Given a function $x(t)$ and a first order low-pass filter $\frac{dy}{dt} = \frac{1}{\alpha} (x(t) - y(t))$, is there a way to choose $\alpha$ analytically such that the derivative of $y(t)$ is bounded, i.e. $\forall t \frac{dy}{dt} \leq c$ for some given constant $c$?

Thanks for your help.

Note that this question is similar to the one here.


1 Answer 1


I will show you two bounds on the derivative of the output signal, both based on the system's impulse response. First, by taking the Laplace transform of the time-domain equation and rearranging you get the Laplace transform of the output signal in terms of the transform of the input signal and of the system's transfer function:


The inverse Laplace transform of $H(s)$ is the system's impulse response


where $u(t)$ is the step function. In the following I will assume that $\alpha>0$ is satisfied, because otherwise the system is not stable.

The transform of the derivative $y^{\prime}(t)$ is


from which it can be seen that the derivative $y^{\prime}(t)$ can be obtained either from filtering the derivative of the input signal $x^{\prime}(t)$ with $h(t)$, or by filtering $x(t)$ with a system with transfer function $sH(s)$. The impulse response of this system is


From these two interpretations we can now derive two bounds on $|y^{\prime}(t)|$. First note that the magnitude of the output of a linear time-invariant system can be bounded as follows:

$$|y(t)|=\left|\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau\right|\le \int_{-\infty}^{\infty}|h(\tau)||x(t-\tau)|d\tau\le |x(t)|_{max}\int_{-\infty}^{\infty}|h(\tau)|d\tau\tag{3}$$

Since $y^{\prime}(t)$ can be obtained from filtering $x^{\prime}(t)$ with $h(t)$ we get the following bound

$$|y^{\prime}(t)|\le|x^{\prime}(t)|_{max}\int_{-\infty}^{\infty}|h(\tau)|d\tau= |x^{\prime}(t)|_{max}\frac{1}{\alpha}\int_{0}^{\infty}e^{-\tau/\alpha}d\tau=|x^{\prime}(t)|_{max}\tag{4}$$

This is the first bound on $|y^{\prime}(t)|$, which is of course only useful if one knows an upper bound on $|x^{\prime}(t)|$.

The other bound is obtained by noting that $y^{\prime}(t)$ can also be interpreted as filtering $x(t)$ with the impulse response $h^{\prime}(t)$ given by (2):


This leads to the following bound


Combining (4) and (5) we obtain


Obviously, both bounds depend on properties of the input signal $x(t)$, and only one of them depends on the system parameter $\alpha$. In sum, without knowing anything about the input signal $x(t)$, there is no way to choose $\alpha$ such that $y^{\prime}(t)$ is bounded for all input signals.

  • $\begingroup$ Thank you very much Matt. Your presentation is very clear. I appreciate the level of detail. It is very instructive. I would vote you up but my score does not admit it yet. $\endgroup$
    – od3
    Jun 9, 2014 at 14:05

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