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
p −pi
,
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)Aiepi⋅nTa =
N
∑
i=1
Aiepi⋅nTa ,
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)z−n =
∞
∑
n=0
(
N
∑
i=1
AiepinTa)z−n =
N
∑
i=1
Ai
∞
∑
n=0
(epiTaz−1)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
1 −epiTaz−1 =
N
∑
i=1
Aiz
z −epiTa = 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 = jω)| 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 kΩand C = 1 nF according to Figure 5.39 shall be replaced by