17-Phys-B4 Signals and Communications · December 2013
Question 5 of 6: Partial-Fraction Inversion of a Right-Sided z-Transform, Two Ways
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B4 Communications, National Examination
December 2013 — a three-hour closed-book examination with one double-sided aid sheet
permitted and an approved calculator. The cover page states any five of the six
questions constitute a complete paper, with only the first five as they appear in the answer
book marked; every question is nonetheless answered in full below so the paper remains a
complete study resource. All six questions carry equal value (20 marks each).
Reference texts. A. V. Oppenheim and A. S. Willsky, Signals and
Systems, 2nd ed. (Fourier series, Fourier transform properties, the sampling theorem,
the unilateral z-transform); S. Haykin and M. Moher, Communication Systems, 5th ed.
(amplitude and angle modulation, transmitted power and sideband power); B. P. Lathi and
Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. (PM/FM instantaneous
phase and frequency); J. G. Proakis and D. G. Manolakis, Digital Signal Processing,
4th ed. (partial-fraction inversion of the z-transform).
Question 5: Partial-Fraction Inversion of a Right-Sided z-Transform, Two Ways (20 marks)
Given.
$X(z)=\dfrac{1}{(1-\frac12 z^{-1})(1-z^{-1})}$, poles at $z=\tfrac12$ and $z=1$, $x[n]$ known to
be right-sided (causal).
Find. (a) $X(z)$ in partial fractions of $z^{-1}$, and $x[n]$; (b) $X(z)$ in
partial fractions of $z$, the resulting $x[n]$, and a demonstration that the two sequences agree.
Approach. Decompose $X(z)$ into first-order partial fractions in whichever
variable ($z^{-1}$ or $z$) the sub-part asks for, then invert each term with the standard
right-sided pair $a^n u[n]\leftrightarrow \dfrac{1}{1-az^{-1}}$ (equivalently
$a^{n-1}u[n-1]\leftrightarrow \dfrac{z^{-1}}{1-az^{-1}}=\dfrac1{z-a}$).
Part (a) — partial fractions in $z^{-1}$. Write
$X(z)=\dfrac{A}{1-\frac12z^{-1}}+\dfrac{B}{1-z^{-1}}$. Clearing denominators,
$A(1-z^{-1})+B(1-\tfrac12z^{-1})=1$; evaluating at the root of each factor
($z^{-1}=2$ and $z^{-1}=1$ respectively) gives $A=-1$ and $B=2$, so
$$X(z)=\boxed{\frac{-1}{1-\frac12z^{-1}}+\frac{2}{1-z^{-1}}}.$$
Both terms are right-sided (the sequence is stated to be right-sided, which picks this ROC over
the alternative left-sided inversion of the same partial fractions), so $\dfrac1{1-az^{-1}}
\leftrightarrow a^n u[n]$ applies term by term:
$$x[n]=\boxed{\big[2-(0.5)^n\big]\,u[n]}.$$
Part (b) — partial fractions in $z$, and the cross-check. Multiplying numerator and denominator by $z^2$ rewrites $X(z)$ as a ratio of polynomials in $z$:
$$X(z)=\frac{z^2}{(z-\frac12)(z-1)}.$$ Since the numerator and denominator have equal degree, long division first extracts a constant
term before the proper-fraction partial expansion:
$$X(z)=1+\frac{1.5z-0.5}{(z-\frac12)(z-1)}=1+\frac{D}{z-\frac12}+\frac{E}{z-1},$$
and evaluating the residues at $z=\tfrac12$ and $z=1$ gives $D=-0.5$, $E=2$:
$$X(z)=\boxed{1-\frac{0.5}{z-\frac12}+\frac{2}{z-1}}.$$ Using $\dfrac{1}{z-a}=\dfrac{z^{-1}}{1-az^{-1}}\leftrightarrow a^{n-1}u[n-1]$ term by term
(and $1\leftrightarrow\delta[n]$),
$$x[n]=\delta[n]-0.5\,(0.5)^{n-1}u[n-1]+2\,(1)^{n-1}u[n-1]
=\delta[n]+\big[2-(0.5)^n\big]u[n-1].$$ At $n=0$ this reduces to $\delta[0]+0=1$, and for every $n\ge1$ the $u[n-1]$ term is identical
to the $u[n]$ term of part (a) — so this is exactly the same sequence,
$$x[n]=\boxed{\big[2-(0.5)^n\big]\,u[n]},$$, which expands $X(z)$ as a power series in
$z^{-1}$ to 8 terms and checks every coefficient against $2-(0.5)^n$ symbolically.
