4.5 Design of Analogue Filters

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133

a bandwidth ∆f = fD2 −fD1 of 20 Hz is to be designed using an RLC element. For this

purpose, the corresponding normalised low-pass filter from the previous example can

be used (cf. Equation 4.32) and then the low-pass-bandstop-transformation according

to Table 4.5 can be performed. This results in:

AnTP(P) =

1

P + 1 ^{bzw. AnTP(jΩ) =}

1

jΩ+ 1 ^.

(4.37)

If the frequencies are expressed in terms of angular frequencies, the associated fre-

quency transformation on denormalisation gives:

Ω= −^{ωB −ω2}

0^/ωB

ω −ω²

0^/ω

,

mit ωB = ωD1 und ^ω2

0

ωB

= ωD2 .

(4.38)

Substituting Equation 4.38 into Equation 4.37, it follows for the bandstop transfer

function:

ABS(jω) =

ω² −ω²

0

ω² −ω²

0 ^{−jω (ωB −ω2}

0^/ωB)

⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

∆ω

.

(4.39)

This transfer function must be identical to that of an RLC element, which is obtained

by applying the equation for a voltage divider (cf. Figure 4.41,left) it follows for

ARLC(jω) = ^U2

U1

=

ω² −1/LC

ω² −1/LC −jωR/L ^.

(4.40)

Since both transfer functions must be identical, it follows by comparison:

ABS(jω) =

ω² −ω²

0

ω² −ω²

0 ^{−jω (ωB −ω2}

0^/ωB)

⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

∆ω

= ARLC(jω) =

ω² −1/LC

ω² −1/LC −jωR/L

(4.41)

and hence:

ω²

0 ^{= 1}

LC

und

∆ω = ^R

L ^.

(4.42)

Because of the relationship ω²

0 ^{= ωD1ωD2 = (2π ⋅50 Hz)2 and ωB −.ω2}

0^{/ωB =}

ωD1−ωD2 = ∆ω = 2π⋅20 Hz one obtains e.g. with capacitance choice of the capacitor C

of 100 μF:

L = ω²

0 ^{⋅C = 101.23 mH}

R = ∆ω ⋅L = 12.73 Ω.

(4.43)

The corresponding frequency response calculated with LTspice according to mag-

nitude and phase is shown in Figure 4.41 (right). The LTspice-simulation of a filtered

ECG signal, which was strongly disturbed by a 50 Hz mains hum signal, is shown in

Figure 4.42.