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How do I visualize this, s, complex angular frequency? Kindly help me.
I think MrAl hit the nail on the head when he mentioned the impulse function. The Laplace transform is an integral and impulses are objects that generate non-zero numbers when integrated. If you start your integration at 0+, you miss the contribution of the impulse function.I'm not still clear about the Q2. Could you please help me with it?
If |exp(-σt)|=1 is incorrect, then what should it be? Please let me know. Thanks.
Thank you, MrAl, Steve.
How do I visualize this, s, complex angular frequency? Kindly help me.
I'm not still clear about the Q2. Could you please help me with it?
If |exp(-σt)|=1 is incorrect, then what should it be? Please let me know. Thanks.
Regards
PG
Re: Q2
Thank you, Steve, MrAl.
As MrAl mentioned impulse function occurs at t=0, so if Laplace transform starts from "0" instead of "0-" then how would the Laplace miss the contribution of the impulse? Kindly help me with it if possible. Thank you.
Regards
PG
The impulse function, despite the fact that it is a weird and almost nonsensical function, starts at zero and then grows to infinity with finite area between 0- and 0+. Hence, we need to start integrating at 0- to capture all information about the function.
Thanks a lot.
In my humble opinion if it starts at zero then there is no need to include 0- which is infinitesimal degree less than zero. So, I would say the impulse function starts at 0- and then grows to infinity with finite area between 0- and 0+. Kindly guide me. Thanks.
Regards
PG
With various transforms, you can imagine that you are applying a particular "test" function to a system. When we use Fourier Transforms, we can imagine that we are applying sine and cosine waves (or complex exponentials) to the system, and we look at how a system responds to that input signal. Laplace Transforms are one step more general than Fourier transforms because you get the Fourier Transform when σ=0.
So, imagine that you are applying signals of the type exp(-σt) sin(ωt) u(t) or exp(-σt) exp(jtω) u(t). In other words, imagine a decaying sine wave for starters. Now, the class of test signals is more general than this because you can have pure sine waves, pure decaying exponentials, exploding exponentials and increasing sine waves etc.
In essence, the σ is a decay (or expand) rate (units of 1/sec are sometimes called rates), and ω is a frequency (units of 1/sec are also sometimes called frequencies). The two are combined and called a complex frequency traditionally, but don't let the word frequency throw you off what the concept is.