# Do Nyquist samples have the same sum as the integral of the signal?

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.

• 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 $$\sum_{n=-\infty}^{\infty}x(nT) \, T \approx \int_{-\infty}^{\infty}x(t) \, \mathrm{d}t$$ ?? – robert bristow-johnson Feb 8 at 21:00
• Yes, you are right. My question is whether the two are equal given the Nyquist criterion. – Peter Kalt Feb 9 at 14:22
• @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. – Matt L. Feb 10 at 11:55

## 1 Answer

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 $$T$$.