5.3 Methods for Analysis and Processing of Discrete Biosignals

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197

the sampling times t = ta = nTa (with n = 1, 2, 3, . . . , and Ta: sampling interval), this

means

gan(t = nTa) = gdi(n) .

(5.88)

To do this, we first assume that the transfer function Gan(p) of the analogue filter is in

a partial fraction form according to

Gan(p) =

N

i=1

Ai

ppi

,

with

pi : pole i

(5.89)

can be represented. In this case, one then obtains for the associated impulse response

gan(t) at the sampling times ta = nTa

gan(nTa) =

N

i=1

σ(nTa)AiepinTa =

N

i=1

AiepinTa ,

σ(t): step function .

(5.90)

Let these values be the impulse response gdigi(n) of the digital filter to be implemen-

ted. The corresponding transfer function G(z) is obtained by the discrete z- transform-

ation:

Gdi(z) =

n=0

gan(nTa)zn =

n=0

(

N

i=1

AiepinTa)zn =

N

i=1

Ai

n=0

(epiTaz1)n

⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

geometrische Reihe

.

(5.91)

The sum signs may be interchanged in the last equation according to the laws of math-

ematics. In this case, with the sum with the upper limit, one obtains a geometric

series whose result can be easily calculated, and the result further follows

Gdi(z) =

N

i=1

Ai

1epiTaz1 =

N

i=1

Aiz

zepiTa = c0 + c1z + c2z2 + ⋅⋅⋅+ cNzN

d0 + d1z + d2z2 + ⋅⋅⋅+ zN

.

(5.92)

From the pole positions pi of the analogue filter, the coefficients ci and di (i

=

0, . . . , N) of the digital filter can be calculated, which can then be realised by a

filter in the 2nd canonical direct form according to Figure 5.34.

Note

So that for small frequencies the magnitude of the transfer function in the continuous-

time domain |Gan(p =)| is approximately equal to the amount of the transfer func-

tion |Gdi(z = ejωTa)| for the discrete-time domain, Gdi(z) is still multiplied by the scal-

ing factor Ta.

Explanatory Example

A simple analogue RC low-pass filter with a cut-off frequency fg of 200 Hz with com-

ponent values R = 800 kand C = 1 nF according to Figure 5.39 shall be replaced by