AEE - Control Systems Most Expected Bits || SAIMEDHA KOTI - HYD

[00:00:06]Hi everyone. Myself Mukesh Bija from Space Engineering Academy and Sai Medha Koti, Hyderabad.

[00:00:13]One control system most important and expected bits that's the bits and bytes series by Vishnu sir videos so choose to watch them.

[00:00:24]Didn't know basics going to be Andy.

[00:00:26]Time response analysis going to be Andy signal flow block diagram going to be Andy frequency response analysis going to be Andy root locus inverse root locus going to be Andy. You only could have already previously released them going to be questions and a questions and you could have already seen them and you might have a response on them.

[00:00:42]And last video state space analysis. Based just going to be ready and going to release this one. This particular video state space analysis based just going to be second video and we're going to show you what's going on. Each and every question would be at the end of the song.

[00:00:58]India's most expert faculty Vishnu sir under the guidance of our management.

[00:01:03]Design challenge going to be but you could question me first we're trying to get Andy.

[00:01:08]I thought that solution you said break down chain Andy. Okay, you want to be bits and bytes series not only with respect to only control systems are going to be comprehensive course low control system okay go to any subject something to most important expected bits and one of the ideas going to be one of the online course low space get up low available going to be and complete latest course and you want to be.

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[00:03:46]States space analysis of bits of.

[00:03:50]Most expected bits that bits and bits explanation by Vishnu sir.

[00:03:54]>> Next question, pick up the correct statement. Once again, controllability and observability. If you see friends here, it is a controllable canonical form, CCF form, because identity matrix is there here. 1 by 1 for a 2 by 2 matrix. Then how to calculate controllability and observability for CCF and observable canonical form is if there is a pole zero cancellation in transfer function, then we can say it is either either matching either not satisfying controllability or not satisfying observability. If there is no pole zero cancellation, it may be controllable and observable. So, this is a shortcut already I have explained during pole zero can uh controllable and observable canonical form. Now, transfer function denominator SS square everything is lost down. This is 2 by 2 matrix A matrix order 2 by 2. Covered SS square. Are they A matrix order 3 by 3 either S cube? That depends on order of the matrix. Write the power of S in denominator. Next.

[00:04:50]Last element is written as first with opposite sign. Plus 3 S plus 2.

[00:04:55]Right?

[00:04:56]Next.

[00:04:58]Corresponding this row, that be constant in the numerator, minus 2. Next.

[00:05:05]Write in reverse form.

[00:05:07]Minus 2 into S power 1 minus 2.

[00:05:12]So, take minus 2 common.

[00:05:15]Minus 2 into minus 2 common, it is S plus 1. Divided by S plus 1 into S plus 2.

[00:05:24]So, it is 4 by S plus 2.

[00:05:27]So, there is a pole zero cancellation.

[00:05:29]There may be either losing controllability or losing observability.

[00:05:33]If no pole zero cancellation, directly we can say controllable and observable.

[00:05:37]Due to pole zero cancellation, we need to check controllability and observability. Controllability matrix is BAB.

[00:05:45]So, engineers, B value is -2 -2.

[00:05:51]AB value, just multiply these two.

[00:05:53]0 into 2, 0. 1 into -2, -2. Next, -2 into -2, +4. -3 into - 6.

[00:06:07]-4 Sorry, +4.

[00:06:12]+6 will be 10. Both the rows are non-symmetrical. Therefore, M is not equal to zero. M is not equal to zero means it is controllable, but not observable. A A though could be missed out in the four, it is controllable, but not observable. Option one will be the correct answer. And then, controllability satisfied and observability satisfied and then meaning directly shortcuts.

[00:06:34]Don't waste the time by solving observability. If you don't believe me, you can shortcut you can do the procedure by long method.

[00:06:41]The concept in the end of pole zero cancellation and then it is not controllable. Either not controllable nor either control either not controllable or not observable. Either one is satisfied at the end. Either one can be found in the end. You would this is equal to zero and then it is not controllable, then 100% is observable.

[00:06:59]Suppose, if it is not equal to zero and then controllable and the second not observable. That is anyone is not satisfying. So, it is not controllable.

[00:07:07]It is controllable, but not observable.

[00:07:09]Therefore, solution is completed engineers for this question.

[00:07:16]Next, another question.

[00:07:19]Which of the following are the properties of state transition matrix?

[00:07:22]Phi of zero is equal to I.

[00:07:26]State transition matrix and then the Phi of T is equal to E power AT.

[00:07:31]When T equal to zero means Phi of zero is equal to E power A into zero. E power A into zero, E power zero, E power zero one. One and then the identity matrix.

[00:07:42]Phi of 0 is equal to I.

[00:07:44]Phi of 0 is equal to I. Phi of T is equal to I is wrong answer. Phi of 0 equal to not I wrong answer. Phi of T inverse Phi of T inverse in right to the knee E power 80 whole inverse. Phi of T and 80 power 80 whole inverse. Therefore, this can be written as E power A into minus T can be E power minus 80 which is also equivalent to minus Phi of minus T.

[00:08:10]So, this is three will be correct answer.

[00:08:12]Next, Phi of T2 minus T1 Phi of T1 minus T0 is going to be Phi of T2 minus T0.

[00:08:19]This answer is Phi of T2 minus T0 will be the correct answer. But, this is also wrong. Only three will be correct.

[00:08:26]So, in this only three is correct. Only three will be correct.

[00:08:30]Remaining all are wrong because I explained all these things here. So, this is the way we have to proceed all these solutions. Next, another question. The minimum number of states necessary to describe the network shown in figure in a state variable. So, how to calculate state variables that depends on storage elements. In the nature only storage elements and a capacitor and then inductor. There are three storage elements. Therefore, minimum number of states required is going to be three. How to calculate number of state variables it depends on storage elements. That's it.

[00:09:02]Next question.

[00:09:05]So, if you see here friends, the control system shown in figure is represented by the matrix like this. Y1 Y2 is equal to matrix G into U1 U2.

[00:09:13]So, if you do this lengthy calculation, Y1 you will get is U1 minus Y2 plus Y2 into 1 by S plus 1.

[00:09:26]And the X1 that is going to be it ain't nothing to X1 dot nothing. It X2 that is going to be it ain't nothing to X2 dot nothing.

[00:09:33]Integrator and integrator differential cancel it X2 nothing to go to the same.

[00:09:38]This is X1.

[00:09:42]Sorry, X1 in X1. X1.

[00:09:45]And again differentiate integrate cancel it also. So, from this Y1 by U1 is equal to 1 by S + 1.

[00:09:53]Y1 by U1 1 by S + 1.

[00:09:56]So, this is going to be wrong. This is wrong.

[00:09:59]And Y1 by U2 also same same symmetrical diagram. Y2 is equal to U2 minus Y1 plus Y1 and I am writing all these things directly by looking the expression. You want to check then apply this summing point concept automatically will be satisfied. So, three will be matches for this question. So, three will be correct answer for this question. Next controllability and observability.

[00:10:23]See for complete controllability there is no pole zero cancellation for the CCF form and OCF form. This is once again CCF form. Denominator S square plus 3S minus 2.

[00:10:39]Numerator this constant 1 into S + 1.

[00:10:45]So, I think right now pole zero cancellation not possible. S + 1 into S minus 2.

[00:10:52]Pole zero cancellation there is no possibility therefore it may satisfies controllability and there is no pole zero cancellation hence it is controllable as well as observable.

[00:11:02]Next question.

[00:11:09]X now X. of T is equal to A into X of T plus B U of T. Y of is C into X of T with U as input impulse with zero initial state. The output Y of T at T is equal to 5.5 seconds and they is given forced response is given forced response means X of T is equal to L inverse of SI minus A whole inverse into B U of T. This is a forced response.

[00:11:36]Because at T is equal to 0.5 second they got it. T is equal to 0 second I think I know for in natural response. L inverse SI minus A inverse into X of 0. That is natural response. But he's asking at T is equal to other than that 0 value.

[00:11:50]Therefore, it is forced response. How to calculate SI minus A inverse? SI and the S 0 0 S. And it said it 2 by 2 last minute because A matrix is given as 2 by 2. Next A matrix minus 1 0 0 minus 2. Therefore, SI minus A S plus 1 0 0 S plus 2. Inverse matrix and the last time D minus B minus C A divided by AD minus BC S plus 1 into S plus 2.

[00:12:19]So, this is inverse matrix. So, after this it is 1 by S plus 1 0 0 1 by S plus 2.

[00:12:27]So, inverse Laplace of this value will give E power minus T 0 0 E power minus 2 state transition matrix obtained multiplied by B.

[00:12:39]State transition matrix is applied multiplied by B.

[00:12:43]So, E power minus T into 0 0 E power minus 2 T multiplied by B value is 1 1.

[00:12:51]So, this is going to be E power minus T and E power minus 2 T. At T is equal to 0.5 seconds friends he's asking.

[00:13:02]At T is equal to 0.5 then Y of T is Y is asking. Y is equal to is given already 0 4 into X. X in the E power minus T E power minus 2 T. So, this Y value is 4 into E power minus 2 T. At T is equal to 0.5 he's asking. So, E E power minus 2 E 0.2 2 cancelled E power minus 1.

[00:13:29]E power minus one and they zero point three six eight e power minus one zero point three six eight that is going to be one point four seven engineers so t is equal to zero point five the response is going to be one point four seven next question the transfer function of the system will be the transfer function of the system so transfer function formula is c into s i minus a inverse into b plus d so this is a matrix this is b matrix this is c matrix this is d matrix d is zero so c matrix is given one zero s i s zero zero s minus a matrix is given zero one zero minus three whole inverse into b matrix is given one zero that's all so this is one zero and this value is going to be s and this is going to be zero minus one minus one and this is zero it is going to be s plus three whole inverse into one zero so how to calculate inverse matrix one zero inverse matrix definition is adjoint of the matrix by determinant s plus three one zero s divided by s into s plus three into one zero so s by s plus three is there s by s plus three means okay first let me complete multiplication of these two first one zero s plus three next zero divided by s into s plus three now one into s plus three s plus 3 by S into S plus 3. Actually, if you stop here, S into S plus 3, sir, only one option is there that is matches, but answer is coming 1 by S.

[00:15:32]So, 1 by S will be the correct answer for this question.

[00:15:35]So, 1 by S will be the correct answer for this question.

[00:15:39]Uh next question.

[00:15:42]Phi of S, the coefficient of matrix is coefficient of matrix A is and the state transition matrix is given, then he's asking coefficient of A matrix. Then again, they S plus 6 by S S squared plus 6 S plus me 5 name right into S plus 2 into partial fraction going to divide the early partial fraction going to divide this into what that apply this value in a coefficient of matrix A.

[00:16:13]So, you will get first option.

[00:16:15]So, do the partial fraction then inverse automatically you will get option A will be the correct answer, friends.

[00:16:22]So, answer is going to be A for this question.

[00:16:31]So, sorry, not A. Fourth will be the correct answer for this question. Do the inverse Laplace and apply all these values and cross checking, so fourth will be matching for this question.

[00:16:42]Next question. Bode plot of a PD controller. PD controller is nothing but high pass filter. PD controller is high pass filter. High pass filter and the characteristics starting constant then rising then constant. This is high pass filter. So, if you check that one, this is low pass filter, wrong answer.

[00:17:02]This is also wrong answer. This is also wrong answer. Fourth will be the high pass filter. So, this is going to be satisfying. Fourth one is going to be satisfying. This is high pass filter.

[00:17:11]17th is going to be fourth one. Next.

[00:17:16]From the given figure state model is given, friends. This is X2 is given.

[00:17:21]Whenever this is X2 means it is X1 dot and this is X1.

[00:17:30]Sorry, it is X2 dot then this is X2.

[00:17:35]And maybe this is X1 dot then it is X1.

[00:17:38]So, X1 dot value is -2 X1 +4 X2.

[00:17:46]So, X1 dot -2 X1 +4 X2. -2 X1 +4 X2.

[00:17:52]So, in the first row, -2 and 4 is there.

[00:17:55]Therefore, this option this option is wrong.

[00:17:57]And input coefficients here 0, here 1. So, input is not there.

[00:18:03]Therefore, directly in this +0 into U.

[00:18:06]So, directly second option is satisfied, third is wrong. Reason, input coefficient is given as 1. Therefore, that is not there. Answer is going to be option two will be the correct answer for this question.

[00:18:17]So, next question will be this question.

[00:18:21]So, if you see this question, friends state transition matrix Sorry, state space representation of the seven signal flow graph I got.

[00:18:29]It is X1 and then definitely this is going to be X1 dot of S.

[00:18:33]And this is X2 and then definitely this is going to be X2 dot of S.

[00:18:37]So, if you see here X1 dot is equal to -10 into X1 plus 1 into R plus it is going to be X2 of S.

[00:18:54]So, it is -10 1 into X plus R is going to be 1 into R.

[00:19:02]So, -10 input is R. -10 1. So, it is wrong. You got the first row of -1 - +1 and then -10 -1 -10 -1 and the first option. In this way, we can save a lot of time by solving this kind of questions with shortcuts only.

[00:19:20]Next.

[00:19:22]This will be the question related to last. That means consider the closed loop transfer function of a system.

[00:19:30]State space model is given, then he's asking There are four forward paths. Okay, let me check. Whenever you are going for signal flow graph discussion, I said already the procedure. Whatever the highest power is there in denominator, divide numerator and denominator equation of the transfer function by same. That is 6s cubed by s cubed plus 4s squared by s cubed plus 3s by s squared s cubed plus 10 by s cubed divided by s cubed by s cubed minus plus 8s squared by s cubed plus 4s by s cubed plus 20 by s cubed.

[00:20:14]So, this is going to be written as 6 plus 4 by s plus 3 by s squared plus 10 by s cubed divided by 1 minus of these two are canceling 1 minus of minus 8 by s minus 4 by s squared minus 20 by s cubed.

[00:20:35]So, there are four forward paths.

[00:20:38]There are three loops.

[00:20:40]There are four forward paths. Correct option. There are three loops. First loop is minus 8 by s minus 4 by s squared minus 20 by s squared.

[00:20:48]So, next.

[00:20:51]Three is There are three forward paths.

[00:20:53]Wrong answer. So, option A also satisfies. How, friends? We have to draw the signal flow graph.

[00:21:00]So, signal flow graph I'm constructing here. Automatically you will understand here.

[00:21:04]So, this is is to be 10.

[00:21:07]Input is U.

[00:21:10]1 by S 1 1 by S 1 Next 1 by S 1. In this way we have to draw.

[00:21:21]Now it is 10 by S cube.

[00:21:25]So first forward path gain is 10 by S cube. Satisfies this one. Second forward path 3 by S square.

[00:21:33]Second forward path 3 by S square.

[00:21:42]So in this way it is going to be 3 by S square. Check the forward path 3 by S square.

[00:21:47]Third forward path 4 by S It is going to be 4 by S. Forward path number three 4 by S.

[00:21:56]Next fourth forward path is going to be 6.

[00:22:01]This is sixth forward path. Oh, sorry.

[00:22:03]Fourth forward path gain is 6.

[00:22:05]That gain I will write in white color.

[00:22:08]Next loop minus 8 by S.

[00:22:11]This is minus 8 by S. Second loop minus 4 by S square.

[00:22:16]Minus 4 by S square. Next third loop minus 20 by S cube.

[00:22:22]Minus 20 by S cube. So let's say this is X1. This is X1 dot.

[00:22:28]This is X2. This is X2 dot. This is X3.

[00:22:32]This is X3 dot. He's asking only matrix A.

[00:22:36]So if you check X1 dot value is friends X1 dot value is minus 8 X1 plus X2.

[00:22:47]And this is going to be another one which is four U.

[00:22:52]So X2 coefficient is one, X1 coefficient is minus 8. You check remaining three also satisfy. Therefore with this first one, second one, third one, all three are correct. Fourth one, three forward paths are included. There are four forward paths. Therefore, all these are complete. These state model questions also completed, dear students. Then we have only one topic called mathematical modeling of control system.