Inverse Z-Transform PDF

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Module 21 InverseZ-Transform Objective:To describe how to obtain inverse z-transform making use of the knowledge of properties of z-Transform and properties of ROC. Introduction: Inverse z-transform maps a function in z-domain back to the time domain. One application is to convert a discrete system response to an input sequence from z-domain back to the time domain. Since discrete system analysis is usually easier in z-domain, the process is to convert the discrete system time domain representation to z-domain (both system and inputs),perform system analysis in z-domain and then convert back to the time domain representation for the response. The reason to do this process in this convoluted way is that due to its properties, the z-transform converts the Linear Constant Coefficient Difference Equations(LCCDE) that describe system behaviour to a polynomial. Also the convolution operation which describes the system action on the input signals is converted to a multiplication operation. These two properties make it much easier to do systems analysis in the z-domain. Inverse z-transform is performed using Long Division Method(Power Series Expansion method), Partial Fraction Expansion and Residue method (Contour Integral Method). Description: Concept of Inverse z-Transform We are aware that the z- transform of a discrete signal x(n) is given by 𝑋𝑧 = ∞ Since z=rejω𝑋𝑟𝑒 𝑗𝜔 =

𝑥 𝑛 (𝑟𝑒 𝑛 =−∞

𝑥 𝑛𝑟 ⟹ 𝑥𝑛 = 𝑟

−𝑛

𝑛

=ℱ

−1

𝑗𝜔 −𝑛

∞

𝑥 𝑛 𝑧 −𝑛

𝑛 =−∞

𝑋𝑟𝑒

𝑗𝜔

𝑥 𝑛𝑟 𝑛 =−∞

∞ =

)

=

𝜋

1 2𝜋

𝜋

1 𝑋𝑟𝑒 𝑗𝜔 𝑒 jω𝑛 𝑑ω 2𝜋 −𝜋

𝑒

−𝑛 −𝑗𝜔𝑛

= ℱ 𝑥 𝑛 𝑟 −𝑛

𝑋𝑟𝑒 𝑗𝜔 𝑒 jω𝑛 𝑑ω

−𝜋

𝜋

1 𝑛 𝑋𝑟𝑒 𝑗𝜔 𝑟𝑒 𝑗𝜔 𝑑ω = 2𝜋 −𝜋

We can recover x(n) from its z-transform evaluated along a contour z=rejω in the ROC, with rfixed and ω varying over a2π interval. Let us now change the variable of integration from ω to z. With z=rejωand r fixed, dz= jrejωdω=jzdω, or dω=(1/j)z-1dz. The integration above is over a 2π interval in ω, which, in terms of z, corresponds to one traversal around the circle |z|=r. Consequently, in terms of an integration in the z-plane the above equation can be written as 𝑥𝑛 =

1 1 𝑋 𝑧𝑧𝑛 𝑧 −1 𝑑𝑧 = 2𝜋𝑗

2𝜋𝑗

𝑋𝑧𝑧 𝑛 −1 𝑑𝑧

where the symbol ↺ denotes integration around a counter clockwise closed circular contour centered at the origin and with radius r. The value of r can be chosen as any value for which X(z) converges – i.e., any value such that the circular contour of integration |z|=r is in the ROC. Long Division Method (Power Series Expansion) Z-transform of the sequence x(n) is given as, 𝑋𝑧 =

∞

𝑥 𝑛 𝑧 −𝑛

= ⋯ + 𝑥 −2𝑧 2 + 𝑥−1𝑧 + 𝑥0 + 𝑥1𝑧 −1 + 𝑥 2𝑧 −2 + ⋯

𝑛 =−∞

From above expansion of z-transform, the sequence x(n) can be obtained as, 𝑥 𝑛 = … , 𝑥 −2, 𝑥 −1, 𝑥 0, 𝑥 1, 𝑥(2, … } The Power series expansion can be obtained directly or by long division method. Illustration Example: Determine inverse z-transform of the following: i) ii) Solution:

𝑋𝑧 =

𝑋𝑧 =

1 1−𝑎𝑧 −1 1 1−𝑎𝑧 −1

, 𝑅𝑂𝐶∶ 𝑧 > |𝑎|

, 𝑅𝑂𝐶∶ 𝑧 < |𝑎|

Partial Fraction Expansion As we know that the rational form of X(z) can be expanded into partial fractions, Inverse z-transform can be taken according to location of poles and ROC of X(z). Following steps are to be performed for partial fraction expansions: Step 1: Arrange the given X(z) as, 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 𝑋(𝑧) = 𝑧 𝑧 − 𝑝1 𝑧 − 𝑝2 … (𝑧 − 𝑝𝑁 ) Step 2:

𝐴1 𝑋(𝑧) = + 𝑧 𝑧 − 𝑝1

𝐴2 + 𝑧 − 𝑝2

𝐴3 + ⋯+ 𝑧 − 𝑝3

𝐴𝑁 𝑧 − 𝑝𝑁

Where Ak for k=1, 2,…N are the constants to be found in partial fractions. Poles may be of multiple order. The coefficients will be calculated accordingly.

Step 3: Above equation can be written as 𝐴1 𝑧

𝐴𝑁 𝑧 𝐴3 𝑧 𝐴2 𝑧 + ⋯ + 𝑧 − 𝑝𝑁 + 𝑧 − 𝑝3 𝑧 − 𝑝2 𝑧 − 𝑝1 𝐴1 𝐴2 𝐴𝑁 𝐴3 = + + ⋯+ + −1 −1 −1 1 − 𝑝3 𝑧 1 − 𝑝1 𝑧 1 − 𝑝2 𝑧 1 − 𝑝𝑁 𝑧 −1 𝑋𝑧 =

+

𝐴

Step 4:All the terms in above step are of the form 1−𝑝 𝑘𝑧 −1 . Depending upon ROC, following 𝑘

standard pairs must be used. 𝑝𝑘 𝑛 𝑢(𝑛)

𝑍

1

1−𝑝 𝑘 𝑧 −1

−𝑝𝑘 𝑛 𝑢(−𝑛 − 1)

𝑍

with ROC: |z| >|𝑝𝑘 |,i.e., causal response 1 1−𝑝 𝑘 𝑧 −1

with ROC: |z| 1

ii)|z| < 0.5

1 1−1.5 𝑧 −1 +0.5 𝑧 −2

iii)0.5 < |z| < 1

Solution: Given 𝑋 𝑧 =

1 1−1.5 𝑧 −1 +0.5𝑧 −2

Which can be written as 𝑋 𝑧 =

𝐴

1−𝑧 −1

𝐵

+ 1−0.5𝑧 −1

After finding the constants as A=2 and B=-1 𝑋𝑧 = i)

ii)

1 2 − −1 1−𝑧 1 − 0.5𝑧 −1

For ROC |z| > 1 i.e., causal or right sided 𝑥 𝑛 = 2𝑢 𝑛 − 0.5𝑛 𝑢(𝑛)

For ROC |z| < 0.5 i.e., non-causal or left sided 𝑥 𝑛 = −2𝑢 −𝑛 − 1 + 0.5𝑛 𝑢(−𝑛 − 1)

iii)

For ROC 0.5 < |z| < 1 i.e., two sided 𝑥 𝑛 = −2𝑢 −𝑛 − 1 − 0.5𝑛 𝑢(𝑛) The ROC is a circular strip between z>0.5 and z |a| using contour integration Solution:

𝑧2

𝑧−𝑎 2

,

Step 1: 𝑋𝑜 𝑧 = 𝑋 𝑧𝑧 𝑛 −1 =

𝑧2

𝑧 𝑛 −1 = 𝑧−𝑎 2

𝑧 𝑛 +1 𝑧−𝑎 2

Step 2:Here the pole is at z=a and it has order m=2. Hence finding the residues at z=a

𝑅𝑒𝑠 𝑋 𝑧= 𝑧=𝑎 𝑜

1 𝑑 2−1 𝑧 − 𝑎2 2 − 1! 𝑑𝑧2−1 =

𝑑 𝑛 +1 𝑧 𝑎𝑡 𝑧 = 𝑎 𝑑𝑧

𝑧 𝑛 +1 𝑎𝑡 𝑧 = 𝑎 𝑧 − 𝑎2

= 𝑛 + 1𝑧 𝑛 𝑎𝑡 𝑧 = 𝑎 = 𝑛 + 1𝑎 𝑛

Step 3: The sequence x(n) is given as 𝑥𝑛 =

𝑅𝑒𝑠 𝑋 (𝑧) 𝑧 = 𝑝𝑖 𝑜 𝑖=1

𝑥(𝑛) = 𝑛 + 1𝑎𝑛 𝑢(𝑛) Since ROC: 𝑧 > |𝑎| Examples: Solved Problems: Problem 1:Find the inverse z-transform of 𝑋 𝑧 = log

1 , 1−𝑎𝑧 −1

|z|>|a|

Solution: 𝑋 𝑧 = log

1 = − log1 − 𝑎𝑧 −1 1 − 𝑎𝑧−1

log(1-p) is expanded with the help of power series. It is given as log1 − 𝑝 = − Therefore, 𝑋 𝑧 = − log1 − 𝑎𝑧 −1 = − − Hence 𝑥𝑛 = i.e., 𝑥𝑛 =

𝑎

𝑛

𝑎𝑛

𝑛

0,

𝑛

,

𝑛≥1

𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

𝑢(𝑛 − 1)

𝑛

𝑎𝑧−1∞ 𝑛=1

𝑛

𝑝𝑛

∞

𝑛 =1

=

𝑛

,𝑝 < 1

𝑛 ∞𝑎 𝑛=1 𝑛

𝑧−𝑛 for |z| > |a|

Problem 2:Find the inverse z-transform of 𝑋 𝑧 =

when ROC is (i) (ii) (iii)

𝑧 using 𝑧+2𝑧−3

partial fraction method

|z| 3 -2 < |z| < 3

Solution: Given𝑋 𝑧 =

𝑧 𝑧+2𝑧−3

Applying partial fractions

i)

ii)

𝑋𝑧 𝑧

=

1 1

⇒ 𝑋𝑧 =

1

𝑥𝑛 =

1

For ROC |z| < -2

−

1

1

5 𝑧+2

𝑧 𝑧 1 − 5 𝑧−3 5 𝑧+2

5

−3𝑛 + −2𝑛 𝑢(−𝑛 − 1)

For ROC |z| > 3 𝑥𝑛 =

iii)

𝑧−3

5

1 5

3𝑛 − −2𝑛 𝑢(𝑛)

For ROC -2 < |z| 1

1 + 2𝑧 −1 + 𝑧 −2

1 − 2 𝑧 −1 + 2 𝑧 −2 3

1

Solution: To arrange X(z) in proper form suitable for partial fraction expansion The highest power of denominator polynomial should be atleast one less than that of numerator polynomial Let us arrange X(z) as follows: 𝑋𝑧 =

𝑧 −2 + 2𝑧 −1 + 1

1 − 2 𝑧 −1 + 2 𝑧 −2 3

1

Now perform one step division so that order of numerator polynomial is reduced by one unit.

Now

−1 + 5𝑧 −1

𝑋𝑧 = 2 + Writing in partial fractions 𝑋𝑧 = 2 +

1 − 2 𝑧 −1 + 2 𝑧 −2 3

1

𝐴 𝐵 + 1 − 𝑧−1 1 − 1 𝑧 −1 2

Finding the constants A =8 and B=-9 𝑋𝑧 = 2 +

9 8 − 1 −1 1−𝑧 1 − 𝑧 −1 2

As ROC is |z| > 1,representing the signal as causal signal which is right sided 𝑥 𝑛 = 2𝛿 𝑛 + 8𝑢 𝑛 − 9

𝑛

1 2

𝑢(𝑛)

Problem 4: An LTI system is characterized by the system function 𝐻𝑧 =

3 − 4𝑧 −1 1 − 3.5𝑧 −1 + 1.5𝑧 −2

Specify the ROC of H(z) and determine h(n) for the following conditions a) The system is causal and unstable b) The system is non-causal and stable c) The system is non-causal and unstable Solution: Given that 𝐻𝑧 =

3 − 4𝑧 −1 3𝑧 − 4 = 1 − 3.5𝑧 −1 + 1.5𝑧 −2 𝑧 − 0.5𝑧 − 3

Using partial fraction expansion, we obtain

𝐻𝑧

𝑧

=

3𝑧 − 4 = 𝑧 − 0.5𝑧 − 3

𝐴 𝐵 +𝑧−3 𝑧 − 0.5

Finding the constants A =1 and B=2 𝐻𝑧 𝑧

1 2 + 𝑧 − 0.5 𝑧 − 3 𝑧 𝑧 +2 𝐻(𝑧) = 𝑧 − 0.5 𝑧−3 =

The system has poles at z=0.5 and z=3

a) For the system to be causal and unstable, the ROC of H(z) is the region in the z-plane outside the outermost pole and it must not include the unit circle. Therefore, the ROC is the region |z| > 3

Hence ℎ 𝑛 = 0.5𝑛 𝑢 𝑛 + 2(3)𝑛 𝑢(𝑛) b) For the system to be non-causal and stable, the ROC of H(z) is the ring in the z-plane and it must include the unit circle. Therefore, the ROC is the region, 0.5 < |z| < 3

Hence ℎ 𝑛 = 0.5𝑛 𝑢 𝑛 − 23𝑛 𝑢(−𝑛 − 1) c) For the system to be non-causal and unstable, the ROC of H(z) is the ring in the zplane inside the inner most pole and it must not include the unit circle. Therefore, the ROC is the region, |z| < 0.5

Hence ℎ 𝑛 = −0.5𝑛 𝑢 −𝑛 − 1 − 23𝑛 𝑢(−𝑛 − 1) Problem 5:Check whether the corresponding LTI system with system function −1 − 0.4𝑧 −1 𝐻𝑧 = 1 − 2.8𝑧 −1 + 1.6𝑧 −2

is stable and causal, If the ROC is i) ii) iii)

|z| > 2 |z| < 0.8 0.8 < |z| 2

Also, find the impulse response for all the three ROCs mentioned above. Solution: Consider the given system function and applying partial fraction expansion 1 2 −1 − 0.4𝑧 −1 = − 𝐻𝑧 = −1 −2 −1 1 − 2.8𝑧 + 1.6𝑧 1 − 0.8𝑧 1 − 2𝑧 −1 The system has poles at z=0.8 and z=2 i)

For the ROC |z| > 2, the system is causal as the ROC is outside the outer most pole and not stable as the ROC does not include unit circle. The impulse response is ℎ𝑛 = 0.8𝑛 𝑢 𝑛 − 22𝑛 𝑢(𝑛)

ii)

For the ROC |z| 1/3 |z| < 1/3, using power series expansion

Solution: Given that 𝑋𝑧 = i)

1 + 𝑧 −1

1−

1

3

𝑧 −1

Since the ROC is |z| > 1/3, we express X(z) as a power series in z -1, so that we obtain a right sided signal. We divide the numerator by the denominator to obtain

We can write, therefore, 𝑋𝑧 = 1 + Hence 𝑥 𝑛 = {1, ii)

4

4

, ,

4

3 9 27

4

4 4 −3 𝑧 −1 + 𝑧 −2 + 𝑧 +⋯ 3 9 27

, … } with x(n) starting from n=0 with value 1

Since the ROC is |z| < 1/3, we express X(z) as a power series in z-1, so that we obtain a left sided signal. We divide the numerator by the denominator to obtain

We can write, therefore, 𝑋 𝑧 = ⋯ − 36𝑧 2 − 12𝑧 − 3

Hence 𝑥 𝑛 = {… ,36, −12, −3} with x(n) ending at n=0 with value 3

Problem 7:Evaluate the inverse z-transform of 𝑥 𝑧 = integral method.

1

1−1.5 𝑧 −1 +0.5𝑧 −2

, |z| > 1 using contour

Solution: Given that𝑥 𝑧 =

1 1−1.5 𝑧 −1 +0.5𝑧 −2

=

𝑧2 (𝑧−0.5)(𝑧−1)

Since ROC is |z| > 1, is the region in the z-plane outside the outermost pole located at z=1, both poles correspond to causal i.e., right sided signals. For n ≥ 0 finding the residues at z=0.5 and z=1

𝑧2

𝑥𝑛 =

𝑧 𝑛−1 𝑑𝑧 =

𝑑𝑧 𝑧 𝑛 +1 (𝑧 − 0.5)(𝑧 − 1)

(𝑧 − 0.5)(𝑧 − 1) 𝑧 𝑛 +1 = 𝑟𝑒𝑠𝑖𝑑𝑢𝑒 𝑜𝑓 𝑎𝑡 𝑧 = 0.5 + 𝑟𝑒𝑠𝑖𝑑𝑢𝑒 𝑜𝑓 (𝑧 − 0.5)(𝑧 − 1) 𝑧 𝑛 +1 + lim𝑧 − 1 (𝑧 − 0.5)(𝑧 − 1) 𝑧→1

= lim 𝑧 − 0.5 𝑧→0.5

𝑧 𝑛 +1 𝑧 𝑛 +1 + lim 𝑧→1 (𝑧 − 0.5) 𝑧→0.5 (𝑧 − 1)

𝑧 𝑛 +1 𝑎𝑡 𝑧 = 1 (𝑧 − 0.5)(𝑧 − 1)

𝑧 𝑛 +1 (𝑧 − 0.5)(𝑧 − 1)

= lim =− =−

0.5𝑛 +1 1𝑛 +1 + 0.5 0.5 𝑛

1 2

+ 2, n > 0

Therefore, 𝑥(𝑛) = 2 −

𝑛

1 2

𝑢(𝑛)

Problem 8:A second order discrete time system is characterized by the difference equation 𝑦𝑛 − 3𝑦𝑛 − 1 + 2𝑦𝑛 − 2 = 𝑥 𝑛 − 2𝑥 𝑛 − 1 Find y(n) for n≥0 when x(n)=u(n) and the initial conditions are given as y(-1)=y(-2)=1 Solution: Given input x(n)=u(n) 𝑋(𝑧) =

1 1 − 𝑧−1

Now consider the given difference equation applying the given input 𝑦𝑛 − 3𝑦𝑛 − 1 + 2𝑦𝑛 − 2 = 𝑢 𝑛 − 2𝑢 𝑛 − 1 Taking unilateral z-transform of the above equation, applying partial fractions, substituting initial conditions and simplifying, we obtain 𝑌𝑧 =

1 + 1 − 𝑧−1 2

1 1−𝑧 −1

Taking inverse z-transform of the above equation for n ≥ 0 yields 𝑦𝑛 = 𝑛 + 2𝑢(𝑛) Problem 9:Determine the inverse z-transform of the function Xz =

z ,𝑧 z−0.5

> 0.5, using long division method

Solution: Given that Xz =

z z−0.5

=

1

1−0.5𝑧 −1

with ROC : 𝑧 > 0.5

Since the ROC is |z| > 0.5, we express X(z) as a power series in z -1 so that we obtain a right sided signal. We divide the numerator by the denominator to obtain

We can write, therefore, 𝑋 𝑧 = 1 + 0.5𝑧 −1 + 0.52 𝑧 −2 + 0.53 𝑧 −3 + ⋯

Hence it is easy to say that 𝑥(𝑛) = 0.5𝑛 𝑢(𝑛)

Problem 10:Given that y(-1)=5 and y(-2)=0, Solve the difference equation, 𝑦𝑛 − 3𝑦𝑛 − 1 − 4𝑦𝑛 − 2 = 0, 𝑛 ≥ 0 Solution: Taking Unilateral z-transform of the given difference equation 𝑌𝑧 − 3𝑧 −1 𝑌 𝑧 + 𝑦−1 − 4𝑧 −2 𝑌 𝑧 + 𝑦−1𝑧 −1 + 𝑦−2 = 0 Substituting the given initial conditions and simplifying 𝑌𝑧 = −

16 1 + −1 1 − 4𝑧 −1 1+𝑧

Therefore, for n≥0𝑦(𝑛) = [−−1𝑛 + 164𝑛 ]𝑢(𝑛) Assignment:

Problem 1: A difference equation of the system is given as 𝑦 𝑛 − 𝑦𝑛 − 1 +

1 4

𝑦𝑛 − 2 = 𝑥 𝑛 +

1 4

𝑥𝑛 − 1 −

1 8

𝑥(𝑛 − 2)

Determine the transfer function and the impulse response of the inverse system. Check whether the inverse system is causal and stable Problem 2: The system function of causal LTI system is given as, 𝐻𝑧 = i) ii)

1 + 2𝑧 −1 + 𝑧 −2 1+

1

2

𝑧 −1 1 − 𝑧−1

Determine unit sample response Determine output if input is 𝑥 𝑛 = 𝑒 𝑗𝑛𝜋 /2

Problem 3:Find the inverse z-transform of 𝑋 𝑧 = log

expansion method.

Problem 4:z-transform of a signal x(n) is 𝑋 𝑧 =

1

1−𝑎 −1 𝑧

, |z| < |a| using power series

1+𝑧 −1

1+3 𝑧 −1 1

Use long division method to determine the values of i) ii)

x(0), x(1) and x(2), assuming the ROC to be |z| > 1/3 x(0), x(-1) and x(-2), assuming the Roc to be |z| < 1/3

Problem 5:Given𝑋 𝑧 =

𝑧

(𝑧−1)3

, find x(n) using contour integration method

Problem 6: Determine the inverse z transform of 𝑋 𝑧 = 𝑒

1

𝑧

with ROC all z except |z| = 0

Problem 7:When the input to an LTI system is 𝑥 𝑛 = corresponding output is 𝑦𝑛 = 5 a) b) c) d)

𝑛

1 3

𝑢𝑛 − 5

𝑛

2

3

𝑛

1

𝑢(𝑛)

3

𝑢 𝑛 + 2𝑛 𝑢(𝑛)and the

Find the system function H(z). Plot the poles and zeros of H(z) and indicate the ROC Find the impulse response h(n) of the system Write a difference equation that is satisfied by the given input and output Is the system stable? Is it causal?

Problem 8: a)Prove that 𝑦𝑛 − 𝑘

𝑈𝑍𝑇

𝑦−𝑘 + 𝑦−𝑘 + 1𝑧 −1 + ⋯ + 𝑦−1𝑧 −𝑘+1 + 𝑧 −𝑘 𝑌(𝑧)

b) Consider the system 𝐻 𝑧 =

𝑧 −2 2 3 −1 2 −2 1− 𝑧 +25 𝑧 5

𝑧 −1 +

1

Determine i) ii)

The impulse response The step response if y(-1)=1 and y(-2)=2

Problem 9:Determine the inverse z-transform of the following 𝑋𝑧 =

𝑧(𝑧 2 − 4𝑧 + 5) (𝑧 − 3)(𝑧 − 2)(𝑧 − 1)

If ROC is given as a) 2 < |z| 3 c) |z| < 1 Problem 10:Using Long division method, determine the inverse z-transform of 𝑋𝑧 =

1+2𝑧 −1 1−2𝑧 −1 +𝑧 −2

if, a) x(n) is causal

b) x(n) is anti-causal

Simulation: In SCILAB,the command one uses now is ldiv, for Long Division Method.The program written is to find the impulse response of Discrete Time system from its Transfer function.

// Discrete time System clc; clear all; close; //Given 2nd order system //y(n)=(3/8)y(n-1)+(2/3)y(n-2)+x(n)+(1/4)x(n-1) z=%z; // first create a variable a=((z^2)+(1/4)*z); b=(z^(2)-(3/8)*z-(2/3)) h = ldiv (a ,b ,10) ; disp (h ,"h(n)=") ; plot2d3(h); xtitle('Impulse response','n-->','response'); OUTPUT: h(n)= 1. 0.625 0.9010417 0.7545573 0.8836534 0.8344082 0.9020054 0.8945242 0.9367835 0.9476432

References: [1] Alan V.Oppenheim, Alan S.Willsky and S.Hamind Nawab, “Signals & Systems”, Second edition, Pearson Education, 8th Indian Reprint, 2005. [2] M.J.Roberts, “Signals and Systems, Analysis using Transform methods and MATLAB”, Second edition,McGraw-Hill Education,2011 [3] John R Buck, Michael M Daniel and Andrew C.Singer, “Computer explorations in Signals and Systems using MATLAB”,Prentice Hall Signal Processing Series [4] P Ramakrishna rao, “Signals and Systems”, Tata McGraw-Hill, 2008 [5] Tarun Kumar Rawat, “Signals and Systems”, Oxford University Press,2011...