One-Sided Z-Transform, Transient/Steady-State, and Stability Triangle

Inverse Z-Transform Methods

1. Partial Fraction Expansion

For rational $X(z) = \frac{N(z)}{D(z)}$, factor denominator and expand:

\[\frac{X(z)}{z} = \frac{A_1}{z-p_1} + \frac{A_2}{z-p_2} + \dots\]

Then use pairs: $\frac{1}{1-az^{-1}} \leftrightarrow a^n u[n]$ (for $

z

>

a

$)

Critical: ROC determines causal vs anti-causal terms!

2. Power Series (Long Division)

Divide $N(z)$ by $D(z)$ to get series in $z^{-1}$:

\[X(z) = \sum_{n} x[n] z^{-n}\]

Coefficients of $z^{-n}$ are $x[n]$. Works for any ROC by choosing division direction.

3. Contour Integration (Cauchy)

\[x[n] = \frac{1}{2\pi j} \oint_C X(z) z^{n-1} dz\]

Where $C$ is counterclockwise contour in ROC. Theoretical; rarely used by hand.

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Relationship to Fourier Transform

\[X(e^{j\omega}) = X(z)\big|_{z=e^{j\omega}} = \sum_{n} x[n] e^{-j\omega n}\]

This is the DTFT (Discrete-Time Fourier Transform).

Condition	Result
ROC includes $	z	=1$	DTFT exists = $X(e^{j\omega})$
ROC excludes $	z	=1$	DTFT does not exist (or is periodic impulse train)

DFT is the sampled DTFT: $X[k] = X(e^{j\omega})\big

_{\omega=2\pi k/N}$

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Why Transform to Frequency Domain?

Time Domain	Frequency Domain
Convolution (hard)	Multiplication (easy)
Difference equations (recursive)	Algebraic equations
Signal “shape”	Spectral content (what frequencies?)
Filter design (trial & error)	Pole/zero placement (systematic)
Stability (check all $n$)	Check pole locations

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Worked Example: System Analysis

System: $y[n] - 0.5 y[n-1] = x[n] + x[n-1]$

Find: $H(z)$, poles/zeros, stability, impulse response

Solution:

1. Z-transform: $Y(z) - 0.5 z^{-1} Y(z) = X(z) + z^{-1} X(z)$
2. $H(z) = \frac{Y(z)}{X(z)} = \frac{1 + z^{-1}}{1 - 0.5 z^{-1}} = \frac{z+1}{z-0.5}$
3. Zero: $z = -1$ (on unit circle → notch at $\omega=\pi$)
4. Pole: $z = 0.5$ (inside unit circle → stable)

5. ROC: $

   z

   > 0.5$ (causal)

6. Inverse: $H(z) = \frac{z}{z-0.5} + \frac{1}{z-0.5} = \frac{z}{z-0.5} + 0.5\frac{1}{1-0.5z^{-1}}$
   • $h[n] = (0.5)^n u[n] + 0.5 (0.5)^{n-1} u[n-1]$
   • $h[0] = 1$, $h[n] = 1.5(0.5)^n$ for $n \ge 1$

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Solving Difference Equations with Initial Conditions

For $y[n] - a y[n-1] = x[n]$ with $y[-1] = y_0$:

Using unilateral Z-transform (sums from $n=0$): \(\mathcal{Z}\{y[n-1]\} = z^{-1}Y(z) + y[-1]\)

\(Y(z) - a(z^{-1}Y(z) + y_0) = X(z)\) \(Y(z) = \frac{X(z) + a y_0}{1 - a z^{-1}}\)

The initial condition appears as an additional input term!

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One-Sided Z-Transform (Proakis 3.6)

For signals with initial conditions, use the one-sided (unilateral) Z-transform:

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

Key Properties:

Property	Time Domain	Z-Domain
Time Advance	$x[n+1]$	$z X_+(z) - z x[0]$
Time Delay	$x[n-1]$	$z^{-1} X_+(z) + x[-1]$
Second-Order Delay	$x[n-2]$	$z^{-2} X_+(z) + z^{-1}x[-1] + x[-2]$

This directly incorporates initial conditions into the algebraic equation.

Worked Example: Second-Order System with Initial Conditions

System: $y[n] - 0.5y[n-1] - 0.25y[n-2] = x[n]$ with $y[-1]=1, y[-2]=2$

Using one-sided Z-transform: \(Y_+(z) - 0.5(z^{-1}Y_+(z) + 1) - 0.25(z^{-2}Y_+(z) + z^{-1} + 2) = X_+(z)\)

\[Y_+(z)(1 - 0.5z^{-1} - 0.25z^{-2}) = X_+(z) + 0.5 + 0.25z^{-1} + 0.5\]

Initial conditions contribute additional terms to the numerator.

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Pole-Zero Cancellations (Proakis 3.5.4)

When a pole and zero coincide, they cancel:

• In $H(z)$: System order reduces
• In $Y(z) = H(z)X(z)$: Input zero suppresses system mode (or vice versa)

Critical: Never rely on pole-zero cancellation to stabilize an unstable system. Finite precision makes exact cancellation impossible → instability in practice.

Example

\(H(z) = \frac{1 - 5z^{-1} + 6z^{-2}}{1 - 2.5z^{-1} + z^{-2}} = \frac{(1-2z^{-1})(1-3z^{-1})}{(1-0.5z^{-1})(1-2z^{-1})}\)

Pole at $z=2$ cancels with zero at $z=2$. Reduced system: \(H_{\text{reduced}}(z) = \frac{1-3z^{-1}}{1-0.5z^{-1}}\)

Original: unstable (pole at 2). Reduced: stable. But quantization makes exact cancellation impossible.

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Transient vs Steady-State Response (Proakis 3.5.2)

For $Y(z) = H(z)X(z)$, partial fraction expansion gives:

\[Y(z) = \sum \frac{A_k}{1-p_k z^{-1}} + \text{(forced response terms)}\]

Component	Time Domain	Behavior
Natural/Transient	$\sum A_k p_k^n u[n]$	Decays if $	p_k	<1$
Forced/Steady-State	Matches input form	Persists as long as input

Example

System: $H(z) = \frac{1}{1-0.5z^{-1}}$, Input: $x[n] = 10\cos(\pi n/4)u[n]$

Transient: $(0.5)^n u[n]$ (decays) Steady-State: $13.56\cos(\pi n/4 - 28.7^\circ)u[n]$ (persists)

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Stability Triangle for Second-Order Systems (Proakis 3.5.6)

For $y[n] = -a_1 y[n-1] - a_2 y[n-2] + b_0 x[n]$, poles are roots of: \(z^2 + a_1 z + a_2 = 0\)

Stability Conditions (all poles inside unit circle): \(|a_2| < 1 \quad \text{and} \quad |a_1| < 1 + a_2\)

These define a triangle in the $(a_1, a_2)$ plane:

a2 = 1
  |
  |   Stable Region
  |  /     \
  | /       \
  |/_________\
a1=-2        a1=2
  a2 = -1

Pole types by region:

• Below parabola $a_2 = a_1^2/4$: Real & distinct
• On parabola: Real & equal (double pole)
• Above parabola: Complex conjugate

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Multiple-Order Poles and Stability (Proakis 3.5.5)

If input has poles coinciding with system poles → multiple-order poles in $Y(z)$.

Output contains terms: $A n^b p^n u[n]$ where $b$ = pole multiplicity - 1

Pole Location	Behavior
$	p	< 1$	Decays to 0 (exponential dominates polynomial)
$	p	= 1$	Unbounded (e.g., step input to integrator $H(z)=1/(1-z^{-1})$ → ramp)
$	p	> 1$	Grows unbounded

BIBO stability requires ALL system poles strictly inside unit circle. Poles on unit circle = marginally stable (only useful for digital oscillators).

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