# How can I determine the order of filter?

This is my first study about signal analysing. I'm very confused about filter order. My problem is how can I know whether its 12-th order, or 2nd order like the book says so? I already knew that slope has 2 choises whether its in n multiples of -6n dB/oct or -20n dB/dec. If the slope is -20 dB/dec then the second order should have -40 dB/dec. For example, An n= 4th order is 24dB/octave slope. But, I saw this example below and pretty confuse about determine the slope of those two line and how can the first line is 12-th order. Its not like the slope reaches -240 dB/dec (-20 x 12) and the second line seems not like -40 dB/dec (-20 x 2). Is there anyone could explain this to me? Thank you.

Note that the given slopes of $6n\textrm{ dB}/\textrm{octave}$ and $20n\textrm{ dB}/\textrm{dec}$, respectively, are asymptotic slopes for large frequencies. So you have to extrapolate the lines that are approximated by the magnitude responses at large frequencies.
For the given figure you could do the following. For the second order filter, look at the attenuations at $f=10^2\textrm{Hz}$ and at the next vertical line, which corresponds to $f=2\cdot 10^2\textrm{Hz}$. The ratio between those two frequencies is an octave, and the difference in attenuation is approximately $-22-(-35)\textrm{dB}=13\textrm{dB}$ (should be $12\textrm{dB}$).
For the $12^{th}$order system you could look at the attenuation values at $30\textrm{Hz}$ and $40\textrm{Hz}$, taking into account that the curve flattens towards its knee, and you need to extrapolate the straight line. With this in mind you would arrive at approximately $-31\textrm{dB}$ at $f=40\textrm{Hz}$, and at $-2\textrm{dB}$ at $f=30\textrm{Hz}$, which corresponds to a difference of $29\textrm{dB}$. Solving
$$\left(\frac{4}{3}\right)^x=2\tag{1}$$
for $x$ gives $x\approx 2.41$, i.e., you have to multiply $29\textrm{dB}$ with $2.41$ to obtain the attenuation per octave: $29\cdot 2.41=69.9\textrm{dB}$ (should be $72\textrm{dB}$).