4.5 Design of Analogue Filters

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131

known, they still have to be selected for AnTP(P) according to Equation 4.18 in such

a way that they all lie in the left P- half plane. Furthermore, the constant factor A0

must still be determined that at the passband cut-off frequency ΩD = 1 the magnitude

square of the transfer function |AnTP(Ω= 1)|² drops to the value

GnTP(jΩ= 1) = |AnTP(jΩ= 1)|² =

1

1 + ϵ²(Ω= 1)^{2n =}

1

1 + ϵ²

(4.26)

Considering now the pole and zero position representation of the normalised transfer

function AnTP(jΩ) analogous to the representation in Equation 4.18, where here there

are no zeros and the poles after selection all lie in the left P- half plane,

|AnTP(jΩ)| =

󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

A0

1

(jΩ−Pp1) ⋅(jΩ−Pp2) ⋅⋅⋅⋅⋅(jΩ−Ppn)

󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

,

(4.27)

so follows Equation 4.26

|A0| =

󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

AnTP(jΩ= 1) ⋅

n−1

∏

l=0

(j −Ppl)

󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

=

1

√1 + ϵ²

󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

⋅

n−1

∏

l=0

(j −Ppl)

󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

.

(4.28)

Substituting the pole places after Equation 4.25 into Equation 4.28 gives:

A0 = ¹

ϵ ^.

(4.29)

Explanatory Examples

Lowpass 1st order

For filtering an ECG, a passive Butterworth-filter 1st order is to be designed with the

aid of an RC element, which has a passband cut-off frequency of 200 Hz and at this an

attenuation of 3 dB.

For this purpose, the corresponding normalised low-pass filter is first determined.

Since at an attenuation of 3 dB the magnitude square of the frequency response has

the value 0.5, it follows from Equation 4.26 that ϵ = 1 must be. From Equation 4.25 it

follows for the pole positions

P±p0 = ^{ej(π/2±(π+0⋅2π)/2⋅1)}

n√1

= ±1

(4.30)

and for |A0|, when all poles are in the left P- half plane, from Equation 4.29:

A0 = ¹

ϵ ^{= 1.}

(4.31)

So the poles are purely real and are at ±1. The pole in the left P- half plane is therefore

P0 = −1 and is selected. The transfer function of the normalised low-pass can now be

given:

AnTP(P) = A0

1

P −Pp0

=

1

P + 1 ^.

(4.32)