常用数学公式30

30.5.1 Analog Integrator
1 H AI (s) = , s 1 hAI (t ) = 0 t ≥0+ t ≤0−
30.5.2 Digital Integrator
y(nTs ) − y(nTs − Ts ) = Y ( z ) − z −1Y ( z ) = Ts [ x(nTs − Ts ) + x(nTs )] 2
30.3.4 Stability
If any pole of HD(z) is located outside the unit circle it can be replaced by their reciprocals without changing the shape of the loss characteristics, although a constant vertical shift will be introduced. If any pole is on the unit circle, its magnitude can be reduced slightly.
30.2 Invariant-Impulse-Response Method
30.2.1 Steps to be taken
1. Deduce hA (t ) = impulse response of the analog filter = L −1{H A (s)}, h(0 + ) = lim s{H A (s)} s→∞ 2. Replace t by nTs in hA (t ) 3. Form the Z-transform of hA (nTs )
∑ 1 − exp( p T )z
Ai
i =1 i s
N
−1
2 2 2 2 +j , p2 = − −j p1 = − , 2 2 2 2
2 s ωc 1 = H A = un-normalized filter = s s (s − ω c p1 )(s − ω c p2 ) ωc ω − p1 ω − p2 c c
Ts −1 [ z X ( z ) + X ( z )] 2
30.5.3 Transfer Function
H DI ( z ) = Y ( z ) Ts z + 1 = X ( z ) 2 z − 1
30.5.4 Bilinear Transformation
H DI ( z ) = H AI (s) s= 2 ( z −1 )
© 1999 by CRC Press LLC
30.4 Matched-Z-Transform Method
30.4.1 Matched-Z-Transform Method
HD ( z ) = ( z + 1) H0
L
∏ (z − e ∏ (z − e
i =1 i =1 N
M
si Ts
) , z = e jωTs )
=
A1 A2 + , s − ω c p1 s − ω c p2
A1 =
2 ωc ωc = , ω c p1 − ω c p2 p1 − p2
A2 =
2 ωc ωc = . ω c p2 − ω c p1 p2 − p1
© 1999 by CRC Press LLC
But Ts = 1 / fs = 1 / 30, 000, ω c Ts = π / 5, and hence H D (z) = = ωc ωc 1 1 + p1 − p2 1 − exp[ω c Ts p1 ] z −1 p2 − p1 1 − exp[ω c Ts p2 ] z −1 ( − j 2 ω c / 2) ( j 2 ω c / 2) + . 1 − exp[ω c Ts p1 ] z −1 1 − exp[ω c Ts p2 ] z −1
© 1999 by CRC Press LLC
30
Recursive Filters (Infinite Impulse Response, IIR)
30.1 Introduction 30.2 Invariant-Impulse-Response Method 30.3 Modified-Invariant-Impulse-Response 30.4 Matched-Z-Transform Method 30.5 Bilinear-Transformation Method 30.6 Digital-Filter Transformations References
30.3.1 Analog Transfer Function
H A (s) = H0
N (s) = H0 D(s)
∏ (s − s )
i
M
∏ (s − p )
i i =1
i =1 N
, M ≤ N;
H A (s) = H0 Conditions:
H A1 (s) , H A2 ( s)
H A1 (s) =
1 , D(s)
H A2 ( s) =
1 N (s) ωs 2
hA1 (0 + ) = 0, hA2 (0 + ) = 0, M, N ≥ 2, H A1 (ω ) = H A2 (ω ) ≅ 0 for ω ≥
30.3.2 Digital Filter
HD ( z ) = H0 ω HD1 ( z ) H (e jωTs ) , HD (e jωTs ) = H0 D1 jωTs ≅ H A (ω ) for ω ≤ s 2 HD2 ( z ) H D 2 (e )
∑
i =1
N
Ai , s − pi
hA (t ) = L−1{H A (s)} =
∑Ae
i i =1
N
pi t
Ai = [(s − pi ) H A (s)] s= p
i
hA (nTs ) =
∑Ae
i i =1
N
pi nTs
H D ( z ) = Z{hA (nTs )} =
∑ ∑
Ai
i =1 n=0
∑
N
N −i
∑
(see 30.5.1)
30.5.6 Discrete-Time Transfer Function
1 ai HDI (s) i=0 HD ( z) = N N −i = H A ( s ) s = 2 ( z −1 ) N 1 1 Ts z +1 bi + HDI ( z ) HDI ( z ) i=0
Hence H D (e jωTs ) =
( − j 2 ω c / 2) ( j 2 ω c / 2) + 1 − exp[ω c Ts p1 ] e − jωTs 1 − exp[ω c Ts p2 ] e − jωTs
30.3 Modified-Invariant-Impulse-Response
i
3. Use Ai’s from step 2 to generate the digital filter system function H(z) = Example The normalized transfer function of a second-order Butterworth analog filter with a 3-dB cutoff frequency at 3000 Hz. The sampling frequency is fs = 30,000 s–1. Solution H A ( s) = 1 1 = normalized analog filter = (s − p1 )(s − p2 ) s2 + 2 s + 1 ω c = 2 π 3000 = 6000 π,
30.3.3 Zeroes and poles of HA(s)
HD1 ( z ) =
∑
i =1
N
Ai z N1 ( z ) , HD2 ( z ) = pi Ts = z−e D1 ( z )
∑ z−e
i =1
Mห้องสมุดไป่ตู้
Bi z
Ts si
=
N2 ( z ) N ( z ) D2 ( z ) , HD ( z ) = H0 1 , D2 ( z ) N2 ( z ) D1 ( z )
Poularikas A. D. “Recursive Filters” The Handbook of Formulas and Tables for Signal Processing. Ed. Alexander D. Poularikas Boca Raton: CRC Press LLC,1999
pi Ts
Values of L
Filter Type All pole Elliptic N odd N even Lowpass N 1 0 Highpass 0 0 0 Bandpass N/2 Bandstop 0
0 for N/2 even
0
30.5 Bilinear-Transformation Method
N
∞
(e piTs z −1 ) n =
∑
i =1
N
Ai = 1 − e piTs z −1
∑ A z−e
z
i i =1
N
pi Ts
30.2.3 Procedure of Impulse-Invariant of IIR Filters
1. Obtain the transfer function HA(s) for the desired analog prototype filter (see Chapter 12) 2. For i = 1, 2,L, N determine the poles of pi and HA(s) and compute the coefficients Ai = [(s − pi ) H A (s)] s= p
Ts z +1
© 1999 by CRC Press LLC
30.5.5 Analog Filter Transfer Function
H A ( s) =
N
∑
i=0
N
ai s N −i
N N −i
s +
∑b z
i i=0 N
1 ai H AI (s) i=0 = N N −i N 1 1 + b i H AI (s) H AI (s) i=0
∑
N −i
∑
by replacing HAI(s) by HDI(z).
30.5.7 Mapping Properties of Bilinear Transformation
2 +s T z= s 2 −s Ts a) The open right-half s-plane is mapped onto the region exterior to the unit circle z = 1 of the zplane; b) The j axis of the s-plane is mapped onto the unit circle z = 1 of the z-plane; c) The open left-half s-plane is mapped onto the interior of the unit circle z = 1; d) The origin of the s-plane maps onto point (1,0) of the z-plane; e) The positive and negative j axes of the s-plane map onto the upper and lower semicircles of z = 1, respectively; jΩT f) The maxima and minima of H A (ω ) are preserved in H D (e s ) ; jΩTs g) If m1 ≤ H A (ω ) ≤ m2 in ω 1 ≤ ω ≤ ω 2 , then m1 ≤ H D (e ) ≤ m2 for a corresponding frequency Ω1 ≤ Ω ≤ Ω 2 ; h) Passbands or stopbands in the analog filter translate into passbands or stopbands in the digital filter; i) A stable analog filter will yield a stable digital filter.
30.2.2 Conditions
If H A (ω ) ≅ 0 for ω ≥ ω s / 2 and h(0 + ) = 0, then Ts H D (e jωTs ) = H A (ω ) for ω ≤ ωs 2
© 1999 by CRC Press LLC
Simple poles
H A ( s) =
30.1 Introduction
30.1.1 Realizable Filter
The transfer function must a) be a rational function of z with real coefficients, b) have poles that lie within the unit circle of the z plane, and c) have the degree of the numerator equal to or less than that of the denominator polynomial.