LINEAR-PHASE FIR FILTERS 1. The amplitude response 2. Why linear-phase? 3. The four types of linear-phase FIR filter 4. Amplitude response characteristics 5. Evaluating the amplitude response 6. Zero locations of linear-phase filters 7. Automatic zeros 8. Design by DFT-based interpolation 9. Design by general interpolation I. Selesnick EL 713 Lecture Notes 1 THE AMPLITUDE RESPONSE If the real and imaginary parts of Hf (!) are given by Hf (!) = R(!) + j · I(!) the magnitude and phase are defined as jHf (!)j = pR2(!) + I2(!) I(!) p(!) = arctan R(!) so that Hf (!) = jHf (!)j · ejp(!): With this definition, jHf (!)j is never negative and p(!) is usually discontinuous, but it can be very helpful to write Hf (!) as Hf (!) = A(!) · ejθ(!) where A(!) can be positive and negative, and θ(!) continuous. A(!) is called the amplitude response. The figure illustrates the difference between jHf (!)j and A(!). I. Selesnick EL 713 Lecture Notes 2 THE AMPLITUDE RESPONSE (2) |H(ω)| = |A(ω)| Phase{H(ω)} 1.4 4 1.2 2 1 0.8 0 0.6 0.4 −2 0.2 0 −4 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 ω/π ω/π A(ω) θ(ω) 1.2 0 1 −5 0.8 0.6 −10 0.4 0.2 −15 0 −0.2 −20 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 ω/π ω/π A linear-phase phase filter is one for which the continuous phase θ(!) is linear. Hf (!) = A(!) ejθ(!) with θ(!) = −M · ! + B: We assume in the following that the impulse response h(n) is real- valued. I. Selesnick EL 713 Lecture Notes 3 WHY LINEAR-PHASE? If a discrete-time cosine signal x1(n) = cos(!1 n + φ1) is processed through a discrete-time filter with frequency response Hf (!) = A(!) · ejθ(!) then the output signal is given by y1(n) = A(!1) cos(!1 n + φ1 + θ(!1)) or θ(!1) y1(n) = A(!1) cos !1 n + + φ1 : !1 The LTI system has the effect of scaling the cosine signal and de- laying it by −θ(!1)=!1. When does the system delay cosine signals with different frequencies by the same amount? θ(!) =) = constant ! =) θ(!) = K! =) The phase is linear The function θ(!)=! is called the phase delay. A linear phase filter therefore has constant phase delay. I. Selesnick EL 713 Lecture Notes 4 WHY LINEAR-PHASE: EXAMPLE Consider an discrete-time filter described by the difference equation y(n) = − 0:1821 x(n) + 0:7865 x(n − 1) − 0:6804 x(n − 2) + x(n − 3) + 0:6804 y(n − 1) − 0:7865 y(n − 2) + 0:1821 y(n − 3): When !1 = 0:31 π, then the delay is −θ(!1)=!1 = 2:45. The delay is illustrated in the figure: 2 1.5 1 0.5 0 −0.5 (n) (INPUT) 1 −1 x −1.5 −2 20 22 24 26 28 30 32 34 36 38 40 2 1.5 1 0.5 0 −0.5 (n) (OUTPUT) −1 1 y −1.5 −2 20 22 24 26 28 30 32 34 36 38 40 Notice that the delay is fractional | the discrete-time samples are not exactly reproduced in the output. The fractional delay can be interpreted in this case as a delay of the underlying continuous-time cosine signal. I. Selesnick EL 713 Lecture Notes 5 WHY LINEAR-PHASE: EXAMPLE (2) Consider the same system given on the previous slide, but let us change the frequency of the cosine signal. When !2 = 0:47 π, then the delay is −θ(!2)=!2 = 0:14. 2 1.5 1 0.5 0 −0.5 (n) (INPUT) 2 −1 x −1.5 −2 20 22 24 26 28 30 32 34 36 38 40 2 1.5 1 0.5 0 −0.5 (n) (OUTPUT) −1 2 y −1.5 −2 20 22 24 26 28 30 32 34 36 38 40 For this example, the delay depends on the frequency, because this system does not have linear phase. I. Selesnick EL 713 Lecture Notes 6 WHY LINEAR-PHASE: MORE From the previous slides, we see that a filter will delay different frequency components of a signal by the same amount if the filter has linear phase (constant phase delay). In addition, when a narrow band signal (as in AM modulation) goes through a filter, the envelop will be delayed by the group delay or envelop delay of the filter. The amount by which the envelop is delayed is independent of the carrier frequency only if the filter has linear phase. (See page 214 in Mitra.) Also, in applications like image processing, filters with non-linear phase can introduce artifacts that are visually annoying. I. Selesnick EL 713 Lecture Notes 7 FOUR TYPES OF LINEAR-PHASE FIR FILTERS Sec 4.4.3 in Mitra Linear-phase FIR filter can be divided into four basic types. Type impulse response I symmetric length is odd II symmetric length is even III anti-symmetric length is odd IV anti-symmetric length is even TYPE I IMPULSE RESPONSE TYPE II IMPULSE RESPONSE 0.4 0.4 0.3 0.3 0.2 0.2 0.1 0.1 0 0 −0.1 −0.1 0 2 4 6 8 10 12 0 2 4 6 8 10 12 TYPE III IMPULSE RESPONSE TYPE IV IMPULSE RESPONSE 0.3 0.3 0.2 0.2 0.1 0.1 0 0 −0.1 −0.1 −0.2 −0.2 −0.3 −0.3 0 2 4 6 8 10 12 0 2 4 6 8 10 12 When h(n) is nonzero for 0 ≤ n ≤ N −1 (the length of the impulse response h(n) is N), then the symmetry of the impulse response can be written as h(n) = h(N − 1 − n) and the anti-symmetry can be written as h(n) = −h(N − 1 − n): I. Selesnick EL 713 Lecture Notes 8 FOUR TYPES OF LINEAR-PHASE FIR FILTERS Important note: If the impulse response h(n) is complex-valued, then to have linear-phase the impulse response should be conjugate- symmetric or conjugate-anti-symmetry. I. Selesnick EL 713 Lecture Notes 9 TYPE I: ODD-LENGTH SYMMETRIC The frequency response of a length N = 5 FIR Type I filter can be written as follows. f −j! −2j! −3j! −4! H (!) = h0 + h1e + h2e + h1e + h0e (1) −2j! 2j! j! −j! −2j! = e h0e + h1e + h2 + h1e + h0e (2) −2j! 2j! −2j! j! −j! = e h0(e + e ) + h1(e + e ) + h2 (3) −2j! = e (2h0 cos (2!) + 2h1 cos (!) + h2) (4) = A(!)ejθ(!) (5) where θ(!) = −2!; A(!) = 2h0 cos (2!) + 2h1 cos (!) + h2: Note that A(!) is real-valued and can be both positive and negative. In general, for a Type I FIR filters of length N: Hf (!) = A(!)ejθ(!) M−1 X A(!) = h(M) + 2 h(n) cos ((M − n)!): n=0 θ(!) = −M! N − 1 M = : 2 I. Selesnick EL 713 Lecture Notes 10 TYPE II: EVEN-LENGTH SYMMETRIC The frequency response of a length N = 4 FIR Type II filter can be written as follows. f −j! −2j! −3j! H (!) = h0 + h1e + h1e + h0e (6) − 3 j! 3 j! 1 j! − 1 j! − 3 j! = e 2 h0e 2 + h1e 2 + h1e 2 + h0e 2 (7) − 3 j! 3 j! − 3 j! 1 j! − 1 j! = e 2 h0(e 2 + e 2 ) + h1(e 2 + e 2 ) (8) 3 − 2 j! 3 1 = e 2h0 cos 2! + 2h1 cos 2! (9) = A(!)ejθ(!) (10) where 3 θ(!) = − !; A(!) = 2h cos 3! + 2h cos 1!: 2 0 2 1 2 In general, for a Type II FIR filters of length N: Hf (!) = A(!)ejθ(!) N −1 X2 A(!) = 2 h(n) cos ((M − n)!) n=0 θ(!) = −M! N − 1 M = : 2 I. Selesnick EL 713 Lecture Notes 11 TYPE III: ODD-LENGTH ANTI-SYMMETRIC The frequency response of a length N = 5 FIR Type III filter can be written as follows. f −j! −3j! −4! H (!) = h0 + h1e − h1e − h0e (11) −2j! 2j! j! −j! −2j! = e h0e + h1e − h1e − h0e (12) −2j! 2j! −2j! j! −j! = e h0(e − e ) + h1(e − e ) (13) −2j! = e (2jh0 sin (2!) + 2jh1 sin (!)) (14) −2j! = e j (2h0 sin (2!) + 2h1 sin (!)) (15) −2j! j π = e e 2 (2h0 sin (2!) + 2h1 sin (!)) (16) = A(!)ejθ(!) (17) where π θ(!) = −2! + ;A(!) = 2h sin (2!) + 2h sin (!): 2 0 1 In general, for a Type III FIR filters of length N: Hf (!) = A(!)ejθ(!) M−1 X A(!) = 2 h(n) sin ((M − n)!) n=0 π θ(!) = −M! + 2 N − 1 M = : 2 I. Selesnick EL 713 Lecture Notes 12 TYPE IV: EVEN-LENGTH ANTI-SYMMETRIC The frequency response of a length N = 4 FIR Type IV filter can be written as follows. f −j! −2j! −3j! H (!) = h0 + h1e − h1e − h0e (18) − 3 j! 3 j! 1 j! − 1 j! − 3 j! = e 2 h0e 2 + h1e 2 − h1e 2 − h0e 2 (19) − 3 j! 3 j! − 3 j! 1 j! − 1 j! = e 2 h0(e 2 − e 2 ) + h1(e 2 − e 2 ) (20) 3 − 2 j! 3 1 = e 2jh0 sin 2! + 2jh1 sin 2! (21) 3 − 2 j! 3 1 = e j 2h0 sin 2! + 2h1 sin 2! (22) 3 π − 2 j! j 2 3 1 = e e 2h0 sin 2! + 2h1 sin 2! (23) = A(!)ejθ(!) (24) where 3 π θ(!) = − ! + ;A(!) = 2h sin 3! + 2h sin 1!: 2 2 0 2 1 2 In general, for a Type IV FIR filters of length N: Hf (!) = A(!) ejθ(!) N −1 X2 A(!) = 2 h(n) sin ((M − n)!) n=0 π θ(!) = −M! + 2 N − 1 M = : 2 I.