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COMPUTING THE INVERSE z-TRANSFORM

Signals and systems (characterized by impulse responses) are functions of (time) index n. Transformation to complex frequency domain z is needed to probe for more characteristics. However, rather frequently we need to take the inverse z-transform to observe the result in time domain . As for other transforms, the inverse z-transform is more difficult than the forward transform . The most standard method for inverse transform is the partial fraction expansion which is discussed in next section. In this section we consider some other methods.

Theoretical inverse z-transform

Officially the inverse z-transform x(n)

x(n) size 12{x \( n \) } {}

of a X(z)

X(z) size 12{X \( z \) } {}

is defined as follows. From the definition of z– transform we multiply both sides with zn−1

zn−1 size 12{z rSup { size 8{n - 1} } } {}

, then take the inlegral of both sides along a closed contour around the origin in the ROC:

∮ C X ( z ) z n − 1 dz = ∮ C ∑ k = − ∞ ∞ x ( k ) z n − 1 − k dz

∮ C X ( z ) z n − 1 dz = ∮ C ∑ k = − ∞ ∞ x ( k ) z n − 1 − k dz size 12{ lInt rSub { size 8{C} } { ital " X" \( z \) z rSup { size 8{n - 1} } ital "dz"} = lInt rSub { size 8{C} } { Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {x \( k \) z rSup { size 8{n - 1 - k} } ital "dz"} } } {}

The direction of C is anticlockwise . Next step is to swap the order of integration and summation:

∮CX(z)zn−1dz=∑k=−∞∞x(k)∮C zn−1−kdz

∮CX(z)zn−1dz=∑k=−∞∞x(k)∮C zn−1−kdz size 12{ lInt rSub { size 8{C} } { X \( z \) z rSup { size 8{n - 1} } ital "dz"} = Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {x \( k \) } lInt rSub { size 8{C} } { ital " z" rSup { size 8{n - 1 - k} } ital "dz"} } {}

(1)

To proceed further we need the Cauchy Integral Theorem which states, for a complex variable z,

12πj∮C zn−1−kdz=1,k=n0,k≠n

12πj∮C zn−1−kdz=1,k=n0,k≠nalignl { stack { size 12{ { {1} over {2πj} } lInt rSub { size 8{C} } { ital " z" rSup { size 8{n - 1 - k} } ital "dz"} =1, matrix { {} # {} } k=n} {} # matrix { matrix { {} # {} # {} # {} } {} # {} # {} } 0, matrix { {} # {} } k <> n {} } } {}

(2)

Applying the theorem, the RHS of Equation 1 becomes just 2πjx(n)

2πjx(n) size 12{2π ital "jx" \( n \) } {}

, hence

x(n)=12πj∮C X(z)zn−1dz

x(n)=12πj∮C X(z)zn−1dz size 12{x \( n \) = { {1} over {2πj} } lInt rSub { size 8{C} } { ital " X" \( z \) z rSup { size 8{n - 1} } ital "dz"} } {}

(3)

To compute the contour integral we call for the Cauchy Residue Theorem . So the direct inverse z – transform is rather cumbersome and is rarely used . One then must resort to other indirect methods.

The power series method

The method is to perform a long division of the rational function and then finding the inverse z-transform the terms of the quotient . This method is also cumbersome and not often used .

Example 1

Find the inverse z – transform of the system function H ( z ) = 1 1 − 3 2 z − 1 + 1 2 z − 2

H ( z ) = 1 1 − 3 2 z − 1 + 1 2 z − 2 size 12{H \( z \) = { {1} over {1 - { {3} over {2} } z rSup { size 8{ - 1} } + { {1} over {2} } z rSup { size 8{ - 2} } } } } {}

When (a) ROC is ∣z∣>1,

∣z∣>1, size 12{ lline z rline >1,} {}

(b) ROC is ∣z∣<0.5

∣z∣<0.5 size 12{ lline z rline <0 "." 5} {}

Solution

First the given system function can be written as H ( z ) = z 2 z 2 − 3 2 z + 1 2 = z 2 ( z − 1 2 ) ( z − 1 )

H ( z ) = z 2 z 2 − 3 2 z + 1 2 = z 2 ( z − 1 2 ) ( z − 1 ) size 12{H \( z \) = { {z rSup { size 8{2} } } over {z rSup { size 8{2} } - { {3} over {2} } z+ { {1} over {2} } } } = { {z rSup { size 8{2} } } over { \( z - { {1} over {2} } \) \( z - 1 \) } } } {}

(a) The system has two poles at z=0.5

z=0.5 size 12{z=0 "." 5} {}

and z=1

z=1 size 12{z=1} {}

. Because the ROC is ∣z∣>1

∣z∣>1 size 12{ lline z rline >1} {}

the signal is causal (right-sided) , we then expand H(z)

H(z) size 12{H \( z \) } {}

as a power series of z−1

z−1 size 12{z rSup { size 8{ - 1} } } {}

by a long division (rather tedious), the result is

H ( z ) = 1 1 − 3 2 z − 1 + 1 2 z − 2 = 1 + 3 2 z − 1 + 7 4 z − 2 + 15 8 z − 3 + 31 16 z − 4 + . . .

H ( z ) = 1 1 − 3 2 z − 1 + 1 2 z − 2 = 1 + 3 2 z − 1 + 7 4 z − 2 + 15 8 z − 3 + 31 16 z − 4 + . . . size 12{H \( z \) = { {1} over {1 - { {3} over {2} } z rSup { size 8{ - 1} } + { {1} over {2} } z rSup { size 8{ - 2} } } } =1+ { {3} over {2} } z rSup { size 8{ - 1} } + { {7} over {4} } z rSup { size 8{ - 2} } + { {"15"} over {8} } z rSup { size 8{ - 3} } + { {"31"} over {"16"} } z rSup { size 8{ - 4} } + "." "." "." } {}

Hence the inverse z – transform of H(z)

H(z) size 12{H \( z \) } {}

, i.e. the impulse respowse h(n)

h(n) size 12{h \( n \) } {}

, is

h ( n ) = [ 1, 3 2 , 7 4 , 15 8 , 31 16 , . . . ]

h ( n ) = [ 1, 3 2 , 7 4 , 15 8 , 31 16 , . . . ] size 12{h \( n \) = \[ 1, { {3} over {2} } , { {7} over {4} } , { {"15"} over {8} } , { {"31"} over {"16"} } , "." "." "." \] } {}

(b) Because ROC is ∣z∣<1

∣z∣<1 size 12{ lline z rline <1} {}

the signal is anticausal (left-sided), we then expand H(z)

H(z) size 12{H \( z \) } {}

as a power series of z instead of z−1

z−1 size 12{z rSup { size 8{ - 1} } } {}

also by a long division the result is

H ( z ) = 1 1 − 3 2 z − 1 + 1 2 z − 2 = 2z 2 + 6z 3 + 14 z 4 + 30 z 5 + 62 z 6 + . . .

H ( z ) = 1 1 − 3 2 z − 1 + 1 2 z − 2 = 2z 2 + 6z 3 + 14 z 4 + 30 z 5 + 62 z 6 + . . . size 12{H \( z \) = { {1} over {1 - { {3} over {2} } z rSup { size 8{ - 1} } + { {1} over {2} } z rSup { size 8{ - 2} } } } =2z rSup { size 8{2} } +6z rSup { size 8{3} } +"14"z rSup { size 8{4} } +"30"z rSup { size 8{5} } +"62"z rSup { size 8{6} } + "." "." "." } {}

Hence

h ( n ) = [ . . . 62 , 30 , 14 , 6,2,0,0 ]

h ( n ) = [ . . . 62 , 30 , 14 , 6,2,0,0 ] size 12{h \( n \) = \[ "." "." "." "62","30","14",6,2,0,0 \] } {}

Transforming equation in z-domain into equation in time domain.

Sometimes we do need to inversely transform an equation in z back to the corresponding equation in n. The delay property of the z-transform is utilized for such situations.

Example 2

Find the inverse transform of the equation 2z 3 Y ( z ) − 3z 2 Y ( z ) + zY ( z ) = 4z − 2 X ( z )

2z 3 Y ( z ) − 3z 2 Y ( z ) + zY ( z ) = 4z − 2 X ( z ) size 12{2z rSup { size 8{3} } Y \( z \) - 3z rSup { size 8{2} } Y \( z \) + ital "zY" \( z \) =4z rSup { size 8{ - 2} } X \( z \) } {}

Solution

Utilizing the delay property directly on the given equation, we have

2y ( n + 3 ) − 3y ( n + 2 ) + 6y ( n + 1 ) = 4x ( n − 2 )

2y ( n + 3 ) − 3y ( n + 2 ) + 6y ( n + 1 ) = 4x ( n − 2 ) size 12{2y \( n+3 \) - 3y \( n+2 \) +6y \( n+1 \) =4x \( n - 2 \) } {}

Because this equation applies for all time , we replace n−3

n−3 size 12{n - 3} {}

for n and get

2y ( n ) − 3y ( n − 1 ) + 6y ( n − 2 ) = 4x ( n − 5 )

2y ( n ) − 3y ( n − 1 ) + 6y ( n − 2 ) = 4x ( n − 5 ) size 12{2y \( n \) - 3y \( n - 1 \) +6y \( n - 2 \) =4x \( n - 5 \) } {}

y ( n ) = 1 . 5y ( n − 1 ) − 3y ( n − 2 ) + 2x ( n − 3 )

y ( n ) = 1 . 5y ( n − 1 ) − 3y ( n − 2 ) + 2x ( n − 3 ) size 12{y \( n \) =1 "." 5y \( n - 1 \) - 3y \( n - 2 \) +2x \( n - 3 \) } {}

The system is a recursive filter .

Example 3

A system has the system function H ( z ) = 1 z ( z − 1 ) ( z − 2 )

H ( z ) = 1 z ( z − 1 ) ( z − 2 ) size 12{H \( z \) = { {1} over { ital "z " \( ital "z " - 1 \) \( ital "z " - 2 \) } } } {}

Find its impulse response .

Solution

Let’s write

H ( z ) = 1 z ( z − 1 ) ( z − 2 ) = Y ( z ) X ( z )

H ( z ) = 1 z ( z − 1 ) ( z − 2 ) = Y ( z ) X ( z ) size 12{H \( z \) = { {1} over { ital "z " \( ital "z " - 1 \) \( ital "z " - 2 \) } } = { {Y \( z \) } over {X \( z \) } } } {}

Crossmultiply the equation :

Y ( z ) [ z ( z − 1 ) ( z − 2 ) ] = X ( z )

Y ( z ) [ z ( z − 1 ) ( z − 2 ) ] = X ( z ) size 12{Y \( z \) \[ z \( z - 1 \) \( z - 2 \) \] =X \( z \) } {}

z 3 Y ( z ) − 3z 2 Y ( z ) + 2 zY ( z ) = X ( z )

z 3 Y ( z ) − 3z 2 Y ( z ) + 2 zY ( z ) = X ( z ) size 12{z rSup { size 8{3} } Y \( z \) - 3z rSup { size 8{2} } Y \( z \) +2 ital "zY" \( z \) =X \( z \) } {}

Take the inverse transform :

y ( n + 3 ) − 3y ( n + 2 ) + 2y ( n + 1 ) = x ( n )

y ( n + 3 ) − 3y ( n + 2 ) + 2y ( n + 1 ) = x ( n ) size 12{y \( n+3 \) - 3y \( n+2 \) +2y \( n+1 \) =x \( n \) } {}

Time-shift the equation :

y ( n ) − 3y ( n − 1 ) + 2y ( n − 2 ) = x ( n − 3 )

y ( n ) − 3y ( n − 1 ) + 2y ( n − 2 ) = x (

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COMPUTING THE INVERSE z-TRANSFORM

Theoretical inverse z-transform

The power series method

Transforming equation in z-domain into equation in time domain.