1. **Problem Statement:** We are given a piecewise function: $$f(x) = \begin{cases} |x|, & -1 \leq x < 1 \\ 0, & \text{otherwise} \end{cases}$$ We need to compute the Fourier transform of $f(x)$ to analyze the frequency components of the signal. 2. **Fourier Transform Formula:** The Fourier transform $F(k)$ of a function $f(x)$ is defined as: $$F(k) = \int_{-\infty}^{\infty} f(x) e^{-i k x} \, dx$$ where $k$ is the frequency variable. 3. **Apply the formula to $f(x)$:** Since $f(x) = 0$ outside $[-1,1)$, the integral limits reduce to: $$F(k) = \int_{-1}^{1} |x| e^{-i k x} \, dx$$ 4. **Split the integral at 0 due to absolute value:** $$F(k) = \int_{-1}^{0} (-x) e^{-i k x} \, dx + \int_{0}^{1} x e^{-i k x} \, dx$$ 5. **Compute each integral separately:** - For the first integral: $$I_1 = \int_{-1}^{0} (-x) e^{-i k x} \, dx = - \int_{-1}^{0} x e^{-i k x} \, dx$$ - For the second integral: $$I_2 = \int_{0}^{1} x e^{-i k x} \, dx$$ 6. **Use integration by parts for $\int x e^{-i k x} dx$:** Let $u = x$, $dv = e^{-i k x} dx$, then $du = dx$, $v = \frac{e^{-i k x}}{-i k}$. So, $$\int x e^{-i k x} dx = \frac{x e^{-i k x}}{-i k} - \int \frac{e^{-i k x}}{-i k} dx = \frac{x e^{-i k x}}{-i k} + \frac{1}{i k} \int e^{-i k x} dx$$ The integral of $e^{-i k x}$ is $\frac{e^{-i k x}}{-i k}$, so: $$\int x e^{-i k x} dx = \frac{x e^{-i k x}}{-i k} + \frac{1}{i k} \cdot \frac{e^{-i k x}}{-i k} + C = \frac{x e^{-i k x}}{-i k} - \frac{e^{-i k x}}{k^2} + C$$ 7. **Evaluate $I_1$:** $$I_1 = - \left[ \frac{x e^{-i k x}}{-i k} - \frac{e^{-i k x}}{k^2} \right]_{x=-1}^{0} = - \left( \left[ \frac{0 \cdot e^{0}}{-i k} - \frac{e^{0}}{k^2} \right] - \left[ \frac{-1 \cdot e^{i k}}{-i k} - \frac{e^{i k}}{k^2} \right] \right)$$ Simplify inside: $$= - \left( - \frac{1}{k^2} - \left( \frac{-1}{-i k} e^{i k} - \frac{e^{i k}}{k^2} \right) \right) = - \left( - \frac{1}{k^2} - \left( \frac{1}{i k} e^{i k} - \frac{e^{i k}}{k^2} \right) \right)$$ $$= - \left( - \frac{1}{k^2} - \frac{1}{i k} e^{i k} + \frac{e^{i k}}{k^2} \right) = \frac{1}{k^2} + \frac{1}{i k} e^{i k} - \frac{e^{i k}}{k^2}$$ 8. **Evaluate $I_2$:** $$I_2 = \left[ \frac{x e^{-i k x}}{-i k} - \frac{e^{-i k x}}{k^2} \right]_{0}^{1} = \left( \frac{1 \cdot e^{-i k}}{-i k} - \frac{e^{-i k}}{k^2} \right) - \left( 0 - \frac{1}{k^2} \right) = \frac{e^{-i k}}{-i k} - \frac{e^{-i k}}{k^2} + \frac{1}{k^2}$$ 9. **Sum $I_1$ and $I_2$ to get $F(k)$:** $$F(k) = I_1 + I_2 = \left( \frac{1}{k^2} + \frac{1}{i k} e^{i k} - \frac{e^{i k}}{k^2} \right) + \left( \frac{e^{-i k}}{-i k} - \frac{e^{-i k}}{k^2} + \frac{1}{k^2} \right)$$ Group terms: $$F(k) = \frac{2}{k^2} + \frac{1}{i k} e^{i k} - \frac{e^{i k}}{k^2} - \frac{e^{-i k}}{i k} - \frac{e^{-i k}}{k^2}$$ 10. **Rewrite terms involving $i k$ and exponentials:** Note that $\frac{1}{i k} = - \frac{i}{k}$ and $\frac{e^{-i k}}{-i k} = - \frac{e^{-i k}}{i k}$. So, $$F(k) = \frac{2}{k^2} - \frac{i}{k} e^{i k} - \frac{e^{i k}}{k^2} + \frac{i}{k} e^{-i k} - \frac{e^{-i k}}{k^2}$$ 11. **Group real and imaginary parts:** $$F(k) = \frac{2}{k^2} - \frac{1}{k^2} (e^{i k} + e^{-i k}) - \frac{i}{k} (e^{i k} - e^{-i k})$$ Recall Euler's formulas: $$e^{i k} + e^{-i k} = 2 \cos k$$ $$e^{i k} - e^{-i k} = 2 i \sin k$$ Substitute: $$F(k) = \frac{2}{k^2} - \frac{2 \cos k}{k^2} - \frac{i}{k} (2 i \sin k) = \frac{2}{k^2} - \frac{2 \cos k}{k^2} - \frac{2 i^2 \sin k}{k}$$ Since $i^2 = -1$: $$F(k) = \frac{2}{k^2} - \frac{2 \cos k}{k^2} + \frac{2 \sin k}{k}$$ 12. **Final expression for the Fourier transform:** $$\boxed{F(k) = \frac{2}{k^2} (1 - \cos k) + \frac{2 \sin k}{k}}$$ 13. **Interpretation:** This Fourier transform shows the frequency components of the signal $f(x) = |x|$ on $[-1,1)$. The terms involving $\sin k$ and $\cos k$ indicate oscillatory components, and the denominators show how the amplitude decays with frequency. --- **Slug:** absolute value transform **Subject:** signal processing **Desmos:** {"latex":"F(k) = \frac{2}{k^2} (1 - \cos k) + \frac{2 \sin k}{k}","features":{"intercepts":true,"extrema":true}} **q_count:** 1