While reading about 1st order low pass discrete implementations I found out, the following 2 equations achieve similar results, but apparently not identical, but I thought they had the same origin. One way is to derive from a filter transfer function and end up with something in the form:
$y[n] = -a_1y[n-1] + b_0x[n] + b_1x[n-1]$$$y[n] = -a_1y[n-1] + b_0x[n] + b_1x[n-1]$$
and the other would be deriving from a differential equation and ending up with something like:
$y[n] = (1-T)*y[n-1] + T*x[n-1]$ , for$$y[n] = (1-T)\cdot y[n-1] + T\cdot x[n-1]$$ for a given $T$ constant.
They seem to achieve very similar results (even though not identical) and I've noticed of course the absence of the second delayed $x$ element on the second equation. However, I thought both should originate from the same differential equation, since the first one I've seen in explanations about 1st order butterworth and the second one I've seen in some 1st order RC or general IIR low pass filters, but I thoguhtthought they should be essentially identical from the mathematics point of view.
What am I missing here? Besides, is there any transformation $f(a_1, b_0, b_1) = T$ such as the second filter would give the exact same response as the first one, even though lacking the second delayed input?
EDIT: So starting with a continuous generic transfer function, I think it would look something like this:
$$H(s) = K\frac{ 1 }{s-a}$$
where $K$ is my gain and $a$ is the position of my pole. In this case I would have a single pole, no zeros, and should represent a low-pass 1st order filter.
Now going to the discrete world, one usually finds something like:
$$H(z) = \frac{b_0 + b_1z^{-1} + ... + b_Nz^{-N} }{1 + a_1z^{-1} + ... + a_Nz^{-N}}$$
which in our case (first order filter) would mean:
$$H(z) = \frac{b_0 + b_1z^{-1} }{1 + a_1z^{-1} }$$
which means having one pole and one zero. And here it starts getting confused. I do not remember anymore how to go from continuous to discrete, but I do remember that if we put a zero at origin on the continuous world, we would have such a transfer function:
$$H(s) = K\frac{ s }{s-a}$$$$H(s) = K\frac{s}{s-a}$$
which I suppose would be a high-pass filter instead, since we block 0Hz, right? On the other hand, we do have a zero on the discrete equation... I do remember they have different meanings on both worlds, but not anymore how.
So getting back to the time discrete equations above, I guess one would originate from:
$$H(z) = \frac{b_0 + b_1z^{-1} }{1 + a_1z^{-1} }$$$$\begin{align} H(z) &= \frac{b_0 + b_1z^{-1}}{1 + a_1z^{-1} } \\ \\ &= \frac{b_0z + b_1}{z + a_1} \\ \end{align}$$
and the other from:
$$H(z) = \frac{b_0 }{1 + a_1z^{-1} }$$$$\begin{align} H(z) &= \frac{b_1z^{-1}}{1 + a_1z^{-1} } \\ \\ &= \frac{b_1}{z + a_1} \\ \end{align}$$
where $b_0=T$$b_1=T$ and $-a_1=(1-T)$
right? If so, I just still don't get what is the difference on the continuous world that imples one less zero on the discrete world (therefore my question was focused on $H(s)$ instead of $H(z)$) and what exactly this zero is causing on the discrete world.
EDIT2: I do remember slowly that poles and zeros on continuous world are this normal 3D plot with Imaginary and Real axis for the position of the poles and zeros, and the Gain or Frequency response axis as the third axis. And, if I'm not mistaken, poles should rather be on the left (negative) side of the Imaginary axis to reach stability, on w=0 for infint oscilation and positive for instability. On the discrete world that was something like a unit circle, where ROC was inside the circle and instability on the outside... still, not sure if same meaning applies from zeros and poles there, because otherwise a zero at origin on the discrete equation would mean blocking DC, or, making a high-pass filter out of it (which is clearly not the case).
