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

Añadido: 2026-05-31

[00:00:02][music] >> Hi everyone. Myself Mukesh Vidya from Sai Medha Coaching Hyderabad and Space Engineering Academy.

[00:00:14]In video time response analysis video number 4 about the now And this is the bits and bytes most important questions by the expert faculty Vishwambhar sir video Okay, the previous year videos so that concepts of stability meter signal flow graph and block diagram meter and basic concepts of control system here already one video release the energy in the the bits and bytes most important questions and the release the energy in the I will utilize this watch and playlist on the videos watch I think that the good videos go when it comes to circulate out to the notification at the bottom of the and you didn't have to go and you didn't have to go AW and you didn't have to go Polytechnic teachers and you didn't have to notification at the bottom of the mission very strong very strong and you the certificate when I want to make a good question is going to come when the notification is going to come I am ready to write it ready to go and the question is going to Okay, let me do what I can do and you didn't have to start preparation and you didn't have to go to the places preparation start changing one by one engineering studies start this engineering studies start changing me preparation help and go when I sign with the coaching Hyderabad make a free go the bits and bytes series at the computer to provide this to me make this a mission come back to start preparing they don't have to go when I sign with the coaching institution and Space Engineering Academy combine nature in the video with the single app Space Gate app download complete course AE and you didn't have AW and you didn't have Polytechnic teachers and you didn't have to go course and design to get in the video when I have app download topic wise video lectures provide this to me topic wise then go transfer this topic wise topic wise topic wise UPSC >> Last minute revision for the revision purpose.

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[00:03:13]And please enjoy the time response analysis bits and bytes video number four.

[00:03:18]For the circuit shown below the value of zeta is asking for RLC all given. Already you know zeta equal to formula R by 2 under root C by L.

[00:03:32]R value 1 kilo means 1,000 by 2.

[00:03:35]C value 0.01 into 10 power minus 6 divided by L value 10 into 10 power minus 3.

[00:03:45]So, 500 into under root is 10 power minus 8 divided by 10 power minus two which is 500 into 10 power minus six. This is going to be 10 power minus three.

[00:04:07]So 500 into 10 power minus three is 0.5.

[00:04:10]Zeta is 0.5. That's all.

[00:04:13]Zeta is 0.5. Two will be the correct answer for this question. Next.

[00:04:20]The system shown in figure A as the time response Y of T R of that is Y of T and input R of T is equal to 10 into U of T yeah shown in figure B.

[00:04:31]U of T being the step input, both K and T are positive. The value of gain K is first of all, what is closed loop transfer function from this system? K by one plus ST plus adding numerator to denominator.

[00:04:45]So from this system is given eight.

[00:04:48]Eight is nothing but final value. How to get final value here? Substituting S tends to zero. K by one plus K is equal to eight is given. K is equal to eight plus eight K.

[00:04:59]From this K is equal to eight plus eight K but very good question actually. Input is given 10 here. Is 10 into U of T. If it is given it as one into U of T directly this value. But if they given 10 into U of T means the value of transfer function become this is 10 into K. That means we have to take 10 into K. This 10 will come this side.

[00:05:33]Therefore, divided by 10.

[00:05:36]So K is equal to eight by 10 is 0.8 and 0.8 K. So 0.2 K is equal to 0.8. K is equal to four. So K is equal to 4 will be the option.

[00:05:50]No options. Therefore, K is equal to 4.

[00:05:52]If they given 1 into U of T, this will be the answer directly. If they given 10 into U of T means, it should be 8 by 10.

[00:06:01]It should be 8 by 10. Like this, we can take directly also. So, K by 1 + K is equal to 8 by 10. Directly take it.

[00:06:08]Instead of this lengthy method also, directly you can take. If for unity it is 8, for 10 it is given, it is going to be 8 by 10. Therefore, K by 1 + K is equal to 8 by 10, which is 0.8. That's all. You can calculate from this, you will get the value of K. Next.

[00:06:28]List one and list two. List one characteristic equation given, list two nature of the unit step response.

[00:06:34]So, calculate anyone. Here, S ^ 2 + 20.25 20.25. Zeta is miss means zeta is zero. Zeta zero means undamped system. C matches with one. When C matches with one, first option is wrong, third option is wrong. Next.

[00:06:53]Take anyone. Let us take this value. S ^ 2 + 20S + 100. 2 zeta omega n is 20.

[00:07:02]Omega n ^ 2 is 100. From this, omega n is 10.

[00:07:07]Omega n is 10. 2 into zeta into 10 is equal to 20. Zeta is 1. Zeta equal to 1 means critically damped system. So, D is 3.

[00:07:16]When D is 3, D is two, wrong. So, this two option also wrong. Fourth will be the correct answer. So, that we need to calculate remaining options, my dear engineers.

[00:07:28]That is advantage here.

[00:07:30]Okay?

[00:07:31]So, these are the advantages if you calculate directly, engineers.

[00:07:35]If you don't calculate directly, that will become disadvantage. Reason, lot of time will be wasted, right? So, if you even you see the solutions, it given lengthy manner, but no need to calculate all the things. If all anyone calculation if our remaining options eliminate means remaining that whatever is left, that will be correct answer.

[00:07:54]The rise time for the system with a transfer function, so it is in standard form. 2 zeta omega n is equal to 4 omega n squared equal to 8.

[00:08:05]Omega n is equal to 2 root 2.

[00:08:09]So, 2 into zeta into 2 root 2 is equal to 4.

[00:08:14]So, zeta is 1 by 2 root 2.

[00:08:16]Rise time formula is pi minus pi by omega d.

[00:08:23]Pi minus pi by omega d. So, omega d formula is omega n under root 1 minus zeta squared. Omega n is 2 root 2 under root 1 minus zeta squared. 1 by zeta squared is 1 by 2.

[00:08:36]So, it is 1 by root 2. Omega d is 2.

[00:08:39]So, pi minus pi So, cos inverse of 1 by root 2 45 degrees. Therefore, pi by 4 divided by 2.

[00:08:48]So, 3 pi by 4 3 pi by 8. 3 pi by 8 will be the correct answer.

[00:08:56]So, rise time is 3 pi by 8. You will get 1.178 seconds.

[00:09:02]So, approx This is asking answer up to two decimals. Up to two decimals means maintain at least two values after the point. That will be the answer for this question.

[00:09:11]Next question.

[00:09:13]Closed loop transfer function of a system given. This is the closed loop transfer function. He's asking what is the type and order. That means product of type and order.

[00:09:21]So, order is going to be four. Order is calculated either for open loop or closed loop, engineers. Highest power of S in denominator of the transfer function is called order.

[00:09:30]So, order is four. Next, how to calculate type? In order to calculate type, we need to convert into open loop because type is determined strictly for open loop transfer function already explained. Therefore, open loop transfer function is numerator 2s + 5 / denominator s ^ 4 + 3s3 7s ^ 2 + 2s + 5 - 2s - 5. This this this this cancel out. So, 2s + 5 / take s ^ common s ^ + 3s + 7. Therefore, number of integrators is two. Therefore, type is two. Type is two, order is four.

[00:10:13]Type is two, order is four.

[00:10:15]Multiplication of 2 into 4 is going to be eight. So, answer is eight will be the correct answer for this question.

[00:10:21]Next question.

[00:10:23]>> [clears throat] >> Unity feedback system it's open loop transfer function given. If you do close loop, this will be omega n ^ 2 / s ^ 2 + 2 zeta omega n s plus adding numerator to denominator that is plus omega n ^ 2.

[00:10:40]If the closed loop pole lie in the shaded region, this is a shaded region.

[00:10:46]This is a shaded region.

[00:10:49]This is shaded region.

[00:10:51]This is shaded region.

[00:10:54]Which one of the following is true?

[00:10:56]Right? So, it is constant zeta locus, friends.

[00:11:00]He given 45° here also 45. Therefore, cos 45 1 / root 2. Next, they given this is 60°.

[00:11:10]This is 60. This is also 60. This means this is also 60. Cos 60 0.5. That means this shaded region starts from cos 45° because cos phi equal to zeta. And this is 60°. Cos phi is equal to zeta. Zeta equal to cos inverse of 60 1 / 2. That means the zeta value is in between 0.5 and 0.707. But, this is exist this side and this side. Zeta must be both the sides.

[00:11:36]Therefore, modulus of zeta is in between 0.0 0.5 and 0.07 0.707. Therefore, C is directly answer and third will be the direct answer. And one more thing, and if you would see the circle, it is constant omega n locus. The shaded region start at 3 radian per second in all the regions.

[00:11:56]And this this is going to be 5 radians.

[00:11:58]So, this shaded region of second is going to be 5. Omega n is going to be in between 3 and 5. So, C will be the correct answer for this question. It's a direct question. It's a beautiful question control system.

[00:12:10]Next, really is a good question. I want to show you where students are generally doing mistakes. Whenever this value is given unity, then this peak overshoot is directly you can calculate. But whenever this value not given unity, peak overshoot MP is equal to percentage peak overshoot MP C of TP minus C of infinity by C of infinity multiplied by 100. So, engineers, C of TP given, so C of TP means at peak time. That is given 2.5 minus C of infinity. Final value given 2 divided by final value 2 into 100. So, this is going to be 0.5 by 2. 0.5 by 2 is 0.0 0.25%.

[00:12:56]Means, peak overshoot is 0.25.

[00:13:00]Don't say peak overshoot is 0.5. 2.5 minus 2 directly we can take, sir. No, that is wrong. That is going to be taken directly whenever they given this value is 1. If other than 1 is given, go to this procedure. So, peak overshoot is given 0.25. Then he's asking what is the damping ratio, zeta. Zeta is equal to So, you know zeta, how to calculate zeta? Peak overshoot formula is e power minus zeta pi by under root 1 minus zeta square. So from this you can get zeta is 0.403.

[00:13:36]0.403 you are getting zeta value.

[00:13:42]37th question.

[00:13:47]Control system is defined by the following mathematical relationship.

[00:13:51]This is the following mathematical equation. The response of the system at t tends to infinity.

[00:13:57]Response of the system at t tends to infinity means they're asking final value. How to calculate final value? By substituting either t tends to infinity or s tends to zero. Before that we need to check system is stable or not. No doubt the given system is stable. Apply Laplace.

[00:14:12]s s square plus 6s plus 5 into x of s is a second order system. All these poles are All these coefficients are same sign in second order. I said already short circuited system is stable which is equal to 10 into 1 by s minus e power minus 5t is 1 by s plus 5.

[00:14:34]So from this x of s value engineers 10 into 1 by s minus 1 by s plus 5 divided by s s square plus 6s plus 5.

[00:14:47]So now you have to calculate t tends to infinity means s tends to zero. That means x of infinity equal to limit s tends to zero s into x of s.

[00:14:58]s into x of s limit s tends to zero s into x of s value is 10 into 1 by s minus 1 by s plus 5 divided by s s square plus 6s plus 5. I think already we solved this question. Maybe it is repeated question I hope.

[00:15:17]So by substituting you will get this value as 2.

[00:15:21]By substituting x is equal to two.

[00:15:23]So, one will be the correct answer for this question.

[00:15:27]Next question. The third peak overshoot and second understood of the system.

[00:15:32]Third peak overshoot. Peak overshoots n is equal to already explained friends.

[00:15:40]These are the second order system like this.

[00:15:44]For peak overshoots n is equal to one, {comma} three, {comma} five. For understoods, it is going to be two, {comma} four like this.

[00:15:54]So, for third peak overshoot n is equal to five. n is equal to five is e power third peak overshoot and second understood of the second order system under damped system is given by. So, third peak overshoot e power minus five zeta pi by under root one minus zeta square. Second understood e four zeta pi by under root one minus zeta square.

[00:16:16]Therefore, answer is going to be fourth, but this value must be minus.

[00:16:23]So, fourth will be the correct answer.

[00:16:25]Right?

[00:16:26]For third overshoot n is equal to five.

[00:16:28]For second understood n is equal to four.

[00:16:32]Next question.

[00:16:35]Consider a unity feedback system with closed loop transfer function. No doubt, he given already the g of s is not open loop, it is closed loop because he said closed loop.

[00:16:44]So, the time required for the response to reach 50% of the final value in first attempt is For 50% final value is that is time is already calculated. We discussed in the classroom, that is delay time. So, delay time formula is one plus point seven zeta by omega n. So, two zeta omega n is eight.

[00:17:05]Omega n square is 64.

[00:17:08]So, from this omega n is eight.

[00:17:11]So, two into zeta into eight is equal to eight. Zeta is one by two.

[00:17:17]Point five.

[00:17:18]When zeta is point substitute here 1 + 0.7 0.5 divided by this omega n is 8.

[00:17:27]So, 0.35 1.35 / 8 1.35 / 8 So, answer is 0.16 8675. So, two will be the correct answer for this question.

[00:17:49]Next question, calculate peak time of the given system. So, dear friends they given closed loop. So, what is the transfer function for unity feedback? 1 by denominator Sorry, numerator by denominator plus numerator.

[00:18:02]So, peak time 2 Zeta omega n is 1 omega n squared is 1 from here. Omega n equal to 1.

[00:18:12]So, when 2 into Zeta into 1 is equal to 1, Zeta becomes 1/2. So, peak time formula is pi by omega d.

[00:18:22]pi by omega d You know, omega d formula is omega n into under root 1 minus Zeta squared directly. 1 under root 1 minus Zeta squared is 1/4. That is 3/4 root 3/2. Omega d is root 3/2.

[00:18:37]Therefore pi by root 3/2. Answer is 2 pi by root 3 is the peak time.

[00:18:44]So, peak time is 2 pi by root 3. So, 2 pi by root 3 first answer is correct answer.

[00:18:52]Next question The block diagram of series RLC circuit is shown below.

[00:18:59]Okay.

[00:19:01]The input to the system is p of t.

[00:19:04]The input to the system is p of t.

[00:19:08]The value of quality factor quality factor q is equal to formula 1 by 2 zeta.

[00:19:15]Already we discussed during control system class, quality factor Q is equal to 1 by 2 zeta. From this, zeta is equal to R by 2 under root C by L. So, Okay, R value, C value, L value not given. Therefore, use the control system concept. Closed loop transfer function numerator by denominator plus numerator.

[00:19:37]So, from this, 2 zeta omega n is 5.

[00:19:42]Omega n squared is 25.

[00:19:45]So, from this, omega n is equal to 5.

[00:19:47]So, 2 into zeta into 5 is equal to 5.

[00:19:51]Zeta is 1 by 2.

[00:19:53]When zeta is 1 by 2, 1 by 2 into 1 by 2, answer is going to be 1. So, quality factor is going to be 1. So, answer for this question is 1, my dear students.

[00:20:04]So, with this, time response questions also completed. Next, in next, we will discuss root locus questions. Thank you.