For a signal u(sint); Fourier series is havinga)Odd harmonics of sine ...
Given x(t) = u(sint)
Drawing waveform

We know for waveform like y(t) shown

DC component is zero and have sine odd harmonics.
Performing upward shift in y(t) by adding dc value of 0.5, then
x(t) = y(t) + 0.5
So, x(t) → DC + sine odd harmonics
For a signal u(sint); Fourier series is havinga)Odd harmonics of sine ...
Understanding the Signal u(sin(t))
The signal u(sin(t)) is a function that is periodic with a fundamental frequency. To analyze its Fourier series representation, we must consider its symmetry and harmonic content.
Fourier Series Representation
- The Fourier series of a periodic function decomposes it into a sum of sine and cosine functions with different frequencies (harmonics).
- A key aspect of the Fourier series is that it can include both even and odd harmonics, as well as a DC component, depending on the signal's symmetry.
Symmetry Analysis
- The function u(sin(t)) is an odd function, meaning it is symmetric about the origin. This characteristic leads to the following implications:
- It contains only sine terms (which are odd functions) in its Fourier series expansion.
- There is no DC component since a DC term would be represented by a cosine term.
Harmonics Present
- The Fourier series will consist of the following:
- Odd Harmonics of Sine: The periodic nature of sin(t) leads to the presence of odd harmonics (3rd, 5th, etc.) in the expansion.
- DC Component: Since the signal is odd, the average value over one period is zero, leading to no DC term.
Conclusion
- Thus, the Fourier series of u(sin(t)) includes:
- Both odd harmonics of sine.
- Zero DC component.
- Therefore, the correct answer is option 'C': DC and odd harmonics of sine.
This understanding highlights the significance of symmetry in determining the harmonic content of periodic signals in the context of Fourier analysis.