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

Añadido: 2026-05-21

[00:00:02][music] [music] Hi everyone myself Mkkesh Bija from space engineering academy and sa so under the guidance of director students reisionceptive design with respect to control system already basic concepts of control systems and questionful response subject learning.

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[00:04:47]Please enjoy the video about the bits and bites questions of signal flow graphs and block diagram. Thank you.

[00:04:56]>> In this topic we will discuss about signal flow graph and block diagram related questions.

[00:05:04]So the first question is consider the following signal flow graph.

[00:05:10]The value of C of S by RF he asked dear friends that means he asked the transfer function. So if you see here denominator delta delta means 1 minus sum of loop gains. The first loop is -4.

[00:05:26]Second loop is 3 into - 5 which is -5 -5.

[00:05:36]Next third loop is 2 into min -5 which is going to be min - 10.

[00:05:42]Next fourth next these two loops are touching or non touching. Which loops are non touching?

[00:05:54]If you see there are three loops.

[00:06:04]First loop is 3 into minus5 is the first loop. Second loop 2 into -5 is the second loop. Minus4 is the self loop.

[00:06:13]Next any other loop 3 into 2 is not the loop. 3 is if you go like this it is repeating.

[00:06:22]This is not a loop.

[00:06:24]And if you see here yeah only three loops are there. All three loops are touching. Sum of loop gains completed. Next numerator how many forward paths? There will be two forward paths. One is going to be two. To that path every loop touching and other forward path is three. For that every loop is touching.

[00:06:58]So if you do if you see this First forward path is 2. Second forward path gain is three. For that every loop is touching 2 + 3 5 divided by 15 + 4 19 19 + 12 5 by 20 which is 1x 4 will be the correct answer. 1x4.

[00:07:22]So 1x4.

[00:07:27]So in this question 1x4 option is not at all there. This will be the solution for this question my dear friends.

[00:07:33]Next second question. The overall transfer function of the following system is first of all calculate delta for this kind of questions 1 minus sum of loop gains. First loop is if you see first loop g of s into 1x g of sum of loop gain g of s into 1x g of s is the first loop gain. Second loop gain G of office. This is both the places given sign as plus. Therefore second loop gain is plus G of S. Both the loops are touching. That's it. How many forward path? There is only one forward path.

[00:08:11]For that gain is G of S. For that every loop is touching 1 minus 0. We need to practice without this step so that you are able to understand all the things.

[00:08:20]Firstly already we discussed all these things in the classroom itself. Dang engineers. These two are cancel outs.

[00:08:27]Now numerator is g of s divided by denominator 1 - 1 + g of s sorry minus g of s because minus into plus minus these two cancel out. So dear friends answer is minus one minus one is going to be first one will be the correct answer.

[00:08:46]Next question. The overall transfer function C by R in the signal flow graph shown in figure A is see dear friends he asked transfer function between C by R.

[00:08:56]R is the input C is the output. C output R is input engine. If you see here whenever defined input node and defined output node both are given. What is the shortcut? I said on the defined input node self loop is invalid loop. On the defined output node self loop is valid loop. Therefore engineers C by R is equal to on the defined input node self loop is invalid loop eliminate that with this how many loops are there only one loop is there minus H2 next how many forward path only one forward path gain is G therefore answer is G by 1 + H2 will be the correct answer g by 1 + h2 three will be the correct answer for this question and the next question if you see they given one system And which one of the following is equivalent signal flow graph or equivalent block diagram. So whenever this kind of questions are coming engine it's very simple no need to do that that siming point before the block after the block and next take a point before the block after block multiplied divided. These are the nonsense techniques already we discussed in the classroom. Simply these two multiplied x into z.

[00:10:11]Therefore, y is equal to x into z. And this is one more uh one more input.

[00:10:17]Sorry, this is one more input. Summing point means many inputs, right? x into z + w.

[00:10:24]xz + w. So, if you see here, y is equal to xz plus w is not given. Therefore, is not matches given given question. And here xz plus wz y is equal to here x into z.

[00:10:41]Here w into g therefore xz plus w z this is also wrong not equal to given question. So here w by z friends and here x both are adding here x + w by z multiplied with g. Therefore x into z plus w matches with given question two will be the correct answer. So no need to check third one automatically we satisfied with the second option engineers. This is the way when they are cascade multiply when they are doing when there there is having summing point just add it that's all. Next question in the feedback system shown in figure the noise component of the output is given by assume high loop gain at high frequencies of interest. Okay, he's asking noise component of the output. That means he's asking what is the transfer function between CFS and NFS. That's the question. So c of s by n of s because he's asking what is the noise component at output. First of all denominator 1 minus sum of loop gains how many loops g of s into minus h1 into minus h2. So sum of loop gains is g of s into minus into minus plus h1 of s into h2 of s.

[00:12:03]Right? Next, there is only one loop.

[00:12:10]There is only one loop. Next, how many forward path from noise to output? There is only one forward path. G minus H2 offs into G1. G of S minus G of S into H2 offs.

[00:12:26]So, this will be the concept my dear friends.

[00:12:33]So he's asking assume high loop gain at frequency of interest means one is far far less than g of s into h1 of s into h2 of s that is a point here that is a point. So whenever this is less c of s by n of s is equal to numerator minus g of s into h2 of s as you said this gain is very very high one is ignored therefore it is minus g of s into h1 of s into h2 of s engineers this this cancel out this this cancel out minus minus cancel out then c of s is equal to n of s divided by h1 of s will be the correct answer. So is there any option? Option two will be the correct answer. So if he did not given assume high loop gain at high frequency of interest then we can't ignore one. So one is ignored depends upon the given statement. That is very very important my dear friends. I hope everyone is catching my points here.

[00:13:48]Next question.

[00:13:51]Signal flow graph is given on the defined input node. Once again invalid loop even though they given that means self loops on the defined input nodes are invalid loops. So transfer function is C by R denominator 1 minus how many loops to one loop that is minus H2.

[00:14:09]How many forward paths? One forward path. The gain of the forward path is G.

[00:14:13]Therefore G by 1 + H2. Answer is going to be 2. G by 1 + H2 will be the correct answer for this question. Next question.

[00:14:24]Already friends in this system shown in figure the to eliminate the effect of disturbance D of S on C of S the function Z DD Z G Z D is inverse of these two values directly explained the shortcut many times. So the gain to eliminate the disturbance Z gd of S is going to be inverse of these two gains that is 1x S + 5 into S + 10 multiplied with 10 by S into S + 5. Therefore answer is S into S + 5 into S + 10 divided by 10 into S + 5 will be the answer engine. So answer is S + 5 S + 5 cancel out. So s into s + 10 by 10. So two will be the correct answer for this question. So option two will be the correct answer for this question. Next question.

[00:15:18]The transfer function between y2 and y1. So y2 and y1 on the defined output node self loop is valid.

[00:15:27]Therefore 1 minus how many loops? There is only one loop. Gain of the loop is c.

[00:15:33]This is only one loop. How many forward path? A and B A + B by 1 - C A + B by 1 - C directly for both the forward path loops are touching. Therefore A into 1 - 0 + B into 1 - 0. That's all my dear engineers. These are the direct shortcuts.

[00:15:52]Next question. The signal flow graph is given that TD of S represent the disturbance of the forward path. The effect of disturbance can be reduced by increasing G1 of S. Many many times we discussed before the disturbance whatever is given directly that gain we have to increase so that noise at the output will be reduced already explained effect of feedback on noise in the control system concept please verify in the notebook of our regular class notebook automatically you'll understand why I'm explaining this before the noise or disturbance whatever the gain is there you have to increase only that don't vary feedback and if you increase Z2 the disturbance also carrying more to the output through Z2 of S. Hence we have to increase only before the disturbance whatever the gain is there that we have to increase ultimately noise will be reduced at the output. Hence increasing G1 of S C will be the correct answer that is third.

[00:16:50]Next they're asking the question friends calculate the return difference. Return difference formula is 1 minus G of S into H of S.

[00:17:00]Here G ofs value is A.

[00:17:03]G ofS value is A and H of S value this is a forward path gain A and H of S value given as minus beta. So the return difference is 1 minus A into minus beta. Answer is 1 + a beta. That is the return difference. 1 + a beta is 2 will be the correct answer.

[00:17:25]Engineers 1 + a beta. Two will be the correct answer for this question. Next the equivalent of the block diagram in given figure S dear friends if you see same concept when they are in cascade multiply that's all and here they missed what this is F and here they have missed input is E. So please note this now how these two blocks E and G1 G2 cascade.

[00:17:52]Therefore, E into G1 into G2. What about F is going to be E into G1 into H.

[00:18:02]Next coming to option one. C is equal to E into G1 is wrong compared to that.

[00:18:07]Therefore, eliminate directly. No need to calculate F. Coming to second option, C is equal to E into G1 into Z2. F is equal to this E into Z1 Z2 into H Z2. This is wrong.

[00:18:22]Not given. Not satisfied the given F even though satisfied C. Next C is equal to E is even wrong option. Third because not satisfied given and automatically fourth option E into G1 Z2 and this is E G1 Z2 by H into Z2. G2 Z2 cancel out F is E into G1 into H. So fourth will be the correct answer. This is the simplest techniques to complete the control system. My dear students, next question is this will be the question they're asking what is the transfer function. So if you see transfer function y of s by r of s equal to denominator delta delta means 1 - sum of loop gains. What is the first loop here? 1x s into 1x s into minus1 will be - 1x s². Second loop 1x s into 1x s into -1 will be - 1x s². Third loop 1x s into -1 into 1x s into minus1. It is not a node. Therefore don't think that both are touching. So third loop is + 1x s² + 1x s². Next how many forward paths?

[00:19:39]All the loops are touching. How many forward path? Three forward paths.

[00:19:42]Sorry, two. First forward path gain is 1x sq. For that every loop is touching.

[00:19:48]Next second forward path gain is 1x s 1x s. Now these two cancel out. So dear friends it is s² + 1 / sq divided by s² + 1x s².

[00:20:09]So this this cancel out. It is going to be 1x s.

[00:20:14]Answer is 1x s. One will be the correct answer. First option will be the correct answer.

[00:20:20]Next question. Same like previous means before one question we discussed from the same as that question engineers. So C is equal to what is the formula? E into G1 into Z2. F is equal to E into G1 H. Same question. I think it is repeated engineers. Now C is equal to E E into G1 E C is E into G1 from this it is eliminated wrong from this E into Z2 it is also eliminated. Now if you see C is equal to E G1 Z2 correct but F is E G1 Z2 into HZ2 this is also wrong. So direct answer is D option C is equal to E into G1 Z2 same as this. Next f is e into z1 z2 into h by g2 z2 z2 cancel out e z1 h therefore fourth will be the correct answer directly this is the shortcut engineers next once again they're asking what is a transfer function so engineers if you see here transfer function 1 minus sum of loop gains how many loops one loop which is -4 by Second loop - 4.

[00:21:40]Third loop - 2x s² - 2x s².

[00:21:49]Next fourth loop s inverse into 1 into -2 will be - 2 by s. Four loops are there. All loops are touching. There are two forward paths. First one is 1x s² for that every loop touching. Second path 1x s for that every loop is touching. This will be the correct one friends. If you simplify this you will get s + 1x 5 s² + 6 s + 2 one will be matching for this question my dear engineers.

[00:22:19]This is the answer for this question.

[00:22:22]Next question. Transfer function of the circuit is given. First of all convert these values into original values. 100 into 10^ - 6.

[00:22:36]Now convert this value into original 10 kilo means 10^ 3. 10 into 10^ 3 10^ 4.

[00:22:43]Next this one is 100 into 10^ - 6. Therefore this value is 10^ -4. This value also 10^ -4. Next what is the shortcut output impedance?

[00:22:57]Output impedance is 10^ 4 + 1x cs 1 by 10^ 4 - 10^ - 4 into s 1x c value is 10^ - 4 divid by total impedance.

[00:23:14]So total impedance is 1x 10^ -4 into s + 10^ 4 + 1x 10^ -4.

[00:23:24]Why I written 1x cs? Because in s domain capacity is written as 1x cs inductor is written as sl. So engineers 10^ 4 common in the numerator 1 + 1 by s denominator also 10^ 4 common 1 + 2x s.

[00:23:44]So this this cancel out answer is s + 1 by s + 2 will be the correct answer for this question. s + 1 by s + 2. Fourth will be the correct answer.

[00:23:56]And this is a question here given electrical electrical circuit and signal flow graph is given. Then he's asking what is the values of G2 and H respectively. What is the value of G2 and H respectively? Okay, this will be the question. First if you see here develop KVL equation for this loop.

[00:24:18]Develop KVL equation for this loop.

[00:24:20]First of all, what is me by node sum of incoming signals to it. So I2 of S first what is I2 of S from the signal flow graph I2 of S is sum of incoming signal speed. So that is G2 of S into I1 of S.

[00:24:37]G2 of S into I1 of S. So we need the relation between I1 and I2. For that develop KVL for second loop. So current flowing through Z2 Z4 Z3 is I2. But Z3 is combination of first loop and second loop. Therefore, with respect to loop 2, current is I2 - I1. I can say I2 - I1 into Z3 of S plus Z2 + Z4 into I2 offs I2 is equal to Z.

[00:25:13]So from this engine is I2 is I2 offs is equal to Z3 of S into I1 of S divided by Z2 + Z3 + Z4.

[00:25:31]So this is going to be G2 office. Z2 office value is Z3 / Z2 + Z3 + Z4.

[00:25:42]So with this three option fourth option eliminated because A and B mean one and two options are satisfied. Next calculate this node. What is this node?

[00:25:53]Sum of incoming signal street. That means I1 of S is equal to V_sub_1 of S into G1 of S.

[00:26:05]Next I2 of S into H of S.

[00:26:12]This is node I1 equation. Next, calculate KVL for this loop. V_sub_1 of S is equal to Z1 + Z3 into I1 of S minus Z3 into I2 of S.

[00:26:33]So write this equation the form of I1.

[00:26:36]I1 is V1 of S by Z1 + Z3 plus Z3 divided by Z1 + Z3 into I2. So compare this equation with above.

[00:26:50]You got Z1 also here G1 is 1x Z1 + Z3 and H of S is going to be Z3 by Z1 + Z3 Z3 by Z1 + Z3 2 will be the correct answer for this question in. So simply develop the node equations and simply calculate KVL equations here. Compare both of them. Automatically you'll get the all necessary answers.

[00:27:13]Next question. Consider the following signal flow graph and they are given these values. And if you see the loops, this is first loop, second loop, third loop, three loops.

[00:27:27]And this is fourth loop and this is fifth loop. This is sixth loop.

[00:27:33]And this is seventh loop. Total number of loops are seven. These are the first loop, second loop, third loop. Next, fourth loop and fifth loop.

[00:27:47]Sixth loop and this is seventh loop.

[00:27:49]There are total seven loops. There is a one combination of three loops which are non- touching to each other.

[00:27:57]This loop and this loop both are non-ouching. This this non-ouching this this non-ouching three non-ouching loops. Therefore, what is the correct answer? Both one and two will be the correct answer for this question. Both one and two. But check the number of loops once again. First loop, second loop, third loop.

[00:28:16]And this is fourth loop. And this is fifth loop.

[00:28:22]And this is sixth loop. And this is seventh loop. The correct both one and two will be the correct answer. These are the questions in signal flow graph.

[00:28:31]Next questions will be stability.