applicationsapproximationfilterslogarithms
I am trying to find derivation (differential attenuation) for frequency's 600 and 2000 Hz for given amplitude characteristic of low pass filter, which look like this:
I assume, that I should transform my approximation (graph) to dB form, but how should I do this (if I am right).
Note that a more general notion of filter is defined on a poset $(P, \le)$ that has meets (two elements $p,q\in P$ have a greatest lower bound $p \land q \in P$; this is called a meet-semilattice) and a minimum $0 \in P$, and then the axioms for a filter $\emptyset \neq F \subseteq P$ become
So an upwards-closed subset of $P$ (condition 3) that is closed under finite meets (2) and is not equal to $P$ itself (the proper part; note that condition (1) is then needed or otherwise $0 \le x$ for all $x$ and all $x$ would be in $F$, which we don't want).
In a powerset of a set $X$ with $\cap$ as the meet and $\subseteq$ as the poset-order and $\emptyset$ as the minimum, this is precisely your definition as a special case. But in topology we also want to talk about filters of closed sets or zero-sets or open sets and open sets/closed subsets/ zero-sets form a meet semi-lattice too, so there this filter notion also makes sense (we can go further and demand just a common lower bound to be in $F$; this is what Wikipedia does, and is also fine).
An ultrafilter $F$ is just a maximal filter on $P$ so if $P \supseteq F' \supseteq F$ is a filter too, then $F=F'$. For the power set $\mathscr{P}(X)$ meet-semilattice (in fact in any Boolean algebra) it is not hard to see that in fact this is equivalent to
$$\forall A \subseteq X: A \in F \lor X\setminus A \in F$$
so it's then not the case that the ultrafilter is all subsets, but exactly "half" of them (a set and its complement cannot both be in $F$ but precisely one of them has to be).
In this poset (and many others that also consist of subsets of some set $X$), a principal filter is one where the intersection of its members is non-empty, so they have at least one common element. In the power set of $X$, if $F$ is a principal ultrafilter it is of the form
$$F_x=\{A \subseteq X\mid x \in A\}$$ for some $x \in X$, as can easily be seen. Such a filter is always maximal and hence an ultrafilter.
The most commonly considered filter on a topological space $X$ is, for any $x \in X$, the set of neighbourhoods of $x$, denoted $\mathcal{N}_x$, which consists of all subsets $N$ of $X$ so that there is an open subset $O$ of $X$, so that $x \in O \subseteq N$. This is clearly a filter (and a principal one, as all members contain $x$; but not any set containing $x$ is in $\mathcal{N}_x$ in general), and the idea of filter convergence is that we define it so that $\mathcal{N}_x \to x$, the neighbourhoods of a point "converge" to that point.
I've added some graphs to make a proper answer of my comments. For the voltage ratio relation you described, the behavior of the function as $ \ f \ $ "tends to infinity" is
$$\frac{U_{out}}{U_{in}} \ = \ \frac{1}{\sqrt{\left(\frac{f}{f_G}\right)^2+1}} \ \ \longrightarrow \ \ \frac{1}{\sqrt{\left(\frac{f}{f_G}\right)^2}} \ \ = \ \ \frac{1}{\left(\frac{f}{f_G}\right)} \ \ = \ \ \frac{f_G}{f} \ \ . $$
So at frequencies "large compared to $ \ f_G \ \ , $ this looks increasingly like a simple inverse relationship between the frequency and the voltage ratio. On a graph, however, this can be rather difficult to read; moreover, a slightly different exponent from $ \ f^{-1} \ $ would be hard to discern. (The graph at left above shows the voltage ratio as a function of frequency with $ \ f_G = 100 \ $ and frequency running out beyond $ \ 25,000 \ \ . $ To its right is a vertical scale enlargement by about a factor of $ \ 30 \ \ $ with frequency out to $ \ 500,000 \ \ . $ The red lines in the first graph indicate the vertical range in the second graph; the red lines in the second graph mark the frequency range in the first one.)
To improve our ability to read the graph to very small voltage ratios, we can apply logarithms to both sides of the above relation to produce a linear equation (mathematicians and some physicists use natural logarithms, but most other disciplines use base-10):
$$ \log_{10} \frac{U_{out}}{U_{in}} \ \ = \ \ \log_{10} \frac{f_G}{f} \ \ \Rightarrow \ \ \log_{10} \frac{U_{out}}{U_{in}} \ \ = \ \ \log_{10} f_G \ - \ \log_{10} f $$ $$ = \ \ (-1)·(\log_{10} f) \ + \ \log_{10} f_G \ \ , $$
which can be rearranged into a "slope-intercept form" with the logarithm of $ \ f \ $ as the independent variable on a line of slope $ \ (-1) \ $ and a $ \ y-$intercept of $ \ \log_{10} f_G \ \ , $ for "sufficiently large" frequencies (as in the graph below, with $ \ f \ $ running from about $ \ 0.001 \ $ to $ \ 400,000 \ ). $ It is possible here to see that beyond a frequency of about $ \ 1000 \ \ , $ the voltage ratio falls by "a decade" for each "decade" increase in frequency.
Any sort of "power-law" function can be treated in the same way to obtain a "log-log" linear equation of the form
$$ \ y \ = \ C·x^n \ \ \longrightarrow \ \ \log_{10} y \ \ = \ \ n· (\log_{10} x) \ + \ \log C \ \ . $$
In experimental practice, data points representing measurements that "follow a power-law" will fall more-or-less along a straight line with a slope that indicates the exponent in the power-law function. (Using natural logarithms instead only changes the numerical values plotted, but has no effect on the observed slope.)
Something similar is done for exponential functions, which also turn up frequently in phenomenological models. There, a "log-linear" transformation is used in which only the ordinate ("vertical coordinate") has a logarithm applied:
$$ y \ = \ C·10^{kx} \ \ \longrightarrow \ \ \log_{10} y \ \ = \ \ \log_{10} [ \ C·10^{kx} \ ] \ \ = \ \ k·x \ + \ \log_{10} C \ \ . $$
Here, there is a straight line in the variable $ \ x \ $ with a slope $ \ k \ \ , $ which gives the "exponential constant" in the relation; the value of this linear function is then $ \ \log_{10} y \ \ . $ (It should be noted that using natural logarithms here will affect the slope, altering it to $ \ (\ln 10)·k \ \ ; $ "natural-logs" would be the proper choice for a model function $ \ C·e^{kx} \ \ . ) $ This last graph shows a full linear and log-linear plot for $ \ 40·10^{-0.02·x} \ \ ; $ the red lines in the graph at right show the extent of the full linear graph.
One last remark that might be made here is that log-log and log-linear plots have "no bottom", since equal steps in the "vertical" direction represent equal factors in $ \ y \ , \ $ so $ \ y = 0 \ $ is transformed to "negative infinity".
[As a historical remark on the "wonderful age in which we live", for quite a long time, one had to purchase specially printed "graph-paper" with "log-log" and "log-linear" grids for these purposes. Once personal computers with graphing utilities became widely available in the 1990's, the market for such items largely evaporated. Not long after that, online graphers also appeared, so it has been quite easy for some while now to make such plots.]
Best Answer
You need to find the slope of the tangent lines to the curve at the frequencies $600$ and $2000$ Hz represented in a Bode plot with two logarithmic scales instead of two linear ones as in your plot. The horizontal axis of the low pass filter should measure the frequency in decades and the vertical axis should measure the amplitude in decibels (dB). The frequencies whose magnitudes differ by a factor of ten such as $1, 10, 100, 1000, 10000$ Hz are equaly spaced and in general two frequencies $f_1,f_2$ are
$$\log_{10}\left(\frac{f_1}{f_2}\right)$$
decades apart. So from $100$ and $600$ Hz are $\log_{10}\left(\frac{600}{100}\right)\approx 0.778$ decades and from $2000$ to $1000$ Hz are $\log_{10}\left(\frac{2000}{1000}\right)\approx 0.301$ decades. To transform the amplitude $A_1$ in a dB change with respect to the amplitude $A_0$ use the conversion formula (if the amplitude in your plot is a voltage and not a power)
$$\text{Gain}_\mathrm{ dB} =20 \log_{10} \left (\frac{A_1}{A_0} \right ), $$
because a change in voltage by a factor of $10$ is a $20$ dB change and means a change in power by a factor of $100=10^2$, assuming that the power is proportional to the square of the voltage. (See Wikipedia). The attenuation is the opposite of the gain.