My problem is how can I know whether its 4-th order, 12-th order, or 2nd order like the book says so? I'd like to know the process behind it.
The order, n of a filter is the number of reactive elements (if all are contributing.)
Using the linear slope (on log-log grid) away from f breakpoint it will be 6dB/octave per order of n.
An n= 4th order is 24dB/octave slope as in both of 1st examples .
I might think it appears to a 10th order filter Butterworth -60dB/oct and 8th order Chebychev -40dB/oct. There is visual ambiguity here from the lack of range after break, to estimate the filter slope when the graph is cutoff near 1 octave above. Also these are filter examples with low&high Q so the breakpoint slopes are very different.
So I agree it is hard to estimate in figure1.12. Whereas Fig 1.11 is easier to measure the slope.
Use a straight edge to go through the Y axis intercept and fit a linear slope to curve. Then measure the slope in n multiples of -6n dB/oct or better if possible -20n dB/dec.
It gets complicate when the Y axis is not big enough.
A decade is 1/10= 20 log 0.1 = -20dB x n order.
An octave is 1/2 = 20 log 0.5 = -6.02dB x n order.
So from Fig 1.11 12th order filter
When you reduce the response of the filter to its transfer function, the order of the differential equation is the order of the filter. See the page:
The order of the filter reflects the number of elements that delay your sampling by one - i.e. a first-order filter needs one sample to produce your desired output, a second-order filter needs two samples, etc.
Here are some examples I'm pulling off google images:
First order low-pass Butterworth filter:
Second order low-pass Butterworth filter:
Most higher-order filters are made of multiple 1st- or 2nd-order filters.
Fourth-order low-pass Butterworth filter: