That looks like a low pass filter, is that what it is?Hey,
I have this bode plot given and I have to determin the time function. The amplitude is 40 dBuVs. The time function will probably be a rectangle, but idk how to determine the time constant τ. Is it like T = 1/f? Thank you very much
View attachment 287008
Hi,Well, there is a certain relaton between the frequency domain and the time domain.
In detail: The step response g(t) is the Laplace transform of the system function H(s) - multiplied with "s".
However, there is a simpler way to find g(t):
* Derive the system function H(s) from the Bode plot. In your case, I can identify two poles and the DC gain.
Therefore, it is easy to write down the function H(s).
* From this 2nd-order function you can derive the quality factor Qp (pole-Q) and the characteristic frequency wo.
* This is all you need to get a rough information about the form of the step response g(t).
(Remember: There is a relation between the pole-Q (frequency domain) and the overshoot (time domain).
Added: The abov applies to a 2nd-order function with a complex pole-pair. In case of two real poles there are some other rules which help to find the step response as a function of the two time constants
It could be that you need to use a function that is more damped than what you are using. That means you may have the wrong values for a and b.@LvW Well K seems to be 100 --> H(s) = 100/(s+0,1)+(s+1). According to this formula g(t9 is determined as following:
View attachment 287109
I used the formula but I got the wrong plot.
View attachment 287110
The time function must have the shape of trapezoid or rectangle. What did I do wrong?
Yes - you are right. Thank you.Sorry but i think you meant the step response of the system is the function H(s) divided by "s" not multiplied. Then of course you use the Inverse Laplace Transform to get the time domain response.
Hi,Yes - you are right. Thank you.
What I wrote was just the opposite: How to derive the transfer function H(s) from the step response.
When we are looking for the step response, of course, we have to apply the inverse Laplace transform to the expression H(s)/s .
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