10

All eigenfunctions of an LTI system can be described in terms of complex exponentials, and complex exponentials form a complete basis of the signal space. However, if you have a system that is degenerate, meaning you have eigensubspaces of dimension >1, then the eigenvectors to the corresponding eigenvalue are all linear combination of vectors from the ...


7

TL;DR: Subspaces are low-dimensional, linear portions of the entire signal space that are expected to contain (or be close to) a large part of the observable and useful signals or transformations thereof, with additional tools that allow us to compute interesting things on the data We are given a set of data. To manipulate them more easily, it is common ...


5

Subspaces are a Linear Algebra concepts. The best representative example I can think of is the relationship of the XY plane to XYZ space, The former is a subspace of the latter. Any vector in the plane also lies in the space. Every vector in space has an orthogonal projection onto the subspace. So a set of vectors in your subspace can only reach vectors ...


4

Yeah, notation is not ideal. It is not - he assumes that each of the $M$ antenna elements is connected to its own RF chain, i.e., there are also $M$ receivers available. If you have fewer receivers you need to modify your $A$, it needs to contain the response your $K$ receivers observe given a wave from a certain direction. He doesn't put it but yeah, $F$ ...


4

That's not true, it's not better. The thing is: the matched filter just implements the projection in the signal vector space, onto the signal vector itself (or a multiple thereof). (You'll find correlation is just an inner product in that space.) The line through that vector is the signal subspace, the plane to which that vector is normal is the noise space. ...


3

Complex exponentials are eigenfunctions of LTI systems because they are eigenfunctions of the convolution operator: $$\begin{align}e^{j\omega_0t}\star h(t)&=\int_{-\infty}^{\infty}h(\tau)e^{j\omega_0(t-\tau)}d\tau\\&=e^{j\omega_0t}\int_{-\infty}^{\infty}h(\tau)e^{-j\omega_0\tau}d\tau\\&=e^{j\omega_0t}H(j\omega_0)\end{align}\tag{1}$$ where $h(t)$ ...


3

My first swing at the answer had some very incorrect claims. I do not have access to the article, so I am inferring some things from the portion posted in the question. NOTA BENE: My arguments assume that the eigenvectors of $\mathbf{R}$ are arranged so that the first $n$ belong to the signal subspace and that the last $m-n$ belong to the noise subspace. ...


3

A subspace is just a vector space that's included in a bigger vector space. Separating a random signal space into two statistically uncorrelated subspaces, a desired signal space and a noise space, yields eigenvectors that are orthogonal to each other. This orthogonality property of those subspaces is used to separate noise from desired isgnal and get a ...


3

For discrete data both are the same - Finding set of orthogonal directions which maximizes the Variance (Energy) of data along them. Sometimes those are called the natural axis of the. Since we're dealing with variance it is only natural both are calculated from the covariance matrix of data. You may encounter places where KL might be even defined on the ...


2

For any convolution matrix (even truncated), your diagonal entries would be $h_0$; hence, the characteristic polynomial, setting $b_0 = h_0-\lambda$, \begin{align}|H-\lambda I| = \begin{vmatrix} b_0 & 0 & 0\\h_1 & b_0 & 0\\h_2 & h_1 & b_0 \end{vmatrix} &= b_0\,\begin{vmatrix} b_0 & 0\\h_1 & b_0 \end{vmatrix} - 0\,\begin{...


2

First, this question is probably better for MATH.SE, but I'll give it a shot. It's been a long long time since I did this stuff. If $N > M$: 1) $C^*$ has rank M. 2) $R_x^{-1}$ has rank M. (or it wouldn't exist as an inverse) 3) Therefore $C^*R_x^{-1}$ has rank M, since $R_x^{-1}$ is a full rank square matrix. 4) $C$ has rank M 5) $C^*R_x^{-1}C$ ...


1

Mathematically (and theoretically), there is no need for the exponential function to be a complex sinusoid. The math is unchanged. The problem is that practical LTI systems are not boundless nor are they acausal. So setting aside those problems, every LTI system has input/output relationship described by the convolution integral (for continuous-time) or ...


1

Complex exponential functions are most generally defined (up to a constant complex or real factor) as $t\mapsto e^{j\omega_0 t}$, $\omega_0\in \mathbb{R}$. Real exponentials are typically of the form $t\mapsto c^{ t}$, $c>0$. The latter are not a subset of the former: one reason is that complex exponentials have a modulus equal to one. This is not the ...


1

I've had issue with this -- please see the errata of the reference. The eigenfunctions are not scaled in the original version of the book. This errata should solve it.


1

Given no formal system model in the question, I will outline in words what each does and the relation between them. Matched Filter: The MF maximizes SNR when the signal is in additive Gaussian noise. You can go back and look at the derivation of the MF, but it does not include any mention of interference. During the derivation, there is a step where we say ...


1

Thanks for your answer Cedron! Taking your same assumption $N \leq M$, and by definition of PD, $x^* R_x^{-1} x > 0\quad \forall x \in \mathbb{R}^M \setminus \{0\}$ and since $C$ is full rank $\dim (\mathcal{R}(C)) = \min(M,N)= N$. By the rank-nulity theorem, $\dim (\mathcal{R}(C)) + \dim(\mathcal{N}(A))=N$, so we have that the null-space is trivial. This ...


1

The eigenvalues belong to the same set of quartic roots of unity verifying $\lambda^4=1$, whatever the order of the DFT. For more details on their multiplicity, you can read: Eigenvectors and Functions of the Discrete Fourier Transform, 1982, Dickinson and Steiglitz (online).


1

You can write $$ R=YY^H $$ where $Y$ is a matrix of size $N\times N_f$ and $N$ is the dimension of $y_k$. $Y$ contains all the measured $y_k$ as its columns. Then, the rank of $R$ is upper bounded by $N_f$. In particular, if $N_f<N$, $R$ will always be a singular matrix. So, if you have too few measurements, you will likely run into the problem you ...


1

The piece I was missing was the distribution of the initial phase values $\varphi_1$ and $\varphi_2$. It is standard to assume that these are uniformly distributed [^]. This leads to: $$ \mathbf{R_x} = \mathbf{E[xx^H]} = \mathbf{S \Lambda S^H} $$ where $\mathbf{\Lambda} = \begin{bmatrix} A_1^2 & 0 \\ 0 & A_2^2 \end{bmatrix}$ and $\mathbf{S} = \begin{...


1

It is an essential component in Principal Component Analysis(PCA) that allows to reduce the number of dimensions by projecting less dimensions of this sort of data, thus narrowing it down faster to finding a pattern. The projection operation characterizes an individual face by a weighted sum of the Eigen faces features and so to recognize a particular face ...


1

I will try giving you some intuition. The SVD says each matrix can be decomposed into 3 operations - Rotation, Stretching (Scaling) and the another Rotation. What matters is which directions are scaled and how. Directions are vectors (Pointing some direction). The SVD has many uses in Linear Algebra. One its most known use is Low Rank Approximation of a ...


1

For any arbitrary LTI sytem, the complex exponential is, to the best of my knowledge, the only known eigensignal. On the other hand, consider the ideal LPF. The $\operatorname{sinc}$ function: $$\operatorname{sinc}(t) \triangleq \frac{\sin(\pi t)}{\pi t}$$ can easily be seen to be an eigen signal. This points to the existence of LTI systems (such as the ...


Only top voted, non community-wiki answers of a minimum length are eligible