Reconstructing a real causal sequence from six transform clues (12 marks): 22-Elec-B1 Digital Signal Processing, May 2015, Q1

Question 1: Reconstructing a real causal sequence from six transform clues (12 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. A purely real sequence $x[n]$ described only through six transform-domain clues; no closed form and no sketch are supplied. The six clues, and the standard property each one invokes, are collected below.

Given data — the six clues
Clue	Statement	Property invoked
(a)	$x[n]$ is causal	definition of causality
(b)	$v[n]=x[n+2]$ has a purely real DTFT	real DTFT $\Leftrightarrow$ conjugate-symmetric sequence
(c)	$\tfrac{1}{2\pi}\int_{-\pi}^{\pi}\left\|X(e^{j\omega})\right\|^{2}d\omega = 24$	Parseval's theorem
(d)	$\lim_{z\to\infty}X(z)=3$	initial-value theorem
(e)	$\tfrac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})e^{j\omega}d\omega = 1$	DTFT synthesis equation at $n=1$
(f)	$x[2] \lt 0$	sign selection

Find. The complete sequence $x[n]$ and its sketch, together with a statement of exactly which feature of $x[n]$ each individual clue pins down — that attribution is what the question awards partial credit for.

Approach. Translate each clue into a constraint on the samples: clues (a) and (b) between them force a finite support of only five samples with a mirror symmetry, which leaves three unknown numbers, and clues (d), (e), (c) and (f) then supply exactly those three numbers.

Causality bounds the support on the left. By definition a causal sequence vanishes before the origin, $$x[n]=0,\qquad n \lt 0 .$$ On its own this clue says nothing about the values; it fixes only where the sequence may live.
A purely real DTFT forces an even sequence. For a real sequence $v[n]$ the DTFT obeys $V(e^{-j\omega})=V^{*}(e^{j\omega})$, so $V(e^{j\omega})$ is purely real if and only if $v[n]$ is even. Since $x[n]$ is real, so is $v[n]=x[n+2]$, and therefore $$v[n]=v[-n]\quad\Longrightarrow\quad x[n+2]=x[-n+2].$$ Writing $m=n+2$ turns this into a mirror symmetry of $x$ about $m=2$: $$x[m]=x[4-m].$$
Symmetry plus causality bounds the support on the right. If $m \gt 4$ then $4-m \lt 0$, and clue (a) makes $x[4-m]=0$; the symmetry then makes $x[m]=0$ as well. The support is therefore the five samples $0\le n\le 4$, and the symmetry leaves only three independent values: $$x[0]=x[4],\qquad x[1]=x[3],\qquad x[2]\ \text{(unpaired)} .$$ Clues (a) and (b) together are what reduce an infinite-dimensional unknown to three numbers.
The initial-value theorem fixes $x[0]$. For a causal sequence every term of $X(z)=\sum_{n\ge 0}x[n]z^{-n}$ except the $n=0$ term vanishes as $z\to\infty$, so $$\lim_{z\to\infty}X(z)=x[0]=3 \quad\Longrightarrow\quad x[0]=x[4]=3 .$$
The synthesis equation at $n=1$ fixes $x[1]$. The DTFT synthesis formula printed in the appendix is $$x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})e^{j\omega n}\,d\omega ,$$ and clue (e) is exactly this integral evaluated at $n=1$. Hence $$x[1]=1 \quad\Longrightarrow\quad x[1]=x[3]=1 .$$
Parseval's theorem fixes the magnitude of the last sample. The appendix gives $$\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|X(e^{j\omega})\right|^{2}d\omega=\sum_{n=-\infty}^{\infty}\left|x[n]\right|^{2}=24 .$$ Substituting the four samples already known, $$2(3)^{2}+2(1)^{2}+x[2]^{2}=18+2+x[2]^{2}=24 \quad\Longrightarrow\quad x[2]^{2}=4 ,$$ so $x[2]=\pm 2$. Parseval alone cannot choose between the two.
The sign clue picks the branch and completes the sequence. Clue (f) states $x[2] \lt 0$, which selects $x[2]=-2$. Assembling all five samples, $$\boxed{\,x[n]=\{\,3,\;1,\;-2,\;1,\;3\,\},\qquad 0\le n\le 4,\ \ x[n]=0\ \text{otherwise}\,}$$

The sketch below shows the answer; the shifted copy confirms clue (b) directly, since $v[n]=x[n+2]$ comes out symmetric about the origin and is therefore guaranteed to have a real transform.

Figure 1.1 — the reconstructed sequence. The single negative sample at n = 2 (red) is the one fixed by clue (f).

Figure 1.2 — the shifted sequence v[n] = x[n+2] is even about n = 0, which is precisely the condition for V(e^jw) to be purely real.

As a numerical check, the energy of the answer is $9+1+4+1+9=24$, matching clue (c) exactly, and $X(z)$ evaluated as $z\to\infty$ returns the leading sample $3$, matching clue (d).

Final results — what each clue determines
Clue	Feature of $x[n]$ it determines
(a) causal	$x[n]=0$ for $n \lt 0$ (left edge of the support)
(b) $V(e^{j\omega})$ real	even symmetry of $x$ about $n=2$; with (a) this also gives $x[n]=0$ for $n \gt 4$
(d) initial value	$x[0]=3$, hence $x[4]=3$
(e) synthesis at $n=1$	$x[1]=1$, hence $x[3]=1$
(c) Parseval	$x[2]^{2}=4$, i.e. $\|x[2]\|=2$
(f) sign	$x[2]=-2$
Complete answer	$x[n]=\{3,1,-2,1,3\}$ on $0\le n\le 4$

Topic. Recovering a sequence from transform-domain constraints. Each clue is a single standard property of the DTFT or the z-transform, read backwards: the exam is testing whether the candidate can name the property rather than manipulate an integral.

Key relations. Symmetry: for real $x[n]$, $X(e^{-j\omega})=X^{*}(e^{j\omega})$, and $X(e^{j\omega})$ is purely real if and only if $x[n]=x[-n]$. Initial value: $\lim_{z\to\infty}X(z)=x[0]$ for causal $x$. Synthesis: $x[n]=\tfrac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})e^{j\omega n}d\omega$. Parseval: $\tfrac{1}{2\pi}\int_{-\pi}^{\pi}|X(e^{j\omega})|^{2}d\omega=\sum_n |x[n]|^{2}$.

Why this works. Causality and even symmetry are contradictory constraints unless the sequence has finite length, so imposing both immediately truncates the signal. That interaction — not either clue by itself — is what makes the problem solvable: it converts an unknown function into three unknown numbers, which the three remaining numerical clues then determine uniquely up to the sign that clue (f) resolves.

Common pitfall. Applying the symmetry to $x$ instead of to $v$. The clue is about $v[n]=x[n+2]$, so the mirror line sits at $n=2$, not at $n=0$; a candidate who writes $x[n]=x[-n]$ concludes that the only causal solution is a single sample at the origin and loses the whole question. The second trap is treating clue (e) as another energy statement: it is the synthesis integral, linear in $X$, so it returns a sample value and not a squared magnitude.

Source. Oppenheim & Schafer, Discrete-Time Signal Processing, 3rd ed., §2.9 (symmetry properties of the DTFT) and §3.4 with Table 3.2 (z-transform properties, including the initial-value theorem); Proakis & Manolakis, DSP, 4th ed., §4.2.

Question 1 of 6: Reconstructing a real causal sequence from six transform clues (12 marks)

Question 1: Reconstructing a real causal sequence from six transform clues (12 marks)