 Hello, I am Milka Jagle, working as assistant professor in Department of Mechanical Engineering, Wolchen Institute of Technology, Solapur. Today, we are going to learn about how to reduce the block diagram problems. Let's see. At the end of this session, students will be able to apply the rules and reduce the block diagrams in the control systems. The content, canonical form, problems on block diagram reduction, and references. Rules for block diagram reduction. We have already studied the 10 rules to reduce the block diagrams. Rule 1, blocks when connected in series that can be algebraically multiplied. Blocks connected in parallel that can be added or subtracted depending on the type of signal, eliminate the feedback loop, associative law for summing point. It shows that in this fourth law, if two summing points are connected one after the other without the presence of any takeoff or any block, then the positions of two summing points can be altered. So, these are the rules in which we can shift the summing point before a block or after a block, shift the takeoff point before a block or after a block and shift takeoff point after a summing point or before a summing point. So, this is the problem in which you need to reduce the block diagram in a single block. So, the problem is given like this. Like some blocks are there, summing point, takeoff points are there. The main aim is to convert this block, this complicated block diagram into a single block. That is we need to find out the transfer function C of s upon R of s that is output upon input. So, let us see how it can be done. So, by observing the problem, we observe that this block that is g2 h2 this take forward path and this backward path forward and reverse path. This forms a standard canonical form. We have already studied in the previous lecture or in the previous video that that can be converted into g2 upon 1 plus g2 h2. Now, let me tell you how this plus sign came. If there is a negative signal or negative feedback, you need to take positive sign over here. So, this block completely dissolve or this block completely converted into a single block that is g2 upon 1 plus g2 h2. So, now if you see here this forms a closed loop. If there was only one summing point present over here, there are two summing points. Like if the if this was not over here, this summing point was not over here, it can be a complete closed loop. So, either I can shift this summing point before this summing point or I can shift this take off point after the block g2 upon 1 plus g2 h2 block. So, you need to decide which step you are going to take. So, in this case, I will shift this take off point after this block. So, this converts into I have shifted the point was here and I have shifted this take off point after this block. So, the rule states that whenever you shift a take off point, you need to add 1 upon g. That is whatever the gain is present over here, you need to add in series with that take off. So, I have added it. So, after doing this step, you observe that this two blocks g1 block and this g2 upon 1 plus g2 h2 block are connected in series. So, this is they are in cascade. So, that can be algebraically multiplied. So, in the next step, we have shown that this g1 g2 upon 1 plus g2 h2, it is algebraically multiplied and can be converted into a single block. Now, if you see this block and this unity feedback, they are connected in parallel. So, we have converted that into a single block and this block and this block, they form canonical form. So, that can be reduced. So, after simplifying this block and this block, you observe that they are connected in a standard canonical form or standard format. So, after that, you have reduced that into single form. That is g2 upon 1 plus g2 g1 g2 upon 1 plus g2 h2 plus g1 g2. So, this is how it is simplified and a single block that is now we get the ratio c of s upon r of s equal to g1 g2 upon 1 plus g1 g2 plus g2 h2 plus g1 h1 plus g1 g2 h1 h2. So, this is how it can be simplified into a simple form. So now, this problem, the main motto is to convert this complicated block diagram into a single block. So, I want you to plan a strategy to solve this problem. So, how it can be solved? If you see over here, it is a closed loop. So, if you see they are connected in series, but one takeoff is present over here. First preference is whether you should check whether the blocks are in series. First step, yes that can be in the g1 and g2 block are said to be in series, but there is a if the presence of g3 is creating a problem. That is why I need to shift to this takeoff point after g2. So, the next step shows that after this takeoff point is shifted, g1 and g2 will be in series. That is why they are converted into g1 and g2. As the takeoff is shifted, we have added a block with 1 by g2 gain. So, now this is over. Now, if you see they are connected in parallel. So, that is why 1 plus g3 upon g2 in the next step it is shown 1 plus g3 upon g2. If you see they are connected in series, that can be algebraically multiplied. But one more step is there. This form a standard form that is g4 upon h1. So, we need to simplify this that is g4 upon 1 plus g4 h1 minus sign is there. So, you need to take plus over here. So, after simplification of all the three blocks that is block 1 g1 g2 upon 1 plus g1 g2 h2, second block 1 plus g3 h2 and the third block g4 upon 1 plus g4 h1. The three blocks are connected in series. So, that can be algebraically multiplied. So, this step shows that all the three blocks are multiplied algebraically. g1 g4 g4 in the bracket g4 plus g3 1 plus g1 g2 h2 1 plus g4 h1. So, this is the block which consists of this term. So, now it can be easily simplified because it has one forward path, one reverse path, one summing point and one gain block. So, this is the standard canonical form. So, that is why we have reduced this into g1 g4 upon g2 plus g3 upon the whole term here the plus sign is indicated just because we have here negative feedback. If the feedback is negative then we need to take the plus sign over here. So, this is how it can be reduced. So, we have found out c of s upon r of s ratio that is output upon input. Quick revision of the two problems which we have studied today. So, these are the blocks which are connected complicated blocks. So, firstly we will reduce this because it is a standard form g2 upon 1 plus g2 h2 and then we can also reduce this but there is a presence of two summing points. So, we need only one summing point that is why it is required to shift the takeoff point. That is why we have shifted this takeoff point after this block as we have shifted we are adding one more block with a gain reverse gain and then we observe that these two are connected in series. So, they are multiplied and then after this multiplication you observe that this is a unity feedback. So, that is why 1 plus g1 g2 upon 1 plus g2 h2 they are added. After simplification the two blocks are connected in parallel. So, canonical form reduction rule is used we receive this. Now, this is reduced by shifting this takeoff point after g2 because first preference we are giving for series connection that is g1 g2. When g1 g2 are simplified as we have shifted this takeoff point reverse gain is added here and then these two are connected in parallel that is it becomes 1 plus g3 upon h2. After that these are connected in parallel. So, it is reduced if you observe three are connected in series and then they are connected into single block. It shows a closed loop these are the this is the simplification. These are the references. Thank you.