Question 1 of 6: Reconstructing a real sequence from six frequency-domain clues

Question 1: Reconstructing a real sequence from six frequency-domain 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 real sequence $x[n]$ described only through the six clues (a)–(f) above: a causality statement about its time-reverse, a purely-imaginary DTFT for the shifted sequence $v[n]=x[n-3]$, an energy value of 28, an initial-value limit of $-1$, one inverse-DTFT sample equal to 2, and a sign condition on $x[-2]$.

Find. Every sample of $x[n]$, together with a statement of which feature each individual clue pins down.

Approach. Convert the two structural clues (a) and (b) into a finite support and an antisymmetry law, which collapses the unknown sequence to three real numbers; then read those three numbers off the initial-value theorem, the DTFT synthesis integral and Parseval's relation, and use the sign clue to resolve the remaining square root.

1. Clue (a) fixes the upper edge of the support. A causal sequence vanishes for negative argument, so $x[-n] = 0$ for $n \lt 0$. Writing $m = -n$, this says $$x[m] = 0 \qquad \text{for } m \gt 0 ,$$ i.e. $x[n]$ is anti-causal: it can be non-zero only at $n \le 0$. Clue (a) alone therefore determines the right-hand edge of the support and nothing else.
2. Clue (b) forces odd symmetry about $n = -3$. For a real sequence the DTFT is purely imaginary if and only if the sequence is odd. (Real and even gives a purely real transform; real and odd gives a purely imaginary one.) Hence $v[-n] = -v[n]$, and substituting $v[n] = x[n-3]$ with $m = n-3$, $$x[-m-6] = -x[m] .$$ The sequence is antisymmetric about the point $n = -3$, and putting $m = -3$ gives $x[-3] = -x[-3]$, so $$\boxed{x[-3] = 0}$$ Combining with Step 1: since $x[m] = 0$ for all $m \gt 0$, the antisymmetry immediately gives $x[k] = 0$ for all $k \lt -6$. The support is therefore exactly $-6 \le n \le 0$.
3. The unknown sequence has collapsed to three numbers. Applying $x[-6-m] = -x[m]$ across the support, $$x[-6] = -x[0], \qquad x[-5] = -x[-1], \qquad x[-4] = -x[-2], \qquad x[-3] = 0 .$$ So only $x[0]$, $x[-1]$ and $x[-2]$ remain free — three real unknowns, which is exactly the number of numerical clues still unused.
4. Clue (d) gives $x[0]$ by the initial-value theorem. Because the support is $n \le 0$, the z-transform is a polynomial in $z$: $$X(z) = \sum_{n=-6}^{0} x[n] z^{-n} = x[0] + x[-1]z + x[-2]z^{2} + x[-3]z^{3} + x[-4]z^{4} + x[-5]z^{5} + x[-6]z^{6}.$$ Letting $z \to 0$ kills every term except the constant, so $\lim_{z\to 0} X(z) = x[0]$ and $$\boxed{x[0] = -1} \qquad\Longrightarrow\qquad x[-6] = +1 .$$ This is the anti-causal mirror of the initial-value theorem printed on page 10 of the paper.
5. Clue (e) is one sample of the synthesis integral. The DTFT synthesis equation gives $x[n] = \frac{1}{2\pi}\int_{-\pi}^{\pi} X(e^{j\omega}) e^{j\omega n}\,d\omega$. The integrand in clue (e) carries $e^{-j\omega} = e^{j\omega(-1)}$, so the integral evaluates the sequence at $n = -1$: $$\boxed{x[-1] = 2} \qquad\Longrightarrow\qquad x[-5] = -2 .$$ No integration is required — recognising the synthesis kernel is the whole step.
6. Clue (c) is Parseval's relation and fixes $|x[-2]|$. Parseval's theorem (page 8 of the paper) equates the two energies: $$\sum_{n=-\infty}^{\infty}|x[n]|^{2} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\left|X(e^{j\omega})\right|^{2} d\omega = 28 .$$ The antisymmetry makes the energy sum pair up, and $x[-3]=0$ contributes nothing, so $$2\left(x[0]^{2} + x[-1]^{2} + x[-2]^{2}\right) = 28 \;\Longrightarrow\; 1 + 4 + x[-2]^{2} = 14 \;\Longrightarrow\; x[-2]^{2} = 9 .$$ Energy is sign-blind, so this clue alone leaves $x[-2] = \pm 3$.
7. Clue (f) resolves the sign. With $x[-2] \gt 0$ the positive root is selected: $$\boxed{x[-2] = +3} \qquad\Longrightarrow\qquad x[-4] = -3 .$$ Every sample is now determined, so the sequence is unique.
8. Assemble and check. The complete sequence is $$\boxed{x[n] = \{\,1,\; -2,\; -3,\; 0,\; +3,\; +2,\; -1\,\} \qquad n = -6,\dots,0 }$$ and $x[n] = 0$ elsewhere. All six clues are satisfied: the support lies at $n\le 0$ (a); $v[n] = x[n-3] = \{1,-2,-3,0,3,2,-1\}$ on $-3\le n\le 3$ is real and odd, so $V(e^{j\omega})$ is purely imaginary (b); the energy is $1+4+9+0+9+4+1 = 28$ (c); $x[0] = -1$ (d); $x[-1] = 2$ (e); and $x[-2] = 3 \gt 0$ (f).

The structure of the answer is worth naming explicitly, because the question invites it: the two qualitative clues do the heavy lifting. Clue (a) bounds the support on the right, clue (b) both bounds it on the left and reduces seven unknown samples to three, and the three quantitative clues then act one unknown at a time. The final sign clue exists only because Parseval's relation cannot see signs.