(a) Sketch a graph of \(u(-t)\) where \(u(t)\) is the unit step function. (b) Sketch a graph of \(f(t)=7 \mathrm{e}^{2 t} u(-t)\). (c) Find the Fourier transform of \(f(t)\).

The unit step function, u(t), is defined as follows: u(t) = 0 , for t < 0 u(t) = 1 , for t ≥ 0 To sketch a graph of u(-t), simply reflect the graph of u(t) about the y-axis.

To sketch the graph of f(t) = 7e^(2t)u(-t), follow these steps: 1. Sketch the graph of the exponential function 7e^(2t) for all t < 0. 2. Multiply the exponential function by the unit step function u(-t) which is 1 for all t < 0 and 0 for all t ≥ 0. This means that the graph of f(t) will coincide with the graph of 7e^(2t) for all t < 0, and will be 0 for all t ≥ 0.

To find the Fourier transform of f(t), we need to use the definition of Fourier transform: \(F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-j\omega t} dt \) Since f(t) = 0 for all t ≥ 0, we only need to compute the integral for t < 0: \(F(\omega) = \int_{-\infty}^{0} 7\mathrm{e}^{2t}\mathrm{e}^{-j\omega t} dt \) Let's define a new function g(t) = 7e^(2t)e^(-jωt), then: \(F(\omega) = \int_{-\infty}^{0} g(t) dt \) To compute the integral of g(t), we can use integration by parts. Let's set: \(u(t) = 7\mathrm{e}^{2t}\) , and \(dv(t) = \mathrm{e}^{-j\omega t} dt\) Then, \(du(t) = 14\mathrm{e}^{2t} dt\) , and \(v(t) = -\frac{1}{j\omega}\mathrm{e}^{-j\omega t}\) By integration by parts, we have: \(F(\omega) = u(t)v(t)\Big|_{-\infty}^0 - \int_{-\infty}^{0} v(t) du(t)\) \(F(\omega) = \Big[-\frac{7}{j\omega}\Big] - \int_{-\infty}^{0} -\frac{14}{j\omega}\mathrm{e}^{(2-j\omega)t} dt\) Now, integrate the remaining integral: \(F(\omega) = -\frac{7}{j\omega} + \Bigg[\frac{14}{(2-j\omega)j\omega}\mathrm{e}^{(2-j\omega)t}\Bigg]_{-\infty}^0 \) \(F(\omega) = -\frac{7}{j\omega} + \frac{14}{(2-j\omega)j\omega} - \lim_{t \to -\infty} \frac{14}{(2-j\omega)j\omega}\mathrm{e}^{(2-j\omega)t} \) Since the limit term goes to 0, the final answer is: \(F(\omega) = \frac{14}{j\omega(2-j\omega)}\) The Fourier transform of f(t) = 7e^(2t)u(-t) is F(ω) = \(\frac{14}{j\omega(2-j\omega)}\).