The Fourier series of f(x) = x on (−π, π) contains:

A constant term and cosine terms

X(z) = z/(z − a). Which statement is correct?

It is aⁿu[n] when |z| > |a| and a different, left-sided sequence when |z| < |a|

It is the transform of aⁿu[n], whatever the region of convergence

It is aⁿu[n] when |z| > |a| and a different, left-sided sequence when |z| < |a|

Its region of convergence is always the whole z-plane

Which of the following are correct? (More than one option may be correct.)

The Fourier series of an even function has no sine terms

For a causal, absolutely integrable signal the Fourier transform is the Laplace transform at s = jω

The z-transform of δ[n] converges for every z

The Fourier series of an even function has no sine terms

For a causal, absolutely integrable signal the Fourier transform is the Laplace transform at s = jω

The z-transform of δ[n] converges for every z

A Fourier series converges to f(x) at every point, including jumps

Transform Theory — Fourier Series, the Fourier Transform and the z-Transform

Check which part is yours before you start. Electrical asks all of this — Transform Theory is a named heading of its mathematics section and its calculus names Fourier series too. Mechanical and Instrumentation ask the first section only: each names Fourier series in its calculus and neither names a transform in its mathematics section, so read section one and stop. Instrumentation does meet the Laplace and z-transforms, but under Signals and Systems, which is a discipline section rather than this one. No other verified paper names any of it in its mathematics section, and if yours is not one of those three, check your own syllabus before spending time here. What unites the three tools is a single idea worth holding above the formulas: a transform trades a hard operation for an easy one. Differentiation becomes multiplication, convolution becomes ordinary multiplication, and a differential equation becomes algebra — which is exactly what the Laplace transform did for initial-value problems in the differential-equations chapter, and these are the same trick in two other settings. A Fourier series writes a periodic function as a sum of sines and cosines, and the single most useful fact about it is that symmetry kills half the coefficients: an even function has only cosine terms, an odd function only sine terms, and noticing which one you have before integrating saves half the work every time. The second most useful fact is what happens at a jump — the series does not pick a side, it converges to the average of the two one-sided limits, so a square wave stepping from −1 to +1 has a series that equals 0 at the step. The Fourier transform is the same decomposition for a function that never repeats, an integral instead of a sum, and it is the Laplace transform evaluated on the imaginary axis: L{e^−atu(t)} = 1/(s + a) becomes 1/(a + jω), which is exactly the Fourier transform of the same signal. The z-transform is the discrete counterpart, X(z) = Σ x[n]z⁻ⁿ, and it carries one thing the others do not make you think about: the region of convergence. z/(z − a) is the transform of a^n u[n] when |z| > |a| and of a quite different, left-sided sequence when |z| < |a| — the algebra is identical and the signal is not, so an answer that gives the formula without the ROC has answered half the question.

Fourier Series — Symmetry First, Integration Second

On (−π, π) a periodic f is written as a₀/2 + Σ [aₙ cos nx + bₙ sin nx], with aₙ = (1/π)∫f cos nx dx and bₙ = (1/π)∫f sin nx dx over the period. Before computing either integral, look at the symmetry — it decides which of them is zero, and it is decided by a glance rather than by work.

Symmetry, and the two worked series
Function on (−π, π)	Symmetry	Coefficients, computed
f(x) = x	Odd	All aₙ = 0, including a₀. bₙ = 2(−1)ⁿ⁺¹/n, so b₁ = 2, b₂ = −1, b₃ = 2/3 — a pure sine series.
f(x) = x²	Even	All bₙ = 0. a₀/2 = π²/3 and aₙ = 4(−1)ⁿ/n², so a₁ = −4 and a₂ = 1 — a pure cosine series.

⚠️ At a jump the series takes the average, not a side

Where f is continuous the Fourier series converges to f. Where it jumps, the series converges to [f(x⁻) + f(x⁺)]/2, the midpoint of the two one-sided limits — it belongs to neither branch. For the square wave that is −1 on (−π, 0) and +1 on (0, π), the series at x = 0 equals (−1 + 1)/2 = 0, which is a value the function itself never takes. The same happens at x = π, where the periodic extension steps back down. A question that asks "to what value does the series converge at x = 0" is asking exactly this, and the options will contain −1, 0 and +1.

The Fourier Transform, and Its Laplace Cousin

The definition. F(ω) = ∫ f(t)e^−jωt dt over the whole line. It is the Fourier series idea for a function that never repeats: a sum over discrete harmonics becomes an integral over a continuum of frequencies.
The two transforms worth knowing by heart. e^−a|t| transforms to 2a/(a² + ω²), and the rectangular pulse that is 1 on [−T, T] transforms to 2 sin(ωT)/ω, which tends to 2T as ω → 0 — the area under the pulse, as the definition requires when the exponential becomes 1.
Narrow in time is wide in frequency. Shrinking T makes the pulse briefer and its transform broader, and the product of the two widths cannot be made small. That is one statement of the uncertainty principle and the reason a short pulse needs a wide channel.
It is Laplace on the imaginary axis. For a signal that is zero before t = 0 and decays, putting s = jω into the Laplace transform gives the Fourier transform: L{e^−atu(t)} = 1/(s + a) becomes 1/(a + jω), and integrating the Fourier definition directly gives the same thing. That is why the differential-equations chapter already did most of the work in this one.

The z-Transform, and Why the ROC Is Half the Answer

The four transforms GATE sets, each with its region of convergence
x[n]	X(z)	Region of convergence
δ[n]	1	The whole plane — one term, nothing to converge.
u[n] (unit step)	z/(z − 1)	\|z\| > 1
aⁿ u[n]	z/(z − a)	\|z\| > \|a\| — right-sided, so the ROC is outside the pole.
−aⁿ u[−n − 1]	z/(z − a) — the same expression	\|z\| < \|a\| — left-sided, so the ROC is inside the pole. Same algebra, different signal.

🎯 Why the region of convergence is not a footnote

The last two rows of that table have identical X(z) and completely different x[n] — one decays forwards in time, the other backwards. The expression alone therefore does not determine the signal, and a question that asks for the inverse z-transform of z/(z − a) without giving an ROC is either incomplete or is testing whether you notice. The ROC also carries the stability answer: a discrete system is stable exactly when its ROC contains the unit circle, which for a causal system means every pole lies strictly inside |z| = 1. The final-value theorem is the other result worth carrying — lim x[n] = lim(z→1) (z − 1)X(z) when the limit exists, so for X(z) = z/(z − 1) − z/(z − ½) it gives 1, matching x[n] = 1 − (½)ⁿ.

Key takeaways

Check which part is yours: Electrical asks all of this, Mechanical and Instrumentation ask the Fourier-series section only, and no other verified paper names any of it in its mathematics section.
A transform trades a hard operation for an easy one — differentiation becomes multiplication and a differential equation becomes algebra. It is the same trick the Laplace transform played.
Look at symmetry before integrating: an even function has only cosine terms, an odd function only sine terms. It halves the work every time.
At a jump the series converges to the average of the one-sided limits, a value the function may never take. For a ±1 square wave that is 0 at the step.
Know two Fourier transforms: e^−a|t| → 2a/(a² + ω²), and the pulse on [−T, T] → 2 sin(ωT)/ω, whose value at ω = 0 is the area 2T.
The Fourier transform is Laplace on the imaginary axis: 1/(s + a) at s = jω is 1/(a + jω), the transform of the same decaying signal.
z/(z − a) is two different signals depending on the ROC — right-sided outside the pole, left-sided inside it. An answer without an ROC is half an answer.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The Fourier series of f(x) = x on (−π, π) contains:
1. Cosine terms only
2. Sine terms only
3. Both sine and cosine terms
4. A constant term and cosine terms
Show answer
Answer: B — Sine terms only
f(x) = x is odd, and an odd function integrated against the even function cos nx over a symmetric interval gives zero — so every aₙ vanishes, a₀ included, and only sine terms survive. Computing them confirms it: bₙ = 2(−1)ⁿ⁺¹/n, so b₁ = 2, b₂ = −1, b₃ = 2/3. The general rule is the one to carry rather than the coefficients: odd → sine only, even → cosine only (plus the constant). Option D is the answer for an even function such as x², whose series is π²/3 + Σ 4(−1)ⁿ cos nx / n². Checking the symmetry first is not a shortcut so much as the method — it decides half the integrals before any of them is done.
A periodic function equals −1 on (−π, 0) and +1 on (0, π). At x = 0 its Fourier series converges to:
1. −1
2. 0
3. +1
4. The series diverges at x = 0
Show answer
Answer: B — 0
At a jump discontinuity the series converges to the average of the one-sided limits: [f(0⁻) + f(0⁺)]/2 = (−1 + 1)/2 = 0. Note that the function never takes the value 0 anywhere, so the series converges to something the function is not — which is the whole point of the question and why both −1 and +1 are offered. Every partial sum of this series is in fact exactly 0 at x = 0, since the series is pure sine and every sine vanishes there. Option D is wrong: the series converges everywhere for a function satisfying the Dirichlet conditions, and a finite jump is allowed. The same average applies at x = ±π, where the periodic extension steps back down.
The Fourier transform of e^−a|t| (a > 0), with the convention F(ω) = ∫ f(t)e^−jωt dt, is:
1. 1/(a + jω)
2. 2a/(a² + ω²)
3. a/(a² + ω²)
4. 2/(a² + ω²)
Show answer
Answer: B — 2a/(a² + ω²)
Split the integral at zero and use the symmetry: the signal is even, so the transform is real and equals twice the integral over t > 0 of e^−atcos ωt, which is 2a/(a² + ω²). Option A is the transform of the one-sided e^−atu(t), which is the same decay without the mirror image on the left — a different signal, and the option a candidate reaches by forgetting the absolute value. A check that costs nothing: at ω = 0 the transform must equal the area under the curve, and ∫e^−a|t|dt = 2/a, which is what B gives and C does not. That check catches both the missing factor 2 and a wrong power in the denominator.
The z-transform of the unit step u[n] is:
1. 1, for all z
2. z/(z − 1), for |z| > 1
3. 1/(z − 1), for |z| > 1
4. z/(z − 1), for |z| < 1
Show answer
Answer: B — z/(z − 1), for |z| > 1
X(z) = Σ_n≥0 z⁻ⁿ is a geometric series with ratio 1/z, so it sums to 1/(1 − 1/z) = z/(z − 1), and it converges exactly when |1/z| < 1, that is |z| > 1. Option D pairs the right expression with the wrong region, and that pairing is not a detail: with |z| < 1 the same expression is the transform of the left-sided sequence −u[−n − 1], a completely different signal. Option A is the transform of δ[n], the one sequence whose transform converges everywhere because the sum has a single term. Option C is what you get by summing from n = 1 instead of n = 0, which drops the sample at the origin.
X(z) = z/(z − a). Which statement is correct?
1. It is the transform of aⁿu[n], whatever the region of convergence
2. It is aⁿu[n] when |z| > |a| and a different, left-sided sequence when |z| < |a|
3. It has no inverse unless a = 0
4. Its region of convergence is always the whole z-plane
Show answer
Answer: B — It is aⁿu[n] when |z| > |a| and a different, left-sided sequence when |z| < |a|
The expression alone does not determine the signal. With ROC |z| > |a| the inverse is the right-sided aⁿu[n]; with ROC |z| < |a| it is the left-sided −aⁿu[−n − 1]. Same algebra, opposite direction in time — which is why an inverse z-transform question is incomplete without an ROC, and why option A is the trap. The ROC also answers stability: a causal system is stable exactly when its ROC includes the unit circle, which means |a| < 1. Option D is false for any a ≠ 0, since the series cannot converge at the pole itself, and it is only δ[n] whose transform converges on the whole plane.
Which of the following are correct? (More than one option may be correct.)
1. The Fourier series of an even function has no sine terms
2. For a causal, absolutely integrable signal the Fourier transform is the Laplace transform at s = jω
3. The z-transform of δ[n] converges for every z
4. A Fourier series converges to f(x) at every point, including jumps
Show answer
Answer: A — The Fourier series of an even function has no sine terms; B — For a causal, absolutely integrable signal the Fourier transform is the Laplace transform at s = jω; C — The z-transform of δ[n] converges for every z
Three hold. A is the symmetry rule — an even f integrated against sin nx over a symmetric interval gives zero. B is the relationship worked in this chapter: 1/(s + a) at s = jω is 1/(a + jω), which is what integrating the Fourier definition returns. C is true because the sum has exactly one non-zero term, so there is nothing to converge and the ROC is the whole plane — the only transform in the table for which that is so. Only D fails, at exactly the points the chapter warns about: at a jump the series converges to the average of the one-sided limits, which is generally not f(x) and may be a value f never takes.
For the Fourier series of f(x) = x on (−π, π), the coefficient b₁ is _____.
Numerical answer — type the value.
Show answer
Answer: 2
bₙ = (1/π)∫_−π^π x sin nx dx, and integrating by parts gives bₙ = 2(−1)ⁿ⁺¹/n, so b₁ = 2. The pattern is worth reading off rather than recomputing each time: b₂ = −1, b₃ = 2/3, b₄ = −1/2 — alternating in sign and decaying as 1/n, which is the slow decay characteristic of a function whose periodic extension jumps. A useful check on any Fourier coefficient: aₙ and bₙ must tend to zero as n grows, so a coefficient that does not decay is wrong whatever the arithmetic said.
A discrete signal has X(z) = z/(z − 1) − z/(z − 0.5), with the region of convergence |z| > 1. By the final value theorem, lim(n→∞) x[n] is _____.
Numerical answer — type the value.
Show answer
Answer: 1
The final value theorem gives lim x[n] = lim(z→1) (z − 1)X(z). The (z − 1) cancels the first pole and kills the second term, leaving z → 1 in z, so the limit is 1. Check it against the sequence itself: the two terms invert to u[n] and (0.5)ⁿu[n], so x[n] = 1 − (0.5)ⁿ, which does tend to 1. The theorem has a condition worth respecting — it applies only when the limit exists, which for a causal signal means every pole of (z − 1)X(z) lies strictly inside the unit circle. Replace 0.5 by 2 and the sequence diverges while the algebra still produces a number, which is how this question is set as a trap in harder years.