# Evaluate the Fourier Transform of a Damped Sinusoidal Wave

• Captain1024
In summary: The cheat link gives a positive result and so do I.In summary, the Fourier Transform of the damped sinusoidal wave g(t)=e^{-t}sin(2\pi f_ct)u(t) is G(f)=\frac{2\pi f_c}{(2\pi f_c)^2 + (1+j2\pi f)^2}. This is equivalent to G(f)=\frac{\omega _c}{\omega ^2 + (\omega _c-1)^2} or G(f)=\frac{\omega _c}{1+\omega ^2-2\omega \omega _c+\omega _c^2
Captain1024

## Homework Statement

Evaluate the Fourier Transform of the damped sinusoidal wave $g(t)=e^{-t}sin(2\pi f_ct)u(t)$ where u(t) is the unit step function.

## Homework Equations

$\omega =2\pi f$
$G(f)=\int ^{\infty}_{-\infty} g(t)e^{-j2\pi ft}dt$
$sin(\omega _ct)=\frac{e^{j\omega _ct}-e^{-j\omega _ct}}{2j}$

## The Attempt at a Solution

$G(f)=\int ^{\infty}_{-\infty}e^{-t}sin(2\pi f_ct)u(t)e^{-j2\pi ft}dt$
$sin(2\pi f_ct)=\frac{e^{j2\pi f_ct}-e^{-j2\pi f_ct}}{2j}$
$G(f)=\frac{1}{2j}\int ^{\infty}_{0}e^{-t}(e^{j2\pi f_ct}-e^{-j2\pi f_ct})e^{-f2\pi ft}dt$
$G(f)=\frac{1}{2j}\int ^{\infty}_{0}e^{-j2\pi ft-t}(e^{j2\pi f_ct}-e^{-j2\pi f_ct})dt$
$G(f)=\frac{1}{2j}\int ^{\infty}_{0}e^{-j2\pi ft-t+j2\pi f_ct}-e^{-j2\pi ft-t-j2\pi f_ct}dt$
$G(f)=\frac{1}{2j}\int ^{\infty}_{0}e^{j2\pi t(f_c-f)-t}-e^{-j2\pi t(f+f_c)-t}dt$

I am not feeling confident on my algebra and I also feel like I should be able to simplify this more before I integrate.

If you're not allowed to cheat (e.g. google 'Table of Fourier Transform Pairs', or 'convolution theorem' ) then all you can do is grin and carry on ...
A possible simplification is to put ##{\bf \alpha} = j2\pi (f_c-f)-1## etc.

New approach:

$sin(\omega _ct)=\frac{e^{j\omega _ct}-e^{-j\omega _ct}}{2j}$
u(t) changes limits of integration (just as before): $G(f)=\int _0^{\infty}e^{-t}(\frac{e^{j\omega _ct}-e^{-j\omega _ct}}{2j})e^{-j\omega t}dt$
Pulling out constants and combining exponentials outside the parentheses: $=\frac{1}{2j}\int _0^{\infty}e^{-(j\omega t+t)}(e^{j\omega _ct}-e^{-j\omega _ct})dt$
Distributing the integrand: $=\frac{1}{2j}[(\int _0^{\infty}e^{t(-j\omega -1+j\omega _c)}dt)-(\int _0^{\infty}e^{-t(j\omega +1+j\omega _c)}dt)]$
Integrating: $=\frac{1}{2j}([\frac{1}{-j\omega -1+j\omega _c}e^{t(-j\omega -1+j\omega _c)}]_0^{\infty}-[\frac{-1}{j\omega +1+j\omega _c}e^{-t(j\omega +1+j\omega _c)}]_0^{\infty})$
$=-\frac{1}{2j}(\frac{1}{-j\omega -1+j\omega _c}+\frac{1}{j\omega +1+j\omega_c})$
$=(\frac{-1}{2j})(\frac{-j\omega +2+j\omega _c}{j(\omega +1+\omega _c)})$
$=\frac{-j\omega +2+j\omega _c}{2(\omega +1+\omega _c)}$
------------------------------------------------------------------------------------------------------------------
Solutions manual gives the following answer: $G(f)=\frac{2\pi f_c}{1+4\pi ^2(f-f_c)^2}$
Which I believe is equivalent to: $G(f)=\frac{\omega _c}{1+\omega ^2-2\omega \omega _c+\omega _c^2}$
Any help would be appreciated.

jddv
I liked your first post and didn't find errors. Here's my bit. Don't have time to do this step by step, so I'll violate PF rules and work it through:
$$G(f)=\frac{1}{2j}\int ^{\infty}_{0}e^{j2\pi t(f_c-f)-t}-e^{-j2\pi t(f+f_c)-t}dt$$ was good. In short: your lack of confidence in your own work was unnecessary!
Use $$\int\limits_0^\infty e^{-\alpha x}dx = {1\over \alpha}$$to get $$G(f) ={1\over 2j}\left [{1\over 1-2\pi j(f_c - f) } - {1\over 1 + 2\pi j(f_c + f)}\right ]$$ so $$G(f) ={1\over 2j}\left [ \left ( 1 + 2\pi j(f_c + f) \right ) - \left ( 1-2\pi j(f_c - f) \right ) \over \left ( 1-2\pi j(f_c - f) \right ) \left ( 1 + 2\pi j(f_c + f) \right ) \right ]$$ and (dividing by ##2j##), the numerator becomes ##2\pi f_c ##, which is good. For the denominator I get $$(2\pi f_c)^2 + (1+j2\pi f)^2$$ which is not in agreement with your book answer. But it does agree with the cheat link I gave (top page 3).

## 1. What is the Fourier Transform of a Damped Sinusoidal Wave?

The Fourier Transform of a Damped Sinusoidal Wave is a mathematical tool used to decompose a signal into its constituent frequencies. It represents the amplitude and phase of each frequency component present in the signal.

## 2. How is the Fourier Transform of a Damped Sinusoidal Wave calculated?

The Fourier Transform of a Damped Sinusoidal Wave can be calculated using the formula: F(ω) = ∫f(t)e^(-jωt)dt, where f(t) is the damped sinusoidal wave and ω is the frequency.

## 3. What is the significance of the Damping Factor in the Fourier Transform of a Damped Sinusoidal Wave?

The Damping Factor in the Fourier Transform of a Damped Sinusoidal Wave represents the rate at which the amplitude of the signal decreases over time. It affects the shape and height of the frequency spectrum.

## 4. Can the Fourier Transform of a Damped Sinusoidal Wave be used to analyze real-world signals?

Yes, the Fourier Transform of a Damped Sinusoidal Wave can be used to analyze real-world signals such as audio, video, and other types of data. It can help identify the frequencies present in the signal and their respective strengths.

## 5. Are there any limitations to using the Fourier Transform of a Damped Sinusoidal Wave?

Some limitations of using the Fourier Transform of a Damped Sinusoidal Wave include the assumption that the signal is periodic and the requirement of a continuous signal. It is also limited by the Nyquist-Shannon sampling theorem, which states that the sampling rate must be at least twice the highest frequency present in the signal.

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