Question

Calculate F.T of (d/dt {u(-2-t)+u(t-2}, sketh (x) lekel the mapitude of each FT

Answer 1

The Fourier Transform (F.T) of
`\( (d)/(dt) \{u(-2-t)+u(t-2)\} \)` results in
`\( X(f) = 2\pi f (1 - e^(-2\pi if)) \)`. The magnitude of the Fourier Transform varies with frequency. At f = 0, the magnitude is 0, and for other frequencies, the magnitude increases proportionally with f.

Step-by-step explanation:

The expression
`\( (d)/(dt) \{u(-2-t)+u(t-2)\} \)` represents the derivative of a piecewise function. It includes two unit step functions, u(-2-t) and u(t-2), which define different intervals for t. The derivative of this function involves finding the derivative of each unit step function within its defined interval.

Taking the derivative of the given piecewise function, the result is
`\( 2\delta(t-2) - 2\delta(t+2) \)`, where
`\( \delta(t) \)` is the Dirac delta function. Applying the Fourier Transform to this result, considering the properties of the Fourier Transform for derivatives and delta functions, yields
`\( X(f) = 2\pi f (1 - e^(-2\pi if)) \)`.

The magnitude of the Fourier Transform, represented by
`\( |X(f)| \)`, is 0 at
`\( f = 0 \)` due to the term involving f. For other frequencies f, the magnitude increases linearly with f, showing a linear relationship between the magnitude and frequency. This linear relationship demonstrates that higher frequencies contribute more significantly to the signal's representation in the frequency domain. Therefore, the magnitude of the Fourier Transform varies proportionally with the frequency f for
`\( f \\eq 0 \)`, following the equation
`\( |X(f)| = 2\pi|f| (1 - e^(-2\pi |f|)) \)`.

Answer 2

Let's find the Fourier transform of the given function \( x(t) = \frac{d}{dt} [u(-2-t) + u(t-2)] \). First, let's express \( x(t) \) in a piecewise form:

\[ x(t) = \begin{cases}

1 & \text{if } -2 < t < 2 \\

0 & \text{otherwise}

\end{cases}

\]

Now, the Fourier transform \( X(f) \) of \( x(t) \) is given by:

\[ X(f) = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt \]

Since \( x(t) \) is a rectangular function with width \( 4 \) centered at \( t = 0 \), its Fourier transform is a sinc function:

\[ X(f) = 4 \text{ sinc}(4f) \]

Now, for the magnitude plot, the sinc function has zero crossings at integer multiples of \( \frac{1}{4} \) (due to the width of the rectangular function). The amplitude of the sinc function is \( 4 \), so the magnitude plot will have peaks at these zero crossings with amplitude \( 4 \). The width of the peaks will be inversely proportional to the width of the rectangular function, which is \( 4 \). Therefore, the peaks occur at \( f = \pm \frac{1}{4} \) and \( f = \pm \frac{3}{4} \).

In summary, the magnitude plot of the Fourier transform will have peaks at \( f = \pm \frac{1}{4} \) and \( f = \pm \frac{3}{4} \) with amplitude \( 4 \), forming a sinc function.