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Eg:

1.Z=1+i*5 (in quadratic form),

2.s(t)=Zexp(iw*t), (i'm multiplying two signals)

now, i want to take second order derivative, yes i know the usual way like this, Equation

but is there any other alternative form to find 2nd order derivative?,

and by doing that i want to extract my original signal ("z", which's given above ) does any one have idea about this ?

I tried this way it's worked but i want reduce it more complex.

After double derivative i will calculate instantaneous angular frequency like this W=(mean((y'')^2)/mean((s(t))^2)) and i will get result as I=2*mean[(s(t)*cos(w*t))] and I=-2*mean[(s(t)*sin(w*t))] this both I and Q are our original signal Z=1+i*5

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  • $\begingroup$ You would get answers more easily if instead of making us read a really long article, you would just explain what you mean by "the usual way". $\endgroup$ – Matti P. Apr 24 '18 at 9:49
  • $\begingroup$ @MattiP. no, you no need to read full article, i'm just referring only the equation of second order derivative (y'') by using this one can calculate the curvature of the signal. But how to find curvature of the signal without 2nd derivative of "s(t)" and at the same time i want extract or demodulate my original signal (Z), u got my question ? $\endgroup$ – user3764118 Apr 24 '18 at 10:11
  • $\begingroup$ So you have measured a signal, right? You have a time series, a long series of numbers? And you want to calculate the second time derivative of that? $\endgroup$ – Matti P. Apr 24 '18 at 10:14
  • $\begingroup$ If you have a time series, your derivatives will anyway be numeric and approximate, and you have to use the values of the time series to calculate the time derivative. The way you calculate it is called a stencil. In the article that you linked, one such stencil is described. Is there a reason why you don't want to use it? $\endgroup$ – Matti P. Apr 24 '18 at 10:15
  • $\begingroup$ @MattiP. yes u r right, i have the signal which is s(t)=Zexp(iw*t) this nothing but original signal (Z) and carrier signal, yes i have time series too, where t=[(T/80)*i], i=0..M-1,M=512; $\endgroup$ – user3764118 Apr 24 '18 at 10:20
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Assuming you have a time series of values $s_k$, there are two ways of approaching this.

  1. Analytical method: If you know exactly that your signal is of the form $$ s(t) = Ze^{i \omega t} = Z \cos{\omega t} + iZ \sin{\omega t} $$ Then you can just calculate the derivative: $$ \frac{d^2 s}{dt^2} = -Z\omega^2 e^{i\omega t} = -\omega^2 s(t) $$ Or, in simple terms, you just have to multiply the original signal by $-\omega^2$. Then the second derivative $y$ in every point is $$ y_k = -\omega^2 s_k $$
  2. Approximate method: If you don't know that your signal is exactly of the mentioned form, you have to indirectly measure the second time derivative, by calculating it at each point. The most straightforward way is the finite difference method, and for live calculation you have to use the backwards difference. If we set the second derivative to be $y$ and use the first-order accurate backwards difference $$ y_k \approx \frac{1}{\Delta t^2}\left(s_k - 2s_{k-1} + s_{k-2} \right) $$ This method uses only the "history" of the signal ($s_{k-1}$ and $s_{k-2}$). It's rather simple but gives an approximate answer. In this equation, $\Delta t$ is the time step used.
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  • $\begingroup$ It's absolutely working fine for derivative but after that i want i want calculate instantaneous angular frequency like this W=(mean((y'')^2)/mean((s(t))^2)) and i will get result as I=2*mean[(s(t)*cos(w*t))] and I=-2*mean[(s(t)*sin(w*t))] this both I and Q are our original signal Z=1+i*5 and i found some way to replace sin and cosine terms. But is there any alternative way to find mean or maybe some other method to reconstruct our Z, $\endgroup$ – user3764118 Apr 24 '18 at 11:43
  • $\begingroup$ I think in general, if you want the frequency, you should take the Fourier transform of the signal. $\endgroup$ – Matti P. Apr 24 '18 at 11:48
  • $\begingroup$ I i'm aware of that but FT is more complected to do in MCU, can we go in the flow of yk=-W^2*sk, from this, is it possible to extract our original signal Z=1+i*5 ? $\endgroup$ – user3764118 Apr 24 '18 at 12:12

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