DSP | Theory | Math | Z-Transform
Calculators Theory Z-Domain Filters
Quantization & Sampling Theorem Math Linear Systems Convolution Fourier Transform
Z-Transform Bilinear Transform
← Back to Digital Signal Processing
Terms
Common terms in z-tramsforms;
Causal
Causal is a sequence (denoted in curly braces) where negative values are \(0\). Thus;
\[f_{n} = \begin{cases} x & n \geq 0 \\ 0 & n < 0 \end{cases}\]Unit Impulse
Denoted by \(\delta_{n}\) and defined as;
\[\delta_{n} = \begin{cases} 1 & n = 0 \\ 0 & n \neq 0 \end{cases}\]Unit Step
Denoted by \(u_{n}\) and defined as;
\[u_{n} = \begin{cases} 1 & n \geq 0 \\ 0 & n < 0 \end{cases}\]Z-Transform Forms
Understanding the Z-Transform
Causal Sequence
For any arbitraty constant \(a\), the z-transform of the causal sequence;
\[f_{n} = \begin{cases} 0 & n = -1, -2, -3, ... \\ a^{n} & n = 0, 1, 2, 3, ... \end{cases}\]is, by definition;
\[F(z) = Z\{f_{n}\} = 1 + az^{-1} + a^{2}z^{-2} + ...\]which is a geometric series with common ratio \(az^{-1}\). Hence, provided \(|az^{-1}| < 1\), the closed form of the z-transform is;
\[F(z) = \frac{1}{1-az^{-1}} = \frac{z}{z-a}\]Examples:
\[\displaylines{ Z\{2^{n}\} = \frac{1}{1-2z^{-1}} = \frac{z}{z-2} & |z| > 2 \\ Z\{(-1)^{n}\} = \frac{1}{1+z^{-1}} = \frac{z}{z+1} & |z| > 1 \\ Z\{e^{-n}\} = \frac{z}{z-e^{-1}} & |z| > e^{-1} \\ Z\{e^{-\alpha n}\} = \frac{z}{z-e^{-\alpha}} & |z| > e^{-\alpha} }\]Multiplication of a Sequence by \(a^{n}\)
Suppose \(f_{n}\) is an arbitrary sequence with z-transform \(F(z)\). Multiplying with \{a_{n}\} would give;
\[\displaylines{ f_{0} + af_{1}z^{-1} + a^{2}f_{2}z^{-2} + ... \\ = \sum_{n=0}^{\infty} a^{n}f_{n}z^{-n} \\ = \sum_{n=0}^{\infty} f_{n}\left(\frac{z}{a}\right)^{-n} }\]That is, multiplying a sequence \({f_{n}}\) by the sequence \({a^{n}}\) does not change the form of the z-transform \(F(z)\). We merely replace \(z\) by \(\frac{z}{a}\) in that transform.
Shifting
by \(a^{n}\)
Delays are essentially the multiplication of \(z\) by \(z^{-1}\). Thus;
\[\displaylines{ Z\{a^{n-1}\} = \frac{zz^{-1}}{z-a} = \frac{1}{z-a} \\ Z\{a^{n-2}\} = \frac{z^{-1}}{z-a} = \frac{1}{z(z-a)} }\]Note the power shift affects the denominator
by \(u_{n}\)
\[\displaylines{ Z\{u_{n-5}\} = \frac{zz^{-5}}{z-1} = \frac{z^{-4}}{z-1} \\ Z\{u_{n} - u_{n-5}\} = \frac{z}{z-1} - \frac{zz^{-5}}{z-1} = \frac{z-z^{-4}}{z-1} \\ Z\{u_{n} - u_{n-1}\} = \frac{z}{z-1} - \frac{zz^{-1}}{z-1} = \frac{z-1}{z-1} = 1 }\]Note the sub-shift affects the numerator