Given the function $x(t) = \sin(10\pi t)$ and the period $T = \frac{1}{3}$, we are to calculate the trigonometric Fourier coefficients.

1. Determining the Fundamental Frequency

The fundamental frequency $f_0$ is given by: $f_0 = \frac{1}{T} = 3 \, \text{Hz}$ The angular frequency $\omega_0$ is: $\omega_0 = 2\pi f_0 = 6\pi \, \text{rad/s}$

2. Express $x(t)$ in terms of $\omega_0$

The given function $x(t) = \sin(10\pi t)$ can be rewritten as: $x(t) = \sin\left(\frac{10\pi}{6\pi} \cdot 6\pi t\right) = \sin\left(\frac{5}{3} \cdot 6\pi t\right)$ This shows that $x(t) = \sin\left(5 \omega_0 t\right)$.

3. Trigonometric Fourier Series

The Fourier series for $x(t)$ is given by: $x(t) = \sum_{n=-\infty}^{\infty} C_n e^{jn\omega_0 t}$ However, since $x(t)$ is purely sinusoidal, it will have non-zero coefficients only for $n = \pm 5$.

4. Fourier Coefficients

For $n = 5$: $C_5 = \frac{1}{2j}$ For $n = -5$: $C_{-5} = -\frac{1}{2j}$

All other Fourier coefficients $C_n$ for $n \neq \pm 5$ are zero.

Final Answer

The trigonometric Fourier coefficients for $x(t) = \sin(10\pi t)$ with $T = \frac{1}{3}$ are: $C_5 = \frac{1}{2j}, \quad C_{-5} = -\frac{1}{2j}, \quad \text{and } C_n = 0 \text{ for all } n \neq \pm 5$

Would you like further details or have any questions?

Related Questions:

1. How do you compute the Fourier series for a more complex periodic function?
2. What happens to the Fourier coefficients if the period $T$ is changed?
3. Can the Fourier coefficients be computed directly from the integral formulas?
4. How would the Fourier series differ if $x(t)$ were a cosine function instead of sine?
5. What is the significance of the magnitude and phase of Fourier coefficients?

Tip: Understanding the symmetry of the function can simplify the calculation of Fourier coefficients. For example, odd functions only have sine terms in their Fourier series.