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Fourier Transform: $$X(\omega) = \int_{t=-\infty}^\infty x(t)e^{-j\omega t}dt$$ Replace $\omega$ with $-\omega$: $$X(-\omega) = \int_{t=-\infty}^\infty x(t)e^{j\omega t}dt$$ Replace $t$ with $-t$: $$X(-\omega) = \int_{-t=\infty}^{-\infty} x(-t)e^{j\omega (-t)}d(-t)$$ $$= \int_{t=-\infty}^{\infty} x(-t)e^{-j\omega t}dt$$ To describe this property intuitively,...

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I think you misunderstand what the book is saying. There is nothing like a "unit-energy" that you can compute. There is, however, a "unit-energy [...] pulse", which is a pulse with energy equal to $1$. So if you have some pulse $p(t)$ with energy $$E_p=\int_{-\infty}^{\infty}|p(t)|^2dt\tag{1}$$ and you want to normalize it such that its ...

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So how come the magnitude of the Fourier transform is still the same? Because the Fourier Transform integrates from $-\infty$ to $+\infty$. There is no finite "observation window" and you always integrate over the entire signal. The time shift doesn't change that. Once you consider a finite interval, you are windowing the signal and that does ...

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If you have a piecewise function, you need to separate the integral over that function into the same number of (sub-)intervals: \begin{align}\int_0^4x(t)e^{-jk\omega_0t}dt&=\int_0^1x(t)e^{-jk\omega_0t}dt+\int_1^2x(t)e^{-jk\omega_0t}dt+\int_2^4x(t)e^{-jk\omega_0t}dt\\&=\int_0^1(-t)e^{-jk\omega_0t}dt+\int_1^2e^{-jk\omega_0t}dt\end{align} with $\... 1 Your method correctly produces the term $$\frac{1}{j\omega}\frac{\sin(\omega/2)}{\omega/2}$$ of the given solution, but you forget that by taking the derivative you lose information about any constant terms in$x(t)$. However, looking at the graph it's easy to figure out that the constant term in$x(t)$is$\frac12\$. The Fourier transform of that constant ...

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