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Circuit Analysis with Laplace Transform
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0:06
Hey friends, welcome to the YouTube channel ALL ABOUT ELECTRONICS. So in this video, we will do the circuit analysis using the Laplace transform. So in the previous video, we have seen that, how to find the Laplace transform, and the inverse Laplace transform. So now with the help of it, now let us see how to do the circuit analysis. So as a first step, we will transform the given electrical circuit into the S domain. That means, we will find the equivalent Laplace domain
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representation for the each circuit element. And after that, using the different circuit analysis techniques, like the nodal or the mesh analysis, we will find the voltages, and the current in the circuit in the S domain. Now while finding these voltages, and the currents, we can even use the other circuit analysis techniques, like the source transformation, and the superposition theorem.
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So once we get the equation of the current and voltage in the S domain, then using the inverse Laplace transform, we will find the equivalent time domain expressions for this currents and the voltage. So in this way, using the Laplace transform, we can do the circuit analysis and by following these steps, we can find the voltage or current across the any circuit element in the time domain. So first of all, let us see how to represent the basic circuit elements,
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like the resistor, capacitor, and the inductor in the S domain. So once we know that, then we can easily represent any circuit in the S domain. So first let us see how to represent the resistor in the S domain. So let's say the voltage across a resistor is equal to V(t). And because of this voltage let's say some current I(t) is flowing through the resistor. So as per the ohms law, we can say that, this voltage V(t) is equal to I(t)* R. So after taking the Laplace transform, let's
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say the voltage and the current in the S domain is equal to Vs and Is. And since the R is constant, so it will remain same in the S domain. So after taking the Laplace transform, we can say that this voltage V(s) is equal to I(s)* R. Or in other words, this R is equal to V(s) / I(s). That means for the resistor the equivalent S domain circuit is same as the time domain circuit. Similarly let
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us see the equivalent S domain representation for the inductor. So let's say the voltage across the inductor is equal to V(t). And the current through the inductor is equal to I(t). Now we know that, the voltage across the inductor can be given as L * dI(t)/dt. So now let us take the Laplace transform of this equation. So after taking a Laplace transform, let's say the voltage and the current are V(s) and the I(s). And using the differentiation property of the Laplace transform,
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this dI/dt can be written as this sI(s)-I(0-), where this I(0-) is the current through the inductor at time t is equal to 0-. That means this V(s) is equal to L*sI(s)-I(0-). So based on the given expression, this is the equivalent S domain circuit for the inductor. Where this Ls presents the inductance and this L * I(0-)is the voltage due to the initial current. So we can say that, this is the equivalent S domain circuit for the inductor. Now using the source transformation,
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this voltage source can also be represented as the current source. So now in this expression, let us bring this term on the right hand side. And now if we divide both sides by the Ls, then this current I(s) can be written as this V(s)/ Ls+I(0-) / s. And this is the equivalent circuit representation for the given expression, where this I(s) is the incoming current to the inductor. And this V(s) is the voltage across the inductor. That means if I(0-) is the initial current through
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the inductor, then these are the equivalent S domain representations for the inductor. So as per your convenience, you can remember any one of these two representations. And using the source transformation, it is possible to derive the other one. So while solving the examples based on the given circuit, we can use any one of the representation. Now in this representations, we have assumed that, there is an initial current through the inductor. And we have denoted as the
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I(0-), but if the initial conditions are 0, or in other words, if this I(0-) is equal to 0, then in the S domain circuits this voltage and current Source will be become 0. That means when the initial conditions are 0, then the equivalent S domain representation for the inductor is equal to Ls. So similarly, now let us see the equivalent S domain representation for the capacitor. So let's
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say the voltage across the capacitor is equal to V(t). And because of that, the current through the capacitor is equal to I(t). So we know that the current through the capacitor can be given as C * dV(t) / dt. So now let us take the Laplace transform of this expression. So after taking the Laplace transform, let's say the current and the voltage are I(s) and the V(s) respectively. And using the differentiation property of the Laplace transform this dV by dt can be written as this
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s * V(s) minus V(0). But this V(0-) is the initial voltage across the capacitor at time t is equal to 0-. So further we can write this expression as the Cs * V(s)minus C * V(0). And based on this expression, this is the equivalent circuit. So as you can see, this I(s) is the incoming current to the capacitor and this V(s) is the voltage across the capacitor. And as you can see this 1 / Cs is
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the impedance of this capacitor. And this current source exist because of the initial voltage across the capacitor. So this is the equivalent S domain circuit representation for the capacitor. Now using the source transformation, this current source can also be represented as the voltage source. So using the source transformation, if we convert this current source into the voltage source, then it will come in series with this capacitor. And its value will be equal to this
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1/ CV * this C * V(0-). Or that is equal to this V (0-) / s. And since this current source is going upwards, so the positive terminal of the voltage source will be on the top side, while the negative terminal will be on the bottom side. So this is the equivalent S domain circuit representation for the capacitor. But suppose if you do not know the source transformation, then from this equation also we can get the same circuit. So if you bring this term on the right hand side,
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then this is how we can write it. And now let us divide both sides by the Cs. That means now this voltage V(s) is equal I(s)/Cs+V(0)/s, where this 1/Cs is the impedance of the capacitor. So based on this expression, this is the equivalent S domain representation for the capacitor. That means these are the two equivalent s domain circuit representations for the capacitor. So according to your convenience you can remember any one of them, and using the source transformation
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it is always possible to find the other one. So like I said earlier, during the circuit analysis, depending on the given circuit we will use any one of the following representations for the capacitor. That means these are the equivalent S domain circuit representation for the capacitor, whenever there is a initial voltage across the capacitor. But suppose, if the initial conditions are 0, or this V (0-) is equal to 0, in that case this current the voltage source
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will become 0. And in that case, the equivalent S domain representation for the capacitor is equal to 1/ Cs. So in this way, we found the equivalent S domain circuit representation for the inductor capacitor, and the resistor. So now, let us take the couple of examples, and through that, let us understand how to do the circuit analysis using the Laplace transform. So in the first example, we have been given this circuit, and we have been asked to find this output voltage across
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the inductor. And here, we have been given that the initial conditions of the circuit are 0. That means there is a no initial current or the voltage across the any element. So as you can see over here, the voltage to the circuit is the unit step function. That means before the time t is equal to 0, the value of this voltage source is equal to 0. And at time t is equal to 0+, it will provide a 1 volt of constant voltage to the circuit. So first of all, let us find the
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equivalent circuit in the S domain. So earlier we have already seen that, how to represent the basic circuit elements in the S domain. But here, since the initial conditions are 0, so we can say that, the equivalent S domain representation for the inductor will be equal to Ls. Similarly, the equivalent S domain representation for the capacitor will be equal to 1/ Cs, and for the resistor that is equal to R. So here in the equivalent S domain circuit, this 1 ohm and
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the 5 ohm resistors will remain as it is, while for this capacitor the equivalent representation is equal to 1/ Cs. And here since the value of C is equal to 1x 3F. So we can say that, that is equal to 3/s. Similarly for the inductor the equivalent representation is equal to Ls. And here since the value of L is equal 1 Henry, so we can say that, the equivalent representation is equal to S. And like I said earlier, here the input is the unit step voltage. That means for
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the t greater than or equal to 0, this input is equal to 1. And for the t less than 0 it's value is equal to 0. So now we know that the Laplace transform of this unit step function is equal to 1/s. So in this way, we got the equivalent S domain representation for the each circuit element. And based on this, this is the equivalent circuit in the S domain. So now what we will do, using the mass analysis, we will find the current in both the loops. And using those currents,
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we will find the output voltage V(s) across the inductor. So let's say the current in both the loops are I1s and the I2s respectively. But here during the mesh analysis, for our simplicity we will represent these two currents as only I1 and I2. So now if we apply the KVL in the first loop, then we can write it as this 1/ s. That is equal to this 1 * I1 + 3/s * I1 - I2. Or in other words,
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this 1/ s that is equal to this 1 + 3 / s * I1 - 3 / s * I2. So let's say, this is our equation number one. Similarly now let us apply the KVL in the second loop. So applying the KVL, we can write this 5 * I2 + S * I2 + 3 / s * this I2 - I1 and that is equal to 0. Or further we can write it as - 3/ s * I1 + 5 + s + 3/ s * I2, and that is equal to 0. Or from this, further we can write it
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as this I1 is equal to s/ 3 * 5 + s+ 3 / s * I2. Or that is equal to this 1/3 * s² + 5 s + 3 * I2. So let's say, this is the equation number two. So now we can put this value of current I1 in the first expression. And by putting that, let us get the expression for the current I2. So now in this expression if you put the value of this current I1, then further we can write it as this 1/ s.
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That is equal to this 1 + 3/ S * 1/3 * s² + 5 S + 3 * I2 - 3 / s * I2. So now if you multiply both sides by the 3s, then we can write it as this 3 is equal to this s + 3 * s² + 5S + 3 * I2 - 9 * I2 or that is equal to s cube+ 5s² + 3s + + 3s² + 15s + 9. And over here, we will also have this - 9. So we can say that, this 3 is equal to this entire term * I2. So here this 9 will get cancel out, and if we further simplify it, then we can say that, this current I2 is equal to 3 /
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s cube + 8s² + 18s. So in this way, we found the expression of this current I2. And we know that, this output voltage V out is equal to s * I2. That means this V out is equal to this s * I2, and that is equal to this 3/ s² + 8s + 18. So now we can further write this expression as this 3/ (s² + 8s + 16 )+ 2. And we know that, this s² + 8 s + 16 is = (s + 4)². So we can say that, this V(s) is equal to 3/(s + 4)² + 2. And this 2 can be written as this root to whole square. So further
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we can write it as, 1/ root2 * 3 * root2 /(s + 4)² + root2². So we know that, the Laplace transform of the sinωt is equal to ω/s² + ω². That means here this ω is equal to root2. That means now if we take the inverse Laplace transform of this term, then we will get the expression in terms of the sin term, where the ω will be equal to root2. And using the frequency shifting property, further we can say that, that is equal to this e^ - 4 t* sin root2t. And here we will also have
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this 3 / root2. That means the V(t) is equal to 3/ root2 * this e^ - 4T * sin root2t. So in this way using the Laplace transform, we found the voltage across this inductor. Now without this Laplace transform, if we had gone for the direct circuit analysis, then it would have been little difficult. But as you can see with the help of the Laplace transform, we can easily do the circuit analysis for this kind of circuits. So now let us take another example. So in this example we have
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been given this circuit, and we have been asked to find the voltage across this 4 ohm resistor. And here once again, the initial conditions in the circuit are 0. So to find this output voltage, first let us find the equivalent circuit in the S domain. So here, since the initial conditions in the circuit are 0. So we can say that, the equivalent circuit for this capacitor will be equal to 1/ Cs. And here since the C is equal to 1/4F. So we can say that, that is equivalent
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to this 4/s. Similarly the equivalent S domain representation for the inductor will be equal to Ls. And here since the L is equal to 1 Henry. So that is equivalent to S. And for the resistor it will remain as it is. That is equal to 4 ohm. And now if we see this current source, then that is equal to 10 * u(t). That is a step function with the amplitude of 10 ampere. So if we take its Laplace transform, then that is equal to 10/s. So in this way, we got the equivalent S domain
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representation for the each circuit element. So we can say that, this is the equivalent S domain circuit for the given circuit. So here to find the output voltage V out, we have a two options. The one is either we can use the nodal analysis, and we can find this node voltage. And from that, we can easily find this V(s). And the second option is using the source transformation we can convert this current source into the voltage source. And then after by applying the KVL,
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we can easily find this output voltage. So here we will use the second approach. So first using the source transformation, let us convert this current source into the voltage source. So as you know in the source transformation, when we convert this current Source into the voltage source, then this impedance will get multiplied with the current source. So in this case, that is equal to 4/s * 10/s. Or that is equivalent to this 40 / s². That means this is the equivalent
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voltage source. And since the current is flowing upwards, so the positive terminal of the voltage source will be at the top side, while the negative terminal will be at the bottom side. And this impedance will come in series with this voltage source. So this is the equivalent circuit after the source transformation. And now by applying the KVL equation in this Loop, we can easily find this
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output voltage. So let's say, the current in the loop is equal to I(s). So applying the KVL in the loop, we can say that, this 40 / s² that is equal to this 4/s + s+ 4 times this current I(s). Or we can say that, this current I(s) is equal to 40 / s² / 4/s + s + 4. So if we further simplify it, then we can write it as, this 40 / s * s² + 4 s + 4. That means this current I(s) is equal to 40
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/s * s² + 4s + 4. And we know that, this output voltage V out is equal to 4 * I(s). So that is equal to this 4 * 40 / s * s² + 4 s + 4. Or we can say that, this output voltage V(s) is equal to 160 / s * s² + 4 s + 4. Now we know that, this s² + 4 s + 4 = (s + 2)². So in this way, we got the expression of this output voltage V(s) in the S domain. So now if we take the inverse Laplace transform of this term, then we will get the time domain expression of the output voltage. And to
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find this inverse Laplace transform, here we will use the partial fraction expansion method. So in the previous video, we have already seen that, how using the partial fraction expansion method, we can easily find the inverse Laplace transform. So here using the partial fraction expansion, we can write this expression as A / s + B / (s + 2) + C / (s + 2)² . And using the residue method, we can easily find these coefficients A B and the C. So here for finding this coefficient A,
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let us multiply the left hand side by the s, and let us evaluate the value of the expression at s is equal to 0. So here, this s in the numerator, and the denominator will get cancel out. And if we put the value of s is equal to 0, then the value of A will come out as 40. Similarly let us find the value of this constant C. So for finding this constant C, what we will do, we will multiply the left hand side by this (s +2)², and we will evaluate the value of this expression
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at s= -2. So after the multiplication, this (s + 2)² in the numerator and the denominator, will get cancel out. And now if you put the value of s = -2, then the value of C will come out as -80. So similarly now let us find the value of the constant B. So for finding this constant B, once again let us multiply the left hand side by this (s +2)². And let us differentiate this term with respect to S. And here since it is a first order differentiation, so we will also divide it
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by the 1 factorial. And after the differentiation, we will evaluate the value of this expression at s = -2. So here this (s +2)² in the numerator and the denominator will get cancel out. And now if we differentiate this 160/ s, then we will get this - 160 / s². So now if you evaluate the value of this expression at s = -2, then the value of B will come out as -40. So in this way, we got the value of the constants A B and the C. That means now the output voltage V(s) can be
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written like this. And now as you can see we can easily take the inverse Laplace transform of the each term. So we know that, the inverse Laplace transform of this 1/ s is equal to u(t). Similarly using the frequency shifting property we can say that, the inverse Laplace transform of this 1/ s + 2 is equal e ^ - 2 T * U of T and likewise we know that the inverse Laplace transform of this
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1/ s² = t * u(t). So we can say that, the inverse Laplace transform of this 1/ (s + 2)² = t* e ^ - 2 t* u(t). So including the coefficients, this is the inverse Laplace transform of the V(s). And if you take this 40 outside, then this is the time domain expression of the output voltage. That means this is the expression of the output voltage for the given circuit. So in this way using the Laplace transform, we can easily find the voltage or current in the given circuit,
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and we can easily find the behavior of the circuit for the different types of input signals. So, so far whatever examples that we have taken, in those examples the initial conditions were 0, but in the next video we will take few more examples, where we do have some initial conditions. And for such case, we will see that how to do the transient analysis, and how to find the behavior of the circuit using the Laplace transform. But I hope in this video,
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you understood the basic idea of doing the circuit analysis using the Laplace transform. So if you have any question or suggestion, then do let me know here in the comment section below. If you like this video, hit the like button and subscribe to the channel for more such videos. videos
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