 Welcome to this session. I am Priyanka Bidla and today we will see about block diagram reduction examples. These are the learning outcomes of this video lecture. At the end of this session students will be able to reduce the block diagram using reduction rules and second outcome will be evaluate the transfer function of the system using reduction rules. These are the contents of this video lecture. You will recall the transformation rules for block diagram reduction. Those are studied in previous lecture. Now pause the video for a minute and you have to recall the rules for block diagram reduction. Transformation rules. So this table represents the all transformation rules. So first rule combine the blocks in cascade. If the blocks are connected in cascade then multiplied with each other. So here G1 and G2 are connected in series. Therefore we take a product of transfer function of that block that is G1 into G2. Next rule combining blocks in parallel. Here G1 and G2 blocks are connected in parallel. Therefore G1 plus or minus G2. If there is positive sign then we can add the blocks G1 plus G2. If there is negative then we can write G1 minus G2. Next rule moving a summing point after a block. So this is a summing point. This summing point is shifted after the block. While shifting a summing point after the block add a block having transfer function as that of the same transfer function of that block that is G. Next rule moving a summing point ahead of a block. Here this summing point is shifted before the block. While shifting a summing point before a block add a block having transfer function as that of the reciprocal of the transfer function of that given block. So here 1 by G. Next rule moving a takeoff point after a block. So here this is a takeoff point and this takeoff point is shifted after the block. For that add a block having transfer function as the reciprocal of the transfer function of that block. That is 1 by G. Then next rule moving a takeoff point ahead of a block. So this takeoff point is shifted before the block with the help of add a block having transfer function as that of the same transfer function of that block. That is G. Then next eliminating a feedback loop. Here this is a summing point and this one is the takeoff point. G is the forward path and H is the feedback path. Forward path is an input to the output and feedback path is nothing but from output to the input. If we want to eliminate a feedback loop use the this rule that is G upon 1 minus plus G into H. Means if there is positive sign then here write down minus sign. If there is minus then write down positive sign. These are the rules for the block diagram reduction. Let us see the procedure to solve the block diagram reduction. There are total six steps. We see one by one step. Step one reduce the blocks those are connected in series. Rule two reduce the blocks connected in parallel. Next step three reduce the minor internal feedback loops. You have to eliminate the feedback loops. Then step four we will try to shift the all takeoff points towards right side and all summing points to the left side. Unless and until is requirement of problem do not use rule nine and ten. These two rules are critical rules. Then step five repeat the above steps that is step one to step four till getting the simplest form. And the step six is nothing but find out the transfer function of whole system after reducing the block diagram. So transfer function is equal to output upon input is nothing but C of S divided by R of S. Consider first example reduce the block diagram to canonical form. This is the given block diagram. Then first step is combine all cascade blocks. In this block diagram there are G1 and G4 are connected in series and G2 and G3 are connected in parallel. So we solve the block diagram step by steps. Step one combine all cascade blocks. So here G1 and G4 are connected in series. So they get multiplied directly G1 into G4. Then step two combine all parallel blocks. Here G2 and G3 are in parallel. So they get added directly G2 plus G3. Then step three eliminate all minor feedback loops. So here H1 is in feedback with the G1 G4 and here is a positive sign. So use a standard formula and reduce the block diagram that is G1 G4 divided by 1 minus G1 into G4 into H1. Step four. Now shifting all summing points to the left side and all takeoff points to the right side. Then step five. Repeat steps one to four until the canonical form has been achieved for the particular input. So here these two blocks are connected in series. So they get multiplied with each other and we get the output. Now here again the H2 is in feedback with this block and here minus sign is there. So use the formula for eliminating the feedback loop. So apply the rule four for this. By reducing the loop containing H2 we obtain this block diagram. So this is the canonical form of the G1 block diagram. Here R of S is the input and C of S is the output. So transfer function is equal to output upon input that is C of S divided by R of S we get this output. This is nothing but this is the transfer function of the system. Now consider second example. Simplify the block diagram and calculate the transfer function. So here R of S is the input and Y of S is the output. First we combine the series blocks then we combine the parallel blocks then eliminate the feedback loops. We see one by one step. So step one is use rule two for blocks G1 and G2 and rule three for blocks G3 and G4. So G1 and G2 are in series so they get multiplied and G3 and G4 are in parallel so they are added. So G3 plus G4. Now in this H1 is in feedback with the G1 into G2 block. So use rule four for blocks G1, G4, G2 and H1. After solving the block we get this block G1, G2 divided by 1 plus G1 into G2 into H1. Now here shift a takeoff point. So next step is use rule A for shifting the takeoff point after the block G5. So here this is a takeoff point now shift a takeoff point after the block. At that time add a block having transfer function as that of the historical of the block. That is instead of H2 here I am writing H2 upon G5. Then next rule use rule two for blocks G3 plus G4 and G5. These blocks are connected in series so they get multiplied G3 plus G4 into G5. H3 is in feedback with this block. So eliminate this feedback loop with the help of rule four again. So use rule four for block G3 plus G4 into G5 and H3. So after simplification we get this output. Here these two blocks are connected in series. So again using rule two they get multiplied directly and we get this output. Now H2 upon G5 block is in feedback with this block. So again eliminate this feedback loop with the help of rule four. So we get the output and this is nothing but this is the canonical form of given block diagram. If we want to calculate the transfer function of this system then write down y of s divided by r of s is equal to this. So this is the transfer function of that block. In this way we have to reduce the block diagram as well as we have to calculate the transfer function of the system. These are the references of this video lecture. Thank you.