Recent content by CEL

k is odd. So, k(n+1) is odd, no matter if (n+1) is odd or even. Since m is even, k(n+1) + m is odd.

The response of a stable system to a ramp input is a ramp, so it tends to infinity. If you have a pole at the origin, the response to a ramp will be a parabola, as your graph shows.

Your system has a pole at the origin. So, it is unstable. With a unit feedback it can be made stable, depending of the gain (your root locus graph).

Your equation is dimensionally wrong! You wrote sV(s)-v(0). You have a voltage (v(0)) subtracted from the derivative of a voltage (sV(s)).

Try (w^4 - 59 w^2 + 630) + jw(14 - w^2) Then divide (w^4 - 59 w^2 + 630) by (14 - w^2)

An nth order differential equation can allways be represented as a set of n first order differential equations.

For a very good analysis of homeopathy, read http://depletedcranium.com/how-homeopathy-supposedly-works-illustrated/

I agree with your answer. There are three observable and four controllable modes.

i + G(s)H(s) = 0 means: s^3+\alpha s^2 + 1 = 0 Since the polynomial is incomplete (there is no term in s^1) there is at least one root in the RHP and the system is unstable. No need to use Routh algorithm.

the main error is in the statement: x(k)=x(k-2).*x(k-1); it should be: x(k)=x(k-2).+ x(k-1);

In your work you made f(t-2) = t-2 so, f(t) = t and F(s) is the Laplace transform of t. F(s) = 1/s^2.

You can use an integrator feeded by a step voltage.

No, the order of the system is the number of poles (at the origin and elsewhere).

Both diodes are forward biased. If they are ideal, you can replace them with short circuits. If not, you can replace them by voltage sources of 0.6 V.

Remember, the column, where the chosen card is, is in the middle of the other two. When you lay the cards the second time, the cards that were formerly in the first column now constitute the first three cards of the first column and the first two cards of the other columns. The cards in the...