The Fourier transform X(jω) of the signal

$\mathrm{x}(\mathrm{t})=\frac{\mathrm{t}}{(1+{\mathrm{t}}^2{)}^2}$

is _______.

Fourier Transform of Signal x(t)

This solution details the process of calculating the Fourier Transform of the specific signal x(t), which is given by $x(t)=\frac{t}{(1+t^2{)}^2}$. We will use standard Fourier Transform properties to find the result.

Identifying the Signal and Key Fourier Transform Pair

The signal we need to analyze is: $x(t)=\frac{t}{(1+t^2{)}^2}$ To compute its Fourier Transform, denoted as X(jω), it's helpful to start with a known Fourier Transform pair that resembles our signal. A common pair involves the function y(t) = \frac{1}{1+t^2}.

Consider the function:

$y(t)=\frac{1}{1+t^2}$

The Fourier Transform of y(t) is a well-known result:

$Y(j\omega )=\mathcal{F}\{y(t)\}=\pi e^{-|\omega |}$

Relating the Signal using Differentiation Property

We can observe that the given signal x(t) is closely related to the derivative of the function y(t). Let's compute the derivative of y(t):

$y^{\prime }(t)=\frac{d}{dt}\left(\frac{1}{1+t^2}\right)$

Using the chain rule for differentiation (specifically, the power rule and the derivative of $1+t^2$):

$y^{\prime }(t)=-1\cdot (1+t^2{)}^{-2}\cdot \frac{d}{dt}(1+t^2)$

$y^{\prime }(t)=-(1+t^2{)}^{-2}\cdot (2t)$

$y^{\prime }(t)=\frac{-2t}{(1+t^2{)}^2}$

Now, by comparing x(t) with y'(t), we can establish a direct relationship:

$x(t)=\frac{t}{(1+t^2{)}^2}$

We see that:

$x(t)=-\frac{1}{2}\left(\frac{-2t}{(1+t^2{)}^2}\right)=-\frac{1}{2}{y}^{\prime }(t)$

This confirms that our signal x(t) is simply a scaled version (by $-\frac{1}{2}$) of the derivative of y(t).

Applying the Fourier Transform Differentiation Property

The Fourier Transform has a useful property related to differentiation in the time domain. The property states:

If $y(t)\leftrightarrow Y(j\omega )$, then $\frac{d}{dt}y(t)\leftrightarrow j\omega Y(j\omega )$.

We can use this property to find the Fourier Transform of y'(t). We already know $Y(j\omega )=\pi e^{-|\omega |}$.

Applying the property:

$\mathcal{F}\{y^{\prime }(t)\}=j\omega Y(j\omega )$

Substitute the known $Y(j\omega )$:

$\mathcal{F}\{y^{\prime }(t)\}=j\omega (\pi e^{-|\omega |})=j\pi \omega e^{-|\omega |}$

Calculating the Final Fourier Transform X(jω)

Now we can find the Fourier Transform of the original signal x(t) using the relationship $x(t)=-\frac{1}{2}{y}^{\prime }(t)$.

Taking the Fourier Transform of both sides:

$X(j\omega )=\mathcal{F}\{x(t)\}=\mathcal{F}\{-\frac{1}{2}{y}^{\prime }(t)\}$

Using the linearity property of the Fourier Transform (scaling):

$X(j\omega )=-\frac{1}{2}\mathcal{F}\{y^{\prime }(t)\}$

Substitute the result from the differentiation property:

$X(j\omega )=-\frac{1}{2}(j\pi \omega e^{-|\omega |})$

Simplifying the expression:

$X(j\omega )=-\frac{j\pi }{2}\omega e^{-|\omega |}$

Rewriting the Result and Comparing with Options

The calculated Fourier Transform is $X(j\omega )=-\frac{j\pi }{2}\omega e^{-|\omega |}$. Let's rewrite this expression to match the format provided in the multiple-choice options. We know that $\frac{1}{j}$ is equal to $-j$.

Therefore, we can rewrite our result as:

$X(j\omega )=\frac{\pi }{2}(-j)\omega e^{-|\omega |}$

$X(j\omega )=\frac{\pi }{2j}\omega e^{-|\omega |}$

Now, let's compare this final expression with the given options:

- Option 1: $\frac{\pi }{2j}\omega e^{-|\omega |}$
- Option 2: $\frac{\pi }{2}\omega e^{-|\omega |}$
- Option 3: $\frac{\pi }{2j}{e}^{-|\omega |}$
- Option 4: $\frac{\pi }{2}{e}^{-|\omega |}$

Our derived Fourier Transform, $\frac{\pi }{2j}\omega e^{-|\omega |}$, perfectly matches the expression in Option 1.

Conclusion

Based on the application of the Fourier Transform differentiation property and comparison with known pairs, the Fourier Transform of the signal $x(t)=\frac{t}{(1+t^2{)}^2}$ is determined to be $\frac{\pi }{2j}\omega e^{-|\omega |}$.