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tentukan deret fourier dari fungsi periodik ini

The function \( f(t) \) provided in the image is periodic with a period of \( T = 2 \), as indicated by the condition \( f(t+2) = f(t) \). The function is defined as a piecewise function:

\[
f(t) = \begin{cases} 
2(1 + t), & \text{for } -1 < t < 0 \\
0, & \text{for } 0 < t < 1 
\end{cases}
\]

To determine the Fourier series of this periodic function, we will use the Fourier series formula:

\[
f(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos \frac{n \pi t}{L} + b_n \sin \frac{n \pi t}{L} \right)
\]

where \( T = 2L = 2 \), which gives \( L = 1 \). The Fourier coefficients are calculated as follows:

### 1. \( a_0 \) (DC term):
\[
a_0 = \frac{1}{T} \int_{-L}^{L} f(t) \, dt = \frac{1}{2} \int_{-1}^{1} f(t) \, dt
\]
We split this into two parts, based on the definition of \( f(t) \):

\[
a_0 = \frac{1}{2} \left( \int_{-1}^{0} 2(1 + t) \, dt + \int_{0}^{1} 0 \, dt \right)
\]

### 2. \( a_n \) (Cosine coefficients):
\[
a_n = \frac{1}{L} \int_{-L}^{L} f(t) \cos \frac{n \pi t}{L} \, dt = \int_{-1}^{1} f(t) \cos(n \pi t) \, dt
\]
We again split this into two parts for the piecewise function.

### 3. \( b_n \) (Sine coefficients):
\[
b_n = \frac{1}{L} \int_{-L}^{L} f(t) \sin \frac{n \pi t}{L} \, dt = \int_{-1}^{1} f(t) \sin(n \pi t) \, dt
\]
As with the previous integrals, we compute this over two regions for the piecewise function.

I will now calculate these integrals.The Fourier series coefficients for the periodic function \( f(t) \) are:

### 1. \( a_0 \) (DC term):
\[
a_0 = 0.5
\]

### 2. \( a_n \) (Cosine coefficients):
\[
a_n = \begin{cases} 
\frac{2(1 - \cos(\pi n))}{\pi^2 n^2}, & \text{if } n \neq 0 \\
1, & \text{if } n = 0 
\end{cases}
\]

### 3. \( b_n \) (Sine coefficients):
\[
b_n = \begin{cases} 
\frac{2(-\pi n + \sin(\pi n))}{\pi^2 n^2}, & \text{if } n \neq 0 \\
0, & \text{if } n = 0 
\end{cases}
\]

Thus, the Fourier series for \( f(t) \) is given by:

\[
f(t) = 0.5 + \sum_{n=1}^{\infty} \left( \frac{2(1 - \cos(\pi n))}{\pi^2 n^2} \cos(n \pi t) + \frac{2(-\pi n + \sin(\pi n))}{\pi^2 n^2} \sin(n \pi t) \right)
\]

Let me know if you need further details or simplifications.

Here are some additional questions to explore:
1. How does the periodicity of the function affect the Fourier series?
2. What happens to the Fourier series when the function has different boundary conditions?
3. Can you find the specific values for the first few Fourier coefficients?
4. How does the convergence of the Fourier series behave for this particular function?
5. What would happen if the function was shifted in time?

**Tip:** When calculating Fourier series, remember that symmetry (even or odd functions) can simplify the calculation of certain coefficients.

Determine the Fourier series for a periodic piecewise function f(t) defined over [-1, 1], with period T = 2. Learn how to compute Fourier coefficients and express the function as a trigonometric series. Suitable for undergraduate students.

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