Because the two–sided (bilateral) z-transform is defined for all time , i.e. −∞<n<∞−∞<n<∞ size 12{ - infinity <n< infinity } {}, it can not be applied to a nonrelaxed system which is described by a difference equation accompanied with initial conditions. In such situations the one-side z- transform is used . We denote X+(z)X+(z) size 12{X rSup { size 8{+{}} } \( z \) } {} for this.

The lower limit of the summation is 0, irrelevant of the causality or noncausality of the signal x(n)x(n) size 12{x \( n \) } {}. This means that X+(z)X+(z) size 12{X rSup { size 8{+{}} } \( z \) } {}does not cantain the past (n<0)(n<0) size 12{ \( n<0 \) } {} information of the signal.

The X+(z)X+(z) size 12{X rSup { size 8{+{}} } \( z \) } {} is only unique for causal signals which don’t exist for n<0n<0 size 12{n<0} {} . Thus for causal signals the one-side or two-sided transforms are the same . The region of convergence (ROC) of causal signals is always outside the circle passing through the largest pole. The one-side transform of anticausaal signals is always equal to zero . For example, all three following signals

x 1 ( n ) = [ 2,3,4,5 ] x 2 ( n ) = [ 1,2,3,2,3,4,5 ] x 3 ( n ) = [ 3,2,1,2,3,4,5 ] x 1 ( n ) = [ 2,3,4,5 ] x 2 ( n ) = [ 1,2,3,2,3,4,5 ] x 3 ( n ) = [ 3,2,1,2,3,4,5 ] alignl { stack { size 12{x rSub { size 8{1} } \( n \) = \[ 2,3,4,5 \] } {} # x rSub { size 8{2} } \( n \) = \[ 1,2,3,2,3,4,5 \] {} # x rSub { size 8{3} } \( n \) = \[ 3,2,1,2,3,4,5 \] {} } } {}

have the same X+(z)X+(z) size 12{X rSup { size 8{+{}} } \( z \) } {}transform which is

X + ( z ) = 2z − 0 + 3z − 1 + 4z − 2 + 5z − 3 = 2 + 3z − 1 + 4z − 2 + 5z − 3 X + ( z ) = 2z − 0 + 3z − 1 + 4z − 2 + 5z − 3 = 2 + 3z − 1 + 4z − 2 + 5z − 3 size 12{X rSup { size 8{+{}} } \( z \) =2z rSup { size 8{ - 0} } +3z rSup { size 8{ - 1} } +4z rSup { size 8{ - 2} } +5z rSup { size 8{ - 3} } =2+3z rSup { size 8{ - 1} } +4z rSup { size 8{ - 2} } +5z rSup { size 8{ - 3} } } {}

The properties of two-sided z-transform (section 4.2) apply as well to one-sided (unilateral) transform , except the time shift . When a signal is delayed its past (n<0)(n<0) size 12{ \( n<0 \) } {} values appear, changing X+(z)X+(z) size 12{X rSup { size 8{+{}} } \( z \) } {}. From the transform of a delayed signal we expand it as

Z + [ x ( n-n 0 ) ] = z − n 0 ∑ k = − n 0 − 1 x ( k ) z − k + ∑ k = 0 ∞ x ( k ) z − k = z − n 0 ∑ k = 1 n 0 x ( -k ) z k + X + ( z ) Z + [ x ( n-n 0 ) ] = z − n 0 ∑ k = − n 0 − 1 x ( k ) z − k + ∑ k = 0 ∞ x ( k ) z − k = z − n 0 ∑ k = 1 n 0 x ( -k ) z k + X + ( z ) alignl { stack { size 12{Z rSup { size 8{+{}} } \[ x \( "n-n" rSub { size 8{0} } \) \] =z rSup { size 8{ - n rSub { size 6{0} } } } left [ Sum cSub {k= - n rSub { size 6{0} } } cSup { - 1} {x \( k \) z rSup { - k} size 12{+ Sum cSub {k=0} cSup { infinity } {x \( k \) z rSup { - k} } }} right ]} {} # size 12{ matrix { matrix { {} # {} # {} } {} # {} # {} } =z rSup { size 8{ - n rSub { size 6{0} } } } left [ Sum cSub {k=1} cSup {n rSub { size 6{0} } } {x \( "-k" \) z rSup {k} size 12{+X rSup {+{}} } size 12{ \( z \) }} right ]} {} } } {}

Thus

For causal signals the second term disappear and we get the known transform. For example , when signals x1(n)x1(n) size 12{x rSub { size 8{1} } \( n \) } {}, x2(n)x2(n) size 12{x rSub { size 8{2} } \( n \) } {}, x3(n)x3(n) size 12{x rSub { size 8{3} } \( n \) } {} mentioned earlier are delayed by 2 samples the transforms become

X 1 ( z ) = ( 2 + 3z − 1 − 4z − 2 + 5z − 3 ) z − 2 X 2 ( z ) = ( 2 + 3z − 1 + 4z − 2 + 5z − 3 ) z − 2 + ( 3z + 2z 2 ) z − 2 X 3 ( z ) = ( 2 + 3z − 1 + 5z − 2 + 5z − 3 ) z − 2 + ( z + 2z 2 ) z − 2 X 1 ( z ) = ( 2 + 3z − 1 − 4z − 2 + 5z − 3 ) z − 2 X 2 ( z ) = ( 2 + 3z − 1 + 4z − 2 + 5z − 3 ) z − 2 + ( 3z + 2z 2 ) z − 2 X 3 ( z ) = ( 2 + 3z − 1 + 5z − 2 + 5z − 3 ) z − 2 + ( z + 2z 2 ) z − 2 alignl { stack { size 12{X rSub { size 8{1} } \( z \) = \( 2+3z rSup { size 8{ - 1} } - 4z rSup { size 8{ - 2} } +5z rSup { size 8{ - 3} } \) z rSup { size 8{ - 2} } } {} # X rSub { size 8{2} } \( z \) = \( 2+3z rSup { size 8{ - 1} } +4z rSup { size 8{ - 2} } +5z rSup { size 8{ - 3} } \) z rSup { size 8{ - 2} } + \( 3z+2z rSup { size 8{2} } \) z rSup { size 8{ - 2} } {} # X rSub { size 8{3} } \( z \) = \( 2+3z rSup { size 8{ - 1} } +5z rSup { size 8{ - 2} } +5z rSup { size 8{ - 3} } \) z rSup { size 8{ - 2} } + \( z+2z rSup { size 8{2} } \) z rSup { size 8{ - 2} } {} } } {}

When a signal is advanced (shifted to the past) there will be less samples with respect to the new origin. Consider signal x(n)x(n) size 12{x \( n \) } {} having transform X+(z)X+(z) size 12{X rSup { size 8{+{}} } \( z \) } {} advanced n0n0 size 12{n rSub { size 8{0} } } {} samples to become x(n−n0)x(n−n0) size 12{x \( n - n rSub { size 8{0} } \) } {} , the new transform is

Z + [ x ( n + n 0 ) ] ∑ n = 0 ∞ x ( n + n 0 ) z − n = z n 0 ∑ k = n 0 ∞ x ( k ) z − k Z + [ x ( n + n 0 ) ] ∑ n = 0 ∞ x ( n + n 0 ) z − n = z n 0 ∑ k = n 0 ∞ x ( k ) z − k size 12{Z rSup { size 8{+{}} } \[ x \( n+n rSub { size 8{0} } \) \] Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {x \( n+n rSub { size 8{0} } \) z rSup { size 8{ - n} } } =z rSup { size 8{n rSub { size 6{0} } } } Sum cSub {k=n rSub { size 6{0} } } cSup { infinity } {x \( k \) z rSup { - k} } } {}

Writing orignal X+(z)X+(z) size 12{X rSup { size 8{+{}} } \( z \) } {} of the signal as

X + ( z ) = ∑ k = 0 ∞ x ( k ) z − k = ∑ k = 0 n 0 − 1 x ( k ) z − k + ∑ k = 0 ∞ x ( k ) z − k X + ( z ) = ∑ k = 0 ∞ x ( k ) z − k = ∑ k = 0 n 0 − 1 x ( k ) z − k + ∑ k = 0 ∞ x ( k ) z − k size 12{X rSup { size 8{+{}} } \( z \) = Sum cSub {k=0} cSup { infinity } {x \( k \) z rSup { size 8{ - k} } = Sum cSub { size 8{k=0} } cSup { size 8{n rSub { size 6{0} } - 1} } {x \( k \) z rSup { size 8{ - k} } } + Sum cSub {k=0} cSup { infinity } {x \( k \) z rSup { - k} } } } {}

We can conclude

The above formula can be checked with the three signals mentioned earlier . Actually , it’d better to take the necessary z- transform directly on the shifted signal rather than using the two formulas above .