noc19-ee36-lec55

In last class, we discussed about the state transition matrix and the state transition equations. So, today we will discuss about the application of state variable analysis in the circuit analysis. So, let us start our discussion of today's lecture. Before going into the application of state variable in circuit analysis, first let us understand what is the relationship between state equations and the transfer functions. So, we know that the linear time invariant system can be described by following the dynamic equation. So, first one is state equation, which we have seen in previous lectures. So, we can say that d x by d t is nothing but a x plus b u, y is the output equation, y t is equal to c x plus d u, where x is considered to be a state vector having n into 1 dimension, u t is input vector having p into 1 as dimension, and y t is the output vector having q into 1 as the dimension. Now, a b c d are the coefficient matrices which correlates the change in the states with respect to its previous state, and similarly c and d define the relationship of a state with the output variables. Now, if you take the Laplace transform on both sides of the state equation, and rearrange the equation in such a way that you get x s that is the Laplace transform of the state vector on the left side, then you get S i minus a inverse x at time t is equal to 0 plus S i minus a inverse b u s as the equation. Similarly, if you take Laplace transform of the output equation, you get y s is equal to c into x s plus d u s. Now, in the output equation if you replace x s with the value which we have just computed for x s, we can write y s is nothing but c into S i minus a inverse into x at time t is equal to 0 plus c into S i minus a inverse d u s plus the component which is there in output equation related to input variable that is d u s. Now, if you see this particular equation, in this you have the value of x at time t is equal to 0. So, by definition the transfer function requires that initial condition to be set to 0. So, this is what we have discussed when we discussed the transfer function in the previous lectures. Now, if you set initial condition to 0, and simplify the above equation you get y s is nothing but c into S i minus a inverse b plus d into u s. And what is y s by u s? y s by u s is nothing but the ratio between output and input which is nothing but the transfer function of the system. So, the transfer function now you get as c into S i minus a inverse b plus d. So, this is the equation which defines the relationship between state equation and the transfer function. So, if you are given the state equations where you can find out the a b c d coefficients or the coefficient matrices, you can simply calculate the transfer function of the system. Let us take one example, so that you can understand what we have discussed. Consider a multivariable system which is described with the help of these two differential equations. So, first is d 2 y 1 by d t 2 plus 4 d y 1 by d t minus 3 y 2 is equal to u 1, second is d y 1 by d t plus d y 2 by d t plus y 1 plus 2 y 2 is equal to u 2. Now, let us define the state variables in the system as x 1 is equal to y 1, so that is one state variable, x 2 is d y 1 by d t and x 3 is y 2. So, when we define these three state variables for the system, we can write these two differential equations in the form of the state variable. So, what you will write for this? So, this is the highest order variable, the component of the differential equation. So, what we will write? We will write d 2 y 1 by d t 2 equal to minus 4 d y 1 by d t plus 3 y 2 plus u 1. Now, if you put the value of the state variable which we have just assigned, so d 2 y 1 by d t 2 can be written as d x 2 by d t or you can simply write x 2 dot equal to minus 4 d y 1 by d t is nothing but x 2 plus 3 y 2, y 2 we have defined as x 3 plus u 1. So, you have got the first differential equation converted into state equation. Similarly, for the second one, let us take d y 2 by d t on the left. So, what you can write? Since y 2 is x 3, you can write x 3 dot equal to minus d y 1 by d t means minus of x 2 minus y 1 that means minus of x 1. So, this is the x 1 minus of 2. What is y 2? y 2 is x 3 plus u 2. Now, you have these two differential equations converted into state equations. When you rearrange, you will get the differential equations can be converted into the state equation. So, two we have got. We have got x 2 dot, we have got x 3 dot, but we need to get x 1 dot also. So, x 1 dot from where you get. So, if you see this equation, this is nothing but x 1 dot equal to x 2. So, now you have three state equations. If you compile it into the matrix form, what you can write? So, that is x 1 dot, x 2 dot and x 3 dot equal to if you take the coefficients using these three equations. So, for x 1 dot, you can write 0 1 0 because only x 2 is the component for x 2 dot. You can write minus 4 x 2 plus 3 x 3. So, you will get minus 4 and 3 as an output. Similarly, for x 3, you will get minus x 1 minus x 2 minus 2 x 3. So, this will become minus 1 minus 1 minus 2 as the coefficients of the matrix and x 1 x 2 x 3 is the state vector. Now, the input, so we do not have any component related to input in this equation. So, we will use 0 for this and for x 2 dot and x 3 dot, we have u 1 and u 2. So, if you arrange it in the matrix form, you can write for u 1 u 2. First equation does not have any input. For second, we have u 1. So, you have 1 0 and for second one, we have u 2. So, you will have 0 and 1. So, this is what you get from the differential equation. Now, you have got the state equation. Similarly, the output equation if you see, the output is y 1 and y 2. So, these are the two variables which we only have in the differential equation. So, we take y 1 and y 2 as an output. So, y 1 and y 2 is the output. Now, y 1 y 2 has to be represented in the form of the state variable that is x 1 x 2 x 3. So, y 1 is nothing but x 1 and y 2 is nothing but x 3. So, this we have defined in our initial assumption. So, we take this y 1 and y 2 as an output variable and compile the state equation. We get the output equation as y 1 y 2 and in the matrix, we get 1 0 0 and 0 0 1 x 1 x 2 x 3. Now, since we do not have any input variable in the output equation, the d coefficient matrix will be 0. So, now this you will get as a, this you will get as b and this is what you have as c matrices, d is of course 0. So, now if we have to find out the transfer function of the system, we will use the derivative which derivation which we have got for the transfer function that is H s is equal to c S i minus a inverse b plus d. Since d is 0, so H s will become c into S i minus a inverse into b. So, now you have transfer function H s is nothing but c into S i minus a inverse b. Now, first we need to find out the value of S i minus a. So, we compile S i minus a and take the inverse. So, when you take the inverse, you need to find the determinant. So, this will be the denominator for the inverse. So, the inverse you will get 1 upon determinant of S i minus a matrix multiplied by the inverse the matrix. Now, you have got the inverse of S i minus a. Now, if you multiply with the c and b matrices in this fashion, you get the transfer function as shown in the slide. So, here you will see since we have the two variable means we are talking about multivariable systems. That is why we are not getting one single transfer function, but the matrix of the transfer functions. Now, let us start our discussion related to state variable method to our circuit analysis. How we will carry out the circuit analysis with the help of state variable method? So, we will discuss. Now, since many engineering system we see have multiple input and multiple outputs. So, that is why the state variable method is very important tool in analyzing and understanding such complex systems, which have multiple input and multiple outputs. So, therefore, the state variable model which we have just discussed is more general than the single input single output model, which we generally see in case of the transfer function. So, what we do in state variable model? We specify a collection of variables that describe the internal behavior of the system, and these variables are known as state variables of the system. These are the variables that determine the future behavior of the system when present state of the system, and the input signals are known to us. So, that means the state variable which you find will give you the future behavior of the system. With respect to our electrical circuit, the state variables we generally consider as inductor current and capacitor voltages, because they collectively describe the energy state of the system. Now, let us see how we will implement it. So, to apply the state variable analysis to our circuit analysis method, we follow the following three steps. First step is that what we will select the inductor current I and capacitor voltage V as a state variable. We have to make sure that they are consistent with the passive sign convention, which we discussed in our initial lectures. Now, let us apply KCL and KVL to the circuit and obtain the circuit variables. So, the circuit variables are voltages and currents in terms of these state variables. This should lead to a set of first order differential equation, because when you talk about the voltage and current, voltage across capacitor and current through inductor, you will get the differential equations, which is necessary and sufficient to determine the state variables. Now, obtain the output equation and put the final result in the state space representation format. So, to see how we can implement this, we will take couple of examples, so that you can understand this concept very clearly. Let us take one example. Here, we have a state space. We need to find the state space representations of the circuit, which is shown in the figure below. We need to determine the transfer function of the circuit also. Here, V s is the input and I x is the output. So, we need to find the value of I x and V s is given as an input. The value of resistance R is 1 ohm. Capacitance is 0.25 Farad. Inductance is 0.5 Henry. Now, in this particular circuit, what we will do? We will first select inductor current I and capacitor voltage V as the state variables. So, when we see them as a state variable, what we can first find? We can find V L that is voltage across inductor is nothing but L into d I by d t. The capacitor current that is I c, you can write as c d V by d t. So, these are the equations, which we have already discussed in our previous lecture. Now, let us apply the KCL at node 1. So, when you apply KCL at node 1, what you can write? You can write I is equal to I x plus I c. What is I x? If you see I x is nothing but the voltage across capacitor, which will be also applied across the resistor. So, I x will be nothing but V by R. So, we have taken I x as V by R and current I, which is unknown. So, what we are doing? We are rearranging this equation. We are taking I c that is c d V by d t on the left. So, what we get on the right is I minus I x, which is minus V by R. If you divide by c, you will get d V by d t is equal to I by c minus V by c R. What is d V by d t? d V by d t is nothing but V dot. So, this is the first equation we have got while applying the KCL at node 1. Now, next we apply the KVL around the outer loop. We get V s is equal to V L plus V. So, when you put the value of V L, you get V s is equal to L into d I by d t plus V. If you put L d I by d t on the left and rearrange and then divide the right side with L, you get d I by d t is nothing but minus V by L plus V s by L. Now, when you get this, you get d I by d t is nothing but I dot. So, you got the second equation for your state equation representation. Now, from the equation 1 and 2, we get the state equation. Now, if we take I x as an output, then I x will be represented as I x. We have just saw I x is nothing but V by R. So, if you compile these equations that is 1 and 2 in the standard form, we get the following state equation. So, on the left, we have V dot and I dot equal to minus 1 by R C 1 by C as the elements of the matrix minus 1 by L and 0, which corresponds to I dot. When you arrange, you get the state equation. Similarly, I x is equal to V by R. So, using this, you get the output equation. Now, you have got A, you have got B and you have got C. Of course, d is 0. So, what we will do? We have A is equal to minus 1 by R C 1 by C minus 1 by L and 0. So, these are the matrix elements. If you put the value of R and C, you get A as minus 4, 4, minus 2 and 0. So, these will be the elements of matrix A. Similarly, for vector B, these are the elements 0 and 1 by L. So, you get B as 0, 2. Similarly, C, you get the elements as 1 by R and 0. If you put the value of R, you will get C as 1, 0. Now, you have got A, B and C. The next task is that we find first the value of S I minus A matrix. Then, we invert it. So, we get this matrix. Finally, we need to find the transfer function of the circuit. So, we get C S I minus A inverse B. If you put the value of all these matrices and simplify, you get the transfer function H S as 8 upon S square plus 4 S plus 8. Now, you can simply say that the transfer function related to the electrical circuit which we have in the example is given by H S equal to 8 upon S square plus 4 S plus 8. Now, let us take another example. So, consider the circuit which is shown in the figure below, which may be regarded as a two input and two output system. In the previous case, we saw the simple single output system. Here, we have two input and two output system. In this case, we need to determine the state variable model and the transfer function of the system. Now, what we will do in this case? We have two inputs. What are those inputs? V S and V I are the two inputs in the circuit and we have two outputs. So, we need to find the value of V naught and I naught. V naught is the voltage across 2 ohm resistance and I naught is the current flowing through the 3 ohm resistance. So, V I and V S are considered as an input in this case and V naught and I naught are the outputs. So, as we did in the previous case, in this case also, we select inductor current and capacitor voltage as the state variables. So, first what we will do? We will apply KVL around the left hand loop. So, this is the left hand loop. If you apply KVL, what you can write? Minus V S plus I 1 into 1 ohm equal to minus I 1, then plus 1 by 6 into d I by d t because voltage across this is L d I by d t. L is 1 by 6. So, you get plus 1 by 6 d I by d t equal to 0. Now, if you rearrange, you can write d I by d t. That is nothing but I dot equal to V S minus I 1 into the value of 6, which you have. Now, we need to eliminate I 1 from this and apply KVL around the loop containing 1 ohm resistor, 2 ohm resistor and 1 by 3 F Farad capacitor. So, if you apply KVL around this, what you get? V S is equal to I 1 plus V naught plus V. So, this is the loop which we are considering for applying the KVL. So, V S you get I 1 plus V naught plus V. So, what is I 1? I 1 you can get from KCL while you apply this KCL at node 1. So, I 1 is nothing but I plus the current flowing through 2 ohm resistance that is V naught by 2. So, you get V naught is equal to 2 into I 1 minus I. Now, from the above two equations, what you can write? You just put the value of V naught in this. So, we will get V S is equal to 3 I 1 plus V minus 2 I or simply you get I 1 as 2 I minus V plus V S by 3. So, you get now I 1 in terms of the other components. So, we can in this way we can eliminate the value the I 1 from the equation. Now, putting the value of I 1 in this equation that is I dot is equal to V S into V S minus I 1 into 6. So, this is V S is V S minus I 1 into 6. So, when you simplify d I by d t is nothing but I dot equal to 2 V plus 4 I minus 4 I plus 4 V S. So, this you get with the help of this. So, when you put this value here and simplify, you get 2 V minus 4 I plus 4 V S. Now, you got one equation. Now, to obtain the second one, what we do? We apply KCL at node 2. So, if you apply KCL at node 2, what you get? The current flowing through 2 ohm resistance is V naught by 2. That will be equal to current flowing through the capacitor. So, you can write 1 by 3 into d V by d t. So, you have got 1 by 3 d V by d t. You can represent as V dot plus the current which is flowing in this equal to I naught. So, when you simplify V dot, you can write equal to 3 by 2 V naught minus 3 I naught. Here again we have to eliminate V naught and I naught because this is the state equation and V naught and I naught are the outputs. So, we need to replace V naught and I naught in the equation which we have V dot equal to 3 by 2 V naught minus 3 I naught. Now, to eliminate what we will do? We get from the right hand loop I naught is nothing but we know that this is the voltage across capacitor is V. So, what we can write? V minus V i by 3 is equal to I naught. So, this is we get from this loop and from the previous discussion that is V naught. We have got as 2 into I 1 minus I and I 1. We have got as 2 I minus V plus V s by 3. So, if you put the value of I 1 in this equation and simplify we get V naught is equal to minus 2 by 3 into V plus I minus V s. Now, we have got the value of I naught and V naught. We put these values in this above equation and simplify. When we simplify we get d V by d t that is V dot is equal to minus 2 V minus I plus V s plus V i. So, now you have got the second state equation. This was the first one. So, now if you compile it in the form of the matrix the state equation which you can now write as V dot I dot. These are the state vector equal to minus 2 minus 1 2 and minus 4 are the elements of the matrix A. Then V i is the state vector plus 1 1 4 0 will be the value of B that is the input vector V s and I s because here we have 2 inputs that is V s and V i. So, that is why you get 4 elements in the B matrix. Then output equation you can compile because V naught and I naught we have just found that is I naught is V minus V i by 3 and V naught is equal to minus 2 by 3 V plus I minus V s. So, using these 2 equations you can compile it that is V naught I naught that is the output vector equal to the C matrix into the state vector that is V i plus the D matrix multiplied by the input vector that is V s and V i. So, with this you can now understand that how you can convert this particular the circuit into a state space representation. Now you have A B C and D. So, the transfer function you can simply calculate that is S i minus A inverse B plus D. So, put the values of A B C D in this matrix and simplify you get the transfer function H s. So, with this we can close our today's discussion. In this session we discussed mainly about the application of state space state variable equations in the circuit analysis. We will continue our discussion related to other aspects of state variable in our next lecture. Thank you.