Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.341:DISCRETE-TIME SIGNAL PROCESSING OpenCourse Ware 2006 Lecture 3 Minimum-Phase and All-Pass Systems Reading:Sections 5.5 and 5.6 in Oppenheim,Schafer Buck(OSB). All-Pass Systems Definition of an all-pass system HAP(z)is as follows: HAP(ej)=A The gain of an all-pass system is a real constant (A doesn't necessarily need to be 1). In order to satisfy the above definition,each pole of HAP(z)should be paired with a conjugate reciprocal zero,as shown in OSB Figure 5.21. A rational all-pass system has the general form given below: (z-1-a) HAP(2)=A (1-akz-1) lak<1 k=1 If a pole is at z=ak then a zero is at z=1/ak,i.e.a pole at ak=rei is paired with a zero at家=e0,Ifhl回is real,,then ax=o OSB Figure 5.24 shows the frequency response for an all-pass system with the pole-zero plot in OSB Figure 5.21.Note that Figure 5.24(b)shows the wrapped phase.The group delay in (c)is largest when w=士π/4andπ，the points on the unit circle that are closest to the poles and zeros.The phase change is greatest around these points. An all-pass system is always stable,since when frequency response characteristics (such as all- pass)are discussed,it is naturally assumed that the Fourier transform exists,thus stability is implied. Example: H(2)=2-1 has a pole at the origin,and a zero at oo,thus it is an all-pass system.In general,any rational function H(z)will have an equal number of poles and zeros (some at ∞ 1
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.341: Discrete-Time Signal Processing OpenCourseWare 2006 Lecture 3 Minimum-Phase and All-Pass Systems Reading: Sections 5.5 and 5.6 in Oppenheim, Schafer & Buck (OSB). All-Pass Systems Definition of an all-pass system HAP (z) is as follows: |HAP (ejω)| = A The gain of an all-pass system is a real constant (A doesn’t necessarily need to be 1). In order to satisfy the above definition, each pole of HAP (z) should be paired with a conjugate reciprocal zero, as shown in OSB Figure 5.21. A rational all-pass system has the general form given below: p (z−1 − a∗ k) HAP (z) = A � (1 − akz−1) |ak| < 1 k=1 If a pole is at z = ak then a zero is at z = 1/a∗ k, i.e. a pole at ak = rejθ is paired with a zero at 1 = 1 ejθ. If h[n] is real, then ak = a∗ a∗ r k. k OSB Figure 5.24 shows the frequency response for an all-pass system with the pole-zero plot in OSB Figure 5.21. Note that Figure 5.24(b) shows the wrapped phase. The group delay in (c) is largest when ω = ±π/4 and π, the points on the unit circle that are closest to the poles and zeros. The phase change is greatest around these points. An all-pass system is always stable, since when frequency response characteristics (such as allpass) are discussed, it is naturally assumed that the Fourier transform exists, thus stability is implied. Example: H(z) = z−1 has a pole at the origin, and a zero at ∞, thus it is an all-pass system. In general, any rational function H(z) will have an equal number of poles and zeros (some at ∞). 1
Minimum-Phase Systems The basic definition of a minimum-phase system is as follows: Stable causal and has a stable causal inverse Stable and causal All poles of H(z)are inside the unit circle. Stable and causal inverse All poles of 1/H(z)are inside the unit circle or equivalently,all zeros of H(z)are inside the unit circle. Thus,to have a minimum-phase system,all poles and zeros of H(z)must be inside the unit circle (no pole or zero at oo).Since the number of poles is always equal to the number of zeros, you have the same number of poles and zeros inside the unit circle.The system in OSB Figure 5.30(a)has four poles and four zeros inside the unit circle,and thus is minimum-phase. Example: H(z)=z-:NOT minimum-phase (a pole at oo) From the definitions,it is clear that an all-pass system cannot be minimum-phase. Spectral Factorization Generally,several different systems can have different phase responses and yet have the same magnitude response.However,for a minimum-phase signal hnl,the frequency response can be uniquely recovered (to within a sign change)from the magnitude alone.This also means that you cannot specify both magnitude and phase independently for a minimum-phase system. For a real rational system: H(e)2=H(e)H*(e)=H(e)H(e-3)=H(2)H(1/2)=. The following example demonstrates the process of recovering H(z)from the given |H(e)2. Example: aeP-蓝-专ce“-最-e-e 5-cosw -ew-e-jm 2
Minimum-Phase Systems The basic definition of a minimum-phase system is as follows: Stable & causal and has a stable & causal inverse Stable and causal ⇔ All poles of H(z) are inside the unit circle. Stable and causal inverse ⇔ All poles of 1/H(z) are inside the unit circle or equivalently, all zeros of H(z) are inside the unit circle. Thus, to have a minimum-phase system, all poles and zeros of H(z) must be inside the unit circle (no pole or zero at ∞). Since the number of poles is always equal to the number of zeros, you have the same number of poles and zeros inside the unit circle. The system in OSB Figure 5.30(a) has four poles and four zeros inside the unit circle, and thus is minimum-phase. Example: H(z) = z − 1 2 : NOT minimum-phase (a pole at ∞) From the definitions, it is clear that an all-pass system cannot be minimum-phase. Spectral Factorization Generally, several different systems can have different phase responses and yet have the same magnitude response. However, for a minimum-phase signal h[n], the frequency response can be uniquely recovered (to within a sign change) from the magnitude alone. This also means that you cannot specify both magnitude and phase independently for a minimum-phase system. For a real rational system: |H(ejω) 2 = H(ejω)H∗(ejω) = H(ejω)H(e−jω) = H(z)H(1/z) z=e | | jω . |H(ejω) 2 The following example demonstrates the process of recovering H(z) from the given | . Example: 5 17 16 − 1 5 17 16 − 1 1 4 ejω e−jω cos ω − H(ejω) 2 = 2 4 | | = 1 1 ejω − e−jω 4 cos ω 4 − 2 − 2 2
We will just replace ei with z(analytic continuation). →H(e)H(1/l=e= Let G(2)=H(2)H(1/z),then we need to factor G(2). Zeros of G(z):z=4,1/4 Poles of G(z):z=2,1/2 1 0.5 0 05 =1 -1.5 Since H(z)is minimum-phase,we know that H(z)must have all its poles and zeros inside the unit circle,thus it has a pole at 1/2 and a zero at 1/4. =4 H/A)=41- 1-方z H(eH0/月l=1e==4e-- 二=A)2=43 From the given H(e)2, IH(e“)Pw=0= This problem of recovering frequency response from the magnitude response is commonly called spectral factorization. 3
e with z =⇒ H(z)H(1/z)|z=e = 17 16 − 1 4 z − 1 4 z−1 5 4 − 1 2 z − 1 2 z−1 |z=e Let G(z H(z)H(1/z G(z). G(z z , 1/4 G(z z , 1/2 −1 0 1 2 3 4 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 1/4 1/2 2 4 We will just replace jω (analytic continuation). jω jω ) = ), then we need to factor Zeros of ) : = 4 Poles of ) : = 2 Since H(z) is minimum-phase, we know that H(z) must have all its poles and zeros inside the unit circle, thus it has a pole at 1/2 and a zero at 1/4. 1 1 z H(z) = A 1 − 4 z−1 H(1/z) = A 1 − 4 1 1 1 − 2 z−1 1 − 2 z 1 4 )2 3 H(z)H(1/z) z=1(ω=0) = A2 (1 − 4 )2 = A2 ( 3 2 )2 = A2( 2 ) 2 1 | (1 − 2 )2 ( 1 |H(ejω) 2 From the given | , 17 1 9 |H(ejω) 2 16 − 2 = 16 9 | |ω=0 = 5 = A = 1 1 4 ⇒ 4 − 1 4 This problem of recovering frequency response from the magnitude response is commonly called spectral factorization. 3
Maximum-Phase Systems The basic definition of a maximum-phase system is as follows: Stable and anti-causal with a stable and anti-causal inverse. All poles and zeros are outside the unit circle and ROC includes the unit circle. OSB Figure 5.30(b)shows the pole/zero plot of a delayed maximum-phase system.The system is stable but causal due to the 4th order pole at z=0.The corresponding sequences associated with OSB Figure 5.30 are shown in OSB Figure 5.31.Note that hon]is the flipped and delayed version of ha[n]as Ho(z)=c.z-4.Ha(1/z). A stable system H(z)can always be expressed as H(2)=HMIN(2)HMAX(a).2-M for some integer M.The factorMallows us to compensate for poles and zeros at the origin or at oo. One example of the factorization is shown below.HMIN(z)takes thethe pole inside the unit circle and HMAx(z)takes the zero outside the unit circle.In order to make the number of poles and zeros equal,we place a zero at the origin for HMIN,and a pole at oo for HMAx.The factor z-1 compensates these zero and pole. H()=日 HMIN=产a HMAX=之-b 之1 4
um-Phase Systems definition of a maximum-phase system is as follows: anti-causal with a stable and anti-causal inverse. oles and zeros are outside the unit circle and ROC includes the unit circle. 5.30(b) shows the pole/zero plot of a delayed maximum-phase system. The system but causal due to the 4th order pole at z = 0. The corresponding sequences associated Figure 5.30 are shown in OSB Figure 5.31. Note that hb[n] is the flipped and delayed ha[n] as Hb(z) = c · z−4 · Ha(1/z). system H(z) can always be expressed as H(z) = HMIN (z) HMAX(z) · z−M integer M. The factor z−M allows us to compensate for poles and zeros at the origin of the factorization is shown below. HMIN (z) takes the the pole inside the unit HMAX(z) takes the zero outside the unit circle. In order to make the number of zeros equal, we place a zero at the origin for HMIN , and a pole at ∞ for HMAX. The 1 compensates these zero and pole. H(z) = z−b z−a ⇓ Maxim ⇔ ∞. factor z− −1 −0.5 0 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 a −1 −0.5 0 0.5 1 1.5 −1 −0.5 0 0.5 1 1.5 INF −1 −0.5 0 0.5 1 1.5 −1 −0.5 0 0.5 1 1.5 INF The basic Stable and All p OSB Figure is stable with OSB version of A stable for some or at One example circle and poles and HMIN = z z−a HMAX = z − b z−1 4
A stable system H(z)can also be expressed as H(z)=HMIN(z)HAP(z). H(a)=-日 8 -t HMIN -1/b 2-a HAP= OSB problem 5.64 illustrates the importance of this concept in compensating the magnitude response of a nonminimum-phase system. xm一H()一H(a)一m H(z)=HMIN(z）HAP(2) Since HMIN(z)has a stable and causal inverse,we can define a stable and causal system Hc(z) as follows: He()=HMIN(2) 1 With this compensating system,the magnitude of the overall frequency response is unity. Y(e)=H(ejw)Hc(e)X(ejw)=HAP(ej)X(ej=X(ej) 5
can also be expressed as H(z) = HMIN (z)HAP (z). H(z) = z−b z−a ⇓ H(z) −1 −0.5 0 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 Real Part Imaginary Part a 1/b −1 −0.5 0 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 1/b b A stable system HMIN = z−1/b z−b z−a HAP = z−1/b OSB problem 5.64 illustrates the importance of this concept in compensating the magnitude response of a nonminimum-phase system. x[n] −→ H(z) −→ Hc(z) −→ y[n] H(z) = HMIN (z) HAP (z) Since HMIN (z) has a stable and causal inverse, we can define a stable and causal system Hc(z) as follows: 1 Hc(z) = HMIN (z) With this compensating system, the magnitude of the overall frequency response is unity. |Y (ejω) = H(ejω) Hc(ejω) jω) = |HAP (ejω) jω)| = X(ejω | | | |X(e | | |X(e | |) 5