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I have found the definition of the following formulas in a paper regarding active vibration control, where they are called sine and cosine modulated integrals.

$y$ is measurement signal with a strong periodic component of frequency $N\Omega$

$$y^{(i)}_{Nc}=\frac{2}{T}\int^{T}_{0}y^{(i)}(ϕ)\cos(Nϕ)dϕ$$

$$y^{(i)}_{Ns}=\frac{2}{T}\int^{T}_{0}y^{(i)}(ϕ)\sin(Nϕ)dϕ$$

where $\phi=\Omega t$.

From these the vector $y_N^{(i)}$ is defined as $y_N^{(i)} = \begin{bmatrix} y_{Nc}^{(1)}\\ y_{Ns}^{(1)}\\\vdots\end{bmatrix}$

The same is done for the control input(s) $u$. Then a quadratic cost function to be minimised at each step is defined using these newly introduced signals in this way:

$$J(k) = y^T_NQy_N+u^T_NRu_N $$

where $Q$ and $R$ are just two weighing matrices.

They should extract the harmonic component considered but is anybody able to explain them a little bit further?

Here the link for the paper

Another question: suppose I have the value $y_N$: how can I invert the relationship to get $y^{(i)}$

Here the link for the paper.

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    $\begingroup$ Welcome to MSE! We'll do our best to address your question on a mathematical level; for a more applied perspective, the Signal Processing Stack Exchange may be helpful. $\endgroup$ – Semiclassical Jul 31 '14 at 14:36
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    $\begingroup$ A few things: 1) The paper you cite is behind a paywall, and so will not be accessible to all readers. So it behooves you to make sure you supply as much context as possible. 2) Your definition of $y_N^{(i)}$ is quite unclear: are the next two elements supposed to be $y^{(2)}_{N,c}$ and $y^{(2)}_{N,s}$? If so, what is the $i$ in $y^{(i)}_N$ supposed to signal? $\endgroup$ – Semiclassical Aug 3 '14 at 14:44
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As someone made me notice, the integrals are just the coefficients of the Fourier Series for that frequency component (when the series is expressed with the cosine and sine parts separately).

The integral is performed on a period much bigger than the period corresponding to the frequency of the disturbance so that a better average is obtained.

To get back the signal in the time domain, it is sufficient to multiply the two components by the cosine and sine respectively. Of course in this way the signal has only one frequency component.

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