In the previous filter tutorials we looked at simple firstorder type low and high pass filters that contain only one single resistor and a single reactive component (a capacitor) within their RC filter circuit design.
In applications that use filters to shape the frequency spectrum of a signal such as in communications or control systems, the shape or width of the rolloff also called the “transition band”, for a simple firstorder filter may be too long or wide and so active filters designed with more than one “order” are required. These types of filters are commonly known as “Highorder” or “n^{th}order” filters.
The complexity or filter type is defined by the filters “order”, and which is dependant upon the number of reactive components such as capacitors or inductors within its design. We also know that the rate of rolloff and therefore the width of the transition band, depends upon the order number of the filter and that for a simple firstorder filter it has a standard rolloff rate of 20dB/decade or 6dB/octave.
Then, for a filter that has an n^{th} number order, it will have a subsequent rolloff rate of 20n dB/decade or 6n dB/octave. So a firstorder filter has a rolloff rate of 20dB/decade (6dB/octave), a secondorder filter has a rolloff rate of 40dB/decade (12dB/octave), and a fourthorder filter has a rolloff rate of 80dB/decade (24dB/octave), etc, etc.
Highorder filters, such as third, fourth, and fifthorder are usually formed by cascading together single firstorder and secondorder filters.
For example, two secondorder low pass filters can be cascaded together to produce a fourthorder low pass filter, and so on. Although there is no limit to the order of the filter that can be formed, as the order increases so does its size and cost, also its accuracy declines.
One final comment about Decades and Octaves. On the frequency scale, a Decade is a tenfold increase (multiply by 10) or tenfold decrease (divide by 10). For example, 2 to 20Hz represents one decade, whereas 50 to 5000Hz represents two decades (50 to 500Hz and then 500 to 5000Hz).
An Octave is a doubling (multiply by 2) or halving (divide by 2) of the frequency scale. For example, 10 to 20Hz represents one octave, while 2 to 16Hz is three octaves (2 to 4, 4 to 8 and finally 8 to 16Hz) doubling the frequency each time. Either way, Logarithmic scales are used extensively in the frequency domain to denote a frequency value when working with amplifiers and filters so it is important to understand them.
Since the frequency determining resistors are all equal, and as are the frequency determining capacitors, the cutoff or corner frequency ( ƒ_{C} ) for either a first, second, third or even a fourthorder filter must also be equal and is found by using our now old familiar equation:
As with the first and secondorder filters, the third and fourthorder high pass filters are formed by simply interchanging the positions of the frequency determining components (resistors and capacitors) in the equivalent low pass filter. Highorder filters can be designed by following the procedures we saw previously in the Low Pass and High Pass filter tutorials. However, the overall gain of highorder filters is fixed because all the frequency determining components are equal.
So far we have looked at a low and high pass firstorder filter circuits, their resultant frequency and phase responses. An ideal filter would give us specifications of maximum pass band gain and flatness, minimum stop band attenuation and also a very steep pass band to stop band rolloff (the transition band) and it is therefore apparent that a large number of network responses would satisfy these requirements.
Not surprisingly then that there are a number of “approximation functions” in linear analogue filter design that use a mathematical approach to best approximate the transfer function we require for the filters design.
Such designs are known as Elliptical, Butterworth, Chebyshev, Bessel, Cauer as well as many others. Of these five “classic” linear analogue filter approximation functions only the Butterworth Filter and especially the low pass Butterworth filter design will be considered here as its the most commonly used function.
The frequency response of the Butterworth Filter approximation function is also often referred to as “maximally flat” (no ripples) response because the pass band is designed to have a frequency response which is as flat as mathematically possible from 0Hz (DC) until the cutoff frequency at 3dB with no ripples. Higher frequencies beyond the cutoff point rollsoff down to zero in the stop band at 20dB/decade or 6dB/octave. This is because it has a “quality factor”, “Q” of just 0.707.
However, one main disadvantage of the Butterworth filter is that it achieves this pass band flatness at the expense of a wide transition band as the filter changes from the pass band to the stop band. It also has poor phase characteristics as well. The ideal frequency response, referred to as a “brick wall” filter, and the standard Butterworth approximations, for different filter orders are given below.
Note that the higher the Butterworth filter order, the higher the number of cascaded stages there are within the filter design, and the closer the filter becomes to the ideal “brick wall” response.
In practice however, Butterworth’s ideal frequency response is unattainable as it produces excessive passband ripple.
Where the generalised equation representing a “nth” Order Butterworth filter, the frequency response is given as:
Where: n represents the filter order, Omega ω is equal to 2πƒ and Epsilon ε is the maximum pass band gain, (A_{max}). If A_{max} is defined at a frequency equal to the cutoff 3dB corner point (ƒc), ε will then be equal to one and therefore ε^{2} will also be one. However, if you now wish to define A_{max} at a different voltage gain value, for example 1dB, or 1.1220 (1dB = 20logA_{max}) then the new value of epsilon, ε is found by:

Transpose the equation to give:
The Frequency Response of a filter can be defined mathematically by its Transfer Function with the standard Voltage Transfer Function H(jω) written as:

Note: ( jω ) can also be written as ( s ) to denote the Sdomain. and the resultant transfer function for a secondorder low pass filter is given as:
To help in the design of his low pass filters, Butterworth produced standard tables of normalised secondorder low pass polynomials given the values of coefficient that correspond to a cutoff corner frequency of 1 radian/sec.
n  Normalised Denominator Polynomials in Factored Form 
1  (1+s) 
2  (1+1.414s+s^{2}) 
3  (1+s)(1+s+s^{2}) 
4  (1+0.765s+s^{2})(1+1.848s+s^{2}) 
5  (1+s)(1+0.618s+s^{2})(1+1.618s+s^{2}) 
6  (1+0.518s+s^{2})(1+1.414s+s^{2})(1+1.932s+s^{2}) 
7  (1+s)(1+0.445s+s^{2})(1+1.247s+s^{2})(1+1.802s+s^{2}) 
8  (1+0.390s+s^{2})(1+1.111s+s^{2})(1+1.663s+s^{2})(1+1.962s+s^{2}) 
9  (1+s)(1+0.347s+s^{2})(1+s+s^{2})(1+1.532s+s^{2})(1+1.879s+s^{2}) 
10  (1+0.313s+s^{2})(1+0.908s+s^{2})(1+1.414s+s^{2})(1+1.782s+s^{2})(1+1.975s+s^{2}) 
Find the order of an active low pass Butterworth filter whose specifications are given as: A_{max} = 0.5dB at a pass band frequency (ωp) of 200 radian/sec (31.8Hz), and A_{min} = 20dB at a stop band frequency (ωs) of 800 radian/sec. Also design a suitable Butterworth filter circuit to match these requirements.
Firstly, the maximum pass band gain A_{max} = 0.5dB which is equal to a gain of 1.0593 (0.5dB = 20log A) at a frequency (ωp) of 200 rads/s, so the value of epsilon ε is found by:
Secondly, the minimum stop band gain A_{min} = 20dB which is equal to a gain of 10 (20dB = 20log A) at a stop band frequency (ωs) of 800 rads/s or 127.3Hz.
Substituting the values into the general equation for a Butterworth filters frequency response gives us the following:
Since n must always be an integer ( whole number ) then the next highest value to 2.42 is n = 3, therefore a “a thirdorder filter is required” and to produce a thirdorder Butterworth filter, a secondorder filter stage cascaded together with a firstorder filter stage is required.
From the normalised low pass Butterworth Polynomials table above, the coefficient for a thirdorder filter is given as (1+s)(1+s+s^{2}) and this gives us a gain of 3A = 1, or A = 2. As A = 1 + (Rf/R1), choosing a value for both the feedback resistor Rf and resistor R1 gives us values of 1kΩ and 1kΩ respectively, ( 1kΩ/1kΩ + 1 = 2 ).
We know that the cutoff corner frequency, the 3dB point (ω_{o}) can be found using the formula 1/CR, but we need to find ω_{o} from the pass band frequency ω_{p} then,
So, the cutoff corner frequency is given as 284 rads/s or 45.2Hz, (284/2π) and using the familiar formula 1/CR we can find the values of the resistors and capacitors for our thirdorder circuit.
Note that the nearest preferred value to 0.352uF would be 0.36uF, or 360nF.
and finally our circuit of the thirdorder low pass Butterworth Filter with a cutoff corner frequency of 284 rads/s or 45.2Hz, a maximum pass band gain of 0.5dB and a minimum stop band gain of 20dB is constructed as follows.
Related products: EMI/RFI Suppression  EMI Filter
Nice.
I’m confused where the 1/10 came from when substituting values into the generalized equation. It doesn’t say why or where those numbers came from. I assume the 10 came from the stop band frequency and the negative sign was omitted because of the square root but that was not explained anywhere and its currently a random assumption for me. The wording is unclear because you say there needs to be a minimum stop band gain of 10 but shouldn’t it say maximum gain instead since we’re trying to attenuate it?
As stated in the example, the minimum gain (Amin) required at the stop band frequency is 20dB. Gain is output/input H(jw) (magnitude function) at the required frequency which is 10 (20dB) at ω = ωs = 800 rads/s.
Minimum or maximum depends on how you look at it. If 0dB is the line, anything above is maximum gain (0.5dB) as we do not want any more, and anything below is minimum gain as ideally we would like more than 20dB.
But….it states it is a Low Pass Filter; so how can you have a gain of 10 at 800 rads/s and 1.0593 at 200 rads/s. That would be High Pass.
You’re looking for H(w) = 20dB = 0.1 (= 1/10) at 800 rads/s and the Numerator should be Ho = 0.5dB = 1.0593.
You get the same answer – almost – and the equation makes sense.
Hi guys,
I’m trying to build a analog filter for my dx7, but I’m struggling to find sources on building a filter that theoretically can change the quality factor by keeping the cutoff freq fixed and viceversa.
Can someone please help me out?
Thank you,
Diogo Gonçalves
Hi wayne, although the butterworth filter has ripple free passband, so why does it need a ripple factor. Please clarify what is the significance of the ripple factor epsilon.
Thankyou.
Please mail the answer if possible.
You are correct that in active filters the Butterworth filters has a flat pass band, but the inclusion of epsilon (e^2) in the standard Butterworth function allows for pass band attenuation to be defined at a frequency relating to the pass band edge rather than the 3dB corner frequency. Clearly, if it is defined at the 3dB point then epsilon = 1 as the pass band edge equals the corner frequency.
In our simple example above, the pass band gain was 1dB then e = 0.5088 resulting in the gain at the pass band edge to be zero. The pass band gain (Ho) can be whatever you choose.
hi! i wanna ask that why butterworth use 2n in his equation? why it can’t be a odd order filter?
The 2n part of the Butterworth transfer function relates to the order of the filter and which can be any positive whole number, 1, 2, 3, 4, . . .etc. The value of 2n will always be an even number representing the pairs of poles at the corner frequency. Then this permits the design of Butterworth filters with four or six poles, or of even higher order.
hi !
I,m a university student from Iran
I need to A Analog Low pass Filter (10 Order, 200dB/Dec)
cut off Frequency = 60KHz
stop Band under 80dB
please Help me
tnks
Hi Wayne, there is a part that says : “and Epsilon ε is the maximum pass band gain, (Amax)”. Should that be changed to Epsilon is a ripple factor, and Amax is the maximum passband linear gain defined by H(0) = Amax = Ho?
Hi again Wayne, just adding a quick note as well – regarding H(jw). The equations with the big square roots should be Magnitude of H(jw), ie. H(jw). I know that it will become obvious for most readers. Just adding – just in case!
Since the Butterworth transfer function does not even involve an epsilon term, then why is this tutorial presenting a socalled Butterworth transfer function with an ‘epsilon’ in there? I’m thinking that the epsilon should just be left out, as it is misleading, and actually wrong.
Hello Kenny, The normalised characteristic response of a Butterworth Filter is one with zero ripple in the passband (maximally flat) and in the stopband making the transition from one to the other very rapidly producing the typical brick wall response. The passband ripple is usually given as the peak to peak variation in dB. Since the maximum ripple value for a Butterworth response is 1, that is 0 dB at DC, this quantity is related to the ripple parameter epsilon (ε) through the above equation. Then epsilon would be simply 1 in the denominator of the generalised Butterworth equation and is therefore omitted. For most designs the passband ripple parameter Ap is always fixed at 3dB 20log(0.7071) for a Butterworth response.
However, in the above example we wanted to set the maximum passband ripple to Amax = 0.5dB which is equal to a gain, Ap of 1.0593 to show that it is possible to have a different ripple parameter within a Butterworth response, similar to Chebyshev.
Hi Wayne! Thanks for clarifying those details. Much appreciated.
Please, how can i design a circuit of fourthorder low pass Butterworth Filter?
Cascade 2 2nd order butterworth filters together.