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Chapter 24: Problem 9

From the definition of the Fourier transform find \(F(\omega)\) when \(f(t)=u(t) t \mathrm{e}^{-3 t}\).

Short Answer

Expert verified
Answer: The Fourier transform of the function f(t) = u(t)t * e^(-3t) is F(ω) = 1/(3 + jω)².

Step by step solution

01

Write the Fourier transform definition

We begin by recalling the definition of the Fourier transform for a continuous-time signal, which is given by: $$ F(\omega) = \int_{-\infty}^{\infty}f(t) \cdot e^{-j \omega t} dt $$
02

Plug in the given function for f(t)

Now, we will substitute the given function f(t) = u(t)t * e^(-3t) into the Fourier transform definition: $$ F(\omega) = \int_{-\infty}^{\infty} u(t)t \mathrm{e}^{-3t} \cdot e^{-j\omega t} dt $$
03

Simplify the integral by exploiting the properties of the unit step function u(t)

Due to u(t) being 0 for t < 0, and 1 for t >= 0, the limits of integration can be changed from -∞ to∞ to 0∞ in order to simplify the integral: $$ F(\omega) = \int_{0}^{\infty} t \mathrm{e}^{-3t} \cdot e^{-j\omega t} dt $$
04

Combine the exponentials in the integrand

Now we will combine the two exponentials in the integrand: $$ F(\omega) = \int_{0}^{\infty} t \mathrm{e}^{-3t - j\omega t} dt $$
05

Apply integration by parts

To evaluate the integral, we need to apply integration by parts. Let's set: $$ u = t \\ dv = e^{-3t - j\omega t} dt $$ By differentiating u, and integrating dv, we get: $$ du = dt \\ v = \frac{-1}{3 + j\omega} e^{-3t - j\omega t} $$ Now applying the integration by parts formula: $$ \int udv = uv- \int v du $$ Substitute u, du, v, and dv: $$ F(\omega) = \left[-\frac{t}{3 + j\omega}\mathrm{e}^{-3t - j\omega t}\right]_0^\infty + \int_0^\infty \frac{1}{3 + j\omega}\mathrm{e}^{-3t - j\omega t} dt $$
06

Evaluate the expression and integral

Upon evaluating the expression and the integral, we get the final answer: $$ F(\omega) = \frac{1}{(3 + j\omega)^2} $$ So, the Fourier transform of the given function f(t) = u(t)t * e^(-3t) is F(ω) = 1/(3 + jω)².

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Most popular questions from this chapter

Find the Fourier series representation of the function with period 1 given by $$ f(t)= \begin{cases}t & 0

(a) Sketch a graph of three cycles of the function with period \(2 \pi\) given by $$ f(t)=1-\frac{|t|}{\pi}, \quad-\pi \leq t<\pi $$ (b) Find its Fourier series representation.

Find, using Table 2.1, the Fourier transform of $$ f(t)=\left\\{\begin{array}{lc} 1 & -1 \leq t \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$

(a) Sketch a graph of $$ f(t)= \begin{cases}\mathrm{e}^{-2 t} & t>0 \\ -\mathrm{e}^{2 t} & t<0\end{cases} $$ (b) Find the Fourier transform of \(f(t)\). (c) Show that this Fourier transform is purely imaginary.

Sketch a graph of \(f(t)=\mathrm{e}^{-|3 t|}\) and find its Fourier transform.
See all solutions

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