22-Elec-A3 Signals and Communications · December 2013

Question 5 of 6: Inverse z-Transform by Partial Fractions, Two Ways

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2013 — 07-Elec-A3, Signals and Communications. Three hours, closed book (one two-sided aid sheet and an approved calculator permitted). Six questions, all of equal value at 20 marks; five constitute a complete paper.

All six questions are solved here. The examination marks only the first five answers submitted, but this document is a study resource, so every question — and every sub-part — is worked in full.

Reference texts.

B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. — Ch. 3 (signal transmission), Ch. 4 (amplitude modulation), Ch. 5 (angle modulation), Ch. 6 (sampling).
B. P. Lathi, Linear Systems and Signals, 2nd ed. — Ch. 6 (Fourier series), Ch. 7 (Fourier transform).
A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. — Ch. 4 (Fourier transform properties), Ch. 7 (sampling), Ch. 10 (z-transform).
S. Haykin, Communication Systems, 5th ed. — Ch. 3 (amplitude modulation), Ch. 4 (angle modulation).
J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. — Ch. 3 (z-transform and inversion).

Question 5: Inverse z-Transform by Partial Fractions, Two Ways (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $X(z) = \dfrac{1}{(1-\frac{1}{2}z^{-1})(1-z^{-1})}$, with $x[n]$ known to be right-sided.

Find. $x[n]$ by partial-fraction expansion in $z^{-1}$ (part a) and again in $z$ (part b), showing the two agree.

Approach. Fix the region of convergence from the right-sidedness, expand in $z^{-1}$ using the cover-up rule to obtain two standard geometric-sequence terms, then repeat with the polynomial-in-$z$ form by first expanding $X(z)/z$ so that the expansion is proper.

Pole–zero map of X(z) in the z-plane: poles (×) at z = 0.5, 1, a double zero (ο) at the origin, and the right-sided ROC |z| > 1 outside the outermost pole (solid blue circle; the dashed circle is the unit circle).

Locate the poles and fix the ROC. $X(z)$ has poles where the factors vanish, at $z = \tfrac{1}{2}$ and $z = 1$, and a double zero at the origin. A right-sided sequence has an ROC that is the exterior of the outermost pole: $$\boxed{\,\text{ROC}: \ |z| \gt 1\,}$$ Because the ROC does not include the unit circle, $x[n]$ is not absolutely summable — expect a non-decaying term, which the pole at $z=1$ supplies.
Expand in powers of $z^{-1}$ (part a). Writing $v = z^{-1}$ and seeking $\dfrac{1}{(1-\frac{1}{2}v)(1-v)} = \dfrac{A}{1-\frac{1}{2}v} + \dfrac{B}{1-v}$, the cover-up rule evaluates each residue at the root of its own factor: $$A = \left.\frac{1}{1-v}\right|_{v=2} = \frac{1}{1-2} = -1, \qquad B = \left.\frac{1}{1-\frac{1}{2}v}\right|_{v=1} = \frac{1}{1-\frac{1}{2}} = 2$$ $$X(z) = \frac{-1}{1-\frac{1}{2}z^{-1}} + \frac{2}{1-z^{-1}}$$
Invert term by term (part a). With the ROC exterior to each pole, the standard right-sided pair is $\dfrac{1}{1-az^{-1}} \leftrightarrow a^n u[n]$ for $|z| \gt |a|$. Applying it to both terms: $$\boxed{\,x[n] = \left[2 - \left(\tfrac{1}{2}\right)^{n}\right]u[n]\,}$$ The first few values are $x[0]=1,\ x[1]=1.5,\ x[2]=1.75,\ x[3]=1.875$, rising monotonically toward the steady value 2 contributed by the unit-circle pole.
Rewrite as a ratio of polynomials in $z$ (part b). Multiplying numerator and denominator by $z^2$: $$X(z) = \frac{z^2}{\left(z-\tfrac{1}{2}\right)(z-1)}$$ This is not proper (numerator and denominator have equal degree), so a direct partial-fraction expansion would need an extra constant term. The standard remedy is to expand $X(z)/z$ instead, which is strictly proper.
Expand $X(z)/z$ and restore the factor of $z$ (part b). With $\dfrac{X(z)}{z} = \dfrac{z}{(z-\frac{1}{2})(z-1)} = \dfrac{A'}{z-\frac{1}{2}} + \dfrac{B'}{z-1}$, the cover-up rule gives $$A' = \left.\frac{z}{z-1}\right|_{z=1/2} = \frac{0.5}{-0.5} = -1, \qquad B' = \left.\frac{z}{z-\frac{1}{2}}\right|_{z=1} = \frac{1}{0.5} = 2$$ Multiplying back through by $z$, $$X(z) = \frac{-z}{z-\frac{1}{2}} + \frac{2z}{z-1}$$
Invert and compare (part b). The pair in this notation is $\dfrac{z}{z-a} \leftrightarrow a^n u[n]$ for $|z| \gt |a|$, so $$x[n] = \left[2 - \left(\tfrac{1}{2}\right)^{n}\right]u[n]$$ which is identical to part (a). This is no accident: $\dfrac{z}{z-a}$ and $\dfrac{1}{1-az^{-1}}$ are the same function written two ways, so the residues $A',B'$ must equal $A,B$.
Verify by long division. Expanding $X(z)$ as a power series in $z^{-1}$ — the Cauchy product of $\sum(\tfrac{1}{2})^k z^{-k}$ and $\sum z^{-k}$ — gives coefficients $1,\ 1.5,\ 1.75,\ 1.875,\dots$, matching the closed form term for term.

Quantity	Result
Poles / zeros	Poles $z = \frac{1}{2},\,1$; double zero at $z=0$
ROC (right-sided)	$\|z\| \gt 1$
(a) Residues in $z^{-1}$	$A = -1$ (pole $\frac12$), $B = 2$ (pole 1)
(b) Residues of $X(z)/z$	$A' = -1$, $B' = 2$ — identical
Sequence $x[n]$	$\left[2 - (1/2)^n\right]u[n]$
First values	1, 1.5, 1.75, 1.875, … → 2

Topic. Inverse z-transform by partial-fraction expansion, in both the $z^{-1}$ and $z$ notations, with the ROC determining the sidedness.

Key relations. $\dfrac{1}{1-az^{-1}} \leftrightarrow a^nu[n]$ for $|z|\gt|a|$ (right-sided) and $\leftrightarrow -a^nu[-n-1]$ for $|z|\lt|a|$ (left-sided); equivalently $\dfrac{z}{z-a} \leftrightarrow a^nu[n]$; expand $X(z)/z$ whenever $X(z)$ is not strictly proper in $z$.

Why this works. Partial fractions decompose the transform into terms whose inverses are memorised, and the algebra is identical in either notation because $z/(z-a)$ and $1/(1-az^{-1})$ are literally the same rational function. The one genuine subtlety is properness: in the $z^{-1}$ variable $X$ is already proper, but in $z$ the numerator $z^2$ matches the denominator degree, so expanding $X(z)$ directly would produce an unwanted constant (an impulse at $n=0$). Dividing by $z$ first, expanding, then multiplying back is the standard device that keeps every term in the recognisable $z/(z-a)$ shape. The ROC is what makes the answer unique — the same algebra with $|z|\lt\frac12$ would give a left-sided sequence instead.

Common pitfall. Expanding $X(z)$ rather than $X(z)/z$ in part (b) and then mis-assigning the leftover constant, which shifts $x[0]$. The other classic error is inverting without stating the ROC, since $1/(1-z^{-1})$ maps to $u[n]$ or to $-u[-n-1]$ depending entirely on which region is intended.

Source. Oppenheim & Willsky, Signals and Systems, 2nd ed., §10.3 and §10.6; Proakis & Manolakis, Digital Signal Processing, 4th ed., §3.4.

← Question 4 Next: Question 6 →

Quantity	Result
Poles / zeros	Poles \(z = \frac{1}{2},\,1\); double zero at \(z=0\)
ROC (right-sided)	\(\|z\| \gt 1\)
(a) Residues in \(z^{-1}\)	\(A = -1\) (pole \(\frac12\)), \(B = 2\) (pole 1)
(b) Residues of \(X(z)/z\)	\(A' = -1\), \(B' = 2\) — identical
Sequence \(x[n]\)	\(\left[2 - (1/2)^n\right]u[n]\)
First values	1, 1.5, 1.75, 1.875, … → 2