Final answer:
The Fourier transform of x(t)=e^{-at}u(t) is computed first, resulting in X(f) = 1/(a + 2πif). Then, applying time-scaling and shift properties of the Fourier transform, the transform for v(t)=x(5t-4) is found to be V(f) = (1/(5a + 2πif))e^{-2πif(4)}.
Step-by-step explanation:
To compute the Fourier transform of x(t)=e⁻ảtu(t), where u(t) is the unit step function and a is a positive constant, we use the integral definition of the Fourier transform:
\(X(f) = \int_{-\infty}^{\infty} e⁻ảtu(t)e⁻²2πift dt\).
Since the unit step function u(t) is 0 for t < 0, we start the integral from 0:
\(X(f) = \int_{0}^{\infty} e⁻ảte⁻²2πift dt\).
Solving this integral yields the Fourier transform of x(t):
\(X(f) = \frac{1}{a + 2πif}\).
Next, to determine the Fourier transform for the signal v(t)=x(5t−4) using Fourier transform properties, we apply the time-scaling property, which states that a scaling of the time variable by a factor A corresponds to a scaling of the frequency axis by 1/A. In this case, the scaling factor is 5. Additionally, due to the time shift by 4, we apply a phase shift in the Fourier domain. Applying these properties results in:
\(V(f) = \frac{1}{5}X\left(\frac{f}{5}\right)e⁻²2πif(4)\).
Substituting the Fourier transform of x(t) into this expression, we obtain:
\(V(f) = \frac{1}{5}\left(\frac{1}{5a + 2πif}\right)e⁻²2πif(4)\).
\(V(f) = \frac{1}{5a + 2πif}e⁻²2πif(4)\).