Question

Derive, from the defining equations, the Fourier transform of the following signal.

x(t)=te−atu(t),a>0.
Hint: ∫te ⁻αᵗdt=−e⁻αᵗ (1+αt)/ /α². This expression is valid even when α is complex-valued.

Answer 1

To derive the Fourier transform of the given signal, we can apply the Fourier transform definition and use the provided hint to evaluate the integral. The final result is (1 + a + ijw) / (a^2 + 2ajw - w^2).

Step-by-step explanation:

To derive the Fourier transform of the given signal, x(t) = te^-atu(t), where a > 0, we can start by applying the Fourier transform definition:
F(w) = Integral[-Infinity to +Infinity] { x(t)e^-jwt } dt

Substituting the given signal into the integral:
F(w) = Integral[-Infinity to +Infinity] { te^-atu(t)e^-jwt } dt

Using the hint provided, we can evaluate the integral:
F(w) = Integral[0 to +Infinity] { te^(-a - ijw)t } dt

Applying the integration rule mentioned in the hint:
F(w) = [-e^(-a - ijw)t (1 + a + ijw) / (a^2 + 2ajw - w^2)] evaluated from 0 to +Infinity

As t approaches +Infinity, the exponential term goes to zero, leaving us with:
F(w) = -(-1)(1 + a + ijw) / (a^2 + 2ajw - w^2) = (1 + a + ijw) / (a^2 + 2ajw - w^2)