Fourier Transform of \(x(t)=\dfrac{t}{(1+t^2)^2}\)
We want to determine the Fourier transform of
$$ x(t)=\frac{t}{(1+t^2)^2} $$using the Fourier transform definition
$$ X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}\,dt $$Step 1 :
Let
$$ x_1(t)=\frac{1}{1+t^2} $$Then
$$ X_1(\omega)=\pi e^{-|\omega|}. $$Proof:
$$ \int_{-\infty}^{\infty}\frac{1}{1+t^2}e^{-j\omega t}\,dt $$Use the Residue Theorem.
$$ \int_{-\infty}^{\infty}\frac{1}{1+z^2}e^{-j\omega z}\,dz $$Or,
$$ \int_{-\infty}^{\infty}\frac{1}{(z+j)(z-j)}e^{-j\omega z}\,dz $$Poles are at \(z=-j\) and \(z=j\).
For the pole at \(z=-j\):
$$ \operatorname*{Res}_{z=-j}\left(\frac{e^{-j\omega z}}{(z+j)(z-j)}\right) = \lim_{z\to -j}(z+j)\frac{e^{-j\omega z}}{(z+j)(z-j)} = \frac{e^{-\omega}}{-2j}. $$Therefore,
$$ -2\pi j\left(\frac{e^{-\omega}}{-2j}\right) = \pi e^{-\omega}. $$For the pole at \(z=j\):
$$ \operatorname*{Res}_{z=j}\left(\frac{e^{-j\omega z}}{(z+j)(z-j)}\right) = \lim_{z\to j}(z-j)\frac{e^{-j\omega z}}{(z+j)(z-j)} = \frac{e^{\omega}}{2j}. $$Therefore,
$$ 2\pi j\left(\frac{e^{\omega}}{2j}\right) = \pi e^{\omega}. $$Step 2 : Differentiate the Time-Domain Function
Step 3 : Apply the Differentiation Property
Step 4 : Multiply by \(-\frac12\)
Step 5 : Simplify
Final Answer
$$ \boxed{ \mathcal F \left\{ \frac{t}{(1+t^2)^2} \right\} = -\frac{j\pi\omega}{2} e^{-|\omega|} } $$Or,
$$ \boxed{ \mathcal F \left\{ \frac{t}{(1+t^2)^2} \right\} = \frac{\pi\omega}{2j} e^{-|\omega|} } $$Complex Plane Integration using Residue Theorem in Fourier Transform
Evaluating Fourier Transform integrals directly is often difficult. Complex Analysis provides a powerful technique called the Residue Theorem, which converts real integrals into contour integrals in the complex plane. This method is frequently used in GATE Signals and Systems.
Why Use the Residue Theorem?
For integrals of the form
$$ \int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt, $$direct integration may be impossible. By replacing the real variable with a complex variable \(z\), the integral becomes easier to evaluate using contour integration.
Residue Theorem
If \(f(z)\) has isolated poles inside a closed contour \(C\), then
$$ \oint_C f(z)\,dz = 2\pi j \sum \text{Residues inside }C. $$- \(\omega>0\) → Lower Half Plane
- \(\omega<0\) → Upper Half Plane
Fourier Transform Example
Find the Fourier Transform of
$$ x(t)=\frac{1}{1+t^2} $$Fourier Transform:
$$ X(\omega) = \int_{-\infty}^{\infty} \frac{e^{-j\omega t}} {1+t^2} dt $$The poles are located at
$$ z=\pm j. $$Choose the contour depending on the sign of \(\omega\), compute the enclosed residue, and apply the Residue Theorem.
The final result is
$$ \boxed{ X(\omega)=\pi e^{-|\omega|} } $$Exam Tips
- Always identify the poles first.
- Choose the contour based on the sign of \(\omega\).
- Only enclosed poles contribute.
- Use Jordan's Lemma for Fourier integrals.
Conclusion
The Residue Theorem is one of the fastest methods for evaluating difficult Fourier Transform integrals. Although contour selection requires practice, mastering this technique is valuable for GATE, ESE, and advanced Signals and Systems courses.