 Today, we are going to study interrelationship between parameters of two port network. Learning outcome at the end of this session students will be able to use two port network parameters to find unknown voltage, current, impedance or admittance. So, before starting you have to pause the video here and you have to answer a question that what are the variables which are dependent and independent while writing equations for two port network parameters like z parameter, y parameter, hybrid parameter or a b c d parameter. Now, we can write equation for one parameter in terms of another parameter. So, first we will see how z parameter can be written in terms of y parameter. So, first we will write the equation for z parameters. So, in case of z parameters independent variables are current that is i 1 and i 2 and dependent variables are v 1 and v 2. So, equation is v 1 equal to z 1 1 i 1 plus z 1 2 i 2. And v 2 is equal to z 2 1 i 1 plus z 2 2 i 2. So, these two equations are related to z parameter ok. Now, we will write equation for y parameters. So, in case of y parameter dependent variables are current i 1 and i 2 and independent variables are v 1 and v 2. So, i 1 is equal to y 1 1 v 1 plus y 1 2 v 2 and i 2 is equal to y 2 1 v 1 plus y 2 2 v 2. So, we will make this as a equation 1 this as a equation or we have to simplify equation number 1 and equation number 2 and these two equations are to be simplified or these two equations are to be converted in such a manner that it should be in the form of the equation related to z parameter ok. So, we will solve it. So, from equation number 1 or we have to solve for equation 1 and we have to find out what should be the value of v 1. So, from equation 1 i 1 minus y 1 2 v 2 is equal to y 1 1 v 1. And if you take this term to the LHS it becomes i 1 by y 1 1 minus y 1 2 by y 1 1 v 2 minus y 1 2 by y 1 1 v 2 minus y 1 2 by y 1 1 is equal to v 1 ok. So, this is the value of v 1 and if we put this value of v 1 in equation number 2. So, i 2 is equal to y 2 1. So, instead of v 1 now we have to put this expression that it is i 1 by y 1 1 minus y 1 2 by y 1 2 minus y 1 1 minus y 1 2 by y 1 1 minus y 1 1 v 2 plus y 2 2 v 2 ok. Now we have to simplify this. So, i 2 is equal to y 2 1 by y 1 1 into i 1 minus y 1 2 y 2 1 by y 1 1 into v 2 minus y 1 2 by y 1 1 minus y 1 1 minus y 1 1 minus plus y 2 2 v 2 ok. So, we have to take terms common related to v 2. So, next step is i 2 is equal to y 2 1 by y 1 1 into i 1 plus in bracket y 2 2 by y 1 1 minus y 1 1 minus ok. We have to cross multiply these two terms. So, it is y 2 2 into y 1 1 minus y 1 2 y 2 1 into v 2 is it yes. Now this term that is y 2 2 into y 1 1 minus y 1 2 into y 2 1 this is del y that is determinant of that y matrix. So, instead of y 2 2 y 1 1 minus y 1 2 y 2 1 I can write del y ok. So, next step is i 2 is equal to y 2 1 upon y 1 1 minus i 1 plus del y upon y 1 1 into v 2 is it. Now value of v 2 we can write it as minus y 2 1 upon y 1 1 into i 1 plus this term should be multiplied by this term. So, this term should be multiplied with y 1 1 upon del y ok. So, v 2 is equal to minus y 2 1 by del y into i 1 plus y 1 1 by del y into i 2. So, if we compare this equation with the equation of z parameter we can say that value of z 2 1 is minus y 2 1 by del y and value of z 2 2 is y 1 1 upon del y. Similarly, we can solve this to find out value of z 1 1 and z 1 2 ok. So, again we write the equation related to y parameters that is i 1 is equal to y 1 1 v 1 plus y 1 2 v 2 and i 2 is equal to y 2 1 v 1 plus y 2 2 v 2 this is equation 1 and this is equation 2. Now in last equation we have got the value of v 1 and we have put that value into equation 2. Now here we have to find out value of v 2 and put that in equation 1 ok. So, from equation number 2 it is i 2 minus y 2 1 v 1 is equal to y 2 2 v 2 ok. So, therefore, value of v 2 is minus y 2 1 by y 2 2 v 1 i 2 by y 2 2 ok. Now we have to put the value of v 2 from above in equation number 1 i 1 is equal to y 1 1 v 1 plus y 1 2 instead of v 2 we have to put this value that is minus y 2 1 by y 2 2 v 1 plus i 2 by y 2 2. So, i 1 is equal to y 1 1 v 1 minus y 1 2 y 2 1 by y 2 2 into v 1 plus y 1 2 by y 2 2 into i 2 ok. So, i 1 is equal to in the equation in the bracket it is y 1 1 y 2 2 minus y 1 2 y 2 1 cross multiplication upon y 2 2 into v 1 plus y 1 2 by y 2 2 i. So, it is i 1 is equal to del y upon y 2 2 v 1 plus y 1 2 by y 2 2 i 2. Now we have to find out value for v 1. So, v 1 is equal to i 1 minus y 1 2 by y 2 2 into i 2 and this is equal to i 1 minus y 2 2 by del y. So, v 1 is equal to y 2 2 by del y i 1 minus y 1 2 by del y into i 2. Now if we compare this equation with the equations of z parameter we can say that z 1 1 is equal to y 2 2 by del y and z 1 2 is minus y 1 2 upon del y is it. So, like this we can find out the z parameter in terms of y parameter. As a reference I have used circuit theory analysis and synthesis by A Chakravarti.