The CTFT of a continuous time signal x(t) = e-A|t|, A > 0 is _______
To find the Continuous Time Fourier Transform (CTFT) of a signal \(x(t)\), we use the definition:
\(X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt\)
The given signal is \(x(t) = e^{-A|t|}\), where \(A > 0\). The absolute value function \(|t|\) means we need to consider two cases: \(t \ge 0\) and \(t < 0\).
- For \(t \ge 0\), \(|t| = t\), so \(x(t) = e^{-At}\).
- For \(t < 0\), \(|t| = -t\), so \(x(t) = e^{-A(-t)} = e^{At}\).
Now, we can split the integral based on these two intervals:
\(X(j\omega) = \int_{-\infty}^{0} e^{At} e^{-j\omega t} dt + \int_{0}^{\infty} e^{-At} e^{-j\omega t} dt\)
Combine the exponential terms in each integral:
\(X(j\omega) = \int_{-\infty}^{0} e^{(A - j\omega)t} dt + \int_{0}^{\infty} e^{(-A - j\omega)t} dt\)
Evaluating the CTFT Integrals
Let's evaluate the first integral:
\(\int_{-\infty}^{0} e^{(A - j\omega)t} dt\)
The antiderivative of \(e^{(A - j\omega)t}\) with respect to \(t\) is \(\frac{1}{A - j\omega} e^{(A - j\omega)t}\). Evaluating the definite integral:
\(\left[ \frac{e^{(A - j\omega)t}}{A - j\omega} \right]_{-\infty}^{0} = \frac{e^{(A - j\omega)0}}{A - j\omega} - \lim_{t \to -\infty} \frac{e^{(A - j\omega)t}}{A - j\omega}\)
Since \(A > 0\), \(\lim_{t \to -\infty} e^{(A - j\omega)t} = \lim_{t \to -\infty} e^{At} e^{-j\omega t} = 0 \times (\text{bounded term}) = 0\).
So, the first integral is \(\frac{e^0}{A - j\omega} - 0 = \frac{1}{A - j\omega}\).
Now, let's evaluate the second integral:
\(\int_{0}^{\infty} e^{(-A - j\omega)t} dt\)
The antiderivative of \(e^{(-A - j\omega)t}\) with respect to \(t\) is \(\frac{1}{-A - j\omega} e^{(-A - j\omega)t}\). Evaluating the definite integral:
\(\left[ \frac{e^{(-A - j\omega)t}}{-A - j\omega} \right]_{0}^{\infty} = \lim_{t \to \infty} \frac{e^{(-A - j\omega)t}}{-A - j\omega} - \frac{e^{(-A - j\omega)0}}{-A - j\omega}\)
Since \(A > 0\), \(-A < 0\). Thus, \(\lim_{t \to \infty} e^{(-A - j\omega)t} = \lim_{t \to \infty} e^{-At} e^{-j\omega t} = 0 \times (\text{bounded term}) = 0\).
So, the second integral is \(0 - \frac{e^0}{-A - j\omega} = - \frac{1}{-A - j\omega} = \frac{1}{A + j\omega}\).
Combining the Results
Add the results of the two integrals to get the total CTFT \(X(j\omega)\):
\(X(j\omega) = \frac{1}{A - j\omega} + \frac{1}{A + j\omega}\)
Combine the fractions by finding a common denominator:
\(X(j\omega) = \frac{(A + j\omega) + (A - j\omega)}{(A - j\omega)(A + j\omega)}\)
Simplify the numerator and the denominator (using the difference of squares formula, \((a-b)(a+b) = a^2 - b^2\)):
\(X(j\omega) = \frac{A + j\omega + A - j\omega}{A^2 - (j\omega)^2}\)
\(X(j\omega) = \frac{2A}{A^2 - (j^2 \omega^2)}\)
Since \(j^2 = -1\):
\(X(j\omega) = \frac{2A}{A^2 - (-1)\omega^2}\)
\(X(j\omega) = \frac{2A}{A^2 + \omega^2}\)
This is the Continuous Time Fourier Transform of the signal \(x(t) = e^{-A|t|}\).
Important Questions from Fourier Series
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