I have two vector norm quantities: $\|\Psi^T(t)\Theta(t)\|$ and $\|\Phi^T(t)\Theta(t)\|$. Here $\Phi^T(t),\Psi^T(t)\in \mathbb{R}^{m\times n}$ and $\Theta(t)\in\mathbb{R}^{n}$. There is a filter relationship between the columns of $\Phi^T$ and $\Psi^T$ (denoted as $\Phi_{ci}^T$ and $\Psi_{ci}^T$ for $i=1...n$) that can be stated as $\Phi_{ci}^T(s)=G(s)\Psi_{ci}^T(s)$. In light of this relationship you might find it more useful to restate the original vector norms as $\|\sum_{i=1}^{n}\Theta_i(t)\Psi_{ci}^T(t)\|$ and $\|\sum_{i=1}^{n}\Theta_i(t)\Phi_{ci}^T(t)\|$.

I would like to make $\|\Phi^T(t)\Theta(t)\|\ge \|\Psi^T(t)\Theta(t)\|$. Is there any design constraint for $G$ that could make this happen?

I have tried creating various bounds in terms of signal and system norms with little success. Any help would be appreciated.

  • $\begingroup$ I think that a transfer function with impulse response $G(t)=a\delta(t)$ and $a\geq 1$ yields the desired inequality ($\delta(t)$ is the Dirac delta function). $\endgroup$ – RTJ Apr 2 '15 at 6:12
  • $\begingroup$ That's a good point. And it may help establish that this isn't feasible. But for now I would need G(s) to be a strictly proper transfer function matrix. $\endgroup$ – KVA Apr 2 '15 at 14:09
  • $\begingroup$ I would also be fine with establishing that the inequality is true below a certain frequency. It would also suffice to show that the two vectors norms are the same order of magnitude. $\endgroup$ – KVA Apr 2 '15 at 15:12

My thoughts on your question:

The condition $\|\Phi^T(t)\Theta(t)\|\geq \|\Psi^T(t)\Theta(t)\|$ is equivalent to $\Theta^T(t)[\Phi(t)\Phi^T(t)-\Psi(t)\Psi^T(t)]\Theta(t)$. This holds true for every $\Theta(t)$ if the matrix $$\Phi(t)\Phi^T(t)-\Psi(t)\Psi^T(t)\geq 0$$ i.e. if it is positive semidefinite. Also it holds true that $$\Phi^T(s)=G(s)\Psi^T(s)$$ and therefore $$\Phi^T(t)=\int_{-\infty}^{\infty}{G(t-\tau)\Psi^T(\tau)d\tau}$$ Now the matrix inequality can be written as $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\Psi(s_1)\bigg[G^T(t-s_1)G(t-s_2)-\delta(t-s_1)\delta(t-s_2)\mathbf{1}\mathbf{1}^T\bigg]\Psi^T(s_2)ds_1ds_2}\geq 0$$ which is obviously true if $G^T(t)=a\delta(t)\mathbf{1}\mathbf{1}^T$ for every $a\geq 1$. For more general conditions my feeling is that you may have to impose extra conditions on $\Psi(t)$.


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