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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.

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  • \$\begingroup\$ Right I get that, but I'm curious as to the reasoning behind it, as in why calculate the area at all? \$\endgroup\$ Oct 2, 2016 at 5:39

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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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  • \$\begingroup\$ ahh, ok so it has to do with creating a continuous signal from a sampled one, like in a DAC? And as a result of the zero order hold, in the limit, we end up with an integral? \$\endgroup\$ Oct 2, 2016 at 5:49
  • \$\begingroup\$ To find the area under a continuous signal you might sample it, add all the samples, then multiply by the sampling increment. That's where the dt comes from in moving from a discrete sum to a continuous integral. \$\endgroup\$
    – Chu
    Oct 2, 2016 at 7:01
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    \$\begingroup\$ In the continuous case, when you apply an input signal to a system you can consider the signal to be divided into series of impulses. Each impulse has a weight equal to the instantaneous value of the signal multiplied by the sampling increment: dt in the limit. These impulses are seen at the system input as a stream. Now, if you add all the contributions of each of the impulse responses at some time, t > 0, you have the overall response of the system at time, t. In the discrete case, the weight of each sample is zero (zero width), and the response signal is a stream of numbers with zero width. \$\endgroup\$
    – Chu
    Oct 2, 2016 at 7:50

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