# First-order low pass filter with bounded derivative

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.

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:

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

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

$$h(t)=\frac{1}{\alpha}e^{-t/\alpha}u(t)$$

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

$$sY(s)=sX(s)H(s)=[sX(s)]H(s)=X(s)[sH(s)]\tag{1}$$

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

$$h^{\prime}(t)=\frac{1}{\alpha}\delta(t)-\frac{1}{\alpha^2}e^{-t/\alpha}u(t)\tag{2}$$

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):

$$y^{\prime}(t)=(x*h^{\prime})(t)=\frac{1}{\alpha}x(t)-\frac{1}{\alpha^2}\int_0^{\infty}e^{-\tau/\alpha}x(t-\tau)d\tau$$

This leads to the following bound

$$|y^{\prime}(t)|\le\frac{1}{\alpha}|x(t)|_{max}+\frac{1}{\alpha^2}|x(t)|_{max}\int_0^{\infty}e^{-\tau/\alpha}d\tau=\frac{2}{\alpha}|x(t)|_{max}\tag{5}$$

Combining (4) and (5) we obtain

$$|y^{\prime}(t)|\le\min\left\{|x^{\prime}(t)|_{max},\frac{2}{\alpha}|x(t)|_{max}\right\}$$

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.

• 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. – od3 Jun 9 '14 at 14:05