Z-Transform Calculator | Transform Pairs, Properties & Inverse Z-Transform
Look up Z-transform pairs from a comprehensive table covering unit impulse, unit step, ramp, exponential, sine, cosine, and more. Apply linearity, time-shift, convolution, and differentiation properties. Compute the inverse Z-transform for causal rational polynomial fractions.
Z-Transform Pair Table
| Sequence x[n] | Z-Transform X(z) | ROC |
|---|---|---|
| δ[n] (unit impulse) | 1 | All z |
| u[n] (unit step) | z / (z − 1) | |z| > 1 |
| n·u[n] (ramp) | z / (z − 1)² | |z| > 1 |
| n²·u[n] | z(z + 1) / (z − 1)³ | |z| > 1 |
| n³·u[n] | z(z² + 4z + 1) / (z − 1)⁴ | |z| > 1 |
| aⁿ·u[n] | z / (z − a) | |z| > |a| |
| n·aⁿ·u[n] | az / (z − a)² | |z| > |a| |
| n²·aⁿ·u[n] | az(z + a) / (z − a)³ | |z| > |a| |
| −aⁿ·u[−n−1] | z / (z − a) | |z| < |a| |
| cos(ω₀n)·u[n] | z(z − cos ω₀) / (z² − 2z·cos ω₀ + 1) | |z| > 1 |
| sin(ω₀n)·u[n] | z·sin ω₀ / (z² − 2z·cos ω₀ + 1) | |z| > 1 |
| aⁿ·cos(ω₀n)·u[n] | z(z − a·cos ω₀) / (z² − 2az·cos ω₀ + a²) | |z| > |a| |
| aⁿ·sin(ω₀n)·u[n] | az·sin ω₀ / (z² − 2az·cos ω₀ + a²) | |z| > |a| |
| x[n − k] (time shift) | z^(−k)·X(z) | ROC of X(z) |
| nCr(n,k)·u[n] | z^k / (z − 1)^(k+1) | |z| > 1 |
| δ[n − k] | z^(−k) | All z ≠ 0 |
| u[n] − u[n − N] | (1 − z^(−N)) / (1 − z^(−1)) | |z| > 0 |
| rⁿ·u[n − k] | z^(−k)·z / (z − r) | |z| > |r| |
| e^(−an)·u[n] | z / (z − e^(−a)) | |z| > e^(−a) |
| 1/n! (causal) | e^(1/z) | |z| > 0 |
| aⁿ/n! (causal) | e^(a/z) | |z| > 0 |
| x[−n] (time reversal) | X(1/z) | 1/ROC |
What Is the Z-Transform Calculator | Transform Pairs, Properties & Inverse Z-Transform?
The Z-transform converts a discrete-time sequence x[n] into a complex-frequency domain representation X(z), analogous to how the Laplace transform works for continuous-time signals. The Region of Convergence (ROC) specifies which values of z make the series converge. For causal signals (x[n] = 0 for n < 0), the ROC is the exterior of a circle in the complex z-plane. Z-transforms are foundational in digital signal processing, discrete control systems, and difference equation analysis. The inverse Z-transform via partial fractions decomposes a rational X(z) into simpler terms whose inverses can be read from standard tables.
Formula
X(z) = Σₙ₌₋∞^∞ x[n]·z^(−n) | Inverse: x[n] = (1/2πj) ∮ X(z)·z^(n−1) dz
How to Use
- 1
Open the "Transform Table" tab and search for a sequence by keyword (e.g. "step", "cos", "ramp") or filter by category.
- 2
Read the Z-transform X(z) and Region of Convergence (ROC) for your sequence from the table.
- 3
Switch to "Properties" to see how linearity, time-shift, scaling, convolution, and time-reversal modify X(z).
- 4
For the inverse Z-transform, switch to "Inverse Z-Transform" and enter numerator [n₀, n₁] and denominator [d₀, d₁, d₂] coefficients.
- 5
Click a preset (e.g. "z/(z−0.5)(z−0.25)") or enter your own coefficients — the denominator must be degree 2 and numerator degree 1 or lower.
- 6
Click "Compute Inverse" to see the pole locations, residues, and the causal inverse x[n] = Σ Aᵢ·pᵢⁿ·u[n].
- 7
Read the x[n] sequence values for n = 0 through 7 displayed as individual tiles.
Use the three tabs: browse the searchable transform pair table, read property formulas, or compute the inverse Z-transform via partial fractions for a rational X(z).
Example Calculation
X(z) = z / (z² − 0.75z + 0.125) = z / (z − 0.5)(z − 0.25). Numerator: [0, 1] (i.e. z). Denominator: [0.125, −0.75, 1]. Poles: p₁ = 0.5, p₂ = 0.25. Residue A₁ = (0 + 1·0.5) / (1·(0.5 − 0.25)) = 0.5/0.25 = 2. Residue A₂ = (0 + 1·0.25) / (1·(0.25 − 0.5)) = 0.25/(−0.25) = −1. x[n] = 2·(0.5)ⁿ·u[n] − 1·(0.25)ⁿ·u[n]. Check: x[0] = 2−1 = 1, x[1] = 1 − 0.25 = 0.75.
Understanding Z-Transform | Transform Pairs, Properties & Inverse Z-Transform
Z-Transform Pairs Quick Reference
| Sequence x[n] | Z-Transform X(z) | ROC | Notes |
|---|---|---|---|
| δ[n] | 1 | All z | Unit impulse |
| u[n] | z/(z−1) | |z|>1 | Unit step |
| n·u[n] | z/(z−1)² | |z|>1 | Unit ramp |
| aⁿ·u[n] | z/(z−a) | |z|>|a| | Causal exponential |
| n·aⁿ·u[n] | az/(z−a)² | |z|>|a| | Damped ramp |
| cos(ω₀n)·u[n] | z(z−cosω₀)/(z²−2z·cosω₀+1) | |z|>1 | Causal cosine |
| sin(ω₀n)·u[n] | z·sinω₀/(z²−2z·cosω₀+1) | |z|>1 | Causal sine |
| aⁿcos(ω₀n)·u[n] | z(z−a·cosω₀)/(z²−2az·cosω₀+a²) | |z|>|a| | Damped cosine |
| δ[n−k] | z^(−k) | |z|>0 | Shifted impulse |
| u[n]−u[n−N] | (1−z^(−N))/(1−z^(−1)) | |z|>0 | Rectangular window |
Z-Transform vs. Laplace Transform Correspondence
| Continuous (Laplace) | Discrete (Z-Transform) | Relationship |
|---|---|---|
| F(s) | X(z) | z = e^(sT) for sampling period T |
| Unit impulse δ(t) → 1 | δ[n] → 1 | Both transforms of impulse = 1 |
| Unit step 1/s | z/(z−1) | Poles at s=0 and z=1 |
| Exponential e^(−at) → 1/(s+a) | aⁿu[n] → z/(z−a) | a = e^(−αT) discretization |
| Left-half plane stable |Re(s)|<0 | Unit disc stable |z|<1 | Stability region maps by z=e^(sT) |
| Imaginary axis jω (Fourier) | Unit circle z=e^(jω) (DTFT) | Stable boundary |
| Integration ÷ s | Accumulation ÷ (1−z^(−1)) | Discrete integration |
Z-Transform in Digital Signal Processing
- ›Transfer function: For a digital filter described by Σ bₖ·x[n−k] = Σ aₖ·y[n−k], the transfer function H(z) = B(z)/A(z) where B and A are polynomials in z^(−1). Poles of H(z) determine stability; zeros determine frequency nulls.
- ›FIR filters: Finite Impulse Response filters have all poles at z=0 (always stable). Their Z-transform is a polynomial in z^(−1).
- ›IIR filters: Infinite Impulse Response filters have poles away from z=0. Butterworth, Chebyshev, and elliptic digital filters are designed by mapping analog prototypes via the bilinear transform z = (2/T)(1−z^(−1))/(1+z^(−1)).
- ›Stability test: All poles inside the unit circle |pᵢ| < 1 is required for BIBO stability. The Jury stability test is the discrete analogue of the Routh-Hurwitz criterion.
- ›Frequency response: H(e^(jω)) = H(z)|_{z=e^(jω)} gives the magnitude and phase response of a digital filter as a function of normalized frequency ω ∈ [0, π].
Frequently Asked Questions
What is the Region of Convergence (ROC)?
The ROC is the set of complex values z for which the Z-transform sum converges absolutely. For causal sequences (starting at n=0), the ROC is |z| > r₀ (outside a circle). For anti-causal sequences, the ROC is |z| < r₁ (inside a circle). For two-sided sequences, the ROC is an annulus. The ROC determines the type of system (stable, causal, anti-causal) and must be specified along with X(z) for uniqueness.
How does the Z-transform relate to the Discrete Fourier Transform (DFT)?
The DFT is the Z-transform evaluated on the unit circle: X(e^(jω)) = X(z)|_{z=e^(jω)}. This is only valid when the unit circle is in the ROC, i.e. the sequence is absolutely summable. The Z-transform generalizes the DFT to arbitrary complex z and is the discrete-time analogue of the Laplace transform.
When can partial fractions be used for the inverse Z-transform?
Partial fractions work when X(z) is a rational function N(z)/D(z) with degree(N) < degree(D). You expand X(z)/z = Σ Aᵢ/(z − pᵢ) for simple poles, multiply by z to get X(z) = Σ Aᵢz/(z − pᵢ), then use the standard pair aⁿ·u[n] ↔ z/(z−a). For repeated poles, include terms Aᵢₖ·z/(z−pᵢ)ᵏ.
What does the time-shift property Z{x[n-k]} = z^(-k)·X(z) mean in practice?
A delay of k samples in the time domain corresponds to multiplication by z^(−k) in the Z-domain. This makes Z-transforms ideal for analyzing difference equations y[n] = Σ aₖ·y[n−k] + Σ bₖ·x[n−k]: each delay operator becomes z^(−k), turning the difference equation into an algebraic equation in z. The transfer function H(z) = Y(z)/X(z) is then read directly.
What is a stable discrete-time system in terms of the Z-transform?
A causal LTI system with transfer function H(z) is BIBO stable (Bounded Input, Bounded Output) if and only if all poles of H(z) lie strictly inside the unit circle: |pᵢ| < 1 for all poles pᵢ. If any pole is outside or on the unit circle, the system is unstable (or marginally stable on the boundary). This is the discrete-time analogue of left-half-plane stability for continuous systems.
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