Can someone verify this statement?

The hold operator is assumed to be a zero order hold

Then the laplace transform of this hold operator has a well known form $(1-e^{-sT})/s$

Let w approach 0, we see that our resulting function is $(1-e^{\alpha T})/\alpha$ where $s = \alpha + j\omega$

What does it mean for the time hold operator to be approximated by time delay of T/2? How can delaying a signal be equivalent to holding that signal

  • $\begingroup$ Me neither, but this claim was used in a high quality paper (which I cannot find for the life of me at the moment) so I am pretty sure it is used elsewhere $\endgroup$
    – Fraïssé
    Commented Jan 29, 2015 at 17:12

1 Answer 1


A zero-order hold can be modeled by a rectangular impulse response of length $T$ (where $T$ is the sampling period) and height $1/T$, starting at $t=0$. The corresponding frequency response is

$$H(f)=\text{sinc}(fT)e^{-j2\pi f T/2}\tag{1}$$

with $\text{sinc(x)}=\sin(\pi x)/(\pi x)$. From (1) it can be seen that $H(f)$ has a linear phase response with a corresponding group delay of $T/2$. Now if the sampling frequency is very high compared to the bandwidth of the sampled signal, then the $\text{sinc}$ function can be approximated by a constant around $f=0$, or, equivalently, the rectangular impulse response can be approximated by an impulse at its center (which is $t=T/2$). So for low-frequency signals (compared to the sampling frequency), the zero-order hold operation can be approximated by a delay of $T/2$.

The following figure illustrates this. The blue line is a signal with a small bandwidth compared to the sampling frequency. The green curve is the result of a zero-order hold operation, where the signal is constant over 1 sampling period $T$. Finally, the red curve is the original signal delayed by $T/2$. If $T$ is very small, the delayed signal is closely approximated by the piecewise constant output of the zero-order hold.

enter image description here


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