Introduction / Context:
Fourier series expansion allows us to represent periodic signals as sums of sine and cosine terms. Depending on whether the function is odd, even, or general, the series simplifies in specific ways. This question tests understanding of symmetry properties and their implications on Fourier coefficients.
Given Data / Assumptions:
- The function is periodic.
- The function is odd: f(−t) = −f(t).
- Standard Fourier series representation is assumed: f(t) = a0/2 + Σ [an cos(nω0 t) + bn sin(nω0 t)].
Concept / Approach:
For odd functions, integration over symmetric intervals causes cosine coefficients an to vanish because cosine is even and product of evenodd is odd, integrating to zero. The constant term a0 is also zero. The only surviving terms are sine terms (bn), which are odd functions themselves. Hence, only sine harmonics remain in the expansion.
Step-by-Step Solution:
Start with Fourier series: f(t) = a0/2 + Σ [an cos(nω0 t) + bn sin(nω0 t)].Odd function: f(−t) = −f(t).Check cosine terms: cos(nω0 t) is even → product with odd f(t) is odd → integral over symmetric interval cancels → an = 0.Check constant term: a0 = 0 by same logic.Check sine terms: sin(nω0 t) is odd → oddodd = even → integral is nonzero → bn survives.Therefore only sine harmonics remain.
Verification / Alternative check:
Take a simple odd function: f(t) = sin(t). Its Fourier series obviously has only sine terms.
Why Other Options Are Wrong:
Odd harmonics only: This applies to half-wave symmetry, not general odd symmetry.Even harmonics only: incorrect, cosine coefficients vanish for odd functions.Cosine harmonics only: cosine terms vanish due to odd symmetry.Both sine and cosine equally: occurs only for general non-symmetric functions.
Common Pitfalls:
Confusing odd symmetry with half-wave symmetry, which specifically removes even harmonics but not all cosine terms.
Final Answer:
Sine harmonics only
Discussion & Comments