Question 24.6: Find the Fourier transform of f(t) = {e^−3t t ≥ 0 e^3t t &lt......

F(\omega)=\int_{-\infty}^{\infty}f(t)\,\mathrm{e}^{-\mathrm{j}\omega t}\,\mathrm{d}t

=\int_{-\infty}^{0}\mathrm{e}^{3t}\,\mathrm{e}^{-\mathrm{j}\omega t}\,\mathrm{d}t+\int_{0}^{\infty}\mathrm{e}^{-3t}\,\mathrm{e}^{-\mathrm{j}\omega t}\,\mathrm{d}t

=\int_{-\infty}^{0}\mathbf{e}^{(3-\mathrm{j}\omega)t}\;\mathrm{d}t\;+\;\int_{0}^{\infty}\mathrm{e}^{-(3+\mathrm{j}\omega)t}\;\mathrm{d}t

=\left[\frac{e^{(3 – j \omega)t} }{3 – j \omega}\right]_{-\infty}^{0} + \left[\frac{e^{-(3 + j \omega)t}}{-(3+j\omega)}\right] _0 ^{\infty}

={\frac{1}{3-\mathrm{j}\omega}}+{\frac{1}{3+\mathrm{j}\omega}}

={\frac{6}{9+\omega^{2}}}

Now

G(\omega)=\frac{6}{10+2\omega+\omega^{2}}=\frac{6}{(\omega+1)^{2}+9}=F(\omega+1)

Then, using the first shift theorem F(ω + 1) will be  {\mathcal{F}}\{\mathrm{e}^{-\mathrm{j}t}f(t)\}; that is, the required function is

g(t)=\left\{\begin{array}{l l}{{\mathrm{e}^{(-3-{j})t}}}&{{t\geqslant0}}\\ {{\mathrm{e}^{(3-j)t}}}&{{t\lt 0}}\end{array}\right.