Fourier Transform Calculation for $e^{-|2t|}$
To find the Fourier Transform of $f(t) = e^{-|2t|}$, we use the definition:
$ F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt $Substitute the given function:
$ F(\omega) = \int_{-\infty}^{\infty} e^{-|2t|} e^{-i\omega t} dt $Handling the Absolute Value
The term $|2t|$ requires splitting the integral based on the sign of $t$:
- For $t \ge 0$, $|2t| = 2t$, so $e^{-|2t|} = e^{-2t}$.
- For $t < 0$, $|2t| = -2t$, so $e^{-|2t|} = e^{-(-2t)} = e^{2t}$.
Splitting and Evaluating the Integral
Split the integral into two parts:
$ F(\omega) = \int_{-\infty}^{0} e^{2t} e^{-i\omega t} dt + \int_{0}^{\infty} e^{-2t} e^{-i\omega t} dt $Combine terms and integrate:
$ F(\omega) = \int_{-\infty}^{0} e^{(2 - i\omega)t} dt + \int_{0}^{\infty} e^{(-2 - i\omega)t} dt $Evaluate the first integral:
$ \int_{-\infty}^{0} e^{(2 - i\omega)t} dt = \left[ \frac{e^{(2 - i\omega)t}}{2 - i\omega} \right]_{-\infty}^{0} = \frac{e^0}{2 - i\omega} - \lim_{t \to -\infty} \frac{e^{(2 - i\omega)t}}{2 - i\omega} = \frac{1}{2 - i\omega} - 0 = \frac{1}{2 - i\omega} $Evaluate the second integral:
$ \int_{0}^{\infty} e^{(-2 - i\omega)t} dt = \left[ \frac{e^{(-2 - i\omega)t}}{-2 - i\omega} \right]_{0}^{\infty} = \lim_{t \to \infty} \frac{e^{(-2 - i\omega)t}}{-2 - i\omega} - \frac{e^0}{-2 - i\omega} = 0 - \frac{1}{-2 - i\omega} = \frac{1}{2 + i\omega} $Combining Results
Add the results of the two integrals:
$ F(\omega) = \frac{1}{2 - i\omega} + \frac{1}{2 + i\omega} $Find a common denominator:
$ F(\omega) = \frac{(2 + i\omega) + (2 - i\omega)}{(2 - i\omega)(2 + i\omega)} $Simplify the numerator and denominator:
$ F(\omega) = \frac{4}{2^2 - (i\omega)^2} = \frac{4}{4 - (-\omega^2)} = \frac{4}{4 + \omega^2} $The Fourier Transform of $e^{-|2t|}$ is $\frac{4}{4 + \omega^2}$.
Important Questions from Fourier Transform
- The FT of $x(t) = e^{4t} u(-t)$ is:
The function f(t) is a periodic function of period 2π. In the range (-π, π), it equals e-t. If f(t) = \(\sum\nolimits_{ - \infty }^\infty {{c_n}{e^{{\mathop{\rm int}} }}}\) denotes its Fourier series expansion, the sum \({\sum\nolimits_{ - \infty }^\infty {\left| {{c_n}} \right|} ^2}\) is
Fourier transform of the unit impulse δ(t) is
Differentiating a signal in the time domain corresponds to _________ its FT in the frequency domain by _________.
The function f(t) has a Fourier transform F(ω). The Fourier transform of F(t) is
