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4 Measurement of Biosignals and Analog Signal Processing

response of the normalised low pass with

|AnTP(jΩ)|² = AnTP(jΩ) ⋅A^∗

nTP^{(jΩ) = AnTP(jΩ) ⋅AnTP(−jΩ)}

(4.17)

and Ω:= 2πF are available. In order to obtain a realisable stable transfer function

AnTP(P) analogous to Equation 4.16, this equation is first extended and the normal-

ised angular frequency Ωis replaced by the normalised complex angular frequency

P := Σ + jΩ. Here the real part Σ is an attenuation part. If one now chooses for the

transfer function AnTP(P) not the polynomial form according to Equation 4.16, but the

representation with the help of the poles and zeros (the polynomials of the numerator

and denominator in Equation 4.16 can each be described by a product of their zeros),

one further obtains:

AnTP(P) ⋅AnTP(−P) = A0

(P −Pn1) ⋅(P −Pn2) ⋅. . . ⋅(P −Pnm)

(P −Pp1) ⋅(P −Pp2) ⋅. . . ⋅(P −Ppn)

⋅A0

(−P −Pn1) ⋅(−P −Pn2) ⋅. . . ⋅(−P −Pnm)

(−P −Pp1) ⋅(−P −Pp2) ⋅. . . ⋅(−P −Ppn) ^.

(4.18)

The product GnTP(P) := AnTP(P) ⋅A^∗

nTP^{(−P) contains twice as many poles and zeros as}

the complex normalised transfer function AnTP(P), which is symmetric in respect to

the imaginary axis jΩ. This expansion of the magnitude square must therefore, given

the magnitude of the normalised low-pass filter, be divided into a symmetrical product

with the same number of poles and zeros in each case in such a way that a realisable

transfer function A(P) can be determined from it, which describes a stable filter. A

filter is stable if its impulse response becomes smaller and smaller with time and tends

towards zero. To achieve this, all poles must lie in the left P half plane. The zeros of

A(P), on the other hand, do not all have to lie in the left P- half plane. Therefore, there

are also different subdivision possibilities in this respect. In a minimum phase system,

which describes the minimum phase change as a function of the frequency, this is the

Tab. 4.5: Overview of the standard-frequency transformations: Let fB denote the reference fre-

quency, f0 := √fD1fD2 the centre frequency, FD, FD1 and FD2 the passband cut-off frequencies and

FS, FS1 and FS2 the stopband cut-off frequencies.

Transformation

transformation equation

cut-off frequencies

nTP ⇒TP

F =

f

fD

fD = fB

nTP ⇒HP

F = −^fD

f

fD = fB

nTP ⇒BP

F =

f−f ²

0 ^/f

fB−f ²

0 ^/fB

fD1 = ^{f 2o}

fB

f ²

0 ^{= fD1fD2}

fD2 = fB

nTP ⇒BS

F = −

fB−f ²

0 ^/fB

f−f ²

0 ^/f

fD1 = fB

f ²

0 ^{= fD1fD2}

fD2 =

f ²

0

fB