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The image shows an example of a **Fourier transform** being applied to a time-domain function \( h(t) \) that is given by:

\[
h(t) =
\begin{cases}
\beta e^{-\alpha t} & t > 0 \\
0 & t < 0
\end{cases}
\]

### The Fourier Transform is:

\[
H(f) = \int_{0}^{\infty} \beta e^{-\alpha t} e^{-j 2 \pi f t} \, dt
\]

### Step-by-step Solution:

1. Combine the exponential terms:

\[
H(f) = \beta \int_{0}^{\infty} e^{-(\alpha + j 2 \pi f)t} \, dt
\]

2. Now, integrate the exponential function:

\[
H(f) = \beta \left[ \frac{-e^{-(\alpha + j 2 \pi f)t}}{\alpha + j 2 \pi f} \right]_{0}^{\infty}
\]

At the limits:

- As \( t \to \infty \), \( e^{-(\alpha + j 2 \pi f)t} \to 0 \).
- At \( t = 0 \), \( e^{-(\alpha + j 2 \pi f) \cdot 0} = 1 \).

Therefore:

\[
H(f) = \frac{\beta}{\alpha + j 2 \pi f}
\]

3. Next, separate the real and imaginary parts:

\[
H(f) = \frac{\beta (\alpha - j 2 \pi f)}{\alpha^2 + (2 \pi f)^2}
\]

This gives us the **real part** \( R(f) \) and the **imaginary part** \( I(f) \) as:

\[
R(f) = \frac{\beta \alpha}{\alpha^2 + (2 \pi f)^2}
\]

\[
I(f) = \frac{-2 \pi f \beta}{\alpha^2 + (2 \pi f)^2}
\]

4. To calculate the **magnitude** \( |H(f)| \) of the Fourier transform:

\[
|H(f)| = \frac{\beta}{\sqrt{\alpha^2 + (2 \pi f)^2}}
\]

5. To calculate the **phase** \( \theta(f) \):

\[
\theta(f) = \tan^{-1} \left( \frac{-2 \pi f}{\alpha} \right)
\]

### Final Results:

- \( R(f) = \frac{\beta \alpha}{\alpha^2 + (2 \pi f)^2} \)
- \( I(f) = \frac{-2 \pi f \beta}{\alpha^2 + (2 \pi f)^2} \)
- \( |H(f)| = \frac{\beta}{\sqrt{\alpha^2 + (2 \pi f)^2}} \)
- \( \theta(f) = \tan^{-1} \left( \frac{-2 \pi f}{\alpha} \right) \)

Would you like further details on any step or clarifications?

Here are 5 related questions to explore this further:

1. How does the value of \( \alpha \) influence the shape of the Fourier transform?
2. What does the magnitude \( |H(f)| \) represent physically in the Fourier domain?
3. How would the Fourier transform change if \( h(t) \) was non-zero for \( t < 0 \)?
4. What is the significance of the real and imaginary parts in the Fourier transform?
5. How does the phase \( \theta(f) \) affect the interpretation of the transformed signal?

**Tip**: In many practical applications, the magnitude of the Fourier transform gives insight into the frequency content of the signal, while the phase provides information about the timing.

Explore the step-by-step solution of the Fourier Transform of the function h(t) = βe^(-αt) for t > 0. This detailed guide covers complex integration, separation of real and imaginary parts, and calculation of magnitude and phase.

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