Consider a continuous-time signal x(t) defined by x(t) = 0 for |t| > 1, and x(t) = 1 - |t| for |t| ≤ 1. Let the Fourier transform of x(t) be defined as$\mathrm{ X}(\omega )={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)e^{-j\omega t}dt}$ . The maximum magnitude of X(ω) is _____

To find the maximum magnitude of the Fourier transform $X(\omega )$ of the signal $x(t)$, we begin by analyzing the properties of the signal $x(t)$.

The signal is defined as:
$x(t)=\{\begin{array}{ll}1-\left|t\right|& \text{for }\left|t\right|\le 1\\ 0& \text{for }\left|t\right|>1\end{array}$
This is a standard triangular pulse centered at $t=0$ with a peak value of $1$ and a base extending from $t=-1$ to $t=1$. Note that $x(t)\ge 0$ for all $t$.

The Fourier transform of $x(t)$ is given by:
$X(\omega )={\int }_{-\infty }^{\infty }x(t)e^{-j\omega t}dt$
Taking the magnitude of $X(\omega )$, we have:
$|X(\omega )|=|{\int }_{-\infty }^{\infty }x(t)e^{-j\omega t}dt|$

Using the triangle inequality for integrals, we can write:
$|{\int }_{-\infty }^{\infty }x(t)e^{-j\omega t}dt|\le {\int }_{-\infty }^{\infty }|x(t)e^{-j\omega t}|dt$
Since $\left|e^{-j\omega t}\right|=1$ and $x(t)\ge 0$ everywhere, the inequality simplifies to:
$|X(\omega )|\ge 0\text{and}|X(\omega )|\le {\int }_{-\infty }^{\infty }x(t)dt$

This upper bound is achieved precisely at $\omega =0$, where:
$X(0)={\int }_{-\infty }^{\infty }x(t)dt$
Since $x(t)$ is non-negative, the maximum magnitude of $X(\omega )$ is equal to the area under the signal $x(t)$.

We calculate this area:
${\int }_{-\infty }^{\infty }x(t)dt={\int }_{-1}^1(1-\left|t\right|)dt$

Since the integrand is an even function, we can compute it as:
$2{\int }_0^1(1-t)dt=2{[t-\frac{t^2}{2}]}_0^1=2(1-\frac{1}{2})=2\times \frac{1}{2}=1$

Alternatively, this represents the area of a triangle with base width $b=2$ (from $t=-1$ to $t=1$) and height $h=1$:
$\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 2\times 1=1$

Therefore, the maximum magnitude of $X(\omega )$ is $1$.