Evaluate the Fourier Transform of a Damped Sinusoidal Wave

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
  • #1
Captain1024
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2

Homework Statement


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

Homework Equations


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

The Attempt at a Solution


[itex]G(f)=\int ^{\infty}_{-\infty}e^{-t}sin(2\pi f_ct)u(t)e^{-j2\pi ft}dt[/itex]
[itex]sin(2\pi f_ct)=\frac{e^{j2\pi f_ct}-e^{-j2\pi f_ct}}{2j}[/itex]
[itex]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[/itex]
[itex]G(f)=\frac{1}{2j}\int ^{\infty}_{0}e^{-j2\pi ft-t}(e^{j2\pi f_ct}-e^{-j2\pi f_ct})dt[/itex]
[itex]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[/itex]
[itex]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[/itex]

I am not feeling confident on my algebra and I also feel like I should be able to simplify this more before I integrate.
 
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  • #2
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.
 
  • #3
New approach:

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

Your book answer must be wrong: it's a real
 

Related to Evaluate the Fourier Transform of a Damped Sinusoidal Wave

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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