InverseZTransform[expr,z,n]
gives the inverse Z transform of expr.
InverseZTransform[expr,{z1,…,zm},{n1,…,nm}]
gives the multiple inverse Z transform of expr.
InverseZTransform
InverseZTransform[expr,z,n]
gives the inverse Z transform of expr.
InverseZTransform[expr,{z1,…,zm},{n1,…,nm}]
gives the multiple inverse Z transform of expr.
Details and Options
- The inverse Z transform of a function
is given by the contour integral
. - The multidimensional inverse Z transform is given by
. - The following options can be given:
-
Assumptions $Assumptions assumptions to make about parameters Method Automatic method to use - In TraditionalForm, InverseZTransform is output using
.
Examples
open all close allBasic Examples (2)
Univariate inverse transforms:
InverseZTransform[z / (z - a), z, n]InverseZTransform[z / (z ^ 2 - 3z + 1), z, n]Multivariate inverse transforms:
InverseZTransform[(z1 z2/(z1 - 1)^2(z2 - 1)^2), {z1, z2}, {n1, n2}]InverseZTransform[E ^ (1 / z1) / z2, {z1, z2}, {n1, n2}]Scope (4)
Constants lead to impulse sequences:
InverseZTransform[1, z, n]InverseZTransform[z ^ -3, z, n]Rational transforms yield exponential and trigonometric sequences:
InverseZTransform[z / (z - a), z, n]InverseZTransform[z / (z - a) ^ 2, z, n]InverseZTransform[1 / ((-3 + z) (-2 + z) (-1 + z) z), z, n]In some cases, additional simplification and transformations are needed:
InverseZTransform[z Sin[w] / (z ^ 2 - 2z Cos[w] + 1), z, n]FullSimplify[%, Element[w, Reals] && Element[n, Integers]]InverseZTransform[(E^I z (4 - 8 E^I z + z^2 + E^2 I (4 + z^2))/(2 z + 2 E^2 I z - E^I (4 + z^2))^2), z, n]//ExpToTrig//FullSimplifyInverseZTransform[E ^ (1 / z), z, n]InverseZTransform[Cos[Sqrt[1 / z]], z, n]InverseZTransform[Log[z / (z + 1)], z, n]InverseZTransform[Sqrt[1 + z ^ -2], z, n]InverseZTransform[PolyLog[-k, c / z], z, n]InverseZTransform[z ^ 3 BesselI[3, 2 / z], z, n]Options (1)
Assumptions (1)
This transform will not evaluate without any constraints on the range of p:
InverseZTransform[z / (z - a) ^ p, z, n]Use Assumptions to limit the range of p:
InverseZTransform[z / (z - a) ^ p, z, n, Assumptions -> p∈Integers && p > 0]Applications (3)
Solve a linear difference equation:
ZTransform[y[n + 1] + 2y[n] == 1, n, z]Add an initial value equation and solve the algebraic equation for the transform:
Solve[% /. {y[0] -> 3}, ZTransform[y[n], n, z]]Get the solution through inverse transformation:
InverseZTransform[ZTransform[y[n], n, z] /. First[%], z, n]Use RSolve:
RSolve[{y[n + 1] + 2y[n] == 1, y[0] == 3}, y[n], n]Solve a linear difference-summation equation:
ZTransform[y[n + 1] == 2y[n] - Sum[2 ^ (n - r)y[r], {r, 0, n}], n, z]Solve[% /. {y[0] -> 1}, ZTransform[y[n], n, z]]Use the inverse transform to get a solution to the original problem:
InverseZTransform[ZTransform[y[n], n, z] /. First[%], z, n]Use RSolve:
RSolve[{y[n + 1] == 2y[n] - Sum[2 ^ (n - r)y[r], {r, 0, n}], y[0] == 1}, y[n], n]A discrete system transfer function:
h = 1 / ((z - 1 / 2)(z - 2))u = ZTransform[DiscreteDelta[n], n, z]InverseZTransform[h u, z, n]u = ZTransform[UnitStep[n], n, z]InverseZTransform[h u, z, n]u = ZTransform[UnitStep[n]n, n, z]InverseZTransform[h u, z, n]Properties & Relations (6)
Use DiscreteAsymptotic to compute an asymptotic approximation:
DiscreteAsymptotic[Inactive[InverseZTransform][1 / (z(-6 + 6 * z ^ 2 + z ^ 3)), z, n], n -> ∞]ZTransform is the inverse operator:
ZTransform[InverseZTransform[F[z], z, n], n, z]InverseZTransform[ZTransform[f[n], n, z], z, n]ZTransform[2 ^ n n, n, z]InverseZTransform[%, z, n]InverseZTransform[a F[z] + b G[z], z, n]InverseZTransform[z ^ (-1) ZTransform[f[n], n, z], z, n]InverseZTransform[z ^ (-2) ZTransform[f[n], n, z], z, n]InverseZTransform[ D[ZTransform[f[n], n, z], z], z, n]InverseZTransform[ D[ZTransform[f[n], n, z], {z, 2}], z, n]InverseZTransform[z / ((z - 1 / 2)(z - 1 / 3)), z, n]Limit[%, n -> 0] == Limit[z / ((z - 1 / 2)(z - 1 / 3)), z -> Infinity]InverseZTransform[z / ((z - 1 / 2)(z - 1 / 3)), z, n]Limit[%, n -> Infinity] == Limit[z / ((z - 1 / 2)(z - 1 / 3)), z -> 0]InverseZTransform is closely related to SeriesCoefficient:
InverseZTransform[E ^ (1 / z), z, n]SeriesCoefficient[E ^ z, {z, 0, n}]Tech Notes
Related Guides
Related Links
History
Introduced in 1999 (4.0) | Updated in 2008 (7.0)
Text
Wolfram Research (1999), InverseZTransform, Wolfram Language function, https://reference.wolfram.com/language/ref/InverseZTransform.html (updated 2008).
CMS
Wolfram Language. 1999. "InverseZTransform." Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2008. https://reference.wolfram.com/language/ref/InverseZTransform.html.
APA
Wolfram Language. (1999). InverseZTransform. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/InverseZTransform.html
BibTeX
@misc{reference.wolfram_2026_inverseztransform, author="Wolfram Research", title="{InverseZTransform}", year="2008", howpublished="\url{https://reference.wolfram.com/language/ref/InverseZTransform.html}", note=[Accessed: 13-June-2026]}
BibLaTeX
@online{reference.wolfram_2026_inverseztransform, organization={Wolfram Research}, title={InverseZTransform}, year={2008}, url={https://reference.wolfram.com/language/ref/InverseZTransform.html}, note=[Accessed: 13-June-2026]}