The Fourier series expansion of the square wave shown below contains:

Question

BlackTom AI · Accepted Answer

To analyze the Fourier series of a square wave, we need to consider the symmetry of the waveform.
Option 1: Only sine terms. This would be true if the square wave were an odd function about the origin (symmetric with respect to the origin), producing only sine (odd) harmonics. However, the plotted square wave sits around a nonzero average level (a dc offset) and is not symmetric about the origin, so sine terms alone cannot capture its shape.
Option 2: Only cosine terms. This would occur if the square wave were an even function (symmetric about the vertical axis) around a chosen origin, yielding only cosine (even) terms. The waveform shown does not exhibit even symmetry about the origin due to the offset and the way the square transitions, so cosine terms alone are insufficient.
Option 3: Both sine and cosine terms. Since the square wave in the figure is not centered at zero and lacks the requisite odd or even symmetry about the origin, its Fourier series typically requires both sine and cosine components to reproduce the waveform accurately, along with a possible constant (dc) term representing the average value. The combination of a nonzero mean and the lack of pure odd or pure even symmetry necessitates both kinds of harmonics.
In summary, the absence of pure odd or pure even symmetry and the presence of a dc offset imply that the Fourier expansion includes both sine and cosine terms (and possibly a constant term).

The Fourier series expansion of the square wave shown below contains:单项选择题

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