# Tag Info

5

Feels like a bit of a trick question. Your answer is indeed correct. $\Psi_1$ ...$\Psi_3$ are an orthonormal basis but it's an incomplete basis. Any signal you can construct with this basis has 4 degrees of freedom, but you only have three basis functions. In order to have a complete basis, you need 4 functions. Turns $x(t)$ (scaled properly) would make the ...

4

An orthogonal matrix has orthogal columns, i.e. the scalar product of two different columns is zero (the case $i\neq j$). For the case $i=j$ you have $a_i^Ta_i=\|a_i\|^2>0$. So, all you can say about an orthogonal matrix with colums $a_i$ is: $$a_i^Ta_j=\begin{cases}c_i>0 & i=j\\ 0 & i\neq j\end{cases}$$ where $c_i>0$ is some constant, ...

4

Let's call your $4$ signals $f_0$, $f_1$, $f_2$ & $f_3$. Just by eyeballing you could have a basis of two function where b0 = f0, for 0 < t < Ts/2, 0 otherwise and b1 = f0, for Ts/2 < t < Ts, 0 otherwise and So you basically have a one first half ramp up in the first half and one second half ramo down. The signals can be expressed simply ...

4

Here is a graphical explanation. Sorry, I have depicted $-x(t)$ in red, and the $\psi_k$ in black. In gray, the area of the product on sub-intervals. Positive when $-x(t)$ and $\psi_k$ have the same sign, negative otherwise. As you can see, the areas sum to zero. So $-x(t)$ and $x(t)$ therefore are orthogonal to the other three functions. Now, you have a ...

3

Wikipedia's entry for the discrete Hartley transform shows states that the $\mathsf{DHT}$ is, up to a scaling, its own inverse. If $x$ is a vector with $N$ entries and $y$ is its discrete Hartley transform, \begin{equation} y = \mathsf{DHT}x, \end{equation} then \begin{equation} x = \frac{1}{N}\mathsf{DHT}y. \end{equation} If $x$ is a vector with $N$ ...

2

Even more simpler than @Hilmar's answer, the signals $f_0$ and $f_2$ are orthogonal just by eyeballing the pictures and recalling basic ideas about integration giving the area under the curve. Thus, the signals are $f_0, -f_0, f_2, -f_2$ with respect to the orthogonal basis $\{f_0, f_2\}$. The basis could be made a orthonormal basis (unit-energy basis ...

2

Add twice 5.65 to 5.64, and let $j = h(0)+h(2)$ and $k = h(1)+h(3)$; then we get j + k = sqrt(2) j*j + k*k = 1 which represents the intersection of a line with a circle, and we get j = k = 1/sqrt(2) Now subtract twice 5.65 from 5.64 and let $m = h(0)-h(2)$ and $n = h(1) - h(3)$; we get m*m + n*n = 1 so for same angle $b$ say we get $m = cos(b)$ and $... 2 After a few quick calculations, it seems to me that the trouble comes from poor notations for the root in your reference. If you read, in the final normalized matrix,$\sqrt{8/64}$and$\sqrt{2/4}$instead of$\sqrt{8}/64$and$\sqrt{2}/4$(along with the$\pm$signs), then the final result is correct. The matrices$V_i$are orthogonal. To normalize them, ... 2 Let$v(x, k)$be an orthonormal set of basis functions such that $$\int_{-\infty}^{+\infty} v(x,a) v^*(x,b) dx = \delta_{a,b}$$ where$\delta_{a,b}$s the Kronecker delta. I believe your question boils down to: for the Fourier Transform, we have $$v(x,a) = e^{jax}$$ and we interpret$a$as the frequency and$x$as time; is there anything special about ... 2 Say you decompose the signal$x(t)$into two components: $$s(t) = x(t) + y(t).$$ As you say, there are myriad ways to do this. The energy of$s(t)is $$E_s = \int_{-\infty}^\infty s^2(t) dt.$$ If we calculate it using the decomposition, we obtain: \begin{align}E_s &= \int_{-\infty}^\infty (x(t) + y(t))^2(t) dt \\ &= \int_{-\infty}^\... 1 Leave time dimension aside for a moment and deal with just one symbol at a time. In a PAM modulation the generated functions are not ortho and not normal. Actually, in a 4-PAM modulation you have four basis pulses:p_0(t)=a_0 \cdot p(t)$,$p_1(t)=a_1 \cdot p(t)$,$p_2(t)=a_2 \cdot p(t)$and$p_3(t)=a_3 \cdot p(t)$. The pulse$p(t)$is explicitly defined ... 1 Suppose you have a set$n$vectors$o_k$in$\mathbb{R}^{m,1}$(length$m$, column-style), pairwise orthonormal, i.e. orthogonal with unit norm. Stack them side by side in a matrix $$A = \left[o_1,\ldots,o_n\right]\,.$$ You have what you call an orthogonal (rectangular) matrix, sometimes called an orthogonal column matrix. The same concept applies row-... 1 The Hartley transform is an involution: it is (up to a scale factor) its own inverse. The classical discrete Hartley transform of order$N$is such that$H_N^{-1} = \frac{1}{N}H_N$. Be careful with your notation, the vector$x$has$N+1$entries, so maybe you are after an$N+1$-order DHT! If$\mathbf{1}$denotes the all-ones vector, then in matrix-vector ... 1 Great question, and a really tricky one to get your head around. I think your dual use of the word distance, and the comparison of vectors in an infinite function space to a vector in a finite-support sequence is a bit confusing. The distance you're describing is the difference between the frequencies$w_1$, and$w_2\$ of two sinusoids. In the time-domain it'...

1

For pairwise stereo calibration try using the Stereo Camera Calibrator app in the Computer Vision System Toolbox. It is much easier to use than Caltech Camera Calibration toolbox. For starters it detects the checkerboard automatically.

1

Use an asymmetric calibration plate. Such as the new circle grid in OpenCV. Check here. You do not need to find the correspondences in the gui, but rather run the calibration directly. Also, do not forget to initialize the optimization problem using the median of the obtained poses. To calibrate multiple view setups, I would also recommend Multi Camera ...

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