Math Problem Statement
Jawablah
Jawablah
Solution
The image shows an example of a Fourier transform being applied to a time-domain function that is given by:
\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.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Fourier Transform
Complex Functions
Integration
Frequency Domain Analysis
Formulas
Fourier Transform formula: H(f) = ∫ h(t) e^(-j2πft) dt
Magnitude of Fourier Transform: |H(f)| = β / √(α^2 + (2πf)^2)
Phase of Fourier Transform: θ(f) = tan^(-1)(-2πf/α)
Theorems
Fourier Transform Theorem
Suitable Grade Level
Undergraduate