How to Handle a Current Source in S Domain for a Loop Analysis

Laplace transforms simplify complex equations.

S-domain is terminology associated with Laplace transforms in the field of mathematics and engineering. Electrical engineers use Laplace transforms to convert complex mathematical time domain representations into S-domain representations. S-domain representations are easier to manipulate during circuit analysis or loop analysis. For example, manipulating the exponential math function e^-at in circuit analysis often require solving complex differential equations and integrals. The Laplace transformation of e^-at is 1/(s + a) where by substituting 1/(s + a) for e^-at in an equation will significantly simplify formula manipulation.

Instructions

1. Identify the reactive components of the circuit. Reactive components are time-dependent components such as inductors and capacitors.

2. Identify and write down the time domain representation for current flowing through the inductors in your circuit. For example, the time domain representation for current flowing through an inductor is:

   I(t) = 1/L * int[Vl(t) from 0 to t dt]

   where I(t) represents current through an inductor as a function of t, L is the inductance value of the inductor in units of henries, and int[Vl(t) from 0 to t dt] represents the mathematical function of integrating the inductor voltage, Vl(t), from 0 to t.

3. Identify and write down the time domain representation for current flowing through the capacitors in your circuit. For example, the time domain representation for current flowing through a capacitor is:

   I(t) = C * [dVc(t)/dt]

   where I(t), in this case, is the current through a capacitor as a function of t, C is the capacitor value in units of farads, and the dVc(t)/dt is the mathematical symbol for differentiating the capacitor voltage, Vc(t), as a function of t.

4. Convert the time domain representation for the current flowing through inductors in Step 2 to S-domain using the "Table of Laplace Transforms" noted in the reference section.

   Time domain from Step 2: I(t) = 1/L * int[V(t) from 0 to t dt];

   Applicable Laplace Transform from Table (sixth row from top): g(t) = int[f(t) from 0 to t dt] converts to G(s) = F(s)/s;

   Apply transform to the Step 2 equation: I(t) = I(s); int[Vl(t) from 0 to t dt] = Vl(s)/s;

   S - domain transformation of Step 2: I(s) = 1/L * Vl(s)/s = 1/Ls Vl(s).

5. Convert the time domain representation for the current flowing through capacitors to S-domain using the "Table of Laplace Transforms" in the reference section:

   Time domain from Step 4: I(t) = C * [dVc(t)/dt];

   Applicable Laplace Transform from Table (fourth row from top): df/dt converts to sF(s) - f(0) where f(0) is the value of the f function at t = 0;

   Apply transform to the Step 4 equation: I(t) = I(s); dVc(t)/dt = sVc(s) - Vc(0);

   S - domain transformation of Step 4: I(s) = C * [sVc(s) - V(0)] = CsVc(s) - CVc(0).

6. Write the loop equation for your circuit by summing all voltages in the circuit and setting the sum equal to zero. For example, if you have a series RLC circuit with one time varying supply voltage source, V(t), a resistor, R, an inductor, L, and a capacitor, C, the S-domain loop equation will be:

   -V(s) + Vr(s) + Vl(s) + Vc(s) = 0

   where V(s) is the S-domain version of V(t), Vr(s) is the voltage across the resistor and the Vl(s) and Vc(s) are represented in Step 4 and 5.

7. Find the current dependent equivalent values of each voltage in the Step 6 loop equation by solving the Step 4 and Step 5 equation for Vl(s) and Vc(s) respectively:

   V(s) is the supply voltage, it has no current dependent equivalent value;

   Vr(s) = I(s) * R based on ohms law where voltage is current times resistance;

   Vl(s) = Ls * I(s) by solving the Step 4 equation for Vl(s);

   Vc(s) = 1/Cs * I(s) + Vc(0)/s by solving the Step 5 equation for Vc(s).

8. Rewrite the loop equation by substituting the equivalent values from Step 7 into the original loop equation from Step 6:

   Loop equation from Step 6:

   -V(s) + Vr(s) + Vl(s) + Vc(s) = 0

   Rewrite by substituting the equivalent current dependent values from Step 7:

   -V(s) + I(s)R + LsI(s) + 1/CsI(s) + Vc(0)/s = 0

9. Solve the loop equation in Step 7 for I(s) to find the final S-domain representation of the current source. For example, in Step 4, you found the S-domain representation of current through an inductor and in Step 5 you found the S-domain representation of current through a capacitor. The total current flowing through the circuit depends on a hybrid of all components interacting in the circuit, including resistors. You see this hybrid in the modified loop equation in Step 8. Now, just solve the equation in Step 8 for I(s):

   -V(s) + I(s)R + LsI(s) + 1/CsI(s) + Vc(0)/s = 0

   I(s)R + LsI(s) + 1/CsI(s) = V(s) - V(0)/s

   I(s)[R + Ls + 1/Cs] = V(s) - Vc(0)/s

   I(s) = [V(s) - Vc(0)/s]/[R + Ls + 1/Cs]

References

• Interactive Mathematics: The Laplace Transformation
• Department of Electrical and Computer Engineering, College of Engineering, University of Wisconsin - Madison: S - Domain Analysis
• Department of Elecrtrical Engineering, Stanford University: Table of Laplace Transforms