n]) + jx[n]*\Im(y[n])$$?
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2 Answers
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Discrete convolution is a linear operation, so yes
$$\begin{align*}x[n]*y[n] &= (x*y)[n]\\ \\ &= \sum_{m=-\infty}^{\infty} x[m]y[n-m] \\ \\ &= \sum_{m=-\infty}^{\infty} x[m]\Re(y[n-m])+ jx[m]\Im(y[n-m]) \\ \\ &= \left[\sum_{m=-\infty}^{\infty} x[m]\Re(y[n-m])\right] + \left[\sum_{m=-\infty}^{\infty} jx[m]\Im(y[n-m])\right] \\ \\ &= x[n]*\Re(y[n])+jx[n]*\Im(y[n])\\ \end{align*}$$
answered Jan 2, 2019 at 17:16
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Yes so you can see this even simpler with distribution of convolution as:
$$ \begin{align} x[n] \star y[n] & = x[n] \star \left( \Re \{ y[n] \} + j ~~ \Im \{ y[n] \} \right) \\ &= x[n] \star \Re \{ y[n] \} + j ~~ x[n] \star \Im \{ y[n] \} \end{align}$$
