digital controls & digital filters lectures 3 & 4 3-4.pdflectures 3 & 4 m.r. azimi, professor...

Inverse z-Transform Representation of LTI systems

Digital Controls & Digital FiltersLectures 3 & 4

M.R. Azimi, Professor

Department of Electrical and Computer EngineeringColorado State University

Spring 2017

M.R. Azimi Digital Control & Digital Filters


Inverse z-Transform Methods

Two methods are covered here.

1-Partial Fraction Expansion (PFE)This method is parallel to partial fraction expansion used in the Laplace

transform, with one minor modification. In this case, we expand X(z)z instead ofX(z), (since unlike L−1{ 1s+a} = e

−at for z-transform Z−1{ 1z+a} does not existin the tables)

Example:

Given X(z) = z2+z

(z−1)2 determine x(n) using PFE.X(z)z =

z+1(z−1)2 =

A1(z−1) +

A2(z−1)2

A1 =ddz{(z − 1)

2X(z)z }|z=1 = 1

A2 = (z − 1)2X(z)z |z=1 = 2Hence:X(z) = zz−1 +

2z(z−1)2

x(n) = Z−1{ zz−1}+ Z−1{ 2z(z−1)2 } = (1 + 2n)us(n)



Inverse z-Transform Methods-Cont.

2-Inversion Formula Method

If {x(n)} has z-transform X(z) =∞∑n=0

x(n)z−n

then x(n) can be recovered from X(z) by the inverse integral formula:

x(n) = 12πj�CX(z)zn−1dz

C: Any closed contour in ROC of X(z).

�C

: Line or contour integral along C in counterclockwise direction.

The inversion integral may easily be evaluated using Cauchy’s residue theorem.



Inversion Formula Method

Cauchy’s Residue Method

Let F (z) be an analytic function in shaded region R (excludes singularities ofF (z)) enclosed by contour C. According to Cauchy theorem the integral ofF (z) over R is zero or, we have

‰c

F (z)dz = 2πj

n∑i=1

‰ci

F (z)dz = 2πj

n∑i=1

ri

where ri =�ciF (z)dz is the ith Residue

associated with z = zi pole of F (z).



Inversion Formula Method-Cont.

Residue Theorem for Inverse z-Transform (IZT)

Given X(z), |z| > R, the corresponding IZT can be found by evaluating theintegral:

x(n) = 12πj�cX(z)zn−1dz =

∑Res of X(z)zn−1 corresponding to the poles

of X(z)zn−1 that lie inside C that encloses |z| = R.

The residue of X(z)zn−1 at a given pole, z = zi with multiplicity m, can becalculated using:

Resz=zi =dm−1

dzm−1

[(z−zi)m(m−1)! X(z)z

n−1] ∣∣∣∣z=zi

For example:m=1: Resz=zi = (z − zi)X(z)zn−1|z=zim=2: Resz=zi =

ddz [(z − zi)

2X(z)zn−1]|z=zi



Inversion Formula Method-Cont.

Example:Find IZT of X(z) = 1(z−1)(z−0.5) using the residue method.

Since X(z)zn−1 = zn−1

(z−1)(z−0.5) has a pole at z = 0 when n = 0 but not for

n > 0, the two cases should be considered separately.Case 1: n=0, X(z)zn−1 = 1z(z−1)(z−0.5)Residue theorem gives: x(0) = Resz=0 +Resz=1 +Resz=0.5Resz=0 = zX(z)z

n−1|z=0 = 2Resz=1 = (z − 1)X(z)zn−1|z=1 = 2Resz=0.5 = (z − 0.5)X(z)zn−1|z=0.5 = −4Thus, x(0) = 0Case 2: n ≥ 1, x(n) = Resz=1 +Resz=0.5Resz=1 = (z − 1)X(z)zn−1|z=1 = 2Resz=0.5 = (z − 0.5)X(z)zn−1|z=0.5 = −2(0.5)n−1Thus, x(n) = 2− 2(0.5)n−1, n ≥ 1

Combine both results: x(n) =

{0 n = 02(1− (0.5)n−1) n ≥ 1



Inversion Formula Method: Special Case

Assume C is confined to the unit circle i.e. z = ejΩ, then dz = jdΩejΩ and theinversion formula becomes

x(n) =1

2πj

‰C

X(z)zn−1dz =1

2π

ˆ π−π

X(ejΩ)ejΩndΩ

which is nothing but the inverse discrete time Fourier transform (DTFT). Thisshows the relation between z-transform and DTFT. The forward transform isthen

X(z)z=ejΩ = X(ejΩ) =

∞∑n=0

x(n)e−jΩn



Representation of Discrete LTI Systems

A discrete-time LTI system can be described by one of the followingformulations:

1 Difference equation-General: Complete solution2 Convolution summation-Only particular solution (or forced response).3 Transfer function-Only particular solution (or forced response).4 State-Space equation-General: Complete solution

1-Difference Equation and Transfer Function RepresentationsA discrete-time LTI system can be represented by the following N th orderconstant coefficient difference equation:

N∑k=0

aky(n− k) =M∑`=0

b`u(n− `), a0 = 1

where y(n): output, u(n): input, aks and bls: coefficients. Alternatively,

y(n)︸︷︷︸present output

=

M∑`=0

b`u(n− `)︸︷︷︸past &present inputs

−N∑k=1

aky(n− k)︸︷︷︸past outputs



Difference Equation-Special Cases

Case a: Consider the case where N = 0 and M 6= 0 then we get

y(n) =M∑̀=0

b`u(n− `)

i.e. no recursion and hence the system is called Nonrecursive system.Taking the z-transform of both sides gives,

Y (z) = (M∑̀=0

b`z−`)U(z), Assuming zero IC’s

which gives transfer function: H(z) = Y (z)U(z) =M∑̀=0

b`z−`

i.e. the transfer function contains all zeros, hence also called All-Zero systems

Comparing H(z) =M∑̀=0

b`z−` with H(z) =

∞∑̀=0

h(`)z−` we note that

h(`) = b` for ` ∈ [0,M ] and h(`) = 0 ∀` > Mi.e. the impulse response is finite duration hence also called Finite ImpulseResponse (FIR). Alternatively we have:

y(n) =M∑̀=0

h(`)u(n− `)

This is called a Moving Average (MA) process.M.R. Azimi Digital Control & Digital Filters


Difference Equation-Special Cases

Case b: Consider the case where N 6= 0 and M = 0, then

y(n) = b0u(n)−N∑k=1

bky(n− k)

i.e. the recursive terms are still there. Taking the z-transform of both sidesgives:

Y (z) = b0U(z)−N∑k=1

akz−kY (z) =⇒ H(z) = Y (z)U(z) =

b0N∑

k=0

akz−k

i.e. an All-Pole system. Compare to H(z) =∞∑n=0

h(n)z−n, it is clear that

h(n) 6= 0 ∀n ≥ 0 i.e. infinite impulse response (IIR).Case c (general): Here N 6= 0 and M 6= 0. Thus, the transfer function becomes,

H(z) = Y (z)U(z) =

M∑̀=0

b`z−`

N∑k=0

akz−k= B(z)A(z)

which has both poles and zeros and is the general case of Recursive or IIRsystems.



Remark:

1 The rational transfer function, H(z), is proper if the order of thenumerator polynomial (in z) is less than or equal to that of thedenominator polynomial (i.e. M ≥ N) and strictly proper if it is less thanthe order of the denominator polynomial (i.e. M > N).

2 Not proper =⇒ Noncausal =⇒ Nonrealizable.

e.g., H(z) = z2+zz−1 =

Y (z)U(z) =⇒ (z − 1)Y (z) = (z

2 + z)U(z)

=⇒ y(n+ 1) = y(n) + u(n+ 2) + u(n+ 1) =⇒ noncausal due to y(n+ 1)depending on u(n+ 2).Alternatively, using long division,

H(z) = z2+zz−1 = z + 2 + 2z

−1 + · · ·

Comparing this with H(z) =∞∑

n=−∞h(n)z−n

it is clear that h(−1) = 1 =⇒ h(n) 6= 0 for n < 0 i.e. noncausal.



Realization of Discrete-Time LTI Systems and SFG

A discrete-time LTI system described by difference equation,

y(n) =M∑̀=0

b`u(n− `)−N∑k=1

aky(n− k)

can be realized on a hardware structure in a variety of forms using three basicelements.

1 Multipliers

2 Delays

3 Adders

A direct form realization of difference equation is shown below.



Realization of Discrete-Time LTI Systems and SFGThe signal flow graph (SFG) version is also shown.

Example 1: Derive the difference equation for a discrete differentiator.

Solution: Analog differentiator is y(t) = du(t)dt . To discretize use backward

difference method y(t) ≈ u(t)−u(t−∆)∆ and let t = nT and ∆ = Ty(nT ) = u(nT )−u((n−1)T )Tor simply y(n) = u(n)−u(n−1)T which is difference equation of discretedifferentiator. Transfer function is

H(z) = Y (z)U(z) =1−z−1T =

z−1zT



Solution of Difference Equations using z-transform

To find the complete solution (i.e. homogenous+particular) of a discrete LTIsystem: (a) take z-transform and impose ICs (b) apply PFE (or residue method)to find y(n) for a given input u(n).

Example 2:Giveny(n+ 2) + 3y(n+ 1) + 2y(n) = u(n), u(n) = δ(n), y(0) = 1, y(1) = −1Find the complete response of the system.Solution:

Recall form properties: Z{x(n+ n0)} = zn0 [X(z)−n0−1∑k=0

x(k)z−k].

Then,z2[Y (z)− y(0)− y(1)z−1] + 3z[Y (z)− y(0)] + 2Y (z) = U(z)

=⇒ (z2 + 3z + 2)Y (z) = U(z) + y(0)z2 + y(1)z + 3zy(0)But U(z) = 1, then

Y (z) = 1z2+3z+2 +z2+2zz2+3z+2 = Yp(z) + Yh(z) =

z2+2z+1z2+3z+2 =

z+1z+2



Solution of Difference Equations using z-transform-Cont.

Using PFE, expand Y (z)z

Y (z)z =

z+1z(z+2) =

1/2z +

1/2z+2

y(n) = 12δ(n) +12 (−2)

nus(n)Note that the particular solution is Yp(z) = H(z)U(z) (here H(z) =

1z2+3z+2 )

while homogeneous solution Yh(z) only depends on IC’s.

If needed, we can find these responses separately, as followsYh(z)z =

z+2z2+3z+2 =

1(z+1) −→ yh(n) = (−1)

nus(n)

WhileYp(z)z =

1z(z2+3z+2) =

1/2z −

1(z+1) +

1/2(z+2)

Thus,yp(n) =

12δ(n)− (−1)

nus(n) +12 (−2)

nus(n)


Inverse z-TransformRepresentation of LTI systems

digital controls & digital filters lectures 3 & 4 3-4.pdflectures 3 & 4 m.r. azimi, professor...

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