Figure 1.52: Location of poles in Example 1 this system be causal and stable? H(z) = 2 + 2 + 3 1 − 0.9e j 4 z−1 1 − 0.9e− j 4 z−1 1 + 2z−1 • Solution: The system has poles at z = e j 4 and z = e− j 4 and z = −2 Ex1(a): Stable system • The location of poles in the z-plane Ex1(a): Stable system For stable system: the ROC must include the unit circle Two inside the unit circle contribute right sided terms
• Find the recursive system is both stable and causal y[n] −y[n − 1] = x[n] with  = 1 + r and r is +ve • H(z) = 1 , a pole is at z =  and is greater than one, the pole is outside the unit circle and hence system can not be both causal and stable system Ex3: Stable and causal system Find the impulse response of a stable and causal system described by a difference equation y[n] + 1 y[n − 1] − 1 y[n − 2] = −2x[n] + − 1] 4 8 4 Solution: find the z-transform from the difference equation and then find the impulse response Ex3: Stable and causal system • z-transform is H(z) = kM=0 bkz−k kN=0 akz−k H(z) = −2 + 45 z−1 1 + 41 z−1 − 81 z−2 • Find the poles of H(z) and Write the denominator in product form Ex3: Stable and causal system • We get 0
• H(z) in terms of partial fraction expansion H(z) = −3 + 1 1 + 21 z−1 1 − 41 z−1 Ex3: Stable and causal system • Now write the impulse response for a causal and stable system h[n] = −3( − 1 )nu[n] + ( 1 )nu[n] Inverse system • Impulse response (hinv) of an inverse system satisfies hinv[n] ∗ h[n] = [n] where h[n] is the impulse response of a system to be inverted • Take inverse z-transform on both sides gives Hinv(z)H(z) = 1 Hinv(z) = 1 H(z) The transfer function of an LTI inverse system is the inverse of the transfer function of the system that we desire to invert If we write the pole-zero form of H(z) as ˜ p M−p 1 H(z) = bz− k=1 (1 − z−l kN=−1l (1 − dkz−1)
˜ = / where b bp al • Then we can write Hinv as z−l N−l (1 − dkz−1) Hinv(z) = k=1 ˜ −p M−p( − −1) bz k=1 1 ckz The zeros of H(z) are the poles of Hinv(z) The poles of H(z) are the zeros of Hinv(z) System defined by a rational transfer function has an inverse system We need inverse systems which are both stable and causal to invert the distortions introduced by the system The inverse system Hinv(z) is stable and causal if all poles are inside the unit circle Poles of Hinv(z) are zeros of (z) Inverse system Hinv(z): stable and causal inverse of an LTI system H(z) exists if and only if all the zeros of H(z) are inside the unit circle The system with all its poles and zeros inside the unit circle is called as minimum-phase system The magnitude response is uniquely determined by the phase response and vice-Vera For a minimum-phase system the magnitude response is uniquely by the phase response and vice-versa
Figure 1.53: Location of poles in a minimum-phase system Ex1: stable and causal system Find the transfer function of an inverse LTI system described by a dif-ference equation y[n] − y[n − 1] + 1 y[n − 2] = x[n] + 1 x[n − 1] − − 2] 4 4 8 Is the system stable and causal? Solution: find the z-transform from the difference equation and then find the impulse response z-transform is M bkz−k H(z) = k=0 Nk=0 akz−k 1 + 1 z−1 − 1 z−2 H(z) = 4 8 1 − z−1 + 14 z−2 • Find the poles and zeros of H(z) and write in the product form
• We can H(z) as (1 − 1 z−1)(1 + 1 z−1) H(z) = 4 2 (1 − 12 z−1)2 • We can write Hinv(z) as Hinv(z) = 1 = (1 − 21 z−1)2 H(z) (1 − 41 z−1)(1 + 21 z−1) The of inverse system are at z = 14 and z = −12 Two poles are inside the unit circle, hence the system can be both stable and causal The zero is also inside the unit circle and the system is system Ex2: inverse system Find the transfer function and difference equation description of the inverse system of discrete LTI system, which describes multipath com-munication channel (two path communication channel) y[n] = x[n] + ax[n − 1] The system is z-transform is M bkz−k H(z) = k=0 Nk=0 akz−k phase
H(z) = 1 + az−1 • The inverse system is Hinv(z) = 1 = 1 H(z) + az−1 1 • The corresponding difference equation description is y[n] + ay[n − 1] = x[n] • The inverse system is both stable and causal when |a| < 1 Ex3: inverse system Find the transfer function of the inverse system and is the inverse sys-tem stable and causal? h[n] = 2[n] + 5 1 n 7 1 n ( − ( − ) u[n] 2 2 2 4
H(z) = 2 + 5 1 − 7 1 2 (1 − 21 z−1) 2 (1 + 41 z−1) • Write H(z) in product form 2(1 − 1 z−1)(1 + 1 z−1)+ 5 (1 + 1 z−1)+ 7 (1 − 1 z−1) H(z)= 2 4 2 4 2 2 (1 − 12 z−1)(1 + 14 z−1) • After simplification we get (1 − 1 z−1)(1 + 2z−1) H(z) = 8 (1 − 12 z−1)(1 + 14 z−1) • We know Hinv(z) = H1(z) , so we can write Hinv(z) = (1 − 12 z−1)(1 + 14 z−1) (1 − 18 z−1)(1 + 2z−1) Poles are at z = 18 and z = −2, so the system can not be both stable and causal 1.8.6 Unsolved examples from [2] Unsolved ex:7.34(a) • Is the system (i)causal and stable (ii)minimum phase H(z) = 2z + 3 z2 + z − 165 H(z)
(ii) All poles and zeros are not inside |z| = 1, the system is not phase. Unsolved ex:7.35(a) Find the transfer function of an inverse system and determine whether it can be both causal and stable H(z) = 1 − 8z−1 + 16z−2 1 − 12 z−1 + 14 z−2 H(z) = (z − 4)2