# Tag Info

5

Let's solve a more general problem (Least Squares with Linear Equality Constraints): \begin{alignat*}{3} \arg \min_{x} & \quad & \frac{1}{2} \left\| A x - b \right\|_{2}^{2} \\ \text{subject to} & \quad & C x = d \end{alignat*} The Lagrangian is given by: $$L \left( x, \nu \right) = \frac{1}{2} \left\| A x - b \right\|_{2}^{2} + {\... 4 One way to interpret the Tikhonov Regularization is using the Maximum A Posteriori (MAP) framework. Lets' say we have a model of the form:$$ \boldsymbol{y} = H \boldsymbol{x} + \boldsymbol{n} $$Where  \boldsymbol{n} \sim N \left( 0, {\sigma}_{n}^{2} \right) , namely Additive White Gaussian Noise, and the prior knowledge about  \boldsymbol{x}  is  \... 4 When dealing with applying a 2D convolution in frequency domain we have to take into account 2 things: Extending the kernel to the dimension of the input data. Dealing with the implicit periodic extension of the frequency domain element wise multiplication. As you wrote in the comments, one way to do it is simple zero padding and fftshift(). Yet this might ... 3 I suggest this reference regarding the comparison between least-squares and Kalman filters : Fundamentals of Kalman Filtering: A Practical Approach by P. Zarchan & H. Mussof Especially Chapter 3 (Recursive Least-Squares Filtering) and Chapter 4 (Polynomial Kalman Filters). In Chapter 4, the authors show that the discrete (time) n-th order polynomial ... 2 This is an example of the Fidelity Term and Prior Term model. In many Inverse Problems we assume some model on the additive noise. This part is modeled by the Fidelity Term ( \mathcal{D} \left(A \boldsymbol{f}, \tilde{\boldsymbol{g}} \right)  in your example). For Gaussian Noise it is given by Least Squares Term:$$ \frac{1}{2} {\left\| A \boldsymbol{f} - \...

2

Note that for a small sampling interval $T$, $\big(d[k+1]-d[k]\big)/T$ is a good approximation for the velocity. So if you fit $au[k]+b$ to a given set of measurements $v[k]$, it is valid to conclude $$d[k+1]=d[k]+T\big(au[k]+b\big)\tag{1}$$ In the text you refer to they might have normalized $T$, so it changes the units without changing the values of $a$ ...

2

Because when a quantity can be complex, and even when it is just real, the absolute squared difference $|f-g|^2$ can be expressed in both domain (complex and real) as: $$|f-g|^2 = (f-g)^H(f-g)$$ and of course this is correct as well for reals. This setting is often related to Hilbert spaces.

1

As explained in Laurent's answer, including the last point, which equals the first point, just gives twice as much weight to that point compared to all the others. This doesn't explain a phase shift in your approximation. If you do things right you actually get an almost perfect fit, even with the last point included: t = 0:0.15:1.5; y = [2.200 1.595 1.031 ...

1

First (wrong) answer (for integrity) The $y$-value of the last point is the same as the first one. As you apparently know the frequency, this point comes in excess of the "fundamental period". It sounds like this additional point comes like an implicit double-weight to the first point of the period. Second take: I have tried to fit the data, with ...

1

Am I understanding this correctly, and if so, will that give me what I need? Yes you are understanding everything correctly. If your "need" is a filter that will match the magnitude response then all choices should work. If you are interested in a very quick and move-on solution, then proceed with Frequency Sampling with a very large number of taps and ...

1

Ideas: 1) Find the minimum eigen vector of $R_1$ and assign this to $w$. This will minimize $J_1$ , doesn't maximize $J_2$. But when $R_1$ and $R_2$ are positive definite or full rank matrices and computation is an issue, this is a decent solution. 2) Form a new objective $J_1 - J_2$ and minimize this with the given constraint 3) Try and formulate as ...

1

I will give you a hint: you can first relax this problem to be a convex optimization problem by editing the second constraint as $$Xw <= \vec{1}$$ where the inequality is elementwise, then form the dual problem or the lagrangian as it is known popularly $$w^TRw + \lambda^T(Xw -1) \tag{1}$$ where $$\lambda <=\vec{0}$$ Differencate (1) with respect ...

1

$$J(h) = \int_{R^+UR-}|e^Th-F_d|^2d\omega\tag{1}\\ = \int(e^Th-F_d)^H(e^Th-F_d)d\omega\\ = \int((e^Th)^H(e^Th) + F_d^HF_d -(e^Th)^HF_d - F_d^He^Th)d\omega\\ = h^H(\int(e^T)^He^Td\omega) h + \int |F_d|^2d\omega -\int (2 Re\{(e^Th)^HF_d\})d\omega$$ The second term in above integral is the integral of $L_2$ norm of $F_d$ in the region $R^+ U R^-$. Since \$F_d(...

1

The reason x[n] must be white is because the solution will effectively spectrally weight the channel response based on the amount of energy present in each spectral frequency location. A white noise source provides equal weight to all frequencies. If energy is not present in any particular frequency bin, a proper solution cannot be found for that frequency. ...

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