Unilateral Z-Transform

Introduction

The unilateral Z-Transform is the workhorse of discrete-time system analysis in engineering practice. While the bilateral Z-Transform sums over all integers from $-\infty$ to $+\infty$, the unilateral version restricts the summation to $n \geq 0$, making it ideally suited for solving initial-value problems. Think of it this way: when you power on a digital filter or start a recursive algorithm, you care about what happens from the starting moment onward, given some known initial state. That is precisely the scenario the unilateral Z-Transform handles elegantly.

The unilateral Z-Transform is the discrete-time counterpart of the unilateral Laplace Transform. Just as the one-sided Laplace Transform naturally incorporates initial conditions when solving ordinary differential equations, the one-sided Z-Transform automatically brings initial conditions into play when solving difference equations.

Formal Definition

The unilateral Z-Transform of a discrete-time sequence $x[n]$ is defined as:

$$X^+(z) = \sum_{n=0}^{\infty} x[n] z^{-n}$$

Notice that the lower limit is $n = 0$ rather than $n = -\infty$. This means the transform ignores any values of the sequence for negative time indices. For causal sequences where $x[n] = 0$ for $n < 0$, the unilateral and bilateral Z-Transforms produce identical results. The distinction matters only when you have non-zero values at negative indices or when initial conditions need explicit handling.

Key Properties with Initial Conditions

The real power of the unilateral Z-Transform emerges in its shifting properties, where initial conditions appear explicitly.

Time Delay Property

For the unilateral Z-Transform, the delay property takes a modified form:

$$\mathcal{Z}^+\{x[n-1]\} = z^{-1}X^+(z) + x[-1]$$

$$\mathcal{Z}^+\{x[n-2]\} = z^{-2}X^+(z) + x[-2] + x[-1]z^{-1}$$

The general formula for a delay of $m$ samples is:

$$\mathcal{Z}^+\{x[n-m]\} = z^{-m}X^+(z) + \sum_{k=1}^{m} x[-k] z^{-(m-k)}$$

The initial conditions $x[-1], x[-2], \ldots$ appear as explicit terms in the expression.

Time Advance Property

For advances, the unilateral Z-Transform yields:

$$\mathcal{Z}^+\{x[n+1]\} = zX^+(z) - zx[0]$$

$$\mathcal{Z}^+\{x[n+2]\} = z^2X^+(z) - z^2x[0] - zx[1]$$

In general:

$$\mathcal{Z}^+\{x[n+m]\} = z^m X^+(z) - \sum_{k=0}^{m-1} x[k] z^{m-k}$$

These formulas show that advancing the sequence requires subtracting out the initial values that get "shifted past" the starting point.

Solving Difference Equations: Worked Example

Problem: Solve the difference equation $y[n] - 0.5y[n-1] = x[n]$ where $x[n] = u[n]$ (unit step) and the initial condition is $y[-1] = 2$.

Step 1: Apply the unilateral Z-Transform to both sides. Using the delay property:

$$\mathcal{Z}^+\{y[n-1]\} = z^{-1}Y(z) + y[-1] = z^{-1}Y(z) + 2$$

So the transformed equation becomes:

$$Y(z) - 0.5[z^{-1}Y(z) + 2] = \frac{z}{z-1}$$

Step 2: Solve for $Y(z)$:

$$Y(z)[1 - 0.5z^{-1}] = \frac{z}{z-1} + 1$$

$$Y(z) \cdot \frac{z - 0.5}{z} = \frac{z}{z-1} + 1 = \frac{z + z - 1}{z-1} = \frac{2z - 1}{z-1}$$

$$Y(z) = \frac{z(2z-1)}{(z-1)(z-0.5)}$$

Step 3: Perform partial fraction expansion. Divide by $z$:

$$\frac{Y(z)}{z} = \frac{2z-1}{(z-1)(z-0.5)}$$

At $z = 1$: $A = \frac{2(1)-1}{1-0.5} = \frac{1}{0.5} = 2$

At $z = 0.5$: $B = \frac{2(0.5)-1}{0.5-1} = \frac{0}{-0.5} = 0$

Therefore: $Y(z) = \frac{2z}{z-1}$, giving $y[n] = 2u[n]$ for $n \geq 0$.

This result makes physical sense: the system reaches a steady state of 2, influenced by both the input step and the initial condition.

Second-Order Example

Problem: Solve $y[n] - 3y[n-1] + 2y[n-2] = 0$ with $y[-1] = 1$ and $y[-2] = 0$.

Applying the unilateral Z-Transform with the delay property:

$$Y(z) - 3[z^{-1}Y(z) + y[-1]] + 2[z^{-2}Y(z) + y[-2] + y[-1]z^{-1}] = 0$$

$$Y(z) - 3z^{-1}Y(z) - 3 + 2z^{-2}Y(z) + 0 + 2z^{-1} = 0$$

$$Y(z)[1 - 3z^{-1} + 2z^{-2}] = 3 - 2z^{-1}$$

$$Y(z) = \frac{3 - 2z^{-1}}{1 - 3z^{-1} + 2z^{-2}} = \frac{3z^2 - 2z}{z^2 - 3z + 2} = \frac{z(3z-2)}{(z-1)(z-2)}$$

Using partial fractions on $Y(z)/z$:

$$\frac{Y(z)}{z} = \frac{3z-2}{(z-1)(z-2)}$$

$A = \frac{3-2}{1-2} = -1$, $B = \frac{6-2}{2-1} = 4$

So $Y(z) = \frac{-z}{z-1} + \frac{4z}{z-2}$, giving $y[n] = (-1 + 4 \cdot 2^n)u[n]$.

When to Use Unilateral vs Bilateral

The choice between unilateral and bilateral Z-Transforms depends on your analysis goal:

Use the Unilateral Z-Transform when:

• Solving initial-value problems with known conditions at $n = -1, -2, \ldots$
• Analyzing transient responses of digital filters during startup
• Implementing recursive algorithms where you track state from a starting point
• Working with causal systems in real-time signal processing

Use the Bilateral Z-Transform when:

• Performing general signal analysis without specific initial conditions
• Studying stability through ROC analysis
• Working with two-sided (non-causal) sequences
• Deriving theoretical relationships between signals and systems

Relationship to the DTFT

When the unilateral Z-Transform converges on the unit circle ($|z| = 1$), evaluating $X^+(z)$ at $z = e^{j\Omega}$ gives the Discrete-Time Fourier Transform of the causal part of the sequence. This connection is important because it links time-domain initial-value analysis to frequency-domain behavior.

Practical Engineering Applications

In digital control systems, the unilateral Z-Transform is used extensively to analyze transient behavior when controllers are activated or when setpoint changes occur. The initial conditions represent the system state at the moment of activation.

In digital audio processing, when a recursive filter begins processing a new audio frame, the initial conditions represent the filter's internal state from the previous frame. The unilateral Z-Transform framework models this scenario precisely.

Key Takeaways

• The unilateral Z-Transform sums from $n = 0$ to $\infty$, assuming or handling causal signals
• Initial conditions $x[-1], x[-2], \ldots$ appear explicitly in the delay property
• The advance property subtracts out initial values: $\mathcal{Z}^+\{x[n+1]\} = zX^+(z) - zx[0]$
• It directly solves difference equations with specified initial states
• For causal sequences, unilateral and bilateral transforms give identical results
• This is the practical engineering tool for digital system transient analysis