
Z-transform should not be confused with Fisher z-transformation.
In mathematics and signal processing, the Z-transform converts a discrete-time signal, which is a sequence of real or complex numbers, into a complex valued frequency-domain (the z-domain or z-plane) representation.[1] [2] [3]
It can be considered a discrete-time counterpart of the Laplace transform (the s-domain or s-plane).[4] This similarity is explored in the theory of time-scale calculus.
While the continuous-time Fourier transform is evaluated on the s-domain's vertical axis (the imaginary axis), the discrete-time Fourier transform is evaluated along the z-domain's unit circle. The s-domain's left half-plane maps to the area inside the z-domain's unit circle, while the s-domain's right half-plane maps to the area outside of the z-domain's unit circle.
In signal processing, one of the means of designing digital filters is to take analog designs, subject them to a bilinear transform which maps them from the s-domain to the z-domain, and then produce the digital filter by inspection, manipulation, or numerical approximation. Such methods tend not to be accurate except in the vicinity of the complex unity, i.e. at low frequencies.
The basic idea now known as the Z-transform was known to Laplace, and it was re-introduced in 1947 by W. Hurewicz[5] [6] and others as a way to treat sampled-data control systems used with radar. It gives a tractable way to solve linear, constant-coefficient difference equations. It was later dubbed "the z-transform" by Ragazzini and Zadeh in the sampled-data control group at Columbia University in 1952.[7] [8]
The modified or advanced Z-transform was later developed and popularized by E. I. Jury.[9] [10]
The idea contained within the Z-transform is also known in mathematical literature as the method of generating functions which can be traced back as early as 1730 when it was introduced by de Moivre in conjunction with probability theory.[11] From a mathematical view the Z-transform can also be viewed as a Laurent series where one views the sequence of numbers under consideration as the (Laurent) expansion of an analytic function.
The Z-transform can be defined as either a one-sided or two-sided transform (similarly to the one-sided Laplace transform and the two-sided Laplace transform).[12]
The bilateral or two-sided Z-transform of a discrete-time signal
x[n]
X(z)
where
n
z
z
z=Aei\phi=A ⋅ (\cos{\phi}+i\sin{\phi})
A
i
\phi
Alternatively, in cases where
x[n]
In signal processing, this definition can be used to evaluate the Z-transform of the unit impulse response of a discrete-time causal system.
An important example of the unilateral Z-transform is the probability-generating function, where the component
x[n]
The inverse Z-transform is:
where
C
C
A special case of this contour integral occurs when
C
X(z)
The Z-transform with a finite range of
n
z
z
The following three methods are often used for the evaluation of the inverse -transform,
This method involves applying the Cauchy Residue Theorem to evaluate the inverse Z-transform. By integrating around a closed contour in the complex plane, the residues at the poles of the Z-transform function inside the ROC are summed. This technique is particularly useful when working with functions expressed in terms of complex variables.
In this method, the Z-transform is expanded into a power series. This approach is useful when the Z-transform function is rational, allowing for the approximation of the inverse by expanding into a series and determining the signal coefficients term by term.
This technique decomposes the Z-transform into a sum of simpler fractions, each corresponding to known Z-transform pairs. The inverse Z-transform is then determined by looking up each term in a standard table of Z-transform pairs. This method is widely used for its efficiency and simplicity, especially when the original function can be easily broken down into recognizable components.
A) Determine the inverse Z-transform of the following by series expansion method,
Solution:
Case 1:
ROC:
\left\vertz\right\vert>1
Since the ROC is the exterior of a circle,
x(n)
Note that in each step of long division process we eliminate lowest power term of .
Case 2:
ROC:
\left\vertz\right\vert<0.5
Since the ROC is the interior of a circle,
x(n)
By performing long division we get
\begin{align}x(n)&=\{30,14,6,2,0,0\}\ & \uparrow\ \end{align}
Note that in each step of long division process we eliminate lowest power term of .
Note:
z
zk
x(-k)
z-k
B) Determine the inverse Z-transform of the following by series expansion method,
Eliminating negative powers if
z
By partial fraction expansion,
Case 1:
ROC:
\left\vertz\right\vert>1
Both the terms are causal, hence
x(n)
Case 2:
ROC:
\left\vertz\right\vert<0.5
Both the terms are anticausal, hence
x(n)
\begin{align}x(n)&=-2{(1)n}u(-n-1)-(-1{(0.5)n}u(-n-1))\ &=((0.5)n-2)u(-n-1)\\ \end{align}
Case 3:
ROC:
0.5<\left\vertz\right\vert<1
One of the terms is causal (provides the causal part) and other is anticausal (provides the anticausal part), hence
x(n)
\begin{align}x(n)&=-2{(1)n}u(-n-1)-1{(0.5)n}u(n)\ &=-2u(-n-1)-(0.5)nu(n)\ \end{align}
The region of convergence (ROC) is the set of points in the complex plane for which the Z-transform summation absolutely converges:
ROC=\left\{z:
| infty | |
| \sum | |
| n=-infty |
\left|x[n]z-n\right|<infty\right\}
Let . Expanding
x[n]
(-infty,infty)
x[n]=\left\{...,(0.5)-3,(0.5)-2,(0.5)-1,1,0.5,(0.5)2,(0.5)3,...\right\}=\left\{...,23,22,2,1,0.5,(0.5)2,(0.5)3,...\right\}.
Looking at the sum
| infty | |
| \sum | |
| n=-infty |
x[n]z-n\toinfty.
Therefore, there are no values of
z
Let
x[n]=(0.5)nu[n]
u
x[n]
(-infty,infty)
x[n]=\left\{...,0,0,0,1,0.5,(0.5)2,(0.5)3,...\right\}.
Looking at the sum
| infty | |
| \sum | |
| n=-infty |
x[n]z-n=
| infty | |
| \sum | |
| n=0 |
(0.5)nz-n=
| infty | ||
| \sum | \left( | |
| n=0 |
| 0.5 | |
| z |
\right)n=
| 1 | |
| 1-0.5z-1 |
.
The last equality arises from the infinite geometric series and the equality only holds if, which can be rewritten in terms of
z
Let
x[n]=-(0.5)nu[-n-1]
u
x[n]
(-infty,infty)
x[n]=\left\{...,-(0.5)-3,-(0.5)-2,-(0.5)-1,0,0,0,0,...\right\}.
Looking at the sum
| infty | |
| \begin{align} \sum | |
| n=-infty |
x[n]z-n&=
| -1 | |
| -\sum | |
| n=-infty |
(0.5)nz-n\\ &=
| infty | ||
| -\sum | \left( | |
| m=1 |
| z | |
| 0.5 |
\right)m\\ &=-
| (0.5)-1z | |
| 1-(0.5)-1z |
\\ &=-
| 1 | |
| 0.5z-1-1 |
\\ &=
| 1 | |
| 1-0.5z-1 |
\\ \end{align}
|(0.5)-1z|<1
z
What differentiates this example from the previous example is only the ROC. This is intentional to demonstrate that the transform result alone is insufficient.
Examples 2 and 3 clearly show that the Z-transform
X(z)
x[n]
In example 2, the causal system yields a ROC that includes
|z|=infty
In systems with multiple poles it is possible to have a ROC that includes neither
|z|=infty
x[n]=(0.5)nu[n]-(0.75)nu[-n-1]
(0.5)nu[n]
The stability of a system can also be determined by knowing the ROC alone. If the ROC contains the unit circle (i.e.,) then the system is stable. In the above systems the causal system (Example 2) is stable because contains the unit circle.
Let us assume we are provided a Z-transform of a system without a ROC (i.e., an ambiguous). We can determine a unique
x[n]
For stability the ROC must contain the unit circle. If we need a causal system then the ROC must contain infinity and the system function will be a right-sided sequence. If we need an anticausal system then the ROC must contain the origin and the system function will be a left-sided sequence. If we need both stability and causality, all the poles of the system function must be inside the unit circle.
The unique
x[n]
| Definition of Z-transform | x[n] | X(z) | X(z)=l{Z}\{x[n]\} x[n]=l{Z}-1\{X(z)\} | r2< | z | a1x1[n]+a2x2[n] | a1X1(z)+a2X2(z) | \begin{align}X(z)&=
(a1x1[n]+a2x
\\ &=a1\sum
x1[n]z-n+a2\sum
x2[n]z-n\\ &=a1X1(z)+a2X2(z)\end{align} | Contains ROC1 ∩ ROC2 | xK[n]=\begin{cases}x[r],&n=Kr\ 0,&n\notinKZ\end{cases} KZ:=\{Kr:r\inZ\} | X(zK) | \begin{align}XK(z)
\\ &=
x[r]z-rK\\ &=
x[r](zK)-r\\ &=X(zK)\end{align} |
| x[Kn] |
X\left(z\tfrac{1{K}} ⋅ e-i{K}p}\right) | ohio-state.edu or ee.ic.ac.uk | - ! Time delay | x[n-k] k>0 x:x[n]=0 \foralln<0 | z-kX(z) | \begin{align}l{Z}\{x[n-k]\}&=
x[n-k]z-n\\ &=
x[m]z-(m+k)&&m=n-k\\ &=
x[m]z-mz-k\\ &=z-k
x[m]z-m\\ &=z-k
x[m]z-m&&x[\beta]=0,\forall\beta<0\\ &=z-kX(z)\end{align} | ROC, except z=0 k>0 z=infty k<0 | - ! Time advance | x[n+k] k>0 | Bilateral Z-transform:Unilateral Z-transform:[14] | - ! First difference backward | x[n]-x[n-1] x[n]=0 n<0 | (1-z-1)X(z) | Contains the intersection of ROC of X1(z) z ≠ 0 | - ! First difference forward | x[n+1]-x[n] | (z-1)X(z)-zx[0] | - ! Time reversal | x[-n] | X(z-1) | \begin{align}l{Z}\{x(-n)\}&=
x[-n]z-n\\ &=
x[m]zm\\ &=
x[m]{(z-1)}-m\\ &=X(z-1)\\ \end{align} | \tfrac{1}{r1}< | z | <\tfrac | - ! Scaling in the z-domain | anx[n] | X(a-1z) | \begin{align}l{Z}\left\{anx[n]\right\}&=
anx[n]z-n\\ &=
x[n](a-1z)-n\\ &=X(a-1z) \end{align} | a | r_2 < | z | < | a | r_1 | - | x*[n] | X*(z*) | \begin{align}l{Z}\{x*(n)\}&=
x*[n]z-n\\ &=
\left[x[n](z*)-n\right]*\\ &=\left[
x[n](z*)-n\right]*\\ &=X*(z*) \end{align} | \operatorname{Re}\{x[n]\} | \tfrac{1}{2}\left[X(z)+X*(z*)\right] | \operatorname{Im}\{x[n]\} | \tfrac{1}{2i}\left[X(z)-X*(z*)\right] | - ! Differentiation in the z-domain | nx[n] | -z
| \begin{align}l{Z}\{nx(n)\}&=
nx[n]z-n\\ &=z
nx[n]z-n-1\\ &=-z
x[n](-nz-n-1)\\ &=-z
(z-n)\\ &=-z
\end{align} | ROC, if X(z) X(z) | x1[n]*x2[n] | X1(z)X2(z) | \begin{align}l{Z}\{x1(n)*x2(n)\}&=l{Z}\left
x1[l]x2[n-l]\right\}\\ &=
\left
x1[l]x2[n-l]\right]z-n\\
x1[l]\left
\right]\\ &=\left
\right]\ | \left [\sum_{n=-\infty}^{\infty} x_2[n]z^ \right ] \\ &=X_1(z)X_2(z)\end | Contains ROC1 ∩ ROC2 |
*x2[n] |
| Contains the intersection of ROC of
X2(z) | - ! Accumulation |
x[k] |
X(z) |
x[k]z-n
(x[n]+ … )z-n\\ &=X(z)\left(1+z-1+z-2+ … \right)\\ &=X(z)
z-j\\ &=X(z)
\end{align} | x1[n]x2[n] |
\ointCX1(v)X
dv | At least r1lr2l< | z | - |
|---|
| infty | |
| \sum | |
| n=-infty |
| * | |
| x | |
| 2[n] |
=
| 1 | |
| 2\pii |
\ointC
| *})v | |
| X | |
| 2(\tfrac{1}{v |
-1dv
x[n]
If the poles of
(z-1)X(z)
Here:
u:n\mapstou[n]=\begin{cases}1,&n\ge0\ 0,&n<0\end{cases}
\delta:n\mapsto\delta[n]=\begin{cases}1,&n=0\ 0,&n\ne0\end{cases}
| Signal, x[n] | Z-transform, X(z) | ROC | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | \delta[n] | 1 | all z | ||||||||
| 2 | \delta[n-n0] |
| z ≠ 0 | ||||||||
| 3 | u[n] |
| z | > 1 | |||||||
| 4 | -u[-n-1] |
| z | < 1 | |||||||
| 5 | nu[n] |
| z | > 1 | |||||||
| 6 | -nu[-n-1] |
| z | < 1 | |||||||
| 7 | n2u[n] |
| z | > 1 | |||||||
| 8 | -n2u[-n-1] |
| z | < 1 | |||||||
| 9 | n3u[n] |
| z | > 1 | |||||||
| 10 | -n3u[-n-1] |
| z | < 1 | |||||||
| 11 | anu[n] |
| z | > | a | ||||||
| 12 | -anu[-n-1] |
| z | < | a | ||||||
| 13 | nanu[n] |
| z | > | a | ||||||
| 14 | -nanu[-n-1] |
| z | < | a | ||||||
| 15 | n2anu[n] |
| z | > | a | ||||||
| 16 | -n2anu[-n-1] |
| z | < | a | ||||||
| 17 | \left(\begin{array}{c}n+m-1\ m-1\end{array}\right)anu[n] | , for positive integer m | z | > | a | ||||||
| 18 | (-1)m\left(\begin{array}{c}-n-1\ m-1\end{array}\right)anu[-n-m] | , for positive integer m | z | < | a | ||||||
| 19 | \cos(\omega0n)u[n] |
| z | >1 | |||||||
| 20 | \sin(\omega0n)u[n] |
| z | >1 | |||||||
| 21 | an\cos(\omega0n)u[n] |
| z | > | a | ||||||
| 22 | an\sin(\omega0n)u[n] |
| z | > | a |
For values of
z
\omega
which is also known as the discrete-time Fourier transform (DTFT) of the
x[n]
2\pi
X(f)
T
x[n]
x[n]
where
T
f
\omega=2\pifT
\omega=2\pi
X(\tfrac{\omega-2\pik}{2\piT})
As parameter T changes, the individual terms of move farther apart or closer together along the f-axis. In however, the centers remain 2 apart, while their widths expand or contract. When sequence
x(nT)
x(nT)
See main article: Bilinear transform. The bilinear transform can be used to convert continuous-time filters (represented in the Laplace domain) into discrete-time filters (represented in the Z-domain), and vice versa. The following substitution is used:
s=
| 2 | |
| T |
| (z-1) | |
| (z+1) |
H(s)
H(z)
z=esT ≈
| 1+sT/2 | |
| 1-sT/2 |
i\omega
i\omega
i\omega
See main article: Starred transform. Given a one-sided Z-transform
X(z)
T
.X*(s)=
| X(z)| | |
| \displaystylez=esT |
The inverse Laplace transform is a mathematical abstraction known as an impulse-sampled function.
The linear constant-coefficient difference (LCCD) equation is a representation for a linear system based on the autoregressive moving-average equation:
| N | |
| \sum | |
| p=0 |
y[n-p]\alphap=
| M | |
| \sum | |
| q=0 |
x[n-q]\betaq.
Both sides of the above equation can be divided by
\alpha0
y[n]=
| M | |
| \sum | |
| q=0 |
x[n-q]\betaq-
| N | |
| \sum | |
| p=1 |
y[n-p]\alphap.
This form of the LCCD equation is favorable to make it more explicit that the "current" output
y[n]
Taking the Z-transform of the above equation (using linearity and time-shifting laws) yields:
Y(z)
| N | |
| \sum | |
| p=0 |
z-p\alphap=X(z)
| M | |
| \sum | |
| q=0 |
z-q\betaq
X(z)
Y(z)
x[n]
Rearranging results in the system's transfer function:
H(z)=
| Y(z) | |
| X(z) |
=
| ||||||||||
|
=
| \beta0+z-1\beta1+z-2\beta2+ … +z-M\betaM | |
| \alpha0+z-1\alpha1+z-2\alpha2+ … +z-N\alphaN |
.
From the fundamental theorem of algebra the numerator has
M
H
N
H(z)=
| (1-q1z-1)(1-q2z-1) … (1-qMz-1) | |
| (1-p1z-1)(1-p2z-1) … (1-pNz-1) |
,
qk
k
pk
k
In addition, there may also exist zeros and poles at
z=0
By factoring the denominator, partial fraction decomposition can be used, which can then be transformed back to the time domain. Doing so would result in the impulse response and the linear constant coefficient difference equation of the system.
If such a system
H(z)
X(z)
Y(z)
y[n]
| style | Y(z) |
| z |
z
Y(z)