Bilateral Z-transform

The bilateral or two-sided Z-transform of a discrete-time signal ${\displaystyle x[n]}$ is the formal power series ${\displaystyle X(z)}$ defined as:

where ${\displaystyle n}$ is an integer and ${\displaystyle z}$ is, in general, a complex number. In polar form, ${\displaystyle z}$ may be written as:

${\displaystyle z=Ae^{j\phi }=A\cdot (\cos {\phi }+j\sin {\phi })}$

where ${\displaystyle A}$ is the magnitude of ${\displaystyle z}$, ${\displaystyle j}$ is the imaginary unit, and ${\displaystyle \phi }$ is the complex argument (also referred to as angle or phase) in radians.

Unilateral Z-transform

Alternatively, in cases where ${\displaystyle x[n]}$ is defined only for ${\displaystyle n\geq 0}$, the single-sided or unilateral Z-transform is defined as:

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 ${\displaystyle x[n]}$ is the probability that a discrete random variable takes the value ${\displaystyle n}$, and the function ${\displaystyle X(z)}$ is usually written as ${\displaystyle X(s)}$ in terms of ${\displaystyle s=z^{-1}}$. The properties of Z-transforms (listed in § Properties) have useful interpretations in the context of probability theory.

Example 1 (no ROC)

Let ${\displaystyle x[n]=(.5)^{n}\ .}$ Expanding ${\displaystyle x[n]}$ on the interval ${\displaystyle (-\infty ,\infty )}$ it becomes

${\displaystyle x[n]=\left\{\dots ,(.5)^{-3},(.5)^{-2},(.5)^{-1},1,(.5),(.5)^{2},(.5)^{3},\dots \right\}=\left\{\dots ,2^{3},2^{2},2,1,(.5),(.5)^{2},(.5)^{3},\dots \right\}.}$

Looking at the sum

${\displaystyle \sum _{n=-\infty }^{\infty }x[n]z^{-n}\to \infty .}$

Therefore, there are no values of ${\displaystyle z}$ that satisfy this condition.

Example 2 (causal ROC)

ROC (blue), |z| = .5 (dashed black circle), and the unit circle (dotted grey circle).

Let ${\displaystyle x[n]=(.5)^{n}\,u[n]}$ (where ${\displaystyle u}$ is the Heaviside step function). Expanding ${\displaystyle x[n]}$ on the interval ${\displaystyle (-\infty ,\infty )}$ it becomes

${\displaystyle x[n]=\left\{\dots ,0,0,0,1,(.5),(.5)^{2},(.5)^{3},\dots \right\}.}$

Looking at the sum

${\displaystyle \sum _{n=-\infty }^{\infty }x[n]z^{-n}=\sum _{n=0}^{\infty }(.5)^{n}z^{-n}=\sum _{n=0}^{\infty }\left({\frac {.5}{z}}\right)^{n}={\frac {1}{1-(.5)z^{-1}}}.}$

The last equality arises from the infinite geometric series and the equality only holds if ${\displaystyle |(.5)z^{-1}|<1,}$ which can be rewritten in terms of ${\displaystyle z}$ as ${\displaystyle |z|>(.5).}$ Thus, the ROC is ${\displaystyle |z|>(.5).}$ In this case the ROC is the complex plane with a disc of radius 0.5 at the origin "punched out".

Example 3 (anti causal ROC)

ROC (blue), |z| = .5 (dashed black circle), and the unit circle (dotted grey circle).

Let ${\displaystyle x[n]=-(.5)^{n}\,u[-n-1]}$ (where ${\displaystyle u}$ is the Heaviside step function). Expanding ${\displaystyle x[n]}$ on the interval ${\displaystyle (-\infty ,\infty )}$ it becomes

${\displaystyle x[n]=\left\{\dots ,-(.5)^{-3},-(.5)^{-2},-(.5)^{-1},0,0,0,0,\dots \right\}.}$

Looking at the sum

${\displaystyle {\begin{aligned}\sum _{n=-\infty }^{\infty }x[n]\,z^{-n}&=-\sum _{n=-\infty }^{-1}(.5)^{n}\,z^{-n}\\&=-\sum _{m=1}^{\infty }\left({\frac {z}{.5}}\right)^{m}\\&=-{\frac {(.5)^{-1}z}{1-(.5)^{-1}z}}\\&=-{\frac {1}{(.5)z^{-1}-1}}\\&={\frac {1}{1-(.5)z^{-1}}}\\\end{aligned}}}$

and using the infinite geometric series again, the equality only holds if ${\displaystyle |(.5)^{-1}z|<1}$ which can be rewritten in terms of ${\displaystyle z}$ as ${\displaystyle |z|<(.5).}$ Thus, the ROC is ${\displaystyle |z|<(.5).}$ In this case the ROC is a disc centered at the origin and of radius 0.5.

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 conclusion

Examples 2 & 3 clearly show that the Z-transform ${\displaystyle X(z)}$ of ${\displaystyle x[n]}$ is unique when and only when specifying the ROC. Creating the pole–zero plot for the causal and anticausal case show that the ROC for either case does not include the pole that is at 0.5. This extends to cases with multiple poles: the ROC will never contain poles.

In example 2, the causal system yields an ROC that includes ${\displaystyle |z|=\infty }$ while the anticausal system in example 3 yields an ROC that includes ${\displaystyle |z|=0.}$

ROC shown as a blue ring 0.5 < |z| < 0.75

In systems with multiple poles it is possible to have a ROC that includes neither ${\displaystyle |z|=\infty }$ nor ${\displaystyle |z|=0.}$ The ROC creates a circular band. For example,

${\displaystyle x[n]=(.5)^{n}\,u[n]-(.75)^{n}\,u[-n-1]}$

has poles at 0.5 and 0.75. The ROC will be 0.5 < |z| < 0.75, which includes neither the origin nor infinity. Such a system is called a mixed-causality system as it contains a causal term ${\displaystyle (.5)^{n}\,u[n]}$ and an anticausal term ${\displaystyle -(.75)^{n}\,u[-n-1].}$

The stability of a system can also be determined by knowing the ROC alone. If the ROC contains the unit circle (i.e., |z| = 1) then the system is stable. In the above systems the causal system (Example 2) is stable because |z| > 0.5 contains the unit circle.

Let us assume we are provided a Z-transform of a system without a ROC (i.e., an ambiguous ${\displaystyle x[n]}$). We can determine a unique ${\displaystyle x[n]}$ provided we desire the following:

• Stability
• Causality

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 ${\displaystyle x[n]}$ can then be found.

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:

${\displaystyle s={\frac {2}{T}}{\frac {(z-1)}{(z+1)}}}$

to convert some function ${\displaystyle H(s)}$ in the Laplace domain to a function ${\displaystyle H(z)}$ in the Z-domain (Tustin transformation), or

${\displaystyle z=e^{sT}\approx {\frac {1+sT/2}{1-sT/2}}}$

from the Z-domain to the Laplace domain. Through the bilinear transformation, the complex s-plane (of the Laplace transform) is mapped to the complex z-plane (of the z-transform). While this mapping is (necessarily) nonlinear, it is useful in that it maps the entire ${\displaystyle j\omega }$ axis of the s-plane onto the unit circle in the z-plane. As such, the Fourier transform (which is the Laplace transform evaluated on the ${\displaystyle j\omega }$ axis) becomes the discrete-time Fourier transform. This assumes that the Fourier transform exists; i.e., that the ${\displaystyle j\omega }$ axis is in the region of convergence of the Laplace transform.

Starred transform

Given a one-sided Z-transform ${\displaystyle X(z)}$ of a time-sampled function, the corresponding starred transform produces a Laplace transform and restores the dependence on ${\displaystyle T}$ (the sampling parameter):

${\displaystyle {\bigg .}X^{*}(s)=X(z){\bigg |}_{\displaystyle z=e^{sT}}}$

The inverse Laplace transform is a mathematical abstraction known as an impulse-sampled function.

Transfer function

Taking the Z-transform of the above equation (using linearity and time-shifting laws) yields:

${\displaystyle Y(z)\sum _{p=0}^{N}z^{-p}\alpha _{p}=X(z)\sum _{q=0}^{M}z^{-q}\beta _{q}}$

where ${\displaystyle X(z)}$ and ${\displaystyle Y(z)}$ are the z-transform of ${\displaystyle x[n]}$ and ${\displaystyle y[n],}$ respectively. (Notation conventions typically use capitalized letters to refer to the z-transform of a signal denoted by a corresponding lower case letter, similar to the convention used for notating Laplace transforms.)

Rearranging results in the system's transfer function:

${\displaystyle H(z)={\frac {Y(z)}{X(z)}}={\frac {\sum _{q=0}^{M}z^{-q}\beta _{q}}{\sum _{p=0}^{N}z^{-p}\alpha _{p}}}={\frac {\beta _{0}+z^{-1}\beta _{1}+z^{-2}\beta _{2}+\cdots +z^{-M}\beta _{M}}{\alpha _{0}+z^{-1}\alpha _{1}+z^{-2}\alpha _{2}+\cdots +z^{-N}\alpha _{N}}}.}$

Zeros and poles

From the fundamental theorem of algebra the numerator has ${\displaystyle M}$ roots (corresponding to zeros of ${\displaystyle H}$) and the denominator has ${\displaystyle N}$ roots (corresponding to poles). Rewriting the transfer function in terms of zeros and poles

${\displaystyle H(z)={\frac {(1-q_{1}z^{-1})(1-q_{2}z^{-1})\cdots (1-q_{M}z^{-1})}{(1-p_{1}z^{-1})(1-p_{2}z^{-1})\cdots (1-p_{N}z^{-1})}},}$

where ${\displaystyle q_{k}}$ is the ${\displaystyle k^{\text{th}}}$ zero and ${\displaystyle p_{k}}$ is the ${\displaystyle k^{\text{th}}}$ pole. The zeros and poles are commonly complex and when plotted on the complex plane (z-plane) it is called the pole–zero plot.

In addition, there may also exist zeros and poles at ${\displaystyle z{=}0}$ and ${\displaystyle z{=}\infty .}$ If we take these poles and zeros as well as multiple-order zeros and poles into consideration, the number of zeros and poles are always equal.

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.

Output response

If such a system ${\displaystyle H(z)}$ is driven by a signal ${\displaystyle X(z)}$ then the output is ${\displaystyle Y(z)=H(z)X(z).}$ By performing partial fraction decomposition on ${\displaystyle Y(z)}$ and then taking the inverse Z-transform the output ${\displaystyle y[n]}$ can be found. In practice, it is often useful to fractionally decompose ${\displaystyle \textstyle {\frac {Y(z)}{z}}}$ before multiplying that quantity by ${\displaystyle z}$ to generate a form of ${\displaystyle Y(z)}$ which has terms with easily computable inverse Z-transforms.