I'm given the formula for describing the voltage ratio bewerten in- and output for a low-pass filter (neglecting physical aspects):
$\dfrac{U_{out}}{U_{in}} = \dfrac{1}{\sqrt{\left(\frac{f}{f_G}\right)^2+1}}$. Here $f$ is the variable frequency, $f_G$ a constant depending on the assemble.
Plotting that "function", gives a graph, decreasing for higher frequency. Don't mind the physics
The mathematical thing I do not understand: if I plot in double-logarithmic scale, say
$\log\left(\dfrac{U_{out}}{U_{in}}\right) = \log\left(\dfrac{1}{\sqrt{\left(\frac{f}{f_G}\right)^2+1}}\right)$ – why does this lead to a linearization with higher frequencies?
I'd just aspect this happening for $\textsf{exponential}$ functions.
Best Answer
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.]