 Welcome to the next lecture which is there on Amplitude Modulations, Transmitted Power and A M Efficiency in Communications. So, we will understand this transmitted power and A M Efficiency is in communication by some examples here. So, these are the learning outcomes. At end of the session, students will be able to illustrate the power and efficiency of an amplitude modulation in communication system here. So, just recall what do you mean by amplitude modulation, frequency modulation and phase modulation here and just recall how much power is required to transmit and total A M signal here and how much bandwidth is required here. So, as we know that the total power required to transmit an A M signal is P T is equals to P C plus P L S B P U S B and bandwidth is 2 FM here. So, now we will see some examples regarding to the amplitude modulations. So, in the main equation of an amplitude modulation, we have seen the three components that is the carrier signal as well as the L S B and U S B. So, depending on that only just we will solve some examples on this. So, this is one example that is prove that in an A M, the maximum average power transmitted by A M is 1.5 times of the carrier power here. So, what we have to prove that is the maximum average power transmitted by A M is 1.5 times of an carrier power here. So, total power required to transmit of an A M signal that is total power transmitted by the A M signal is P C plus P L S B plus P U S B. So, these are the three components are there here. So, individually we have to solve for how much power is required to transmit and carrier signal L S B and U S B here. But by solving all these, so we know that that is P T is equals to P C bracket 1 plus M square by 2. So, this is the total power required to transmit an A M signal here P C 1 plus M square by 2 here. So, when it will be the maximum power, so as we know that the modulation index M which lies 0 to 1 here. So, at 0 it will be lowest power required and 1 will be the maximum power will required. So, M equals to 1 will be the maximum power will be required. So, substituting the value of an M in this equation that is P C 1 plus M square by 2 here. So, if you solve this that is P C 1 plus 1 upon 2. So, it will be P C into 1.5 that is nothing but P T equals to therefore, P T equals to 1.5 P C here. So, this is the total power will be 1.5 into P C here. So, that is the maximum average power transmitted by A M will be 1.5 times of an carrier power here total. If you substitute the value of an M equals to 1 in this equation that is 1 plus M square by 2 in that. So, you will get the answer that is P C 1.5. So, now we will understand one more problem that is based on the antenna current here. So, the antenna current of an A M broadcaster broadcaster transmitter modulated to the depth of 40 percent here. So, it is modulated to how much that is the modulated to M equals to 40 percent means it will be 0.4 by an audio sine wave is 11 amperes here. So, what is this 11 ampere it is an the total current or IC current that is the power carrier current. So, antenna current of an A M broadcast transmitter. So, it will be in total. So, it will be I T equals to 11 amperes here. Now, second statement is what it increases to 12 amperes as a result of an simultaneous modulation by another audio sine wave. So, whenever there is an 11 ampere it increases to the 12 amperes when simultaneous modulation by another sine wave. So, one more modulating signal is there if you make use of that signal. So, it will be increased to the 12 amperes here. So, at that time I T will be 12 amperes here. Find the modulation index due to this second sine wave not to the first. So, try to understand this. So, this is modulation index due to this second sine wave here. So, we will see. So, in this example. So, when there will be more than one sine wave. So, how modulation index is calculated for a multiple sine wave here. So, before that first we have to solve for the IC because we required for the second in second equation here. So, we know that formula from the previous lecture that is I T square is equals to IC square bracket 1 plus M square by 2 that is I T total current equals to IC current into 1 plus M square by 2 it is in the form of an modulation index in the form of current here. So, when this will be the case 1 this will be the case 1. So, when M equals to 40 percent. So, it will be 0.4 and I T is equals to will be an 11 amperes here. So, if you solve this. So, that is IC equals to I T under root it will be 1 plus M square by 2 now substituting these values that is M and I T into this equation. So, it will be IC equals to 11 upon under root 1 plus 0.4 square upon 2. So, which gives 10.58 amperes. So, this is total IC current that is the power carrier current that is IC equals to 10.58 amperes here. So, now in the case 2 what is there? So, your case 2 is what it increases to I T increases to 12 amperes. Now, we have to find out the modulation index here that is the M 2 here. So, if you donate this M 1. So, what will be your M 2 here? So, we know that relation that is power is directly proportional to I square. So, that is why we have to take this equation that is the I T square is equals to IC square plus 1 plus M square it is same like an P T is equals to P C 1 plus M square by 2 because power is directly proportional to the I square here. So, now, if you solve for case 2 that is the I T when increases to 12 amperes and M 2 will be what? So, before that so, we have to solve for the modulation index M here. So, it will be same like IC equals to I T under root 1 plus M square by 2. So, we know that IC from this IC will be 10.58 10.58 and I T will be 12 under root 1 plus M square by 2. So, we will solve here now after that it will be 12 upon 58 equals to 1 plus M square by 2. So, after solving this M comes to be an 0.75. So, this M comes to be an 75. Now, if you have the multiple sine wave if you have an multiple sine wave. So, at that time you are going to solve for this M square is equals to M 1 square plus M 2 square here. So, total modulation index M. So, this will be M of T. So, it will be M 1 square plus M 2 square here. So, now, we have to solve for the M 2 here find the modulation index due to the second sine wave. So, we have solved for the M here that is we have taken as an M 2 here. So, it will be substituting the values in this equation that is M 2 square is equals to M T minus M 1 square here. This is M T now if you donate this. So, it will be 0.75 square minus 0.4 square. So, it comes to 0.64 here M 2 will comes to 0.64 here. So, if you have a multiple sine wave. So, we have to make use of this formula M total T that is modulation total modulation will be of M 1 square plus M 2 square plus like this will be M n square here. This will be the modulation index here. So, there is one more problem if you solve for that. So, determine the efficiency and percentage of total power carried by the side bands of an AM view for the modulation index M equals to 1.5 and 0.3. So, we will see the efficiency and percentage of total power carried by the side bands here. So, how power efficiency is calculated? So, we know that by using the formula that is M square by 2 plus M square. So, now for the case 1 if you see M equals to 1 here by putting here. So, it will be 1 upon 1 plus 2. So, 1 by 3. So, it will be 33 percentage for case 2 for M equals to 0.5. So, if you substitute this 0.5 square upon 2 plus 0.5 square. So, it comes to be 0.11 that will be 11.11 percentage here. So, for case 3 if you see M equals to 0.3 square upon 2 plus 0.3. So, which comes to when 0.43. So, which comes to when 4.3 percentage here. So, what we have to analyze? So, we have calculated the efficiency. Now, the percentage of total power carried by the side band. So, how much will be the side band side band power will be carried here it will be 67 percent. So, here the side band will carry how much that is 88.89 percentage here. So, here this the total power carried by the side bands of the wave that is the carrier power will be 95.7 here. Okay. So, this is the carrier power will be there. So, this is the 11.11 percentage. So, it will be an 88.89 here. So, this is the determination of an efficiency and the percentage of total power carried by the side band here. So, this will be the carrier signal. So, this will be a carrier side. So, we will complete the lecture here. Thank you. Thank you.