Question

How to choose the bandwidth of the Butterworth, 2 poles filter for ASK modulation? What bandwidth should be if symbol duration is 3 ms?

Answer 1

The Butterworth filter is the best compromise between attenuation and phase response. It has no ripple in the pass band or the stop band, and because of this is sometimes called a maximally flat filter. The Butterworth filter achieves its flatness at the expense of a relatively wide transition region from pass band to stop band, with average transient characteristics The normalized poles of the Butterworth filter fall on the unit circle (in the s plane). The pole positions are given by: (2K-1) -sin (2K-1)T +j cos K-1,2 n 2n 2n where K is the pole pair number, and n is the number of poles The poles are spaced equidistant on the unit circle, which means the angles between the poles are equal Given the pole locations, o, and a (or Q) can be determined. These values can then be use to determine the component values of the filter. The design tables for passive filters use frequency and impedance normalized filters. They are normalized to a frequency of 1 rad/sec and impedance of 1 . These filters can be denormalized to determine actual component values. This allows the comparison of the frequency domain and/or time domain responses of the various filters on equal footing. The Butterworth filter is normalized for a -3 dB response at oo = 1.

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 roll-off also called the "transition band", for a simple first-order 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 "High-order" or "nth-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 roll-off and therefore the width of the transition band, depends upon the order number of the filter and that for a simple first-order filter it has a standard roll-off rate of 20DB/decade or 6dB/octave Then, for a filter that has an nth number order, it will have a subsequent roll-off rate of 20n dB/decade or 6n dB/octave. So a first-order filter has a roll-off rate of 20dB/decade (6dB/octave), a second-order filter has a roll-off rate of 40dB/decade (12DB/octave), and a fourth-order filter has a roll-off rate of 80dB/decade (24DB/octave), etc, etc. High-order filters, such as third, fourth, and fifth-order are usually formed by cascading together single first-order and second-order filters. For example, two second-order low pass filters can be cascaded together to produce a fourth-order low pass filter, and so on. Although there is no limit to the order of the filter

Decades and Octaves 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. Logarithmic Frequency Scale One Decade One Octave f 2000 10 20 1000 50 100 200 500 Frequency, (Hz)

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 OHz (DC) until the cut-off frequency at -3dB with no ripples. Higher frequencies beyond the cut-off point rolls-off 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

Ideal Frequency Response for a Butterworth Filter Note that the higher the Butterworth Gain dB 10 90° Corner 1st-order filter order, the higher the number of Maximally flat 0 2nd-order cascaded stages there are within the 3rd-order etc. filter design, and the closer the filter -10 becomes to the ideal "brick wall" -20 response n 1 Ideal "Brick wal Response -30 In practice however, Butterworth's -40 ideal frequency response is n 2 unattainable as it produces excessive -50 n3 6 5 4 passband ripple. -60 F (Hz) 30 0.1 0.3 1 3 10 Normalised Frequency Where the generalised equation representing a "nth" Order Butterworth filter, the frequency response is given as: S. CO

Find the order of an active low pass Butterworth filter whose specifications are given as: Amax 0.5dB at a pass band frequency (wp) of 200 radian/sec (31.8Hz), and Amin -20dB at a stop band frequency (ws) of 800 radian/sec. Also design a suitable Butterworth filter circuit to match these requirements. Firstly, the maximum pass band gain Amax 0.5dB which is equal to a gain of 1.0593, remember that: 0.5dB 20*log(A) at a frequency (wp) of 200 rads/s, so the value of epsilon E is found by: 1+ 1.0593 8=0.3495 and =0.1221 Secondly, the minimum stop band gain Amin = -20DB which is equal to a gain of 10 (-201B 20 log(A)) at a stop band frequency (ws) of 800 rads/s or 127.3Hz. Substituting the values into the general equation for a Butterworth filters frequency response gives us the following:

V810.811 42 28.475 log 28.475 2.42 log 4 Since n must always be an integer (whole number) then the next highest value to 2.42 is n 3, therefore a "a third-order filter is required" and to produce a third-order Butterworth filter, a second-order filter stage cascaded together with a first-order filter stage is required. From the normalised low pass Butterworth Polynomials table above, the coefficient for a third-order filter is given as (1+s)(1+s+s2) and this gives us a gain of 3-A 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 1k and 1k respectively as: ( 1k2/1k)+1 = 2 We know that the cut-off corner frequency, the -3dB point (wo) can be found using the formula 1/CR, but we need to find wo from the pass band frequency w, then,

We know that the cut-off corner frequency, the -3dB point (wo) can be found using the formula 1/CR, but we need to find wo from the pass band frequency wp then, Ho H(jo) 3dB 1.414 at = » 1 1 1.414 2n 1 2n 2 1 1

We know that the cut-off corner frequency, the -3dB point (w) can be found using the formula 1/CR, but we need to find wo from the pass band frequency w, then, Н. Н jо) 2n 1+ 1.414 at o 3dB 1 1 1.414 2n 1 2n 2 18 P