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For a first order system like an RC filter you have something like $$y(t)=1-e^{-\frac{t}{\tau}}\\ y'(t)=\frac{1}{\tau}e^{-\frac{t}{\tau}}\\ y'(0)=\frac{1}{\tau}$$

But for a second order system there are 4 different cases. What I'm interested in is y'(0) of any second order low-pass filter in terms of its frequency and damping.

Taking any random image from the web, it can be seen that for all different damping modes, it appears to start slow (low derivative) ad then ramp up. Is y'(0)=0, or some fixed relation like in the first order case? You can't really tell from the graphs, and I've been unable to find any confirmation about what it is.

step response

Doing it the time domain way, you get into pesky situations where some things are imaginary or not, leading to those 4 cases. Is there a way to get a general answer?

Using the frequency approach I did some questionable math to arrive at the following. Taking a generic second order low-pass system $$H(s)=\frac{\omega^2}{s^2+\frac{\omega}{Q}s+\omega^2}$$ The step response is $$C(s)=\frac{\omega^2}{s(s^2+\frac{\omega}{Q}s+\omega^2)}$$ Derivative in the time domain is multiplication by s in the frequency domain $$c'(t) \rightarrow sC(s)=\frac{\omega^2}{s^2+\frac{\omega}{Q}s+\omega^2}$$ Now apply initial value theorem $$c'(0) = s^2C(\infty)=\frac{s\omega^2}{s^2+\frac{\omega}{Q}s+\omega^2}$$ Apply L'Hôpital's rule $$c'(0) = s^2C(\infty)=0$$ Does that make sense at all? Can this be seen directly in time domain as well?

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Starting from any transfer function \$H(s)\$, applying a step function will multiply it by \$\frac{1}{s}\$, then deriving it is the same as multiplying by \$s\$ in the Laplace domain. You find

$$\mathcal{L}\left\{ y'(t) \right\} = H(s)\cdot \frac{1}{s}\cdot s$$

We can then use the initial value theorem to find the value at t=0:

$$\lim_{t\to 0^+} y'(t) = \lim_{s\to +\infty} s\cdot H(s)$$

For a second-order transfer function, we have then two cases, one with and one without a zero (two zero's means that the transfer function can be reduced).

  1. Without a zero

$$H(s) = \frac{b_0}{1 + a_1s + a_2s^2}$$

Applying the initial value theorem will yield

$$\lim_{t\to 0^+} y'(t) = \lim_{s\to +\infty} s\cdot \frac{b_0}{1 + a_1s + a_2s^2} = 0$$

  1. With a zero

$$H(s) = \frac{b_0 + b_1s}{1 + a_1s + a_2s^2}$$

Applying the initial value theorem will yield

$$\lim_{t\to 0^+} y'(t) = \lim_{s\to +\infty} s\cdot \frac{b_0 + b_1s}{1 + a_1s + a_2s^2} = \frac{b_1}{a_2}$$

So there you have it. The derivative is always zero at \$t=0\$ unless there is a zero in the transfer function.

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The under-damped (and frequency normalized) low-pass filter is this: -

$$H(s) = \dfrac{1}{s^2+2\zeta s+1}$$

It has a step response of: -

$$\dfrac{1}{s}\cdot\dfrac{1}{s^2+2\zeta s+1}$$

And, for the convenience of being able to perform an inverse Laplace transformation, it is made equal to this: -

$$\dfrac{1}{s[(s+a)^2+b]}$$

Where \$a=\zeta\$ and \$b = \sqrt{1-\zeta^2}\$

Using Laplace tables this converts to: -

$$1-\dfrac{1}{\sqrt{1-\zeta^2}}\cdot e^{-\zeta t}\cdot \sin(t\cdot\sqrt{1-\zeta^2}+\phi)$$

Where \$\phi = \arccos(\zeta)\$

Examining the formula you can see that when \$\zeta\$ approaches zero, the time derivative also approaches zero at t = 0. See the lower half of the picture: -

enter image description here

Picture source.

For other values of \$\zeta\$ just differentiate the formula above the picture but, what you will find is that the derivative at t = 0 is always zero because if you look at a typical 2nd order low- pass filter: -

enter image description here.

And you apply a step at the input, the rate of change of output voltage of the capacitor is proportional to current BUT the current doesn't change instantly because of the inductor. Hence the derivative at t = 0 is always zero.

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Because a certain parameter in the time domain is wanted, the most direct method would be to find the function g(t) for the step repsonse and calculate the first derivative. This is, however, a rather involved procedure.

Hence, it is more convenient not to leave the frequency domain and apply the "initial slope theorem" (which immediately gives the slope g′(0) of the step response at t=0):

Similar to the initial value theorem (find (Hs) for s>>>infinity) the initial slope theorem requires to find the value s*H(s) for infinite frequencies.

This calculation step is identical to the last part of the original question leading to g′(0)=0.

Comment: The initial slope theorem can be extended also to higher derivatives (Ref: Active Network Design, Clause S. Lindquist)


Addendum (Time domain):

For the given system function H(s) the corresponding step response function can be found to be

g(t)=1 - exp(Dwot)*{cos(wnt) + [D/SQRT(1-D²)]sin(wnt)}

with the natural frequency wn=wo*SQRT(1-D²).

and the first derivative at t=0 can be found to be

g′(t=0)=woD-Dwn/SQRT(1-D²)]

Replacing wn by wo we get (as expected):

g′(t=0)=0

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  • \$\begingroup\$ Are you aware you can use LaTeX notation for equations? That would make your time domain solution much easier to read. \$\endgroup\$ – Pepijn Jun 17 '18 at 7:43
  • \$\begingroup\$ Yes - I agree that my formulas are not good to read. However, I am not familiar with LaTex, sorry. \$\endgroup\$ – LvW Jun 17 '18 at 8:03

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