This is my first study about signal analysing. I'm very confused about filter order. My reference book says that this graph shows 4-th order filter.

And, there are also 2th-order and 12-th order

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.

  • \$\begingroup\$ This question already has been answered here. \$\endgroup\$ – Matt L. Aug 15 '18 at 16:51

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.

Graphical Method

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

The graphical method has some uncertainty but is closest to 12th order. enter image description here

  • \$\begingroup\$ Hmmmm do quartz tuning forks count? Like as a really specific frequency pass filter? so when you set one up to ring with a capacitor that would be a second order filter for the feed voltage? \$\endgroup\$ – K H Aug 15 '18 at 4:17
  • \$\begingroup\$ For BP filters with high Q a different method using linear asymptotes are used on the skirts far away from resonance , even necessary for high Q Chebychev with extreme 12 dB ripple in the pass band \$\endgroup\$ – Sunnyskyguy EE75 Aug 15 '18 at 5:01
  • \$\begingroup\$ @TonyEErocketscientist Sorry. From your explaination, I make conclusions 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? Am I right? And, Im confuse with part " decade is 10x. An octave is 2x"? \$\endgroup\$ – Evangelina Tessia P Aug 15 '18 at 5:43
  • \$\begingroup\$ That is right . For x axis of f you know octaves from music being 1/4f, 1/2f , f , 2f etc and a decade are 1/10 or x10 in frequency about f. \$\endgroup\$ – Sunnyskyguy EE75 Aug 15 '18 at 6:17
  • \$\begingroup\$ @TonyEErocketscientist for figure 1.11. Am I right that filter graph shows about -20n dB/dec and then the second order is -20 x 2 = -40 dB/dec? If its right, then how does the next line become 12th order? Thank you. \$\endgroup\$ – Evangelina Tessia P Aug 15 '18 at 9:45

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:

enter image description here

Second order low-pass Butterworth filter:

enter image description here

Most higher-order filters are made of multiple 1st- or 2nd-order filters.

Fourth-order low-pass Butterworth filter:

enter image description here


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