The Inverse z-Transform
4.3 The Inverse z-Transform
In this section we consider how to find x[n] for a given z-transform X [z] with its ROC. From the complex variable theory, the inverse z-transform formula can be derived (see [O-2], Sect. 2.2) as
x [n] =
X[z]z −n dz
C means the complex integral along a closed contour C within the ROC of X[z] encircling the origin of the z-plane in the counterclockwise direction. It is,
however, difficult to directly evaluate this integral and therefore we make resort to alternative procedures for obtaining the inverse z-transform.
4.3 The Inverse z-Transform 219
4.3.1 Inverse z-Transform by Partial Fraction Expansion
Noting that the complex variable z appears in the numerator of almost every basic z -transform listed in Table B.9, we apply the same procedure as with the inverse Laplace transform (Sect. A.4) to get the partial fraction expansion on X [z]/z and then multiply both sides by z so that we can directly use the z-transform table to get the inverse z-transform.
More specifically, let X [z]/z be rational as
X [z]
Q 1 [z]
b M z M +...+b 1 z+b 0 (4.3.2)
z = P [z] = a N z N +...+a 1 z+a 0
where M and N are the degrees of the numerator and denominator polynomials, respectively. If M ≥ N , we divide Q 1 [z] by P[z] starting with the highest powers
of z to produce the remainder polynomial of degree less than N :
Q [z] z = P [z] +c M−N z M−N +...+c
X [z]
1 z+c 0 (4.3.3)
If M < N , we have only the first term on the RHS where c i = 0 for all i. Now, for the purpose of illustration, we assume that all the poles of Q[z]/P[z] are simple except one multiple pole of order L at z = p so that we can write Q[z]/P[z] in the following form:
Q [z] N −L r
r N P [z] =
r N −L+1
L +K (4.3.4)
(z − p) where
i = 1, 2, . . . , N − L (4.3.5a)
P [z] z=p i
1 d l L Q r [z] N −l =
(4.3.5b) l ! dz l (z − p) P [z] l = 0, 1, . . . , L − 1
z=p
Now, substituting Eq. (4.3.4) into Eq. (4.3.3), multiplying the both sides by z, and using the z-transform Table B.9, we can obtain the inverse z-transform of X [z] as
N −L
x [n] = n r
i p n i +r N −L+1 p +r N −L+2 np n−1 +...
c i δ [n + i + 1] (4.3.6) (L − 1)!(n − L + 1)!
p n−L+1 u s [n] +
i =0
220 4 The z-Transform Example 4.7 The Inverse z-Transform by Partial Fraction Expansion
(a) Let us find the inverse z-transform of
X [z] =
2 (E4.7.1) − 2z + (5/4)z − 1/4
(z − 1)(z − 1/2) with ROC R = {z : |z| > 1}
We first divide this by z and then expand it into partial fractions:
X [z]
z−1 2 z − 1/2 (z − 1/2)
where the coefficient of each term can be found from Eq. (4.3.5) as
(E4.7.3a)
z=1
d 1 r 2 = dz (z − 1/2)
(4.3.5b) d 2 X [z]
l=1
z=1/2
dz z−1 z=1/2
(E4.7.3b) (z − 1) z=1/2
(z − 1/2) z
= −2 (E4.7.3c)
z−1 z=1/2 Now, moving the z (which we have saved) from the LHS back into the RHS
Then we can use Table B.9(3), (5), and (10) to write the inverse z-transform as
x [n] = 4−4
2 − 4n
u s [n] (E4.7.5)
where the right-sided sequences are chosen over the left-sided ones since the given ROC is not the inside, but the outside of a circle. We can use the MATLAB command ‘residue( )’ or ‘residuez( )’ to get the par- tial fraction expansion and ‘iztrans( )’ to obtain the whole inverse z-transform.
4.3 The Inverse z-Transform 221 It should, however, be noted that ‘iztrans()’ might not work properly for high-
degree rational functions.
>>Nz=[0 1]; Dz=poly([1 1/2 1/2]), [r,p,k]=residue(Nz,Dz); [r p],k %(E4.7.2)
r= 4.0000 p=
4/(z-1) -4.0000
% (E4.7.3a)
% (E4.7.3b) -4/(z-0.5) -2.0000
% (E4.7.3c) -2/(z-0.5)ˆ2 k = [] >>syms z, x=iztrans(z/(zˆ3-2*zˆ2+1.25*z-0.25)) % (E4.7.1) x = 4-4*(1/2)ˆn-4*(1/2)ˆn*n
% (E4.7.5)
(b) Let us find the inverse z-transform of
(z − 1/2)(z + 1/4) with one of the following three ROCs:
R 3 = z : < |z| < (E4.7.7)
We first divide this by z and then expand it into partial fractions:
X [z]
−4 (E4.7.8) z
; X [z] = 4
z + 1/4 Now, depending on the ROC, we use Table B.9(5) or (6) to write the inverse
z -transform as follows:
4 (E4.7.9a)
(ii) R 2 = z : |z| <
x [n] = −4
u s (E4.7.9b)
2 [−n − 1] + 4 − 4 [−n − 1]
(iii) R 3 = z : <
u s [n] (E4.7.9c)
>>syms z, x=iztrans(3*z/(zˆ2-0.25*z-1/8)) % (E4.7.6) x = 4*(1/2)ˆn-4*(-1/4)ˆn % (E4.7.9a) just a right-sided sequence
222 4 The z-Transform Example 4.8 The Inverse z-Transform by Partial Fraction Expansion
Let us find the inverse z-transform of
2z 2 2z 2
X [z] =
(E4.8.1) − z + 1/2
(z − 0.5 − j0.5)(z − 0.5 + j0.5) with ROC R = {z : |z| > 1}
We first divide this by z and then expand it into partial fractions:
X [z] r R + jr I r R − jr I z =
z − 0.5 − j0.5
z − 0.5 + j0.5
z − 0.5 + j0.5 Now, moving the z (which we have saved) from the LHS back into the RHS yields
z − 0.5 − j0.5
z − 0.5 + j0.5 We can use Table B.9(5) to write the inverse z-transform as
z − 0.5 − j0.5
[n] = (1 − j)(0.5 + j0.5) n u s [n] + (1 + j)(0.5 − j0.5) u s [n]
= . 2 e 2 e +e 2 e − jnπ/4 u s [n] √ −n+3
√ - − jπ/4 √ −n j nπ/ 4 j π/ 4 √ −n
π = 2 cos((n − 1) )u s [n]
(E4.8.4)
As a nice alternative for X [z] having complex conjugate poles like (E4.8.1), we can decompose it into the following form, which can be matched exactly with some element of the z-transform table:
X [z] = 2z(z − 1/2) 2 2 +
2(1/2)z
2 (z − 1/2) 2 + (1/2) (z − 1/2) + (1/2)
2 (E4.8.5) (z − r cos Ω 1 ) +r sin Ω 1 (z − r cos Ω 1 ) 2 +r 2 sin Ω 1
2, and Ω 1 = π/4. Then we can use B.9(18) and (17) to obtain the same result as (E4.8.3):
where r = 1/
2, cos Ω 1 = 1/
2, sin Ω 1 = 1/
x [n] = 2 n cos(Ω
)+r n sin(Ω
2 cos( n )+ 2 sin( n ) u s [n]
√ −n+3
√ −n+3
π = 2 cos((n − 1) )u s 4 [n] = 2 sin((n + 1) )u s [n]
(E4.8.6)
4.3 The Inverse z-Transform 223 (cf) It seems that the MATLAB command ‘iztrans()’ does not work properly for
this problem:
>>syms z, x=iztrans(2*zˆ2/(zˆ2-z+1/2)) % (E4.8.1) >>n=1:10; xn=2.ˆ(-(n-3)/2).*cos((n-1)*pi/4); stem(n,xn), hold on %(E4.8.6) >>[r,p,k]=residuez([2 0 0],[1 -1 1/2]) % Partial fraction expansion >>xn1=real(r.’*[p(1).ˆn; p(2).ˆn]); stem(n,xn1,’r’) % (E4.8.3) Alternative
4.3.2 Inverse z-Transform by Long Division
Noting that the inverse z-transform can rarely be found in an elegant form like Eq. (E4.8.6), we may think of it as an alternative to expand X[z] = Q 1 [z]/P[z]
into a polynomial in powers of z −1 and equate each coefficient of z −n to x[n]. More specifically, starting with the highest/lowest powers of z depending on the shape of
the ROC (for a right/left-sided sequence), we divide Q 1 [z] by P[z] to expand X[z] into the power series form of the z-transform definition (4.1.1). For example, let us consider X[z] given by Eq. (E4.7.6) in Example 4.7.
3z
3z
X[z] = z 2 =
(z − 1/2)(z + 1/4) (Case 1) If the ROC is given as {z : |z| > 1/2}, we perform the long division as 3z −1 + (3/4)z . −2 + (9/16)z −3 +···
− (1/4)z − 1/8
z 2 − (1/4)z − 1/8 3z 3z − 3/4
− (3/16)z −2 − (3/32)z (9/16)z −1 − (3/32)z −2 ...............................
Then each coefficient of the quotient polynomial in z −n is equated with x [n], yielding
n=0
x [n] = [ 0 3 3/4 9/16 · · · ] : the same as Eq. (E4.7.9a) (4.3.8a)
224 4 The z-Transform (Case 2) If the ROC is given as {z : |z| < 1/4}, we perform the long division as
. −24z + 48z − 288z 2 +··· − 1/8 − (1/4)z + z 3z
.................................. Then each coefficient of the quotient polynomial in z n is equated with
x [−n], yielding
x [n] = [ −288 48 −24
0 · · · ] : the same as Eq. (E4.7.9b) (4.3.8b)
(Case 3) If the ROC is given as {z : r − = 1/4 < |z| < r + = 1/2}, X[z] should
be separated into two parts, one having the poles on or inside the circle of radius r − and the other having the poles on or outside the circle of radius r + . Then, after performing the long division as in case 1/2 for the former/latter, we add the two quotients and equate each coefficient of the resulting polynomial in z ±n with x[∓n].