The explanation is in your own posts.Steve10 said:Welcome back, but I don't know what makes you think I am hostile. That sounds funny. Care to explain?
irfan1 said:I would like to raise a similar question. What does the zeros of a function in fourier domain mean in the time domain?
zorro said:Steve10, you misunderstood at the beginning the question posed by Highlander-SP: you mistook what was “the function”, you confused whether the transfer function was f(t) or F
(s). This is not grave. Perhaps you are not familiar with some terms of circuit analysis employed in electrical engineering. This is not grave. The point is that you don’t
have reason for being so provocative and treat other people as they were ingorant. It’s funny for you? Not for me. It is pathetic.
Now, let’s go back to the question about zeroes.
....
zorro said:Steve10,
Analyze the problem, and you will see your mistake. See from the first post. Pay attention at the first one, what is called the ‘function’.
Highlander-SP said:The function
F(s) = 12*exp(-2s) / [(s-4)(s-3)]
has the poles: (4,0);(3,0)
and what is the zero of the function? How do i analize it?
Thanks
zorro said:F(s) has two finite poles (at 4+j0 and 3+j0), two zeroes at infinity, and a time shift of 2.
Suppose H(s) is a transfer function expressed as a quitient of polynomials in the variable s.
If H(s) has no zero at infinity, then its impulse response includes an impulse at t=0+.
If H(s) has a single zero at infinity, then its impulse response is finite (but not zero) at t=0+.
If H(s) has a double zero at infinity, then its impulse response is zero at t=0+, and its first derivative is finite (but not zero) at t=0+.
… and so on.
...
zorro said:F(s) has an essential singularity at Infinity because of the term exp(-2s). I agree with that.
But, for circuit analysis, F(s) is better specified saying that it is composed by the product (the cascade) of two transfer functions:
1) a function that has poles at s=4 and s=3, and unity gain at d.c.;
2) a function that, in time domain, represents a pure time delay of 2.
The first has the two poles (and in consequence two zeroes, that are at Inf).
The second does not contribute with poles nor zeroes to F(s), but with the essential singularity at Inf.
For circuit analysis purposes (response in frequency domain, transient response, stablity analysis, etc.) it is not useful to describe F(s) as a function with an essential singularity at Inf. Saying that, the only conclusion would be that standard methods of circuit analysis could not be applied.
Instead, describing F(s) in the other way, those methods are appropriate.
For instance, the doble zero at Inf says the behaviour of F(s) at “high frequencies” (in the imaginary axis): for frequencies high enough, the amplitude of the transfer function decays at 40 dB/decade, and the phase is +/-Pi. The additional term exp(-sT) doesn’t modify the amplitude response, and adds a linear phase term.
steve10 said:If you want to take this chance to make a speech about "circuit analysis", you can. But why do I have to respond?
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