Bilateral Z-Transform¶
The Z-transform is the time-discrete equivalent of the Laplace transform: It extends the discrete Fourier transform to also analyse dynamic behavior.
For a time series $x[n]$, the bilateral (or two-sided) Z-transform is defined by the following equation:
$$ X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} $$
$z$ is a complex number with the radius $r$:
$$z = r \ e^{j \omega}$$
For $r=1$, the Z-transform becomes the (Discrete-Time Fourier Transform) DTFT . This is similar to the relation Laplace transform, that becomes the Fourier transform for $\sigma=0$:
$$\begin{eqnarray} X(e^{j \omega}) & = & \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n} \\ \end{eqnarray}$$
For all $r\neq1$, the Z-transform is:
$$\begin{eqnarray} X(z) = X(r e^{j \omega}) & = & \sum_{n=-\infty}^{\infty} x[n] \left(r e^{j\omega}\right)^{-n} \\ & = & \boxed{\sum_{n=-\infty}^{\infty} x[n] r^{-n} e^{-j\omega n}} \\ \end{eqnarray}$$
Similar to the Laplace transform, the last equation shows two separate terms:
- The decay-term $r^{-n}$
- The frequency-term $e^{-j\omega n}$
Visualisation¶
The following plots visualize the influence of the decay term and the frequency term for the z-transform.
Changing $r$¶
$r$ is the decay factor.
- For $r<1$, the transform increases with time until infinity.
- For $r>1$, the transform decreases and converges towards $0$.
- For $r=1$ we see the sine and cosine of the DTFT.
Changing $\omega$¶
The following plots show different values for $\omega$ with a fixed decay variable $r$. This means different frequencies are compared with the same decay pattern.
The Z-Plane¶
The Z-plane is the discrete time equivalent to the S-plane for the Laplace transform. Applying Euler, $z$ can be expressed as a combination of real and imaginary part:
$$ z = r e^{j\omega} = r \cos \varphi + r j \sin \varphi $$
The frequency $\omega$ is the angle of the complex number $z$, with the radius $r$. A radius of $r=|z|=1$ marks the unit circle.
Comparison to the S Plane¶
The z-plane is the discrete equivalent of the s-plane. More precisely, the discrete z-plane it is a bended version of the continuos case:
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The imaginary axis of the s-plane corresponds to the unit circle.
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Vertical lines in the s-plane correspond to circles around the origin (0,0) in the z-plane.
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Vertical lines on the left half of the s-plane are circles around the origin inside the unit circle.
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For the Laplace transform, the dashed line shows the Fourier transform with $\sigma = 0$.
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For the Z-transform, the dashed line shows the (Discrete-Time Fourier Transform) DTFT with $r = 1$.
Region of Convergence¶
The Region of Convergence (ROC) of a z-transform inlcudes all $z$ for that $X(z)$ converges:
$$ \sum_{n=−\infty}^{\infty} |x[n]r^{-n}| < ∞ $$
- If x[n] is a finite duration sequence, then the ROC is the entire z-plane: IT MUST CONVERGE
- The ROC cannot contain any poles (that will make more sense later).
Causality¶
We can observe the causality by observing the left- and right-sided part of a transform:
$$\begin{eqnarray} X(z) & = & N(z) + P(z) \\ &=& \sum_{n=-\infty}^{-1} x[n] \left(r e^{j\omega}\right)^{-n} + \sum_{0}^{\infty} x[n] \left(r e^{j\omega}\right)^{-n} \end{eqnarray}$$
Causal Signal¶
A causal signal is a right sided sequence - it consists only of $P(Z)$.
The ROC of causal signals lies outside a radius, including $|z|=\infty$.
An example for a causal function is the unit step function function:
$$ u[n] = \begin{cases} 1, n\geq 0 \\ 0, n < 0 \\ \end{cases} $$
$$\begin{eqnarray} U(z) & = & \sum_{n=0}^{\infty} u[n] z^{-n} \\ & = & \sum_{n=0}^{\infty} z^{-n} \\ & = & \sum_{n=0}^{\infty} \left( z^{-1} \right) ^n \\ \end{eqnarray}$$
This can be solved with a geomeric series:
$$\begin{eqnarray} U(z) & = & \frac{1}{1-z^{-1}} \\ U(z) & = & \frac{1}{z-1} \\ \end{eqnarray}$$
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The series converges for all $|𝑧| > 1$.
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We can also see that the denominator has a root at $z = 1$ - this is referred to as a pole.
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The ROC of the Z-transform of the unit step sequence expands from this pole towards infinity.
Anti-Causal Signal¶
An anticausal signal is a left sided sequence - it consists only of $N(Z)$.
The ROC of such signals lies inside a radius, including $|z|=0$.
An example is the following signal:
$$\begin{eqnarray} x[n] = -a^n u[-n-1] \end{eqnarray}$$
The z-transform is:
$$\begin{eqnarray} X(z) & = & \sum_{n=-\infty}^{-1}-a^n z^{-n} \\ & = & \sum_{n=-\infty}^{-1}-a^n z^{-n} \\ & = & - \sum_{n=1}^{\infty}a^{-n} z^{n} \\ & = & 1 - \sum_{n=0}^{\infty} \left( a^{-1} z \right)^{n} \\ &=&1-\frac{1}{1-a^{-1}z} \\ &=&1\frac{1}{1-az^{-1}} \\ \end{eqnarray}$$
To have the series converge, we need to have:
$$\begin{eqnarray} |a^{-1}z| & < & 1 \\ |z| & < & |a| \\ \end{eqnarray}$$
- There is a pole at $|z| = |a|$.
- The ROC extends from this pole inwards.
Mixed Causality¶
A mixed-causal signal is defined for positive and negative $n$.
$$ x[n] = a^{|n|} $$
The ROC of such signals is a ring:
