Signals and Systems

10.3 The Inverse Z-Transform

Imron Rosyadi

Formal Inverse Z-Transform

Derivation from Inverse Fourier Transform:

Recall from Section 10.1 that \(X(r e^{j\omega}) = \mathfrak{F}\{x[n]r^{-n}\}\).

Applying the inverse Fourier transform: \[ x[n]r^{-n} = \frac{1}{2\pi} \int_{2\pi} X(r e^{j\omega}) e^{j\omega n} d\omega \] Multiplying by \(r^n\): \[ x[n] = \frac{1}{2\pi} \int_{2\pi} X(r e^{j\omega}) (r e^{j\omega})^n d\omega \] Let \(z = r e^{j\omega}\). Then \(dz = j r e^{j\omega} d\omega = jz d\omega\), so \(d\omega = \frac{1}{jz} dz\).

The integration over \(2\pi\) for \(\omega\) corresponds to a counterclockwise closed circular contour integration for \(z\) around the origin.

Formal Inverse Z-Transform

The Inverse Z-Transform Integral: \[ x[n]=\frac{1}{2 \pi j} \oint X(z) z^{n-1} d z \]

• The contour integral is taken around a counterclockwise closed circular path \(|z|=r\) within the ROC.
• This is the formal definition, often evaluated using the Cauchy’s Residue Theorem from complex analysis.

Formal Inverse Z-Transform

Practical Approaches for Inverse Z-Transform:

While the contour integral is the formal definition, it’s rarely used in practice for typical engineering problems. We primarily use two main methods:

1. Partial-Fraction Expansion:
   • Decompose \(X(z)\) into simpler terms.
   • Use known Z-transform pairs and ROC properties to find the inverse of each term.
   • Sum the inverse transforms.
2. Power Series Expansion (Long Division):
   • Expand \(X(z)\) into a power series of \(z\) or \(z^{-1}\).
   • The coefficients of the series directly correspond to the values of \(x[n]\).

Important

The Region of Convergence (ROC) is absolutely critical for determining the unique \(x[n]\) from \(X(z)\)!

The formal inverse Z-transform involves a complex contour integral. While mathematically rigorous, it’s not the most intuitive or direct method for finding \(x[n]\) in engineering contexts. Instead, we rely on more practical methods.

The two main methods we’ll explore are partial-fraction expansion, which you might recall from Laplace transforms, and power series expansion, which directly leverages the definition of the Z-transform. Remember, the ROC is paramount in both these methods because it disambiguates the time-domain signal.

Method 1: Partial-Fraction Expansion

Procedure:

1. Express \(X(z)\) as a rational function: \(X(z) = \frac{Numerator(z)}{Denominator(z)}\)
2. Multiply by \(z\) (optional but often convenient): Consider \(X(z)/z\) for partial fraction expansion, then multiply back. This avoids constant terms for \(z^{-1}\) forms.
3. Perform Partial-Fraction Expansion: Decompose \(X(z)\) (or \(X(z)/z\)) into a sum of simpler terms, typically of the form \(\frac{A}{1-az^{-1}}\) or \(\frac{A}{z-a}\).
4. Determine Inverse Transform of Each Term:
   • For each term, use known Z-transform pairs.
   • Crucially, use the overall ROC of \(X(z)\) to determine the ROC for each individual term. This dictates whether the inverse is right-sided (\(a^n u[n]\)) or left-sided (\(-a^n u[-n-1]\)).
5. Sum the Inverse Transforms: The original signal \(x[n]\) is the sum of the inverse transforms of the individual terms.

Method 1: Partial-Fraction Expansion

Common Z-Transform Pairs for PFE:

• Right-sided: \(a^n u[n] \stackrel{\mathcal{Z}}{\longleftrightarrow} \frac{1}{1-a z^{-1}}, \quad |z|>|a|\)
• Left-sided: \(-a^n u[-n-1] \stackrel{\mathcal{Z}}{\longleftrightarrow} \frac{1}{1-a z^{-1}}, \quad |z|<|a|\)

Tip

For terms with repeated poles or higher order poles, specific formulas are available, similar to Laplace transforms.

Partial-fraction expansion is a workhorse for finding inverse Z-transforms of rational functions. The general idea is to break down a complex fraction into a sum of simpler fractions, each of which has a known inverse Z-transform.

The critical step is to correctly assign the ROC for each individual term based on the overall ROC of \(X(z)\). This determines whether you’re dealing with a right-sided or a left-sided exponential. Without considering the ROC, the inverse transform is ambiguous.

Example 10.9: Partial-Fraction Expansion (Right-Sided)

Given \(X(z)\) and ROC: \[ X(z)=\frac{3-\frac{5}{6} z^{-1}}{\left(1-\frac{1}{4} z^{-1}\right)\left(1-\frac{1}{3} z^{-1}\right)}, \quad|z|>\frac{1}{3} \] Poles: \(z=1/4\) and \(z=1/3\). ROC: \(|z|>1/3\). This implies \(x[n]\) is a right-sided sequence.

Partial-Fraction Expansion: \[ X(z)=\frac{A}{1-\frac{1}{4} z^{-1}}+\frac{B}{1-\frac{1}{3} z^{-1}} \] Solving for \(A\) and \(B\): \(A = \left. \frac{3-\frac{5}{6} z^{-1}}{1-\frac{1}{3} z^{-1}} \right|_{1-\frac{1}{4} z^{-1}=0 \implies z^{-1}=4} = \frac{3-\frac{5}{6}(4)}{1-\frac{1}{3}(4)} = \frac{3-\frac{10}{3}}{1-\frac{4}{3}} = \frac{-\frac{1}{3}}{-\frac{1}{3}} = 1\) \(B = \left. \frac{3-\frac{5}{6} z^{-1}}{1-\frac{1}{4} z^{-1}} \right|_{1-\frac{1}{3} z^{-1}=0 \implies z^{-1}=3} = \frac{3-\frac{5}{6}(3)}{1-\frac{1}{4}(3)} = \frac{3-\frac{5}{2}}{1-\frac{3}{4}} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2\)

So, \(X(z)=\frac{1}{1-\frac{1}{4} z^{-1}}+\frac{2}{1-\frac{1}{3} z^{-1}}\)

Example 10.9: Partial-Fraction Expansion (Right-Sided)

Inverse Transform with ROC:

• For the first term: \(\frac{1}{1-\frac{1}{4} z^{-1}}\). Since the overall ROC is \(|z|>1/3\), which is also \(|z|>1/4\), this term has an ROC of \(|z|>1/4\). \(\Rightarrow x_1[n] = \left(\frac{1}{4}\right)^n u[n]\)
• For the second term: \(\frac{2}{1-\frac{1}{3} z^{-1}}\). The overall ROC is \(|z|>1/3\), so this term has an ROC of \(|z|>1/3\). \(\Rightarrow x_2[n] = 2\left(\frac{1}{3}\right)^n u[n]\)

Result: \[ x[n]=\left(\frac{1}{4}\right)^{n} u[n]+2\left(\frac{1}{3}\right)^{n} u[n] \]

Note

The ROC \(|z|>1/3\) indicates a right-sided sequence, so both individual terms are also right-sided.

This example shows a straightforward application of partial-fraction expansion. Because the overall ROC is an exterior region (\(|z| > 1/3\)), it tells us that the resulting time-domain signal must be right-sided. Consequently, both terms in the partial-fraction expansion also correspond to right-sided exponentials. This is a consistent application of the ROC properties.

Example 10.10: Partial-Fraction Expansion (Two-Sided)

Given \(X(z)\) and ROC: \[ X(z)=\frac{3-\frac{5}{6} z^{-1}}{\left(1-\frac{1}{4} z^{-1}\right)\left(1-\frac{1}{3} z^{-1}\right)}, \quad\frac{1}{4}<|z|<\frac{1}{3} \] Poles: \(z=1/4\) and \(z=1/3\). ROC: \(1/4<|z|<1/3\). This implies \(x[n]\) is a two-sided sequence.

Partial-Fraction Expansion (same as before): \[ X(z)=\frac{1}{1-\frac{1}{4} z^{-1}}+\frac{2}{1-\frac{1}{3} z^{-1}} \]

Example 10.10: Partial-Fraction Expansion (Two-Sided)

Inverse Transform with ROC:

• For the first term: \(\frac{1}{1-\frac{1}{4} z^{-1}}\). The overall ROC (\(1/4<|z|<1/3\)) implies that for this term, the ROC is \(|z|>1/4\). \(\Rightarrow x_1[n] = \left(\frac{1}{4}\right)^n u[n]\)
• For the second term: \(\frac{2}{1-\frac{1}{3} z^{-1}}\). The overall ROC (\(1/4<|z|<1/3\)) implies that for this term, the ROC is \(|z|<1/3\). \(\Rightarrow x_2[n] = -2\left(\frac{1}{3}\right)^n u[-n-1]\)

Result: \[ x[n]=\left(\frac{1}{4}\right)^{n} u[n]-2\left(\frac{1}{3}\right)^{n} u[-n-1] \]

Important

The ROC is crucial! The term associated with the pole at \(z=1/3\) is now a left-sided signal because the overall ROC is inside that pole.

Here, we use the exact same algebraic expression for \(X(z)\), but with a different ROC. This time, the ROC is an annular region (\(1/4 < |z| < 1/3\)), indicating a two-sided signal.

This means that for the pole at \(z=1/4\), the ROC is outside, so it corresponds to a right-sided exponential. But for the pole at \(z=1/3\), the ROC is inside, so it corresponds to a left-sided exponential. This clearly demonstrates how the ROC dictates the form of the inverse Z-transform.

Example 10.11: Partial-Fraction Expansion (Left-Sided)

Given \(X(z)\) and ROC: \[ X(z)=\frac{3-\frac{5}{6} z^{-1}}{\left(1-\frac{1}{4} z^{-1}\right)\left(1-\frac{1}{3} z^{-1}\right)}, \quad|z|<\frac{1}{4} \] Poles: \(z=1/4\) and \(z=1/3\). ROC: \(|z|<1/4\). This implies \(x[n]\) is a left-sided sequence.

Partial-Fraction Expansion (same as before): \[ X(z)=\frac{1}{1-\frac{1}{4} z^{-1}}+\frac{2}{1-\frac{1}{3} z^{-1}} \]

Inverse Transform with ROC:

• For the first term: \(\frac{1}{1-\frac{1}{4} z^{-1}}\). The overall ROC (\(|z|<1/4\)) implies that for this term, the ROC is \(|z|<1/4\). \(\Rightarrow x_1[n] = -\left(\frac{1}{4}\right)^n u[-n-1]\)
• For the second term: \(\frac{2}{1-\frac{1}{3} z^{-1}}\). The overall ROC (\(|z|<1/4\)) implies that for this term, the ROC is \(|z|<1/3\). \(\Rightarrow x_2[n] = -2\left(\frac{1}{3}\right)^n u[-n-1]\)

Example 10.11: Partial-Fraction Expansion (Left-Sided)

Result: \[ x[n]=-\left(\frac{1}{4}\right)^{n} u[-n-1]-2\left(\frac{1}{3}\right)^{n} u[-n-1] \]

Note

Both terms are now left-sided because the overall ROC is inside both poles.

In this final scenario with the same \(X(z)\) algebraic form, the ROC is an interior region (\(|z| < 1/4\)), meaning the signal must be left-sided. Consequently, both terms in the partial-fraction expansion now correspond to left-sided exponentials. This further reinforces the idea that the ROC is not just a mathematical detail but fundamentally changes the time-domain signal.

Method 2: Power Series Expansion (Long Division)

Procedure:

1. Express \(X(z)\) as a rational function.
2. Perform Long Division:
   • Divide the numerator polynomial by the denominator polynomial.
   • Important: The direction of division (ascending or descending powers of \(z\) or \(z^{-1}\)) depends on the ROC.
     • If ROC is an exterior region (\(|z|>R\)), divide to get a series in negative powers of \(z\) (\(z^{-1}, z^{-2}, \dots\)). This corresponds to a right-sided sequence.
     • If ROC is an interior region (\(|z|<R\)), divide to get a series in positive powers of \(z\) (\(z, z^2, \dots\)). This corresponds to a left-sided sequence.
3. Identify Coefficients: The coefficients of the power series directly give the values of \(x[n]\) for corresponding powers of \(z\). \(X(z) = \dots + x[-2]z^2 + x[-1]z^1 + x[0]z^0 + x[1]z^{-1} + x[2]z^{-2} + \dots\)

Method 2: Power Series Expansion (Long Division)

Example 10.12: Finite Sequence (Trivial Case)

\(X(z)=4 z^{2}+2+3 z^{-1}, 0<|z|<\infty\).

By inspection, the coefficients are:

\(x[-2]=4\)

\(x[0]=2\)

\(x[1]=3\)

\(x[n]=0\) otherwise.

This gives \(x[n]=4 \delta[n+2]+2 \delta[n]+3 \delta[n-1]\).

Tip

This method is particularly useful for non-rational Z-transforms or when you need the first few samples of \(x[n]\).

The power series expansion method is more direct than partial fractions if you’re comfortable with polynomial long division. The key insight is that the Z-transform definition itself is a power series.

The direction of your long division is crucial and is determined by the ROC. For right-sided signals (ROC is exterior), you want terms with negative powers of \(z\). For left-sided signals (ROC is interior), you want terms with positive powers of \(z\).

Example 10.13: Power Series for \(X(z) = \frac{1}{1-az^{-1}}\)

Case 1: ROC \(|z|>|a|\) (Right-Sided)

We need a series in negative powers of \(z\).

Perform long division of \(1\) by \((1-az^{-1})\):

        1   + az⁻¹ + a²z⁻² + ...
      ___________________
1 - az⁻¹ | 1
          -(1 - az⁻¹)
          _________
                az⁻¹
              -(az⁻¹ - a²z⁻²)
              ____________
                    a²z⁻²

Result: \(X(z) = 1 + az^{-1} + a^2z^{-2} + \dots = \sum_{n=0}^{\infty} a^n z^{-n}\)

By inspection: \(x[n] = a^n u[n]\). (Consistent with Example 10.1)

Example 10.13: Power Series for \(X(z) = \frac{1}{1-az^{-1}}\)

Case 2: ROC \(|z|<|a|\) (Left-Sided)

We need a series in positive powers of \(z\).

Rearrange \(X(z)\) to \((az^{-1})^{-1} \frac{1}{ (az^{-1})^{-1} - 1} = a^{-1}z \frac{1}{a^{-1}z - 1}\).

Or, divide by \((az^{-1}-1)\) instead of \((1-az^{-1})\):

      -a⁻¹z - a⁻²z² - ...
    _________________
az⁻¹-1 | 1
      -(1 - a⁻¹z)   (dividing by -1+az⁻¹)
      _________
            a⁻¹z
          -(a⁻¹z - a⁻²z²)
          ___________
                a⁻²z²

Result: \(X(z) = -a^{-1}z - a^{-2}z^2 - \dots = -\sum_{k=1}^{\infty} a^{-k} z^k = -\sum_{n=-\infty}^{-1} a^n z^{-n}\)

By inspection: \(x[n] = -a^n u[-n-1]\). (Consistent with Example 10.2)

This example clearly illustrates how the ROC dictates the direction of the power series expansion. For a right-sided sequence (ROC \(|z|>|a|\)), we divide to get negative powers of \(z\). For a left-sided sequence (ROC \(|z|<|a|\)), we divide to get positive powers of \(z\). The coefficients of these series directly give the time-domain samples \(x[n]\).

Example 10.14: Power Series for Non-Rational \(X(z)\)

Given \(X(z)\) and ROC: \[ X(z)=\log \left(1+a z^{-1}\right), \quad|z|>|a| \] Condition for Taylor Series:

The ROC \(|z|>|a|\) implies \(|az^{-1}| < 1\). This is the condition for the Taylor series expansion of \(\log(1+v)\) where \(v = az^{-1}\).

Taylor Series Expansion:

Recall \(\log(1+v) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} v^n}{n}, \quad |v|<1\).

Substitute \(v = az^{-1}\): \[ X(z)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1} (az^{-1})^n}{n} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} a^n z^{-n}}{n} \]

Example 10.14: Power Series for Non-Rational \(X(z)\)

Identifying \(x[n]\):

Comparing with \(X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n}\), we match coefficients:

• For \(n \ge 1\): \(x[n] = \frac{(-1)^{n+1} a^n}{n}\)
• For \(n \le 0\): \(x[n] = 0\)

Result: \[ x[n]=\left\{\begin{array}{ll} \frac{(-1)^{n+1} a^{n}}{n}, & n \geq 1 \\ 0, & n \leq 0 \end{array}\right. \] This can also be written as \(x[n]=\frac{-(-a)^{n}}{n} u[n-1]\).

Note

This method is particularly powerful for non-rational Z-transforms where partial-fraction expansion is not applicable.

This example highlights the versatility of the power series expansion method. It’s not limited to rational Z-transforms. Here, we use a known Taylor series expansion for the logarithm function. The condition for the Taylor series to converge perfectly aligns with the given ROC. By matching the powers of \(z^{-n}\) to the definition of the Z-transform, we directly extract the time-domain sequence \(x[n]\).

Conclusion: Inverse Z-Transform

Key Methods for Inverse Z-Transform:

1. Partial-Fraction Expansion:
   • Best for rational \(X(z)\).
   • Decomposes \(X(z)\) into simpler terms.
   • Crucially relies on ROC to determine if terms are right-sided (\(a^n u[n]\)) or left-sided (\(-a^n u[-n-1]\)).
2. Power Series Expansion (Long Division):
   • Applicable to rational and some non-rational \(X(z)\).
   • Directly extracts \(x[n]\) coefficients.
   • Direction of division (ascending/descending powers) is determined by ROC.
   • Useful for finding first few terms or for non-rational transforms.

Conclusion: Inverse Z-Transform

Importance in ECE:

• System Response: If you have the Z-transform of the output of an LTI system, the inverse Z-transform allows you to find the actual time-domain output signal.
• Filter Design: After designing a filter in the Z-domain, the inverse Z-transform gives you the difference equation or impulse response for implementation.
• Control Systems: Analyzing the transient and steady-state behavior of discrete-time control systems often involves finding the inverse Z-transform of system variables.

Important

The ability to move back and forth between the time domain (\(x[n]\)) and the Z-domain (\(X(z)\)) is fundamental for analyzing, designing, and understanding discrete-time systems.

In summary, the inverse Z-transform is the bridge back to the time domain. You have two main tools: partial-fraction expansion for rational functions, and power series expansion, which is more general. In both cases, the ROC is your guide, telling you about the causality and behavior of the resulting time-domain signal. Mastering these techniques is essential for any ECE student working with discrete-time signals and systems.