Math Problem Statement

Considerando la función de periodo T=8, obtenga su serie compleja. La función f(t) está definida por: f(t) = 0, para -4 < t < -2 f(t) = 3, para -2 < t < 2 f(t) = 0, para 2 < t < 4

Solution

We are tasked with finding the complex Fourier series for a periodic function f(t)f(t) with period T=8T = 8 and defined as follows:

0, & -4 < t < -2 \\ 3, & -2 < t < 2 \\ 0, & 2 < t < 4 \end{cases}$$ The general form of the complex Fourier series for a periodic function $$f(t)$$ is: $$f(t) = \sum_{n=-\infty}^{\infty} c_n e^{i \frac{2\pi n}{T} t}$$ where the coefficients $$c_n$$ are given by: $$c_n = \frac{1}{T} \int_{T} f(t) e^{-i \frac{2\pi n}{T} t} \, dt$$ Given that $$T = 8$$, let's compute the Fourier coefficients $$c_n$$ by integrating over one period, from $$-4$$ to $$4$$. ### Step 1: Compute the Fourier Coefficients The expression for $$c_n$$ becomes: $$c_n = \frac{1}{8} \int_{-4}^{4} f(t) e^{-i \frac{\pi n}{4} t} \, dt$$ We break this integral into three parts based on the piecewise definition of $$f(t)$$: $$c_n = \frac{1}{8} \left( \int_{-4}^{-2} 0 \cdot e^{-i \frac{\pi n}{4} t} \, dt + \int_{-2}^{2} 3 e^{-i \frac{\pi n}{4} t} \, dt + \int_{2}^{4} 0 \cdot e^{-i \frac{\pi n}{4} t} \, dt \right)$$ Since $$f(t) = 0$$ for $$t \in (-4, -2) \cup (2, 4)$$, only the integral from $$-2$$ to $$2$$ contributes to $$c_n$$: $$c_n = \frac{3}{8} \int_{-2}^{2} e^{-i \frac{\pi n}{4} t} \, dt$$ ### Step 2: Solving the Integral The integral we need to solve is: $$\int_{-2}^{2} e^{-i \frac{\pi n}{4} t} \, dt$$ This is a standard integral for the exponential function, and its result is: $$\int e^{ax} \, dx = \frac{1}{a} e^{ax}$$ Here, $$a = -i \frac{\pi n}{4}$$, so: $$\int_{-2}^{2} e^{-i \frac{\pi n}{4} t} \, dt = \frac{4}{-i \pi n} \sin\left( \frac{\pi n}{2} \right)$$ ### Step 3: Final Expression for $$c_n$$ Now, substituting the result back into the expression for $$c_n$$: $$c_n = \frac{3}{8} \cdot \frac{4}{-i \pi n} \sin\left( \frac{\pi n}{2} \right)$$ Thus, the Fourier coefficients are: $$c_n = \frac{3}{2 \pi n i} \sin\left( \frac{\pi n}{2} \right)$$ This gives us the coefficients for the complex Fourier series. For $$n = 0$$, we handle the case separately by evaluating the integral directly, since $$c_0$$ represents the average value over one period. Would you like the explicit Fourier series or further details on any part of this process? Here are 5 related questions you might explore: 1. How is the Fourier series connected to signal processing? 2. What happens to the Fourier coefficients if we change the period $$T$$? 3. Can we derive the real Fourier series from this complex series? 4. What are some applications of Fourier series in physics or engineering? 5. How do the harmonics in the Fourier series represent the function graphically? **Tip:** To handle discontinuous functions, the Fourier series converges to the average of the left-hand and right-hand limits at points of discontinuity.

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Math Problem Analysis

Mathematical Concepts

Fourier Series
Complex Fourier Series
Piecewise Functions
Integration

Formulas

c_n = \frac{1}{T} \int_T f(t) e^{-i \frac{2\pi n}{T} t} dt
f(t) = \sum_{n=-\infty}^{\infty} c_n e^{i \frac{2\pi n}{T} t}

Theorems

Fourier Series Theorem

Suitable Grade Level

Undergraduate (Calculus, Advanced Engineering Mathematics)