To find the Fourier series of the given periodic function $f(t)$, defined as:

• $f(t) = 0$ for $-\pi \leq t < -\frac{\pi}{2}$
• $f(t) = 1$ for $-\frac{\pi}{2} \leq t \leq \frac{\pi}{2}$
• $f(t) = 0$ for $\frac{\pi}{2} < t \leq \pi$

we first need to identify the coefficients of the Fourier series, which are given by:

$a_0 = \frac{1}{T} \int_{-T/2}^{T/2} f(t) \, dt$

$a_n = \frac{1}{T} \int_{-T/2}^{T/2} f(t) \cos\left(\frac{2\pi nt}{T}\right) \, dt$

$b_n = \frac{1}{T} \int_{-T/2}^{T/2} f(t) \sin\left(\frac{2\pi nt}{T}\right) \, dt$

where $T = 2\pi$ for our function.

Step 1: Calculate $a_0$

$a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) \, dt = \frac{1}{2\pi} \left( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dt \right)$ Calculating the integral:

$= \frac{1}{2\pi} \left( t \Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \right) = \frac{1}{2\pi} \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) = \frac{1}{2\pi} \cdot \pi = \frac{1}{2}$

Step 2: Calculate $a_n$

$a_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) \cos(t) \, dt$ Since $f(t) = 1$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$:

$= \frac{1}{2\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(t) \, dt$

Calculating this integral:

$= \frac{1}{2\pi} \left( \sin(t) \Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \right) = \frac{1}{2\pi} \left( \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) \right) = \frac{1}{2\pi} \left( 1 - (-1) \right) = \frac{1}{2\pi} \cdot 2 = \frac{1}{\pi}$

Step 3: Calculate $b_n$

$b_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) \sin(t) \, dt$

Since $f(t) = 1$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$:

$= \frac{1}{2\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(t) \, dt$

Calculating this integral:

$= \frac{1}{2\pi} \left( -\cos(t) \Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \right) = \frac{1}{2\pi} \left( -\cos\left(\frac{\pi}{2}\right) + \cos\left(-\frac{\pi}{2}\right) \right) = \frac{1}{2\pi} (0 - 0) = 0$

Step 4: Assemble the Fourier Series

Thus, the Fourier coefficients are:

• $a_0 = \frac{1}{2}$
• $a_n = \frac{1}{\pi}$ for $n = 1, 2, \ldots$
• $b_n = 0$

The Fourier series is given by:

$f(t) = \frac{1}{2} + \sum_{n=1}^{\infty} \frac{1}{\pi} \cos(nt)$

Conclusion

The Fourier series for the given function $f(t)$ is:

$f(t) = \frac{1}{2} + \frac{1}{\pi} \sum_{n=1}^{\infty} \cos(nt)$

If you have any questions or need further details, let me know! Here are some related questions you might find interesting:

1. How would the Fourier series change if the function were defined differently in one of the intervals?
2. Can you explain the significance of the Fourier coefficients in signal processing?
3. What is the convergence behavior of this Fourier series?
4. How would you compute the Fourier series for a piecewise linear function?
5. What role does the period of the function play in the Fourier series representation?

Tip: When working with Fourier series, always check the function's continuity and periodicity in the defined intervals, as this affects convergence.