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Prototype Analog Filter Design

Module by: Douglas L. Jones. E-mail the author

Analog Filter Design

Laplace transform: Hs= h a te(st)d t H s t h a t s t Note that the continuous-time Fourier transform is Hjλ H λ (the Laplace transform evaluated on the imaginary axis).

Since the early 1900's, there has been a lot of research on designing analog filters of the form Hs= b 0 + b 1 s+ b 2 s2+...+ b M sM1+ a 1 s+ a 2 s2+...+ a M sM H s b 0 b 1 s b 2 s 2 ... b M s M 1 a 1 s a 2 s 2 ... a M s M A causal IIR filter cannot have linear phase (no possible symmetry point), and design work for analog filters has concentrated on designing filters with equiriplle ( L L ) magnitude responses. These design problems have been solved. We will not concern ourselves here with the design of the analog prototype filters, only with how these designs are mapped to discrete-time while preserving optimality.

An analog filter with real coefficients must have a magnitude response of the form |Hλ|2=Bλ2 H λ 2 B λ 2

HjλHjλ¯= b 0 + b 1 jλ+ b 2 jλ2+ b 3 jλ3+...1+ a 1 jλ+ a 2 jλ2+...Hjλ¯= b 0 b 2 λ2+ b 4 λ4+...+jλ( b 1 b 3 λ2+ b 5 λ4+...)1 a 2 λ2+ a 4 λ4+...+jλ( a 1 a 3 λ2+ a 5 λ4+...) b 0 b 2 λ2+ b 4 λ4+...+jλ( b 1 b 3 λ2+ b 5 λ4+...)1 a 2 λ2+ a 4 λ4+...+jλ( a 1 a 3 λ2+ a 5 λ4+...)¯= b 0 b 2 λ2+ b 4 λ4+...2+λ2 b 1 b 3 λ2+ b 5 λ4+...21 a 2 λ2+ a 4 λ4+...2+λ2 a 1 a 3 λ2+ a 5 λ4+...2=Bλ2 H λ H λ b 0 b 1 λ b 2 λ 2 b 3 λ 3 ... 1 a 1 λ a 2 λ 2 ... H λ b 0 b 2 λ 2 b 4 λ 4 ... λ b 1 b 3 λ 2 b 5 λ 4 ... 1 a 2 λ 2 a 4 λ 4 ... λ a 1 a 3 λ 2 a 5 λ 4 ... b 0 b 2 λ 2 b 4 λ 4 ... λ b 1 b 3 λ 2 b 5 λ 4 ... 1 a 2 λ 2 a 4 λ 4 ... λ a 1 a 3 λ 2 a 5 λ 4 ... b 0 b 2 λ 2 b 4 λ 4 ... 2 λ 2 b 1 b 3 λ 2 b 5 λ 4 ... 2 1 a 2 λ 2 a 4 λ 4 ... 2 λ 2 a 1 a 3 λ 2 a 5 λ 4 ... 2 B λ 2
(1)
Let s=jλ s λ , note that the poles and zeros of Bs2 B s 2 are symmetric around both the real and imaginary axes: that is, a pole at p 1 p 1 implies poles at p 1 p 1, p 1¯ p 1 , p 1 p 1 , and p 1¯ p 1 , as seen in Figure 1.

Figure 1
s-plane
s-plane (fig1.png)

Recall that an analog filter is stable and causal if all the poles are in the left half-plane, LHP, and is minimum phase if all zeros and poles are in the LHP.

s=jλ s λ : Bλ2=Bs2=HsHs=H(jλ)H((jλ))=H(jλ)Hjλ¯ B λ 2 B s 2 H s H s H λ H λ H λ H λ we can factor Bs2 B s 2 into HsHs H s H s , where Hs H s has the left half plane poles and zeros, and Hs H s has the RHP poles and zeros.

|Hs|2=HsHs H s 2 H s H s for s=jλ s λ , so Hs H s has the magnitude response Bλ2 B λ 2 . The trick to analog filter design is to design a good Bλ2 B λ 2 , then factor this to obtain a filter with that magnitude response.

The traditional analog filter designs all take the form Bλ2=|Hλ|2=11+Fλ2 B λ 2 H λ 2 1 1 F λ 2 , where F F is a rational function in λ2 λ 2 .

Example 1

Bλ2=2+λ21+λ4 B λ 2 2 λ 2 1 λ 4 Bs2=2s21+s4=(2s)(2+s)(s+α)(sα)(s+α¯)(sα¯) B s 2 2 s 2 1 s 4 2 s 2 s s α s α s α s α where α=1+j2 α 1 2 .

Note:

Roots of 1+sN 1 s N are N N points equally spaced around the unit circle (Figure 2).

Figure 2
Figure 2 (fig2.png)

Take Hs=LHP H s LHP factors: Hs=2+s(s+α)(s+α¯)=2+ss2+2s+1 H s 2 s s α s α 2 s s 2 2 s 1

Traditional Filter Designs

Butterworth

Bλ2=11+λ2M B λ 2 1 1 λ 2 M

Note:

Remember this for homework and rest problems!
"Maximally smooth" at λ=0 λ 0 and λ= λ (maximum possible number of zero derivatives). Figure 3. Bλ2=|Hλ|2 B λ 2 H λ 2

Figure 3
Figure 3 (fig3.png)

Chebyshev

Bλ2=11+ε2 C M 2λ B λ 2 1 1 ε 2 C M λ 2 where C M 2λ C M λ 2 is an Mth Mth order Chebyshev polynomial. Figure 4.

Figure 4
(a)
Figure 4(a) (fig4a.png)
(b)
Figure 4(b) (fig4b.png)

Inverse Chebyshev

Figure 5.

Figure 5
Figure 5 (fig5.png)

Elliptic Function Filter (Cauer Filter)

Bλ2=11+ε2 J M 2λ B λ 2 1 1 ε 2 J M λ 2 where J M J M is the "Jacobi Elliptic Function." Figure 6.

Figure 6
Figure 6 (fig6.png)

The Cauer filter is L L optimum in the sense that for a given MM, δpδp, δsδs, and λpλp, the transition bandwidth is smallest.

That is, it is L L optimal.

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