z-transform formulas

Defined only for $z$ in the ROC, the set of $z$ where the sum converges absolutely. Always quote $X(z)$ together with its ROC.

Ignores $x[n]$ for $n<0$; it equals the bilateral transform of $x[n]u[n]$. When it exists, its ROC is the exterior of a circle, $|z|>r$ (including $z=\infty$), so it is usually not written. For a causal sequence the two transforms are identical.

Valid only when the ROC contains the unit circle $|z|=1$. $X(e^{j\Omega })$ is periodic in $\Omega$ with period $2\pi$.

With $a=1$: $u[n]\leftrightarrow \frac{z}{z-1}$, $|z|>1$. With $a=0$ (limit): $\delta [n]\leftrightarrow 1$ for all $z$.

Same algebraic transform as $a^{n}u[n]$; only the ROC differs. Note the minus sign and the $u[-n-1]$ (it starts at $n=-1$).

Follows from differentiation in $z$. Useful variants: $nu[n]\leftrightarrow \frac{z}{(z-1{)}^2}$ and $(n+1)a^{n}u[n]\leftrightarrow \frac{z^2}{(z-a{)}^2}$, both right-sided.

Here $r>0$ and ${\Omega }_0$ are real. ROC $|z|>r$; poles at $r{e}^{\pm j{\Omega }_0}$. Put $r=1$ for the undamped $\cos ({\Omega }_{0}n)u[n]$ and $\sin ({\Omega }_{0}n)u[n]$, whose poles lie on the unit circle (ROC $|z|>1$).

ROC unchanged except possibly for $z=0$ or $z=\infty$. A delay of $k$ samples multiplies by $z^{-k}$. For the unilateral version with initial conditions see Section 5.

The ROC inverts: $r_1<|z|<r_2$ becomes $1/r_2<|z|<1/r_1$. A pole at $p$ moves to $1/p$.

ROC scales to $|a|r_1<|z|<|a|r_2$; every pole and zero moves from $p$ to $ap$. With $a=e^{j{\Omega }_0}$ (modulation) the pole-zero plot rotates by ${\Omega }_0$.

Same ROC. Apply it twice for $n^{2}x[n]$.

ROC contains $R_1\cap R_2$ (it can be larger if a pole cancels against a zero).

ROC contains $R\cap \{|z|>1\}$. The inverse operation, the first difference $x[n]-x[n-1]$, gives $(1-z^{-1})X(z)$.

Same ROC. For a real sequence $X(z^{\ast })=X^{\ast }(z)$, so complex poles and zeros come in conjugate pairs.

For a causal $x[n]$ ($x[n]=0$ for $n<0$), or for any $x[n]$ with the unilateral transform. The next sample is $x[1]={\lim }_{z\to \infty }z\left(X(z)-x[0]\right)$. A rational $X(z)$ whose numerator degree in $z$ exceeds the denominator degree cannot be the transform of a causal sequence, and the theorem does not apply. For an anticausal $x[n]$ ($x[n]=0$ for $n>0$) the mirror result is $x[0]={\lim }_{z\to 0}X(z)$.

For a causal $x[n]$ (or the unilateral transform of any $x[n]$), and only when the limit exists: every pole of $(z-1)X(z)$ must lie strictly inside the unit circle. $X(z)$ itself may have one simple pole at $z=1$.

Equivalent form in $z^{-1}$: $X(z)={\sum }_k\frac{A_k}{1-p_k{z}^{-1}}$ with $A_k=(1-p_k{z}^{-1})X(z){|}_{z=p_k}$. That form needs the numerator degree in $z^{-1}$ to be less than the denominator degree; otherwise divide out the polynomial part first (it gives impulses).

$C$ is a counter-clockwise circle centred on the origin and lying in the ROC. For a causal $x[n]$ it encloses every pole. For small $n$, check whether $z^{n-1}$ adds a pole at $z=0$ (for example at $n=0$ when $X(z)$ has no zero at the origin) and include its residue.

For a simple pole ($m=1$) this is just $(z-p)X(z)z^{n-1}$ evaluated at $z=p$.

Take the transform term by term with $x[n-k]\leftrightarrow z^{-k}X(z)$ and zero initial conditions. Sign convention: every output term is on the left. If the equation is written $y[n]=0.5y[n-1]+x[n]$, the feedback coefficient changes sign when moved across: $H(z)=\frac{1}{1-0.5z^{-1}}$.

Exists when the ROC contains the unit circle (a stable system). $\Omega =0$ is the point $z=1$ (DC) and $\Omega =\pi$ is $z=-1$ (the highest digital frequency). $H(e^{j\Omega })$ has period $2\pi$.

For a stable LTI system with real $h[n]$. It is exact for an input present for all $n$; for an input switched on at $n=0$ it is the response once the transients have died out. A constant input $A$ gives the output $AH(1)$.

The cases used most: $x[n-1]\leftrightarrow z^{-1}{X}^{+}(z)+x[-1]$ and $x[n-2]\leftrightarrow z^{-2}{X}^{+}(z)+z^{-1}x[-1]+x[-2]$.

For $k=1$: $x[n+1]\leftrightarrow z{X}^{+}(z)-zx[0]$. For $k=2$: $z^2{X}^{+}(z)-z^{2}x[0]-zx[1]$.

$T$ is the sampling period. The digital frequency is $\Omega =\omega T=2\pi f/f_s$ rad/sample. A continuous-time pole at $s=p$ appears at $z=e^{pT}$ in the transform of the sampled signal.

For real $a$. The pole $s=-a$ maps to $z=e^{-aT}$, inside the unit circle when $a>0$.

Exact at the sampling instants: $G(z)$ is $(1-z^{-1})$ times the transform of the sampled step response of the plant.

For $a\ne 0$. From $\frac{K}{s(s+a)}=\frac{K}{a}\left(\frac{1}{s}-\frac{1}{s+a}\right)$. Without the hold, $\mathcal{Z}\{G_p(s)\}=\frac{Kz}{z-e^{-aT}}$, a different answer.

When a sampler separates $G_1(s)$ and $G_2(s)$, their pulse transfer functions multiply. When none does, multiply in $s$ first and then transform; in general $G_1{G}_2(z)\ne G_1(z)G_2(z)$. Example with $G_1(s)=G_2(s)=\frac{1}{s+1}$ and $T=1$: with a sampler between them, $\frac{z^2}{(z-e^{-1}{)}^2}$; without, $\mathcal{Z}\left\{\frac{1}{(s+1{)}^2}\right\}=\frac{e^{-1}z}{(z-e^{-1}{)}^2}$.

$G(s)$ is the forward path after the sampler (hold included) and $H(s)$ the continuous feedback path. Because $G(s)$ and $H(s)$ are not separated by a sampler, they are multiplied before transforming. The loop is stable when every root of $1+GH(z)=0$ lies inside the unit circle.

Unity feedback, the sampler on the error, $G(z)$ the pulse transfer function of the forward path (hold included), and a stable closed loop (otherwise the final value theorem does not apply). From $E(z)=\frac{R(z)}{1+G(z)}$ and the final value theorem, with $R(z)=\frac{z}{z-1}$ for a step and $R(z)=\frac{Tz}{(z-1{)}^2}$ for the sampled ramp. The constants are often written $K_p={\lim }_{z\to 1}G(z)$ and $K_v=\frac{1}{T}{\lim }_{z\to 1}(z-1)G(z)$, so that $e_{ss}=\frac{1}{1+K_p}$ for a step and $\frac{1}{K_v}$ for a ramp. Each pole of $G(z)$ at $z=1$ plays the part of an integrator.

For real coefficients, all three conditions together are necessary and sufficient for both roots to lie strictly inside the unit circle (the Jury conditions for order 2). For $a_2{z}^2+a_{1}z+a_0$ with $a_2>0$ the first condition becomes $|a_0|<a_2$. At any order $n$, with a positive leading coefficient, $P(1)>0$ and $(-1{)}^{n}P(-1)>0$ are necessary. For a full test at higher orders, use the Jury table, or substitute $z=\frac{1+w}{1-w}$ and apply the Routh test to the polynomial in $w$.