Tying Convolution with Laplace and Fourier Transforms

Discussion in 'Homework Help' started by jp1390, Nov 6, 2011.

  1. jp1390

    Thread Starter Member

    Aug 22, 2011
    45
    2
    Hi, this post is really to put my understanding to a test as I am not fully confident with my conclusions, so if anyone can point me in the correct direction that would be awesome!

    Convolution: the resulting steady-state output of an input passed into a LTI system.

    Laplace Transform: general form of Fourier transform, s = σ + jw. Displays frequency spectrum with decay (σ ≠ 0)

    Fourier Transform: special case of Laplace transform, s = jw. Non-periodic and periodic signals. Displays frequency spectrum without decay (σ = 0). Phasors?

    Fourier Series: Displays frequency spectrum of period signals

    Okay, I understand that the Fourier Transform is a special case of the Laplace where there is no σ term, but when would you use either?

    For instance, to produce a Bode plot of a system, you end up using the Laplace Transform, but could you not have just used the Fourier Transform? What is the difference in information that you receive for either case?

    Also, for periodic signals, the Fourier Series states that it can be broken up into sinusoids, does this just tell which frequencies are going to be affected when passed into a system?

    Thanks in advance,
    JP
     
  2. Peytonator

    Active Member

    Jun 30, 2008
    105
    3
    Hi JP,

    I'm still also in the process of learning all this stuff, but from my limited experience here are a couple answers to your questions.

    Often a system is given as a differential equation. The laplace transform can easily be used to transform from the time to s-domain. The reason for doing this is because differentiation and integration in time correspond to multiplication and division by s in the laplace domain. So its much easier to work with, and once you've solved the problem in s, you can transform back to t.

    By a simple substitution (s = jw), you get the fourier transform (frequency domain). So if you have a transfer function in s, and you want to plot the bode diagram, just substitute s = jw and plot. Sure, you could have used the fourier transform directly (using either the tables or integration). Since the Laplace transform is a special case of the Fourier transform, both will give you the same result eventually.

    The fourier TRANSFORM is usually used to tell you what frequencies will be affected when passed into the system, since it gives you the frequency information of the signal. A fourier SERIES is merely an approximation of the actual waveform using sine waves. The more sine waves you add, the more the fourier series will match the actual signal. Ultimately adding an infinite number of sine waves (with varying amplitudes) gives you the original signal. In a sense it does reveal frequency info due to the presence of sinusoids at different frequencies.
     
    Last edited: Nov 8, 2011
    jp1390 likes this.
  3. jp1390

    Thread Starter Member

    Aug 22, 2011
    45
    2
    That was an amazingly clear explanation! Thank you!
     
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