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I want to know the amplitude of the first-order hold filter at the Nyquist frequency (the roll-off amplitude/gain).

I know that the Fourier transform of the reconstruction is given by:

$$\sum^\infty_{k=-\infty}F\left(\nu - \frac{k}{T}\right)\ \mathrm{sinc}^2(\nu T)$$

which corresponds to low-pass filtering of the spectrum in the frequency domain by a $\mathrm{sinc}^2$ function. This causes some aliasing due to the sinc's sidelobes. However at the Nyquist frequency $\nu = 1/2T$, approximately we have the amplitude:

$$\mathrm{sinc}^2\left(\frac{T}{2T}\right)\approx0.4053$$

Is this close to the correct value? Also, how should I display this gain as a dB reading? MATLAB's mag2db function gives -7.8, not sure if that's right.

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    $\begingroup$ @OlliNiemitalo: I guess the OP means a triangular impulse response for linear interpolation. $\endgroup$
    – Matt L.
    Commented Apr 10, 2017 at 14:11
  • $\begingroup$ @MattL. Thanks, that's right. I forgot it's called first-order hold, and was thinking of zeroth order hold with a rectangular impulse response. $\endgroup$
    – Olli Niemitalo
    Commented Apr 10, 2017 at 14:44
  • $\begingroup$ @OlliNiemitalo: yes, there are even n'th order holds (and some controversy about them), as discussed in this question. $\endgroup$
    – Matt L.
    Commented Apr 10, 2017 at 15:04
  • $\begingroup$ the only controversy is simply about what the definitions are of these $n$th-order holds. at Wikipedia i have seen 3 different definitions for 1st-order hold. the last one seems a little goofy to me. $\endgroup$
    – robert bristow-johnson
    Commented Apr 11, 2017 at 6:18

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That's correct, $0.4053$ is the approximate magnitude of the frequency response of linear interpolation at the Nyquist frequency. With the definition:

$$\operatorname{sinc}(x) = \frac{\sin(\pi x)}{\pi x}$$

the exact value is:

$$\operatorname{sinc}^2\left(\frac{1}{2}\right)= \left(\frac{\sin\left(\frac{\pi}{2}\right)}{\frac{\pi}{2}}\right)^2 = \left(\frac{1}{\frac{\pi}{2}}\right)^2 = \left(\frac{2}{\pi}\right)^2 = \frac{4}{\pi^2}$$

or in dB:

$$20\log_{10}\left(\frac{4}{\pi^2}\right) = 20\frac{\log\left(\frac{4}{\pi^2}\right)}{\log(10)}\approx-7.844795081\text{ dB}$$

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