A continuous-time signal $x(t)$ is defined as $x(t)=0$ for $|t|1$, and $x(t)=1-|t|$ for $|t|\le 1$. Let its Fourier transform be $X(\omega)=\int_{-\infty}^{\infty} x(t)e^{-j\omega t}\,dt$. The maximum magnitude of $X(\omega)$ is $\underline{\hspace{2cm}}$.

Question

A continuous-time signal $x(t)$ is defined as $x(t)=0$ for $|t|>1$, and $x(t)=1-|t|$ for $|t|\le 1$. Let its Fourier transform be $X(\omega)=\int_{-\infty}^{\infty} x(t)e^{-j\omega t}\,dt$. The maximum magnitude of $X(\omega)$ is $\underline{\hspace{2cm}}$.

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At $\omega = 0$: $$ X(0) = \int_{-1}^{1} (1 - |t|)\,dt $$ Compute area (triangle of height 1, base 2): $$ X(0) = 1 $$ Since Fourier magnitude decreases away from $\omega=0$, the maximum occurs at zero frequency: $$ \max |X(\omega)| = |X(0)| = 1 $$ $$ \boxed{1} $$

A continuous-time signal \(x(t)\) is defined as \(x(t)=0\) for \(|t|>1\), and \(x(t)=1-|t|\) for \(|t|\le 1\). Let its Fourier transform be \(X(\omega)=\int_{-\infty}^{\infty} x(t)e^{-j\omega t}\,dt\). The maximum magnitude of \(X(\omega)\) is \(\underline{\hspace{2cm}}\).

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