# First order low pass filter

I am trying to better understand the first-order low pass filter:

Summary:

Per wikipedia, a first order low pass filter yields the following in discrete time:
$$\frac{Y(s)}{U(s)}= \frac{\omega_{c}}{s+\omega_{c}}$$ yields $$y[k] = \left(\frac{\omega_c T_{s}}{1+\omega_{c} T_{s}} \right) u[k]+\left(\frac{1}{1+\omega_c T_{s}}\right) y[k-1]$$ or $$y[k] =\alpha \, u[k] \ + \ (1 - \alpha)y[k-1]$$

where

$\begin{array}{cclc} \omega_{c} &:& \text{Cutoff angular frequency of filter} & [\frac{rad}{s}] \\ T_{s} &:& \text{Sampling period} & [s] \\ \end{array}$

Question 1:

Even though this filter is in discrete time, does it still model an analog ($s$-plane) filter?

• If I wanted to use a discrete computing system to filter in real-time,
would I need to use the digital ($z$-plane) equivalent?
• If so, what is the general process for performing this?
My best guess is:
1. Determine digital cutoff frequency $\omega_d$.
2. Convert to analog cutoff frequency $\omega_a = \frac{2}{T} \cdot \mathrm{tan}(\omega_d \cdot \frac{T}{2})$.
3. Determine transfer function for (first-order low-pass) analog filter
using analog cutoff frequency $\omega_a$.
4. Transform to transfer function for digital filter
using bilinear transform $z = \frac{2}{T} \cdot \frac{z-1}{z+1}$

Relation to exponential smoothing:

On the same page, exponential smoothing is referenced.
The exponential smoothing page describes an exponential weighted average as: $$y[k] =\alpha u[k]+\left(1 - \alpha\right)y[k-1]\quad\text{where}\quad\alpha = 1 - e^{-\omega_{c} \cdot T_{s}}$$

Question 2:

How is it possible to relate the first-order low-pass filter alpha
with the exponential smoothing alpha?

• appears to me that the two $\alpha$'s are exactly the same. – robert bristow-johnson Aug 11 '17 at 17:29
• @robertbristow-johnson Can you demonstrate that $\frac{1}{1+x} = 1 - e^{-x}$? – kando Aug 11 '17 at 17:31
• @Fat32 : Can you provide some examples? I was not able to find basic-level descriptions to the former, or anything involving $e$ with respect to exponential smoothing. – kando Aug 11 '17 at 17:32
• oh I'm extremely sorry I mean bilinear transform of course... Search with bilinear transform filter design please... – Fat32 Aug 11 '17 at 17:53
• @Fat32 : Haha, no worries - I actually noticed that, but feared that you meant what you said ; j (I'm still not having much luck in either case, but have seen some stellar answers that you've made. Nice job, Mr. ~6K!) – kando Aug 11 '17 at 17:55

The filter in the digital domain of course has a correspondence in the analog domain. However, the analog domain behavior differs from the analog filter corresponding via bilinear transform due to frequency warping: attenuation goes to $-\infty$ as the signal approaches the Nyquist frequency rather than as the signal approaches infinite frequencies.