Impulse and step response differences
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Hello I'm trying to find the inverse laplace transform of this function:
num = [5];
den = [1 2 5];
>Z(s) = tf(num,den);
I did it manually by hand and got,
>Z(t) = (5/2)*e^(-t)*sin(2t)
It appears that the 'impulse' function matches my inverse laplace function, although my Dr asked for the overshoot, corresponding time peaks, and decaying characteristics. Using 'impulse' function the graph when right clicked there aren't features to show the overshoot or the time peaks. Unless I used a 'step' function it shows these features when I right cliick the graph.
Is it possible that I needed to assume a step input to get these values? Like assume :
Z(s)=(1/s)*(5/s^2+2s+5);
Otherwise is there a function to find the overshoot of the impulse on Matlab?
采纳的回答
The differential equation in the Question is ambiguously written. Your solution for the impulse response assumed that the differential equation is really:
D2x(t) + 2Dx(t) + 5x(t) = 5u(t), u(t) = delta(t), x(0-) = Dx(0-) = 0.
where u(t) is the generic symbol for the input to the system, not the unit step. But, the equation as written doesn't show the "u(t)" explicitly on the right hand side. Based on that, and the context of the question about overshoot, the fact that the question doesn't explicitly define the input but needs to be solved anyway, and that some treatments might be a bit loose in their notation for step inputs, I'm going to guess that the differential equation is really:
D2x(t) + 2Dx(t) + 5x(t) = u(t), u(t) = 5*step(t), x(0-) = Dx(0-) = 0.
where step(t) is the unit step function. If my guess is correct, you do really want that second form of Z(s), though more formally it would look like:
H(s) = 1/(s^2 + 2s + 5) % transfer function
U(s) = 5/s % Laplace of 5*step(t)
Z(s)= H(s)U(s) = (1/s^2+2s+5)*(5/s);
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