# Circuit laws in S-domain

1. Apr 2, 2013

### MotoPayton

I am looking for a proof to why we can apply kirchoff's laws and ohm's law in the s-domain.
If anyone has an intuitive explanation or possibly knows of a link explaining it that would be awesome. Thanks

2. Apr 2, 2013

### marcusl

I think it's because multiplying by exp(-st) and integrating are both linear operations, so the transformation to the s domain is also linear.

3. Apr 2, 2013

### AlephZero

I agree. For example, write down Ohm's law for time dependent current and voltage, $V(t) = R\,I(t)$. Take the Laplace transform of both sides, and you get an equation that looks exactly the same as "Ohm's law" but in the s domain.

4. Apr 3, 2013

### rbj

it's because the Laplace transform of a sum of functions is the sum of the Laplace transforms.

there are 3 physical principles we bring to lumped-element circuit analysis:

1. Kirchoff current law applied to every node
2. Kirchoff voltage law applied to every loop
3. The volt-amp description or characteristic of every device in the circuit

now when you apply the Laplace transform to KCL for every node and KVL for every loop, you simply get an equivalent expression, but where $i_n(t)$ is replaced by $I_n(s)$ and $v_m(t)$ is replaced by $V_m(s)$.

now if (and only if) the volt-amp characteristic of each device (what relates $i_n(t)$ to $v_n(t)$) can also be transformed to a corresponding relationship in the Laplace domain, then we can restate the whole circuit description as some set mathematical equations in the time domain as a counterpart set of equations in the frequency domain. the neat thing about the Laplace transform is that it converts the volt-amp characteristics of capacitors and inductors (that are described with derivatives or integrals) into simple algebraic expressions that simply scale voltage by some function of s to become current, or the reverse. that is the same kind of volt-amp characteristic that resistors have. when we do that, we don't call the capacitors and inductors "resistance", but we call them "impedance" and treat them like resistors, but, instead of $R$, we use $sL$ or $1/(sC)$ in the expression.

what that means is that these techniques we learn with resistive networks, usually called the Node-voltage method or the Loop-current method, we can apply these techniques to circuits with resistors, capacitors, and inductors, but we have to do that in the s-domain. if you get good at it, you should be able to write the matrix equation directly from observation of the circuit.

where this all breaks down is if there are non-linear elements in the circuit (like diodes, or transistors not biased into their linear region). then the Laplace transform does not help you. but KCL, KVL, and the raw V-I characteristics are still valid descriptions, in the time domain, of the circuit and the circuit behavior.

Last edited: Apr 3, 2013
5. Apr 3, 2013

### MotoPayton

I think I may need to learn more about linear operations and how they affect their arguments

6. Apr 3, 2013

7. Apr 4, 2013

### rbj

the first sentence speaks to why KVL and KCL are still applicable in the s-domain. it seems obvious to me.

for example, KCL says that the sum of all of the currents going into a node add to zero:

$$i_1(t) + i_2(t) + i_3(t) +... + i_n(t) = 0$$

if you apply the Laplace transform to both sides, you get

$$\mathcal{L} \left\{ i_1(t) + i_2(t) + i_3(t) +... + i_n(t) \right\} = \mathcal{L} \left\{ 0 \right\}$$

or

$$\mathcal{L} \left\{ i_1(t)\right\} + \mathcal{L} \left\{i_2(t)\right\} + \mathcal{L} \left\{i_3(t)\right\} +... + \mathcal{L} \left\{i_n(t) \right\} = \mathcal{L} \left\{ 0 \right\}$$

or

$$I_1(s) + I_2(s) + I_3(s) +... + I_n(s) = 0$$

where

$$X(s) = \mathcal{L} \left\{x(t)\right\} \ = \ \int_0^{\infty} e^{-st} x(t) \,dt$$

that just is a restatement of KCL, but it is applied to the Laplace Transforms of all of the currents, rather than directly to the currents (in the time domain).

nothing more than that.

Last edited: Apr 4, 2013