The Initial Value Theorem of the laplace transform states as follows:
This is useful for finding the initial conditions of a function needed when we perform the transform of a differentiation operation (see above).
Final Value Theorem
Similar to the Initial Value Theorem, the Final Value Theorem states that we can find the value of a function f, as t approaches infinity, in the laplace domain, as such:
This is useful for finding the steady state response of a circuit. The final value theorem may only be applied to stable systems.
Transfer Function
If we have a circuit with impulse-response h(t) in the time domain, with input x(t) and output y(t), we can find the Transfer Function of the circuit, in the laplace domain, by transforming all three elements:
In this situation, H(s) is known as the "Transfer Function" of the circuit. It can be defined as both the transform of the impulse response, or the ratio of the circuit output to its input in the Laplace domain:
Transfer functions are powerful tools for analyzing circuits. If we know the transfer function of a circuit, we have all the information we need to understand the circuit, and we have it in a form that is easy to work with. When we have obtained the transfer function, we can say that the circuit has been "solved" completely.
Convolution Theorem
Earlier it was mentioned that we could compute the output of a system from the input and the impulse response by using the convolution operation. As a reminder, given the following system:
x(t) = system input h(t) = impulse response y(t) = system output
We can calculate the output using the convolution operation, as such:
Where the asterisk denotes convolution, not multiplication. However, in the S domain, this operation becomes much easier, because of a property of the laplace transform:
Where the asterisk operator denotes the convolution operation. This leads us to an English statement of the convolution theorem:
Convolution in the time domain becomes multiplication in the S domain, and convolution in the S domain becomes multiplication in the time domain.[1]
Now, if we have a system in the Laplace S domain:
X(s) = Input
H(s) = Transfer Function Y(s) = Output
We can compute the output Y(s) from the input X(s) and the Transfer Function H(s):
Notice that this property is very similar to phasors, where the output can be determined by multiplying the input by the network function. The network function and the transfer function then, are very similar quantities.
Resistors
The laplace transform can be used independently on different circuit elements, and then the circuit can be solved entirely in the S Domain (Which is much easier). Let's take a look at some of the circuit
[Transform of Resistors]
[Transform of Ohm's Law]
[Tranform of Resistor]
[Transform of Capacitor]
[Transform of Inductor]
elements:
Resistors are time and frequency invariant. Therefore, the transform of a resistor is the same as the resistance of the resistor:
Compare this result to the phasor impedance value for a resistance r:
You can see very quickly that resistance values are very similar between phasors and laplace transforms.
Ohm's Law
If we transform Ohm's law, we get the following equation:
Now, following ohms law, the resistance of the circuit element is a ratio of the voltage to the current. So, we will solve for the quantity , and the result will be the resistance of our circuit element:
This ratio, the input/output ratio of our resistor is an important quantity, and we will find this quantity for all of our circuit elements. We can say that the transform of a resistor with resistance r is given by:
Capacitors
Let us look at the relationship between voltage, current, and capacitance, in the time domain:
Solving for voltage, we get the following integral:
Then, transforming this equation into the laplace domain, we get the following:
Again, if we solve for the ratio , we get the following:
Therefore, the transform for a capacitor with capacitance C is given by:
Inductors
Let us look at our equation for inductance:
putting this into the laplace domain, we get the formula:
And solving for our ratio , we get the following:
Therefore, the transform of an inductor with inductance L is given by:
Impedance
Since all the load elements can be combined into a single format dependent on s, we call the effect of all load elements impedance, the same as we call it in phasor representation. We denote impedance values with a capital Z (but not a phasor ).
RCL circuit with zero capacitance and zero initial current.
In the network shown, determine the character of the currents , , and assuming that each current is zero when the switch is closed.
Solution[2]
Current flow at a joint in circuit
Since the algebraic sum of the currents at any junction is zero, then ...(182)
Voltage balance on a circuit
Applying the voltage law to the circuit on the left we get
... (182-1)
Applying again the voltage law to the outside circuit, given that E is constant, we get
... (182-2)
Laplace Transforms of current and voltage equations
Transforming (182), (182-1) and (182-2), we get ...(182-3)
... (182-4)
... (182-5)
Review on implementing Laplace Transformation
The three Laplace transformed equations (182-3), (182-4), and (182-5) show the benefits of integral transformation in converting differential equations (http://www.amazon.com/Advanced-Mathematics-Personal-Study-Notes/dp/1461084423/ref=tmm_pap_title_0?ie=UTF8&qid=1366661958&sr=8-1) into linear algebraic equations that could be solved for the dependent variables (the three currents in this case), then inverse transformed to yield the required solution.
In equation (182-3), we utilized the sum property of Laplace transforms.
In equation (182-4), we utilized the transform of differential derivative as follows.
...(182-4.1)
Since, substituted by the given initial condition:
In equation (182-5), we also utilized the transform of differential derivative
...(182-5.2)
Again, we substituted by the given initial condition:
The fact that the applied voltage was constant, implied the use of Laplace transform of constant, as follows:
...(182-5.3)
Solution linear simultaneous equations
The three linear simultaneous equations (182-3), (182-4), and (182-5) have the three unknown , , and and can be solved by Cramer’s rule of matrices among other simple methods of elimination, as follows.
... (182-6)
Where, the determinant ∆ for the matrix is determined as follows
... (182-6.1)
Since we are interested in the factors of Δ, we consider the equation Δ =0. Since all coefficients of this equation are positive, hence it cannot have any positive roots. Its discriminant is ... (182-6.1.1)
which can be written
... (182-6.1.2)
which is positive. Hence the equation Δ = 0 has two negative distinct roots and , say.
Therefore,
... (182-6.2)
Where, and are the roots of the quadratic equation (182-6.1) as follows
... (182-6.2.1) The inverse Laplace transform of (182-7) is therefore,
...(182-8)
The remaining variables and and the corresponding voltages are determined by equations (182), (182-1) and (182-2)
Analysis of circuit dynamics
The electric current in equation (182-8) shows a time-independent component and two decay terms, which reach asymptotic values as t reaches ∞. In other words, the currents in the three circuits lack sinusoidal osculation, mainly because: (1) the applied voltage is constant and (2) the circuit does not have capacitance components.
Notes: This example could be is modified in various ways[3] to involve voltage impulse, sinusoidal voltage source, capacitance,and various boundary and initial conditions of charges and currents.