2. For a standard second order system given by this transfer...

2. For a standard second order system given by this transfer function: \(G(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_ns + \omega_n^2}\) where \(\zeta = 0.6\) and \(\omega_n = 5\). Answer the following questions: • Find the: rise time, peak time, maximum overshoot, and settling time (the two criterion we discussed in class) if the system input is a unit step function. • Show a plot of how \(T_r\), \(T_p\), and %OS all vary with respect to different values of \(\zeta\) and \(\omega_n\). Ideally, you should do that on MATLAB.

Transcript

00:01 So, here in the first part of question for the unit step response of the first order system we are considering we have to find the transfer function of the system.

00:09 This curve is given here in x axis it is representing the time in second and it is in y axis it is representing the responses.

00:18 So, this is the term which we are given here.

00:21 So, here to find out the transfer function we need to analyze the unit step responsible for the transfer function.

00:26 So, the transfer function of the system can be written as g of s that become equals to k which is divided by s plus a where we are having the value of the k that is the steady state k is the steady state value and a from here is the time constant a is the time constant.

00:46 So, for the graph we can estimate the steady state value as 1 and the time constant as approximately 0 .1.

00:52 Therefore the transfer function that is g s become equals to 1 which is divided by s plus 0 .1 this is the value of the transfer function hence the answer to the first part of the question.

01:02 Now we are considering about the second part where we have to find out the overshoot for the overshoot we are considering here settle time rise time and peak time for the system with the transfer function.

01:11 So, here we are considering about t s that is equals to 14 .145 which is divided by s raise to the power 2 plus 1 .204 of s plus 2 .829 which is multiplied by the s plus 5.

01:25 So, this is the term which we are given here.

01:27 So, from here we can say that to find out the overshoot settling time rise time and the peak time for the system with the given transfer function we need to analyze the transfer function.

01:36 So, the overshoot is the maximum percentage by which the response exceed the steady state value.

01:41 So, in this case we see that that there is a no overshoot since the response reaches a steady state value of 1.

01:49 So, the settling time is the time it takes for the response to reach and stay within a certain pressure percentage let us say that percentage is 2 percentage of the steady value.

01:59 So, to find out the settling time settling time to find out the settling time we can use the formula for the settling time time from here is equals to 2 which is divided by tau omega n where tau is the damping ratio omega n is the natural frequency.

02:14 So, for the given transfer function we can use a natural frequency omega n that is equals to 5.

02:20 So, the settling time can be calculated as settling time become equals to 4 which is divided by tau that is multiplied by the omega n where tau is the damping ratio and omega n is the natural frequency.

02:34 So, from the given transfer function we can see that the natural frequency is approximately that is represented by omega n is approximately equals to 5...

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