Lecture: 31

So, in this week, we will see how we can utilize the Laplace transform into circuit analysis. So, let us start our discussion. Before going into the detail, let us first understand what we discussed when we were discussing about the Laplace transform. So, basically the Laplace transform allows a signal represented by a time domain function to be analyzed in the S domain. So, basically if you have a signal which has some function in time domain that can be represented in S domain for analysis. Now, properties of the Laplace transform and the Laplace transform of basic common functions were discussed in the previous classes. Then we discussed the inverse Laplace transform and we said that the inverse Laplace transform can be found using partial fraction expansion or using Laplace transformation paired technique. And then we also discussed that real poles lead to exponential functions and complex poles lead to damped sinusoid. And we also discussed the initial value and final value theorems also through which we can find out the initial and final conditions required for circuit analysis. So, these are the key points which we discussed when we were discussing about the Laplace transform. Now, let us talk about how we will do the circuit analysis using Laplace transform. So, the Laplace transform is considered to be one of the best technique for analyzing a electrical circuit. Actually involves three major steps, one is that you transform the circuit from time domain to S domain, it means that whatever the circuit elements you have in the electrical circuit all will be first transformed into S domain. And then we will solve the circuit using the circuit analysis techniques which we discussed in the previous classes, those are like nodal analysis, mesh analysis or may be source transformation, superposition or any circuit analysis technique which we discussed, this includes your Thevenin's and Norton's equivalence also. So, the second step will be that whatever the circuit we have formed in S domain, we will utilize the circuit analysis technique to solve the circuit in S domain. And finally, when we get the answer in S domain, we will use inverse Laplace transform for the solution. And finally, we will obtain the solution in time domain. So, these are the three major steps we will follow when we will do the circuit analysis. So, first thing which we have to understand is that how we can convert the various circuit elements from time domain into S domain. So, let us first see the three key elements which we generally have in our circuit, those are resistors, inductors and capacitors. So, how we can transform the three elements into the S domain. So, for resistor the voltage current relationship in time domain, we know that it is given by V is equal to I into R that is as per Ohm's law. Now, if you take the Laplace transform, we get the time domain equation getting converted into S domain. So, V t will become V s, R is a constant quantity, there is no change in the R and I t will be converted into I s. So, V t V at any time t is equal to I at any time t into R will be converted into S domain as V s is equal to I into I s into R. So, this we get when we convert resistor into frequency or S domain. Now, for inductor as we know that the voltage across inductor can be given by this equation that is L di by dt. Now, we will use time differentiation technique which we discussed in the previous classes when we were discussing about the various properties of Laplace transform. So, we will convert this time domain equation for inductor into Laplace domain that is S domain. So, V t will become V s. Now, L will remain same being a constant quantity, but in case of di by dt, we will use time differentiation technique. So, we will get the Laplace transform of di by dt as S into I s minus I at value of I at 0. So, we can say that the Laplace transform of inductor voltage can be given by S into L I s minus L I at time t is equal to 0. So, this we get in the S domain. Next is how we can represent that equation in the circuit. So, we know that if you have V s is equal to S L I s minus L I naught L I at time t is equal to 0. At the same time, you can also calculate the value of current I s with the help of the same equation. So, you will rearrange the equation in such a way that you get I s is nothing but 1 upon S L into V s plus I at time t is equal to 0 divided by S. So, now we have the inductor voltage as well as inductor current in S domain. Now, what we will do? We will convert these equations into a equivalent circuit. So, if you see in time domain, we have an inductor which is carrying some current I at any time t with initial current as I at 0 and voltage across inductor is V at any time t. So, this is the time domain representation of inductor voltage and current. Now, if we have to convert them into S domain, V t will become V s, I t will become I s. Now, for the inductor L, what we will do? We know that V s is nothing but S into L I s minus L I 0. So, that is what we got from the Laplace transform. So, minus sign added to L I at 0 term, what will happen? To represent the minus sign, the polarity of a fictitious voltage source that will represent the initial condition for inductor will become L at time t is equal to 0 and the polarity will be opposite. So, the equation for voltage in S domain for the inductor will become as follows. S L that is the inductor in series with voltage source equal to L at L into I at 0 and with negative polarity on top and positive in the bottom. So, this you can see that will represent the equation which we just derived while we were transforming time domain equation into S domain. Similarly, for current, now you know that if the voltage across inductor is V at any time t, it will be represented as V in S domain and similarly, current I t will be represented as I s. Now, if you see this equation, so from this you can see that current I s is converted is divided into two parts. So, we will use that equation to find out the equivalent circuit in case of we want to find out the current I s. So, from first the value what we get is I 0 by S. So, this will be the value of a current source that will represent the initial condition that is initial current flowing through the inductor. So, I 0 by S, the direction will be downward because it is positive. So, current I s will give you the same direction for the fictitious current source which you will add for representing the initial condition of current flowing through the inductor. Now, next the value of current flowing through the inductor. So, I 0 by S will be equal to so this would be nothing but V s upon S L. So, this is what is represented. So, it means that this value of inductor will be S L. So, in this way we can represent the inductor either for in the form of voltage V s or in the form of current I s. So, these are required whenever you are solving the circuit equation. This would be usually utilized when you are using the Kirchhoff voltage law and this can be utilized when you are utilizing the Kirchhoff current law for converting the circuits. Now, next let us talk about the capacitor. So, in case of capacitor we know that current at any time t can be given by C d V by d t. So, using the equation in time domain we will transform this into S domain. So, when we will convert it into S domain it will become I s equal to C and then d V by d t will be converted again into S into V s minus V at time t is equal to 0. So, finally, the current I s we will get is S into C V s minus C V at time t is equal to 0. Now, in the same way if you rearrange this equation or this expression in terms of voltage V s. So, you can simply write V s is nothing but 1 upon S C into I s plus V at time t is equal to 0 divided by S. So, these two would be the governing equations for capacitor in S domain. So, we can represent as we did in case of inductor we can do in case of capacitor also and represent these two equation in the circuit. So, let us see how we will represent the equivalent of capacitor in S domain. So, this is the equivalent representation of capacitor in time domain where you have voltage across capacitor as V t initial voltage across capacitor is V 0 that is V at time t is equal to 0 and then a current I is flowing through the capacitor at any time t. Now, when we are asked to find out the voltage V s in S domain across the capacitor. So, we will use this equation and you can simply see that it contains two terms and both are since it is a voltage. So, both term can be a voltage terms and since it is a positive it means that a positive voltage source that is plus polarity on the top and negative at the bottom will be the value. So, this will be the fictitious voltage source that is V at time t is equal to 0 by S which will represent the initial condition that is the voltage across capacitor at time t is equal to 0. Now, plus 1 upon S c will be the value of the capacitor here because when you say the current I s is flowing through the circuit. So, I s into 1 by S c will be the drop across capacitor plus the voltage that is V 0 that is V 0 at V at time t is equal to 0 by S and when you sum up you will get the value of V at in S domain. So, this will be required when you are having the Kirchhoff voltage law to be applied. Now, next you can also represent the current I s in terms of these two terms. So, when you see these two terms these are nothing but the current terms because I s is the current. So, means I s is divided into two parts. So, when we represent it in circuit the first term you will see the value is 1 upon S c and when you divide 1 upon S c by V s you will get S c into V s which will be nothing but the current flowing through this lag of the circuit. Next is c at c into V at time t is equal to 0 this is again a current term which is a constant value because this will represent the voltage across capacitor at time t is equal to 0 and since this is having a negative sign the direction of current would be opposite of the direction of the current I s. So, in this way you can represent equivalently the equation for I s and this you can utilize whenever you see the circuit to be analyzed with the help of Kirchhoff current law. So, these voltage and current equations you can use whenever it is required to solve the circuit according to your convenience. Now, from these equations we observe that the initial conditions are now part of the transformation. So, we need not to worry about how we will incorporate the initial conditions in S domain because the S domain equations take cares of initial conditions also. Now, this is the one of the biggest advantage of you will be using Laplace transform in circuit analysis. Now, another advantage is that this will give you a complete response which will include transient as well as steady state response of the circuit. So, if you remember when we were discussing about the step response of may be first order second order circuits we converted the solution into natural response and the first response and we found both of the responses separately and finally, we sum them up to get the final response. But in this case we need not to worry about having two solutions analyzed separately and then summing up because the Laplace transform and the circuit analysis using Laplace transform will directly give you the complete response of the circuit. Now, if you see closely these two equations that is V s is equal to S L I s minus L I at time t is equal to 0 and I s is equal to S into C V s minus C V at time t is equal to 0. So, this is for inductor and this is for capacitor. So, when you observe then you can confirm the duality because if you see both of the equations you will see that L is dual of C, I s is dual of V s and V at time t is equal to 0 is dual of current I at time t is equal to 0. So, these creates the dual pairs. Now, when we talk about the impedance in S domain. So, impedance would be nothing, but the ratio of voltage Laplace transform and current Laplace transform under 0 initial conditions. So, we can write the impedance Z s in S domain would be nothing, but V s upon I s. So, the impedance of the three circuit elements which we discussed will become Z s for resistance will be only R for inductor it will be Z s is equal to S L for capacitor the Z s would be 1 upon S C. Now, if you want to find out the admittance, admittance would be the reciprocal of the impedance. So, in case of Z s if you want to find out the admittance that is Y s. So, it will become I s by V s and similarly the values of like conductance and the inverse of inductor and capacitor will also change. Now, the use of Laplace transform in circuit analysis will facilitate the use of various signal sources. So, like impulse response, step response, ramp and exponential source response can be utilized for analyzing the circuit. So, now let us take some examples. So, that we can understand how we will utilize the circuit elements and their Laplace transform in the overall circuit analysis. So, let us take one example where we need to find out V naught at time any time t across the inductor and we assume that there is no initial current through the inductor or initial voltage across the capacitor. So, they it will have the 0 initial condition. Now, what we have to do first? First we have to transform the circuit from time domain into S domain. So, the source which is unit step function when you convert into S domain it will become 1 by S. Inductor will become the inductor is S L and value of L is 1. So, it will become S. The 1 by 3 Farad capacitor will have the value as 3 by S in S domain. So, now we have all these elements in S domain. We can represent this circuit in S domain now. So, source will now become 1 by S. There will be no change in the resistances. So, this will these will remain same. Capacitor has now become 3 by S and inductor has become S and we need to find out first the voltage across inductor in S domain and then we will convert it into time domain. Now, if you see the circuit you can utilize the mesh analysis. So, let us say that the current I 1 S is in loop 1, I 2 S is in loop 2. So, for mesh 1 you can write the equation as given in the slide that is 1 by S equal to 1 plus 3 by S that is the total sum of impedance of the circuit that is 1 plus 3 by S into I 1 that is the current I 1 flowing in the mesh minus 3 by S I 2 because for the capacitor I 1 is in this direction and I 2 is in opposite direction. So, you will get another term minus 3 by S for I 2. For mesh 2 what will happen you will sum up all the impedances. So, it will become S plus 5 that is the value of resistance and then plus 3 by S that is the capacitance value. So, you will get for I 2 you will get S plus 5 plus 3 by S and then since I 1 is in opposite direction of I 2 for capacitor you will get another term for capacitor that is minus 3 by S into I 1. So, now when you simplify this you will get the value of I 1 as 1 by 3 into S square plus 5 S plus 3 into I 2. Now, you have I 1 in terms of I 2 you will put the value of I 1 in terms of I 2 in this equation. So, you will get 1 by S into 1 plus 3 1 plus 3 by S into 1 by 3 into this term into I 2 minus 3 by S I 2. Now, if you simplify what you will get you will get this expression that is 3 is equal to S cube plus 8 S square plus 18 S into I 2. So, from this you will get I 2 as 3 by S cube plus 8 S square plus 18 S. So, you can take S out and when you multiply S at both sides you will get S I 2. Now, S I 2 is what if you see the circuit S into I 2 will be V naught S. So, you can write V naught S is nothing but, S I 2 is equal to 3 divided by S square plus 8 S plus 18. Now, to solve it you will utilize the creating the square technique. So, you will first create the square of these terms. So, it will become S plus 4 square plus since you will have S square plus 8 S plus 16 for this term. So, S plus 4 square you will left with term 2. So, you can say it as root 2 square. Now, in the numerator you can multiply by root 2 and in denominator you can have root 2. So, if you see this particular component what you can see you can see this is nothing but, the Laplace transform of e to the power minus 4 t into sin of root 2 t because the Laplace transform for sin omega t is nothing but, omega upon S square plus omega square. So, you can use this the output of the Laplace transform of sin omega t you can say that this is nothing but, the Laplace transform of sin of root 2 t. So, this with this you will get the voltage response across the inductor. Now, let us see another case when you have the initial condition that is voltage V across capacitor and the value is 5 volt. So, we will first transform the circuit into S domain. So, we will see the complete circuit in next slide, but we have to first see that the initial condition which is given will have the current source that is C V naught because if you see these all are connected in parallel it means that Kirchhoff current law would be the most suitable option. So, now when you see Kirchhoff current law means you have to convert this capacitor into equivalent S domain. So, now this particular term will be converted into one upon S C and in parallel with you will have current source value of C V naught, V naught is given as 5 volt, C is given as 0.1 f. So, you will get C V naught is nothing but, 0.5 ampere current source. Now, using this particular circuit you can transform the complete circuit in S domain. So, it will be converted into S domain as shown in the figure. The unit step function multiplied by the exponential means you are getting the time shift. So, the Laplace transform of the voltage source will become 10 upon S plus 1 and then resistances will remain same. The capacitor will now be converted into the 1 upon S C would be nothing but, 10 by S and in parallel with you will have one current source having 0.5 ampere as value because this will represent the initial condition. And then you have current source another current source of value 2 delta T when you convert into Laplace or transform it will become 2 ampere. Now, what you have to do you have to apply the Kirchhoff current law at the top node. So, you will see 1, 2 and 3 are the currents which are going inside the node and 2 are coming out of the node. So, when you compile incoming and outgoing currents you can say that 10 upon S plus 1 minus the node voltage V naught divided by 10 will be the value of this current plus 2 and 0.5 are the current sources will be equal to outgoing currents. So, here the outgoing current is V naught by 10 and the here outgoing current will be V naught upon 10 by S. Now, if you multiply both side by 10 and rearrange you will get 10 upon S plus 1 plus 25 equal to V naught into S plus 2 or if you rearrange again you will get V naught would be 25 S plus 35 divided by S plus 1 into S plus 2. Now, you can represent this as partial fraction you can utilize it will become A upon S plus 1 plus B upon S plus 2. When you solve you can get the value of A as 10 B is 15 and your V naught S will become 10 upon S plus 1 plus 15 upon S plus 2 which you can easily write in terms of inverse Laplace transform. You will get V naught t is nothing but 10 into e to the power minus t plus 15 into e to the power minus 2 t and then you multiply by unit step function to represent that the time this will be valid for time t greater than 0. So, now with this we can close our today's session. So, in this we discussed about the various circuit elements and how you can convert them into Laplace transform and we will talk about now the transfer function in the next class. Thank you.