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📡Bioengineering Signals and Systems Unit 8 Review

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8.3 Inverse Z-transform

8.3 Inverse Z-transform

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
📡Bioengineering Signals and Systems
Unit & Topic Study Guides

The inverse Z-transform is a crucial tool in digital signal processing. It allows us to convert Z-domain representations back into discrete-time signals, enabling analysis and manipulation of digital systems.

This topic covers methods for finding the inverse Z-transform, including partial fraction expansion, power series, and residue methods. These techniques help break down complex Z-domain expressions into simpler terms for easier conversion to time-domain signals.

Inverse Z-Transform

Concept of inverse Z-transform

  • Converts Z-domain representation X(z)X(z) back to discrete-time signal x[n]x[n]
    • Reverse process of Z-transform which converts discrete-time signal to Z-domain representation
  • Essential for analyzing and manipulating signals in Z-domain and converting results back to discrete-time domain
    • Allows design and analysis of discrete-time systems (digital filters, control systems)

Partial fraction expansion for Z-transforms

  • Technique used to decompose rational Z-transform expressions into sum of simpler terms
    • Rational Z-transform expression is ratio of two polynomials in zz: X(z)=P(z)Q(z)X(z) = \frac{P(z)}{Q(z)}
  • Process involves following steps:
    1. Factorize denominator polynomial Q(z)Q(z) into product of linear and irreducible quadratic factors
    2. Determine form of partial fraction expansion based on factors of Q(z)Q(z)
      • Distinct linear factors: A1zp1+A2zp2++Anzpn\frac{A_1}{z-p_1} + \frac{A_2}{z-p_2} + \cdots + \frac{A_n}{z-p_n}
      • Repeated linear factors: A1(zp1)1+A2(zp1)2++Am(zp1)m\frac{A_1}{(z-p_1)^1} + \frac{A_2}{(z-p_1)^2} + \cdots + \frac{A_m}{(z-p_1)^m}
      • Irreducible quadratic factors: B1z+C1(zα1)(zβ1)++Bkz+Ck(zαk)(zβk)\frac{B_1z+C_1}{(z-\alpha_1)(z-\beta_1)} + \cdots + \frac{B_kz+C_k}{(z-\alpha_k)(z-\beta_k)}
    3. Solve for unknown coefficients (AiA_i, BiB_i, CiC_i) by equating original expression with partial fraction expansion and comparing coefficients or evaluating at specific points
  • Simplifies process of finding inverse Z-transform by breaking down expression into easier to handle terms
Concept of inverse Z-transform, Frontiers | TSI-SD: A time-sequence-involved space discretization neural network for passive ...

Methods for Finding the Inverse Z-Transform

Power series method for inverse Z-transform

  • Used to find inverse Z-transform of simple Z-domain expressions (single term, geometric series)
  • For Z-domain expression of form X(z)=azkX(z) = az^{-k}, inverse Z-transform is:
    • x[n]=aδ[nk]x[n] = a\delta[n-k], where δ[n]\delta[n] is unit impulse function
  • For Z-domain expression representing geometric series, such as X(z)=a1bz1X(z) = \frac{a}{1-bz^{-1}}, inverse Z-transform is:
    • x[n]=abnu[n]x[n] = ab^nu[n], where u[n]u[n] is unit step function
  • Involves expanding Z-domain expression into infinite series and identifying corresponding discrete-time signal based on series coefficients

Residue method for inverse Z-transform

  • Powerful technique for finding inverse Z-transform of rational Z-domain expressions
  • After applying partial fraction expansion to decompose rational Z-transform expression into simpler terms, residue method can be used to find inverse Z-transform of each term
  • For simple pole at z=piz = p_i, residue is:
    • Res[X(z)]z=pi=limzpi(zpi)X(z)\text{Res}[X(z)]_{z=p_i} = \lim_{z \to p_i} (z-p_i)X(z)
  • Inverse Z-transform of simple pole term Aizpi\frac{A_i}{z-p_i} is:
    • xi[n]=Aipinu[n]x_i[n] = A_ip_i^nu[n]
  • For pole of order mm at z=piz = p_i, residue is:
    • Res[X(z)]z=pi=1(m1)!limzpidm1dzm1[(zpi)mX(z)]\text{Res}[X(z)]_{z=p_i} = \frac{1}{(m-1)!} \lim_{z \to p_i} \frac{d^{m-1}}{dz^{m-1}} [(z-p_i)^mX(z)]
  • Inverse Z-transform of repeated pole term Aj(zpi)j\frac{A_j}{(z-p_i)^j} is:
    • xij[n]=Ajnj1(j1)!pinu[n]x_{ij}[n] = A_j\frac{n^{j-1}}{(j-1)!}p_i^nu[n]
  • Total inverse Z-transform is sum of inverse Z-transforms of all individual terms obtained from partial fraction expansion
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