 Hello, I am Milka Jagle working as assistant professor in Department of Mechanical Engineering Vulture Institute of Technology, Solapur. Today, we are going to learn how to solve problems on block diagrams. Learning outcome. At the end of the session, students will be able to apply the rules and reduce the block diagrams in control systems. So, these are the content, canonical form or standard feedback loop of a control system, problems on block diagram reduction and references. So, let us see what is canonical form? A canonical form consists of one forward path, one feedback path, one summing point and one takeoff point. So, in the last video we have studied how to reduce this canonical form into simple transfer function that is C of s upon R of s. So, this is how the canonical form looks like. It has one forward block, one reverse block, summing point, takeoff point that is input signal and this is output signal. This is error signal that is E of s, B of s is feedback signal. So, to reduce any control system or any block diagrams, so a complicated control system is given which consists of n number of blocks. So, we need to reduce these kind of problems by using the rules. So, what are those rules? Blocks which are connected in cascade or series that can be converted into by algebraically multiplying them. Now, blocks in parallel that is blocks in one above the other or one below the other that can be added or subtracted depending on the feedback given to them. Now, associative law for summing point, associative law means the position of a summing point can be changed if there is no any takeoff point or block between the two summing points. So, we are using this laws so as to reduce the block diagrams. So, these are the rules, total 10 rules we have discussed in the last. So, it is just a revision. Shifting the summing point before a block, shifting summing point after a block, shifting takeoff point before a block, shifting takeoff point after a block, shifting takeoff point after a summing point or shifting takeoff point before the summing point. So, now let us see. So, this is the problem given to you. So, the question is using block diagram reduction technique, find the transfer function from each point to the output C from the system shown in the figure below. So, this is R of S and this is C of S. So, in control systems, so this is a complicated or very critical control system, we need to find out the transfer function that is C of S upon R of S that can be written only if one block is in between input and output. Our main motive is to reduce this block diagram into a single block by using block diagram reduction rules. Now, let us see how it can be done. So, now if you see in this diagram, G1 and G4 blocks are connected in series. So, by using the rule 1, they can be algebraically multiplied that is G1 and G4. The block, these two blocks are converted into single block by using rule 1 that is when blocks are connected in series, the gains of the block can be multiplied that is G1 and G4. Now, if you observe this block that is G1, G4 block and H1 block, they form a standard canonical form that is it has one takeoff, one forward path, one reverse path and summing point. This can be reduced by using rule 3 that is G upon 1 plus or minus G into H. I also wanted to tell you before that going to the reduction, here G3 and G2 blocks are connected in parallel. So, the reduction of two blocks is converted into G2 plus G3. So, the sign is depending upon the whether it is a positive signal or negative signal. So, if you see here, these two can be converted into one block and these two are added because they are connected in parallel. Now, if you see here, it forms a closed loop as we have discussed earlier. So, the result of this is G1, G4 upon 1 minus G1, G4 H1, why minus because here it is a positive signal, a positive feedback. If a positive feedback, then you need to take negative sign. If there is a negative feedback, you need to take positive sign. So, if you see in the next step, G upon 1 minus G1 G4 H1, there is a minus sign is taken because here we have positive feedback. So, it is the simplification of this block. This complete loop can be simplified into a single block that is this one. Now, the further it can be seen from the figure that these two blocks are connected in series. The block G1 G4 and 1 minus G1 G4 H1 are connected in series. So it is observed that the blocks G1 G4 and 1 minus G1 G4 H1, so it can be multiplied. So the next step shows that the two blocks that are multiplied. So now, it is shown that G1 G4 G2 plus G3 1 minus G1 G4 H1 this block and this H2 block are they form a standard loop that is that can be converted into G upon 1 plus or minus H2 that is if minus sign we need to take plus sign. Here it is a minus sign, so we need to take in the reduction plus sign. So this is the solution of our problem that is C of S upon R of S. So next problem, next problem in the same way it shows that it is we will follow the same steps this is a control loop that is feedback loop. So G2 upon 1 plus G2 H1 this is if this can be simplified then G1 this block and G3 they are connected in series. So that can be algebraically multiplied. So in the next step it is shown that G1 this block that is closed loop and then G3 these three are connected in series. So they are algebraically multiplied in the next step. Now if you see here it is a closed loop a standard loop G1 G2 G3 upon 1 plus G2 H. So this is the standard loop that can be simplified in the next step that is G5 plus G1 G2 G3 upon 1 plus G2 H1. Now if you see the diagram these two blocks are connected in series. So that can be algebraically multiplied in the next step and the final step is that this block and H2 block are in a closed loop format or a standard loop that is why they can be reduced in a single block that is and after solving this we can get C of S upon R of S ratio which is G1 G2 G3 G4 plus G4 G5 plus G2 G4 G5 H1 upon 1 plus G2 H1 plus H2 G1 G2 G3 G4 plus G4 G5. So this is our final problem. So these are the references which are referred. Thank you.