Fourier Transform of product of heaviside step function and another function

In summary, the conversation discusses using Fourier transforms and inverse transforms to find the response of a damped harmonic oscillator driven by a specific force. The student is having trouble with the integration of the Heaviside step function in the Fourier transform. The expert suggests using a trick to avoid integration by parts and simplifying the integral.
  • #1
Dextrine
102
7

Homework Statement


A damped harmonic oscillator is driven by a force of the form f(t)=h(t) t^2 Exp(-t), where h(t) is a Heaviside step function. The Oscillator satisfies the equation x''+2x'+4x=f(t). Use pencil-and-paper methods involving Fourier transforms and inverse transforms to find the
response of the oscillator, x(t), assuming that x(0)=0 and x'(0)=1.

Homework Equations


The Fourier Transform F[f(t)]
The Inverse Fourier Transform F^(-1)[f(ω)]
Integration by parts

The Attempt at a Solution


First thing I did was take the Fourier transform of the left hand side, which I'm sure I got correct. The part I'm stuck on is taking the Fourier transform of f(t). I used integration by parts using
u=t^2/e^t dv=h(t)(Exp[i ω t])
du=(2t-t^2)/e^t v=πδ(ω)+i/ω

but now when I try to complete the integration, i get an integral that does not converge because of the i/ω. There have been multiple heaviside problems that I have been working on that all end up the same, and I'm not sure where to go from here.
 
Last edited:
Physics news on Phys.org
  • #2
In general actually, how would one go about solving the Fourier transform of a function with the form f(t)H(t)

for example, how could I do the Fourier transform of tH(t)? Again, I've tried doing it by parts, but nothing good comes of it
 
  • #3
The way you split the integrand up is unusual. The step function allows you to write the integral as
$$\int_{-\infty}^\infty t^2e^{-t}h(t) e^{i\omega t}\,dt = \int_0^\infty t^2e^{(i\omega-1)t}\,dt,$$ which you can integrate by parts. A trick you could use to avoid integration by parts is to note that ##t^2 e^{i\omega t} = -\frac{\partial^2}{\partial \omega^2}e^{i\omega t}##, so you can say
$$\int_0^\infty t^2e^{(i\omega-1)t}\,dt = -\frac{\partial^2}{\partial \omega^2}\int_0^\infty e^{(i\omega-1)t}\,dt$$
 
  • Like
Likes Dextrine
  • #4
Thank you so much you are a life saver. My homework became super easy after your post!
 
  • #5


It seems like you are on the right track with using integration by parts. However, instead of using the delta function in your calculation for v, try using the Fourier transform of the Heaviside step function, which is 1/(iω) + πδ(ω). This will give you a convergent integral and allow you to complete the calculation. Keep in mind that you will also need to use the convolution theorem to simplify the Fourier transform of the product of the Heaviside step function and the exponential function.

Once you have the Fourier transform of f(t), you can then use the inverse Fourier transform to find the solution for x(t). Remember to also use the initial conditions given in the problem to solve for the constants in your solution. Hope this helps!
 

1. What is the Fourier Transform of a product of a Heaviside step function and another function?

The Fourier Transform of a product of a Heaviside step function and another function is the convolution of the Fourier Transform of the Heaviside step function and the Fourier Transform of the other function.

2. How do you calculate the Fourier Transform of a product of a Heaviside step function and another function?

To calculate the Fourier Transform of a product of a Heaviside step function and another function, you can use the convolution theorem, which states that the Fourier Transform of a convolution of two functions is equal to the product of their individual Fourier Transforms.

3. What is the significance of the Heaviside step function in the Fourier Transform?

The Heaviside step function acts as a switch in the time domain, turning the signal on or off at a specific time. In the frequency domain, it appears as a constant value, representing the DC component of the signal.

4. Can the Fourier Transform of a product of a Heaviside step function and another function be simplified?

Yes, depending on the specific functions involved, the Fourier Transform of a product of a Heaviside step function and another function can be simplified using properties such as linearity, time shifting, and frequency shifting.

5. What applications does the Fourier Transform of a product of a Heaviside step function and another function have?

The Fourier Transform of a product of a Heaviside step function and another function is commonly used in engineering and physics to analyze and process signals and systems. It has applications in areas such as signal processing, image processing, and control systems.

Similar threads

  • Advanced Physics Homework Help
Replies
2
Views
809
  • Advanced Physics Homework Help
Replies
0
Views
211
  • Advanced Physics Homework Help
Replies
11
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
932
  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
742
  • Calculus and Beyond Homework Help
Replies
5
Views
326
  • Advanced Physics Homework Help
Replies
2
Views
728
  • Advanced Physics Homework Help
Replies
17
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
819
Back
Top