Continuous Time Active Filter Design (Delyannis 1999, Crc Press) Fixed

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©1999 CRC Press LLC 1.6 Continuous-Time Filter Functions As was mentioned in Section 1.2, the response of a continuous-time filter to the continuous- time excitation e(t) is a continuous-time function r(t) given as follows: (1.41) where h(t) is the impulse response (see Section 1.6.3) of the filter. In the frequency domain this equation is written as follows: (1.42) where R(s), H(s), E(s) are the Laplace transforms of the time functions r(t), h(t), and e(t) respectively, and s is the complex frequency variable. H(s) is the filter function, transfer or driving-point impedance, or admittance function. These are shown in Fig. 1.19 in block dia- gram form. For the filters we are concerned with, H(s) will be a rational function of s, i.e., the ratio of two real and finite polynomials in s. It is written in the following form: (1.43) where N(s) and D(s) are the numerator and denominator polynomials, respectively, with m n, a i , b i real and b i positive (for stability reasons explained below). If z i , i = 1,2,...n are the roots of N(s), i.e., the zeros of H(s) and p i , i = 1,2,....,m are the roots of D(s), i.e., the poles of H(s), then Eq. (1.43) can be written as follows: (1.44) If the signal is sinusoidal of frequency ω, in Eqs. (1.43) and (1.44) s is substituted by jω. Function H(jω) obtained this way is in fact the continuous-time Fourier transform of h(t). It can then be written in the following form: (1.45) i.e., in terms of the magnitude and phase of H(jω). rt () ht τ ( ) e τ ()τ d 0 t = Rs () Hs () Es () = FIGURE 1.19 Block diagram representation of the filter (a) in the time domain and (b) in the frequency domain. Hs () Ns () Ds () ----------- a n s n a n 1 s n 1 a 1 s a o + + + + s m b m 1 s m 1 b 1 s b o + + + + -------------------------------------------------------------------------- = = Hs () a n s z i ( ) i 1 = n s p j ( ) j 1 = m -------------------------- = Hj ω ( ) Hj ω ( ) e j ϕω ( ) =

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Transcript of Continuous Time Active Filter Design (Delyannis 1999, Crc Press) Fixed

  • 1999 CRC Press LLC

    1.6 Continuous-Time Filter Functions

    As was mentioned in Section 1.2, the response of a continuous-time lter to the continuous-time excitation e(t) is a continuous-time function r(t) given as follows:

    (1.41)

    where h(t) is the impulse response (see Section 1.6.3) of the lter.In the frequency domain this equation is written as follows:

    (1.42)

    where R(s), H(s), E(s) are the Laplace transforms of the time functions r(t), h(t), and e(t)respectively, and s is the complex frequency variable. H(s) is the lter function, transfer ordriving-point impedance, or admittance function. These are shown in Fig. 1.19 in block dia-gram form.

    For the lters we are concerned with, H(s) will be a rational function of s, i.e., the ratio oftwo real and nite polynomials in s. It is written in the following form:

    (1.43)

    where N(s) and D(s) are the numerator and denominator polynomials, respectively, withm n, ai, bi real and bi positive (for stability reasons explained below).

    If zi, i = 1,2,...n are the roots of N(s), i.e., the zeros of H(s) and pi, i = 1,2,....,m are the rootsof D(s), i.e., the poles of H(s), then Eq. (1.43) can be written as follows:

    (1.44)

    If the signal is sinusoidal of frequency , in Eqs. (1.43) and (1.44) s is substituted by j.Function H(j) obtained this way is in fact the continuous-time Fourier transform of h(t). Itcan then be written in the following form:

    (1.45)

    i.e., in terms of the magnitude and phase of H(j).

    r t( ) h t ( )e ( ) d0

    t

    =

    R s( ) H s( )E s( )=

    FIGURE 1.19Block diagram representation of the lter (a) in the time domain and (b) in the frequency domain.

    H s( ) N s( )D s( )------------

    ansn

    an 1 sn 1

    a1s ao+ + + +

    sm bm 1 s

    m 1 b1s bo+ + + +

    ---------------------------------------------------------------------------= =

    H s( ) ans zi( )

    i 1=

    n

    s p j( )

    j 1=

    m

    --------------------------=

    H j( ) H j( ) e j ( )=

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    Clearly,

    (1.46)

    It is usual practice to present the magnitude of H(j) in the form

    (1.47)

    and thus express it in dB. This gives the lter gain in dB.However, in most cases, we talk about the lter attenuation or loss, A(), also in deci-

    bels. In some cases, the attenuation is given in nepers obtained as follows:

    Attenuation in nepers: (1.48)

    In most lter design cases, H(s) represents the ratio of the Laplace transform of the out-put voltage to the Laplace transform of the input voltage to the lter being thus dimen-sionless. However, it may also represent ratio of currents, when it will again bedimensionless, or ratio of output voltage to input current (transimpedance) or output cur-rent to input voltage (transadmittance) having the dimensions of impedance or admit-tance, respectively. Finally, it may represent a driving point function, i.e., the ratio of thevoltage to the current in one port of the lter network or vice versa. In these cases, H(s)will represent either an impedance function or an admittance function, again not beingdimensionless.

    1.6.1 Pole-Zero Locations

    The roots of N(s), which are the zeros zi of H(s) (because for s = zi H(s) becomes zero), canbe real or complex conjugate, since all of the coefcients of N(s) are real. Each of these zeroscan be located at a unique point in the complex frequency plane as shown in Fig. 1.20. Incase of a multiple zero, all of them are located at the same point in the s-plane.

    On the other hand, the roots of D(s), which are the poles pi of H(s) (because for s = pi , H(s)becomes innite) can be real or complex conjugate, since D(s) also has real coefcients.However, their real part can only be negative for reasons of stability. Also, for a network tobe useful as a lter, its transfer function H(s) should not have poles with real part equal tozero. Thus, the poles of function H(s) should all lie in the left half of the s-plane (LHP)excluding the j-axis, while its zeros can lie anywhere in the s-plane, i.e., in the left-half andin the right-half s-plane (RHP).

    ( ) H j( )[ ]arg=

    A ( ) 20 H j( )log=

    ( ) H j( )ln=

    FIGURE 1.20(a) Possible zero and (b) possible pole locations in the s-plane.