Question

Derive, from the defining equations, the Fourier transform of the following signal. x(t)=te−ᵃᵗu(t),a>0. Hint:∫te−ᵃᵗdt=e−ᵃᵗ(1+αt)/a².This expression is valid even when α is complex-value

Answer 1

The Fourier transform of
`\(x(t) = te^(-\alpha t)u(t)\), where \(\alpha > 0\)`, is given by
`\(X(f) = (1)/((\alpha+j2\pi f)^3) + (1)/((\alpha+j2\pi f)^2)\)`.

Step-by-step explanation:

To derive the Fourier transform of the given signal
`\(x(t) = te^(-\alpha t)u(t)\)` where
`\(\alpha > 0\)`, we can use the definition of the Fourier transform and employ the provided hint. The Fourier transform of
`\(x(t)\)`, denoted as
`\(X(f)\)`, is given by:

`\[ X(f) = \int_(-\infty)^(\infty) x(t) e^(-j2\pi ft) \, dt \]`

Now, let's substitute the expression for
`\(x(t)\)` into this integral and simplify using the given hint.

`\[ X(f) = \int_(0)^(\infty) te^(-\alpha t)e^(-j2\pi ft) \, dt \]`

**Detailed Calculation:**

1. **Apply the hint:**

`\[ X(f) = \int_(0)^(\infty) te^(-\alpha t)e^(-j2\pi ft) \, dt = \int_(0)^(\infty) te^(-(\alpha+j2\pi f)t) \, dt \]`

2. **Integration:**

`\[ = \int_(0)^(\infty) te^(-st) \, dt \]`

`\[ \text{where } s = \alpha + j2\pi f \]`

3. **Use the provided expression in the hint:**

`\[ = \int_(0)^(\infty) (1)/(s^2)e^(-st)(1+st) \, dt \]`

`\[ = (1)/(s^2) \int_(0)^(\infty) (1+st) e^(-st) \, dt \]`

4. **Integrate each term separately:**

`\[ = (1)/(s^2) \left[ \int_(0)^(\infty) e^(-st) \, dt + \int_(0)^(\infty) ste^(-st) \, dt \right] \]`

5. **Evaluate each integral:**

`\[ = (1)/(s^2) \left[ (1)/(s) + (s)/(s^2) \right] \]`

`\[ = (1)/(s^3) + (1)/(s^2) \]`

6. **Substitute back for \(s\):**

`\[ = (1)/((\alpha+j2\pi f)^3) + (1)/((\alpha+j2\pi f)^2) \]`

Therefore, the Fourier transform
`\(X(f)\)` of the given signal
`\(x(t)\)` is:

`\[ X(f) = (1)/((\alpha+j2\pi f)^3) + (1)/((\alpha+j2\pi f)^2) \]`