Assume I sample a signal according to the Nyquist criterion. Then I perform a simple summation / integral over a linearly interpolated signal. Is this equivalent to the integral over the continuous signal, i.e.?

$$\sum_{n=-\infty}^{\infty}x(nT) T = \int_{-\infty}^{\infty}x(t)\text{d}t$$

I hope my representation is correct for what I mean. My question relates to whether a reconstruction, e.g. Whittaker–Shannon interpolation, is necessary in practical application. I would also appreciate any literature pointers for discussions on how this might affect practical cases with finite time and imperfect filters.

  • 1
    $\begingroup$ what is "$B$"? is it the signal bandwidth. is $T=\frac{1}{2B}$? if so, the summation is a Riemann summation for the integral and they might come out close if $T$ is sufficiently small. why not just say $$\sum_{n=-\infty}^{\infty}x(nT) \, T \approx \int_{-\infty}^{\infty}x(t) \, \mathrm{d}t$$ ?? $\endgroup$ – robert bristow-johnson Feb 8 at 21:00
  • $\begingroup$ Yes, you are right. My question is whether the two are equal given the Nyquist criterion. $\endgroup$ – Peter Kalt Feb 9 at 14:22
  • $\begingroup$ @robertbristow-johnson: As long as $T$ is chosen such that the sampling theorem is satisfied, the sum and the integral don't just come close, they're identical. $\endgroup$ – Matt L. Feb 10 at 11:55

The equation in your question is correct. If we assume that $x(t)$ is an ideally band-limited signal with bandwidth $B$, then, according to the sampling theorem, we have

$$x(t)=\sum_{n=-\infty}^{\infty}x(nT)\frac{\sin\left[\pi (t-nT)/T\right]}{\pi (t-nT)/T}\tag{1}$$

where the sampling rate $1/T$ satisfies $1/T>2B$.

If we assume that the integral over $x(t)$ exists, using $(1)$ we obtain

$$\begin{align}\int_{-\infty}^{\infty}x(t)dt&=\int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}x(nT)\frac{\sin\left[\pi (t-nT)/T\right]}{\pi (t-nT)/T}dt\\&=\sum_{n=-\infty}^{\infty}x(nT)\int_{-\infty}^{\infty}\frac{\sin\left[\pi (t-nT)/T\right]}{\pi (t-nT)/T}dt\\&=\int_{-\infty}^{\infty}\frac{\sin(\pi t/T)}{\pi t/T}dt\cdot \sum_{n=-\infty}^{\infty}x(nT)\\&=T\sum_{n=-\infty}^{\infty}x(nT)\tag{2}\end{align}$$

where the last equality follows from the fact that the integral in the one but last line just equals the DC value of an ideal low pass filter with gain $T$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.