Recent content by Karl Karlsson

They come to the conclusion that there are 2 zeros by ##wi## becoming dominant for large values of w and each ##wi## in the denominator gives a phase shift of -90° and each ##wi## in the numerator gives a phase shift of +90°. Thus there must be 2 zeros because you see that ##G(iw)## phase shift...

The Nyquist curve for a fifth order system with transfer function G(s) is given in the attached photo. Here we have plotted G(iω) for both positive and negative frequencies ω. The transfer function G(s) has no poles or zeros strictly on the right side of the imaginary axis Point −1 is marked...

Thanks!

I want to solve the heat equation below: I don't understand where the expression for ##2/R\cdot\int_0^R q\cdot sin(k_nr)\cdot r \, dr## came from. The r dependent function is calculated as ##sin(k_nr)/r## not ##sin(k_nr)\cdot r##. I don't even know if ##sin(k_nr)/r## are orthogonal for...

No, since we should be able to write ##\begin{bmatrix} 0,3 & 0 & 0,5 & 0 & 1\\ 0 & 0,2 & 0 & 0,6 & 0\\ 0,5 & 0 & 0,5 & 0 & 0\\ 0 & 0,8 & 0 & 0,4 & 0\\ 0,2 & 0 & 0 & 0 & 0 \end{bmatrix}\cdot\begin{bmatrix} a & b & c & d & e\end{bmatrix}^T = \begin{bmatrix} 0,3 & 0 & 0,5 & 0 & 1\\ 0 & 0,2 & 0 &...

Right, I had not thought about that. So the answer should be ##\frac {a+c+e} {11}\cdot\begin{pmatrix}5\\5\\1\end{pmatrix}##, right?

Yeah, right, a+c+e<=1. I don't know how i am supposed to know which multiple of the eigenvector ##\lim_{n\to\infty}\begin{pmatrix}0.3 & 0.5 & 1\\0.5 & 0.5 & 0\\ 0.2 & 0 & 0\end{pmatrix}^n\begin{pmatrix}a\\c\\e\end{pmatrix}## goes towards, unless i diagonalize the matrix...

Isn't p(t) always a vector where the sum of its entries are 1? So how could the expression go towards anything other than ##\frac {1} {11}\cdot\begin{pmatrix}5\\5\\1\end{pmatrix}##?

Yes I see now that what I wrote before is wrong. I calculated the eigenvector corresponding to eigenvalue 1, which was ##\frac {1} {11}\cdot\begin{pmatrix}5\\5\\1\end{pmatrix}## . But how do I know when the matrix A that was given will go towards ##\frac {1}...

Ok, so if ##\begin{bmatrix}j & g & r\end{bmatrix}^T## is the eigenvector for the 3x3 matrix and ##\begin{bmatrix}s & t\end{bmatrix}^T## is the eigenvector for the 2x2 matrix which are corresponding to ##\lambda = 1##, is ##lim_{t\rightarrow∞}\vec p(t) = \frac {1} {j+g+r+s+t} \begin{bmatrix}j &s&...

I get it because in a steady state ##A\cdot p(t) = p(t+1)##. But if there are multiple eigenvectors corresponding to ##\lambda = 1##, as there are in this case, how do I know which one ##\lim_{n\to\infty}A^nx_0## converges to?

That's smart, I did not think about that. But it still takes a while to find the eigenvectors for every eigenvalue and then I have to find the inverse of the eigenvectors matrix in order to be able to calculate ##A^n = BD^nB^{-1}##. Would that really be the fastest way of solving this?

I have already in a previous task shown that A is not irreducible and not regular, which I think is correct. I don't know if I can use that fact here in some way. I guess one way of solving this problem could be to find all eigenvalues, eigenvectors and diagonalize but that is a lot of work and...

Yeah, I realized that afterwards. So i can choose h = 1 on 0 <= t <= 1/3 and h=3t on 1/3 < t < 2/3 and h=2 on 2/3 <=t <= 1. An eigenvector to eigenvalue 1 could be a function that is t-1/3 on the intervall 0<=t<=1/3 and 0 on 1/3<t<=1. An eigenvector to eigenvalue 2 could be a function that is 0...

Ooh, right! You mean if I for example said h(t)=1 in the intervall 0<t<0.5 and h(t)=2 in the intervall 0.5<t<1. One eigenvector for the eigenvalue 1 could be the function f(t) which is ##t-0.5## on the intervall 0<t<0.5 and f(t)=0 on 0.5<t<1 ? And for for the eigenvalue 2 an eigenvector could be...