The equation governing a leaky integrator (according to Wikipedia at least) is

$\frac{d\mathcal{O}}{dt} + A\mathcal{O}(t) = \mathcal{I}(t)$.

Is a continuous-time leaky integrator thus the same thing as a low pass filter with time-constant $A$, up to some scaling of the input?

  • 2
    $\begingroup$ Yes, but be sure to check the definition of time constant. $\endgroup$ Aug 22 '12 at 21:52

A so-called leaky integrator is a first-order filter with feedback. Let's find its transfer function, assuming that the input is $x(t)$ and the output $y(t)$:

$$ \frac{dy(t)}{dt} + Ay(t) = x(t) $$

$$ \mathcal{L}\left\{\frac{dy(t)}{dt} + Ay(t)\right\} = \mathcal{L}\left\{x(t)\right\} $$

where $\mathcal{L}$ denotes application of the Laplace transform. Moving forward:

$$ sY(s) + AY(s) = X(s) $$

$$ H(s) = \frac{Y(s)}{X(s)} = \frac{1}{s + A} $$

(taking advantage of the Laplace transform's property that $\frac{dy(t)}{dt} \Leftrightarrow sY(s)$, assuming that $y(0) = 0$).

This system, with transfer function $H(s)$, has a single pole at $s = -A$. Remember that its frequency response at frequency $\omega$ can be found by letting $s=j\omega$:

$$ H(j\omega) = \frac{1}{j\omega + A} $$

To get a rough view of this response, first let $\omega \to 0$:

$$ \lim_{\omega \to 0} H(\omega) = \frac{1}{A} $$

So the system's DC gain is inversely proportional to the feedback factor $A$. Next, let $w \to \infty$:

$$ \lim_{\omega \to \infty} H(\omega) = 0 $$

The system's frequency response therefore goes to zero for high frequencies. This follows the rough prototype of a lowpass filter. To answer your other question with respect to its time constant, it's worth checking out the system's time-domain response. Its impulse response can be found by inverse-transforming the transfer function:

$$ H(s) = \frac{1}{s+A} \Leftrightarrow e^{-At}u(t) = h(t) $$

where $u(t)$ is the Heaviside step function. This is a very common transform that can often be found in tables of Laplace transforms. This impulse response is an exponential decay function, which is usually written in the following format:

$$ h(t) = e^{-\frac{t}{\tau}}u(t) $$

where $\tau$ is defined to be the function's time constant. So, in your example, the system's time constant is $\tau = \frac{1}{A}$.

  • $\begingroup$ Thanks for the answer! So it seems as transfer functions $\frac{1}{1+i \omega \tau}$ and $\frac{1}{\tau + i \omega}$ are different... $\endgroup$
    – Kris
    Aug 23 '12 at 2:45

The frequency response is the same, yes, but the application is different:

  • With a low-pass filter, your signal is in the passband. The cutoff frequency of the filter is set above the highest frequency you want to keep in your signal.
  • With a leaky integrator, your signal is in the stopband. The cutoff frequency of the filter is set below the lowest frequency in your signal.

enter image description here

Also, integrators are always first-order, while low-pass filters can be any order.

  • 3
    $\begingroup$ Same response except for the DC-gain... $\endgroup$
    – Arnfinn
    Sep 16 '16 at 8:24

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