Why Does a Fourier Series Overshoot by About 9% at a Jump, However Many Terms You Add?

Studyloaf Team•October 10, 2026•Last verified October 10, 2026•7 min read

Because adding terms makes the overshoot narrower, not lower. For a square wave that jumps between −1 and +1, every partial sum peaks just after the jump, and as the number of terms grows that peak settles at 1.179, an overshoot of 0.179, which is 8.95% of the jump of 2. This is the Gibbs phenomenon: the series still converges in energy, but not uniformly near a discontinuity.

What you'll learn

  • Which square wave and series this post uses, with every convention stated
  • Why the peak of the partial sum settles near 1.179 instead of falling to 1
  • A worked example: the peak of the five-term partial sum, computed term by term
  • What does shrink as you add terms: the width of the ringing and the mean-square error
  • What the series gives at the jump itself

Which square wave and series are we using?

Take the square wave x(t) = +1 for 0 < t < T/2 and −1 for T/2 < t < T, repeating with period T. It is an odd function with zero average, so its Fourier series has only sine terms, and only at odd harmonics: x(t) = (4/π)[sin(ω0t) + sin(3ω0t)/3 + sin(5ω0t)/5 + ...], where ω0 = 2π/T. Write SN(t) for the partial sum that keeps every odd harmonic up to N. So S9 has five terms (k = 1, 3, 5, 7, 9), and S49 has 25.

Amplitudes of the square wave's harmonicsBar chart of the Fourier series amplitudes 4 divided by pi k for the odd harmonics k equals 1, 3, 5, 7 and 9: 1.273, 0.424, 0.255, 0.182 and 0.141. Even harmonics are zero. The amplitudes fall only as 1 over k, which is why a jump needs so many terms.135790.00.40.81.2harmonic number kamplitude 4/(πk)1.2730.4240.2550.1820.141
The harmonic amplitudes 4/(πk) for k = 1 to 9. They fall only as 1/k, so a sharp jump needs many terms, and the slow decay is what makes the overshoot persist.

Why doesn't the peak fall to 1?

Each partial sum has its first maximum after the jump at t = T/(2(N + 1)), so as N grows the peak moves towards the jump. Near the jump the sum behaves like a scaled copy of the same curve, squeezed into a narrower window, so the height of its first peak barely changes. In the limit the peak value works out to (2/π) times the sine integral Si(π), which is 1.1790. Computing the peaks directly shows the levelling off: 1.273 with one term (N = 1), 1.200 with N = 3, 1.182 with N = 9, 1.180 with N = 25, and 1.179 for N = 101 and N = 1001.

Peak of the partial sum against the number of harmonicsComputed peak values of the partial sum for highest harmonic N equals 1, 3, 9, 25, 101 and 1001: 1.273, 1.200, 1.182, 1.180, 1.179 and 1.179. They level off at the Gibbs limit of 1.179, an overshoot of about 9 percent of the jump, instead of falling to 1.1392510110011.151.201.251.30highest harmonic Npeak value1.2731.2001.182limit 1.179
Computed peak of the partial sum against the highest harmonic N. The values level off at the limit 1.179 instead of falling towards 1.

Worked example: the peak of the five-term sum

For N = 9 the first peak is at t = T/(2 × 10) = T/20, where ω0t = 2π/20, so the k-th harmonic's angle is k × 18°. The five terms sin(k × 18°)/k are: sin 18° = 0.3090; sin 54°/3 = 0.2697; sin 90°/5 = 0.2000; sin 126°/7 = 0.1156; sin 162°/9 = 0.0343. They add to 0.9286, and multiplying by 4/π = 1.2732 gives S9(T/20) = 1.1823. The partial sum overshoots +1 by 0.182, or 9.1% of the jump. With 25 terms (N = 49) the peak is 1.179 at t = T/100: five times closer to the jump, almost exactly as high.

Partial sums of the square wave near the jump at t equals 0The ideal square wave jumps from minus 1 to plus 1 at t equals 0. The partial sum with harmonics up to 9 rings and peaks at 1.182 at t over T equals 0.05. The partial sum with harmonics up to 49 hugs the jump more tightly but still peaks at 1.179, at t over T equals 0.01. Both pass through 0 at the jump itself.-0.10.00.10.20.3-101time t/Tpartial sumN = 9: peak 1.182N = 49: peak 1.179dashed N = 9, solid N = 49thin line: ideal
The partial sums with N = 9 and N = 49 near the jump at t = 0, computed at 501 points. The N = 49 sum hugs the jump far more tightly, but its first peak is almost as high.

What does get better with more terms?

  • The width of the overshoot: the first peak sits at T/(2(N + 1)), so ten times as many terms squeeze the ringing into a window about ten times narrower.
  • The mean-square error over a period: 0.0404 with N = 9, 0.0156 with N = 25 and 0.0040 with N = 101. These match Parseval's theorem exactly, since the error power equals the power in the harmonics left out.
  • The value at any fixed point away from the jump: S9(T/4) = 1.063 while S101(T/4) = 1.006, both heading to 1.

What does the series give at the jump itself?

At t = 0, and at every other jump, every sine term is zero, so every partial sum equals 0: the midpoint of −1 and +1. That is the general rule for Fourier series at a jump discontinuity (under the usual Dirichlet conditions): the series converges to the average of the left and right limits, not to either side.

Why does it matter outside the exam?

Truncating a Fourier series is the same as passing the signal through an ideal low-pass filter that keeps the first few harmonics. So the same overshoot appears whenever a signal with sharp edges is band-limited too abruptly: ringing on a filtered step, or ripple near the edges of a filter designed by cutting off an ideal frequency response. Windowing methods in FIR filter design taper the cutoff for exactly this reason, trading a wider transition band for much smaller ripple.

Common mistakes

  • Saying the overshoot is 9% of the amplitude. It is about 9% of the jump; for a jump from −1 to +1 that is 0.179, about 18% of the amplitude 1.
  • Expecting more terms to remove the overshoot. They narrow it; its height tends to 1.179, not 1.
  • Claiming the series fails to converge at a jump. It converges to the midpoint there, and converges in mean square everywhere.
  • Counting terms and harmonics as the same thing. Up to the 9th harmonic is five terms, because the even harmonics are zero.
  • Using the peak location T/(2N) instead of T/(2(N + 1)) for this square wave's partial sum.

Check your understanding

  • Q1. A square wave switches between −5 V and +5 V. Roughly what peak will a long partial sum reach? Answer: 5 × 1.179 = 5.89 V, an overshoot of about 0.89 V.
  • Q2. Where is the first peak after the jump for the partial sum with N = 49? Answer: t = T/(2 × 50) = T/100.
  • Q3. Going from N = 101 to N = 1001 changes the peak height by how much? Answer: almost nothing; both peaks are 1.179 to four significant figures, while the peak moves ten times closer to the jump.

Related

Official source

Last verified

The GATE 2027 EC syllabus lists Fourier series and Fourier transform under continuous-time signals in Section 2, Networks, Signals and Systems (checked against the official PDF from IIT Madras on 10 October 2026). Every partial sum, peak and error figure here was computed in code, with the peak finder cross-checked against a direct sum.

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