Evaluate the Fourier Transform of following continuous time signal:[i] x(t) = u(-t) * eA(-at) [ii] x(t) = 6 (t)

To evaluate the Fourier Transform of the given continuous-time signals, we will follow the steps outlined.

1. Break Down the Problem

We have two signals to analyze:

1. $x(t) = u(-t) e^{-at}$
2. $x(t) = 6 \delta(t)$ (assuming $6(t)$ is a delta function)

2. Relevant Concepts

The Fourier Transform $X(f)$ of a continuous-time signal $x(t)$ is defined as:

$X(f) = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt$

We will apply this definition to both signals.

3. Analysis and Detail

For the first signal $x(t) = u(-t) e^{-at}$:

• The unit step function $u(-t)$ is equal to 1 for $t < 0$ and 0 for $t \geq 0$. Thus, the expression simplifies to: $x(t) = e^{-at} \quad \text{for } t < 0$
• The Fourier Transform becomes: $X(f) = \int_{-\infty}^{0} e^{-at} e^{-j2\pi ft} dt$

We can simplify this integral: $X(f) = \int_{-\infty}^{0} e^{-(a + j2\pi f)t} dt$ Evaluating the integral results in: $X(f) = \left[ \frac{e^{-(a + j2\pi f)t}}{-(a + j2\pi f)} \right]_{-\infty}^{0} = \frac{1}{a + j2\pi f}$

For the second signal $x(t) = 6 \delta(t)$:

• The Fourier Transform of the delta function is given by: $X(f) = 6 \cdot \int_{-\infty}^{\infty} \delta(t) e^{-j2\pi ft} dt$ Using the sifting property of the delta function: $X(f) = 6$

4. Verify and Summarize

We have evaluated the Fourier Transforms for both signals:

1. For $x(t) = u(-t) e^{-at}$: $X(f) = \frac{1}{a + j2\pi f}$

2. For $x(t) = 6 \delta(t)$: $X(f) = 6$

Final Answer

• For $x(t) = u(-t) e^{-at}$: $X(f) = \frac{1}{a + j2\pi f}$
• For $x(t) = 6 \delta(t)$: $X(f) = 6$