Electrical – Derivation of convolution integral from the discrete convolution sum

control systemconvolutionsignal-theory

I was wondering if anyone can provide a derivation of the continuous-time convolution integral
$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau$$
from the discrete-time convolution sum
$$y[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]$$

The discrete sum makes sense to me conceptually- as sum of amplitudes, but an integral means area… which is where I start to get lost, at least intuitively.

Best Answer

A discrete signal does not exist between the sampling increments, so it doesn't have an area under the signal. If you wanted to create a continuous signal from a discrete one you might use a zero order hold which gives it area by extending each sample into a pulse of width \$\Delta t\$, or \$dt\$ in the limit.

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