This page describes how to form the homogeneous equation of a second-order circuit and then obtain its response. Equation (1) presents the second-order linear homogeneous differential equation with c1 and c2 as the constant coefficients. The chapter on differential equations has described how to obtain the the homogeneous response of a second-order system and you may need to read that chapter, if you cannot understand what is presented in this page and the subsequent pages.
Equation (1.1) can be presented differently, as shown by equation (1.2). Coefficients k and ωd can be expressed in terms of ξ and ωn , as shown by equation (1.3). We call ξ as the damping coefficient and ωn as the natural frequency of oscillation. Hence we obtain equation (1.4) from equation (1.1).
Depending on the value of ξ , the system represented by equation (1.4) can be classified as follows:
The roots of characteristic polynomial obtained from equation (1.4) are the poles of the system represented by equation (1.4) and Fig. 1 displays their location for different values of ξ.
The zero-input response of the system depends on the location of its poles. When a system has zero-input, its response is only due to its initial conditions and this response is often called the natural response. The process of obtaining the zero-input response is explained with the help of a number of examples. Once a differential equation is formed to describe the behaviour of the system, the solution is obtained as explained in the chapters on differential equation and Laplace transforms. We start off with a series RLC circuit.
A series RLC circuit is presented in Fig. 2. The differential equation that defines this circuit is obtained as follows. For the series RLC circuit,
Equation (1.5) is obtained by applying Kirchoff’s voltage law. Since the current is leaving the positive terminal of capacitor, the capacitor voltage falls and the rate of change of capacitor voltage is negative. When the capacitor current marked in Fig. 2 is to be a positive value, the capacitor current and the rate of change of capacitor voltage have opposing signs. From equations (1.5) and (1.6), we obtain equations (1.7) and (1.8).
The dependent variable in equation (1.8) is vC, the capacitor voltage reflecting energy stored in the capacitor. It is possible to express the differential equation with current as the dependent variable, and it is also acceptable since current reflects energy stored in the inductor.
Equation (1.10) is obtained by differentiating equation (1.8) and expressing it in terms of the current. Though both equations are formed for a variable representing energy stored in the circuit, the differential equation in equation (1.8) is usually the preferred form. The reason for preferring (1.8) over equation (1.10) can be understood, when the circuit is excited by a source. This aspect is explained in the next page.
Since we have a second-order equation in equation (1.8), we need two initial conditions to solve this equation. The initial conditions are vC(0) and i(0) . Note that vC reflects the energy stored in the capacitor and i the energy stored in the inductor. It needs to be stressed that the variable of the differential equation should be a variable reflecting energy stored in a network. Then only it is continuous, and its derivative exists and it has a finite value at any instant. Given this network, it is preferable to use vC over current. For the differential equation in equation (1.8), the auxiliary equation is presented by equation (1.11). The roots are presented by equation (1.12).
The condition for the circuit to be over-damped is presented by equation (1.13). It has been stated previously that the system is over-damped if ξ > 1. When ξ > 1, we get the condition stated by equation (1.13).
Let us take a specific example of an over-damped system. The values of components and the initial conditions are stated in equation (1.16). The differential equation, obtained earlier, is presented again. By substituting the values of components into equation (1.8), we get equation (1.17).
The solution to equation (1.17) is obtained as shown below.
The plots of the capacitor voltage and the current through the circuit are presented below. The script used for getting the plots is presented below.
% Plots of voltage and current % Step size =0.05 s
for n = 1:501; x(n) = (n-1)/200; v(n) = 240*exp(2*(1-n)/200) - 140*exp(3*(1-n)/200); i(n) = 80*exp(2*(1-n)/200) - 70*exp(3*(1-n)/200); end;
subplot(211)
plot(x,v)
title('Capacitor Voltage')
ylabel('Volt')
axis([0 2.5 0 120])
grid
subplot(212)
plot(x,i)
title('Current')
ylabel('Amp')
xlabel('Time in second')
axis([0 2.5 0 20])
grid
The solution using Laplace transform is presented next. Using the formula for the derivative of a function, equations (1.24) and (1.25) are obtained.
From equation (1.25), equation (1.26) is obtained. Equation (1.27) presents the Laplace transform expression for the capacitor voltage. The capacitor voltage is expressed by equation (1.28).
We can use the transformed network to get the Laplace transform function for the capacitor voltage.
From the circuit in Fig. 3, we can form the following equations.
We can get an expression for the capacitor voltage, as shown below.
Next the solution is obtained using the state-variable approach. We get the state-transition matrix first and then the solution.
Equation (1.34) presents the state equation. Find the eigenvalues next.
Next the coefficients are obtained.
Then the state transition matrix is obtained.
Finally the solution is obtained.
We can use the Laplace transforms for the state-variable approach and get the solution, as shown below.
The auxiliary equation of a second-order circuit is presented by equation (1.11). For the critically-damped circuit, the roots of this circuit are obtained from equations (1.47) and (1.48).
Given the component values and the initial values as stated in equation (1.50), the differential equation (1.51) of the critically-damped circuit is obtained from equation (1.8).
The time-domain expressions for the capacitor voltage and the current through the circuit can be obtained as shown below.
To use the Laplace transforms approach, we can use the transformed circuit to get the solution. From the circuit in Fig. 3, we get the following equations.
We can get an expression for the capacitor voltage, as shown below.
Next the solution is obtained using the state-variable approach. We get the state-transition matrix first and then the solution.
Equation (1.63) presents the state equation. Find the eigenvalues next.
Next the coefficients are obtained.
Then the state transition matrix is obtained.
Finally the solution is obtained.
We can use the Laplace transforms to get the solution. We start with the (sI - A) matrix.
The auxiliary equation of a second-order circuit is presented by equation (1.11). For the under-damped circuit, the roots of this circuit are obtained from equations (1.76), (1.77), (1.78) and (1.79).
Given the component values and the initial values as stated in equation (1.80), the differential equation (1.81) of the under-damped circuit is obtained from equation (1.8).
The time-domain expressions for the capacitor voltage and the current through the circuit can be obtained as shown below.
We can get the same answer using the Laplace transforms. From the circuit in Fig. 3, we get the following equations.
We can get an expression for the capacitor voltage, as shown below.
Next the solution is obtained using the state-variable approach. We get the state-transition matrix first and then the solution.
Equation (1.91) presents the state equation. Find the eigenvalues next.
Next the coefficients are obtained.
Then the state transition matrix is obtained.
Finally the solution is obtained.
We can use the Laplace transforms to get the solution. We start off with the (sI - A) matrix.
It is easy to obtain the inverse transform of equation (1.100) and then obtain expressions for the capacitor voltage and the inductor current. It can be verified that the same results are obtained.
For the undamped circuit, the roots of this circuit are obtained from equations (1.101) and (1.102).
The solution can be obtained as shown below.
We can get the same answer using the Laplace transforms. From the circuit in Fig. 3, we get the following equations.
We can get an expression for the capacitor voltage, as shown below.
Next the solution is obtained using the state-variable approach.
Equation (1.112) presents the state equation. Find the eigenvalues next.
Next the coefficients are obtained.
Then the state transition matrix is obtained.
Finally the solution is obtained.
We can use the Laplace transforms to get the solution.
It is easy to obtain the inverse transform of equation (1.121) and then obtain expressions for the capacitor voltage and the inductor current. It can be verified that the same results are obtained.
It is possible to obtain the solution using a numerical routine. To illustrate this procedure, we use equation (1.8) which is presented below for convenience.
Let
Equations (1.122) and (1.123) can be converted to a routine as shown below. The increments are calculated as shown below:
The flow chart shows how a program can be developed using equations (1.124) and (1.125). The plot obtained using the program is presented below. The Matlab script is presented after the plot. It can be seen that the capacitor voltage exhibits the characteristic of an under-damped system.
% Solution for series RLC circuit:Zero-input Response
R=1; % Value in Ohm
L=1; % Value in Henry
C=0.1; % Value in Farad
v1=100; % Initial condition
v2=-100; % Initial condition
step=0.01; % Step size is 0.01 sec
for n=1:401; % Calculated over 4 secs
zeit(n)=(n-1)/100; % Time computed stored in an array
vcap(n)=v1; % Capacitor voltage stored in an array
incrV1=v2*step; % Increments computed
incrV2=-step*(R/L*v2+v1/L/C);
v1=v1+incrV1; % v1 updated
v2=v2+incrV2; % v12 updated
end;
plot(zeit,vcap)
axis([0 4 -80 100])
ylabel('output voltage')
xlabel('time in seconds')
grid
For the RLC circuit shown in Fig. 4, obtain the differential equation for the capacitor voltage and the inductor current.
Solution:
From Fig. 4a, we get the following equations.
The capacitor current is the sum of the resistor current and the inductor current. It is possible to express the resistor current and the capacitor current, as shown by equation (2.1). The differential equation expressed by equation (2.2) can then be obtained. For the parallel-connected RLC circuit, it is preferable to use the inductor current as the variable of the differential equation.
From the differential equation, the auxiliary equation described by equation (2.3) can be obtained.
The parallel-connected RLC circuit can be over-damped, critically-damped or under-damped, depending on the values of R, L and C. The condition to be satisfied for the circuit to be over-damped has been presented by equation (2.4). Equation (2.5) presents the condition for the circuit to be critically-damped and , equation (2.6) presents the condition for the circuit to be under-damped.
It is possible to express the differential equation in terms of the damping coefficient ξ, and the natural frequency of oscillation ωn, as shown by equation (2.8).
From the circuit in Fig. 4b, we can obtain an expression for the capacitor voltage and the inductor current in terms of Laplace transforms.
Equation (2.9) expresses the capacitor current in terms of its voltage. The capacitor current is the sum of the resistor current and the inductor current and hence we get equation (2.10). The capacitor voltage is also the inductor voltage, and hence we get equation (2.11).
Replace the capacitor voltage in equation (2.10) by the expression for the capacitor voltage, displayed by equation (2.11). The resultant equation is expressed by equation (2.12). Re-arrange equation (2.12) and obtain equation (2.13). The Laplace transform expression for the inductor current is expressed by equation (2.14). This expression can be obtained from the differential equation also, as shown below.
We can replace the derivatives of inductor current by the corresponding expressions displayed by equations (2.15) and (2.16). Hence we obtain equation (2.16) from equation (2.2). From equation (2.16), it is easy to obtain equation (2.14).
The easiest technique to obtain a Laplace transform for the inductor current is to use the principle of superposition. It is shown below how the principle of superposition theorem can be applied.
After some algebraic manipulation, an expression for the inductor current can be obtained, as shown below.
A numerical example is presented now.
The solution can be obtained as shown below.
Using the Laplace transforms, the solution can be obtained as shown below.
The solution using the state-variable approach is not presented, since it is similar to the solution presented for the series-connected RLC circuit. Examples for the critically-damped and the under-damped parallel-connected RLC circuit are also not presented, since these solutions are similar to what has been presented for the series connected RLC circuit.
For the circuit in Fig. 5, determine i(t) and vC(t) for t > 0. Both the switches remain closed prior to t = 0 and they both are open for t ≥ 0.
Solution:
The steps involved are as follows.
a. Determine the initial conditions.
b. Re-draw the circuit for t ≥ 0 and then solve.
The circuit has a dc excitation signal prior to t = 0 and the assumption is that the circuit is in a steady-state just before the switches are thrown open at t = 0. To determine the steady-states for an RLC circuit with a dc excitation, the technique is as follows.
With a dc excitation, the responses in the circuit become fixed values. Hence the rate of change of inductor current is zero. It means that the voltage across the inductor is zero. Hence consider the inductor to be a short-circuit and find the current flowing through the inductor.
The rate of capacitor voltage is zero and hence the current through the capacitor is zero. Treat the capacitor as an open-circuit and find the capacitor voltage. We can draw a circuit to determine the initial conditions, and this circuit is shown in Fig. 6.
From the circuit in Fig. 6, it can be seen that the inductor current at t = 0 is the current through the 2 Ω resistor and the capacitor voltage at t = 0 is the voltage across the 2 Ω resistor.
For t ≥ 0, the circuit in Fig. 5 can be simplified and the circuit that needs to be considered for t ≥ 0 is presented in Fig. 7.
For the circuit in Fig. 7,
Based on the initial conditions, we can obtain the values of A and B.
From the transformed network in Fig. 7, we get that
For the circuit in Fig. 8,
vS(t) = 2 - 2.u(t).
Find expressions for vC(t), and iL(t). Sketch these functions.
Solution:
The solution is obtained using the following techniques;
a. Using the differential equations
b. Using the Laplace Transforms, and
c. Using the state-variable approach.
The zero-state responses of vC(t), and iL(t) are to be obtained. Hence it is necessary to determine the initial conditions at t = 0.
The input voltage can be sketched as shown in Fig.9 and the initial conditions are obtained as shown. The voltage across resistor R2 is vC(0) and the current through R2 is iL(0). From Fig 9, we get that
Differential equation approach
The circuit that needs to be considered for the solution using the differential equation is presented in Fig. 10.
The differential equation for the inductor current is obtained as follows.
We know the value of iL(0). As shown below, the rate of change of current across inductor at t = 0 is zero, since the inductor voltage is zero at t = 0.
Solution using the Laplace Transforms
The Laplace transform function for the inductor current is obtained using the principle of superposition. Find out the contribution due to vC(0) alone. Determine the current through the capacitor. Use current division rule and get the the contribution by vC(0) to the inductor current. Next determine the inductor current due to the impulse voltage. Add the two components.
State variable approach
Next the coefficients are obtained.
Then the state transition matrix is obtained.
Finally the solution is obtained.
We can use the Laplace transforms to get the solution.
The time-domain expressions can be obtained from the inverse transform of the corresponding functions.
The plots of the capacitor voltage and the inductor current are shown below.
For the circuit in Fig. 12, vS(t) = 60 - 60.u(t). Form a differential equation for the circuit using R1, R2, L1 and L2. Find the condition for the circuit to be over-damped, critically damped or under-damped. For the given values of R1, R2, L1 and L2, find i1(t) and i2(t) for t ≥ 0.
Solution:
For t ≥ 0, the input voltage is zero. From the circuit in Fig. 13, we get the following equations with the input voltage set to zero.
The initial conditions are obtained from the circuit shown below.
We get that i1(0) = 10 A, and i2(0) = 0 A.
Hence the solution is obtained as follows.
The solution using the Laplace transforms can be obtained as follows.
We can use the state-variable approach as shown below.
We get the same solution obtained earlier.
For the circuit in Fig. 16, vS(t) = 3 - 3.u(t). Find v2(t) for all t.
Solution:
When vS(t) = 0, current through R1 is obtained as follows.
Hence the dependent current source of 10i(t) can be replaced by a 1 Ω resistor.
For t < 0, the circuit is in steady-state and hence the inductor voltage v2 is zero. For t < 0, i(t) = 3 A and the inductor current is 10i(t), which is equal to 30 A. The transformed circuit is shown below.
For the circuit in Fig. 17, we can form the following equations.
To obtain the solution using differential equations approach, the circuit below is used.
For the parallel connected RLC circuit, the differential equation obtained earlier is presented below. First, an expression for the inductor current is obtained and then an expression for the capacitor voltage can be obtained.
For the circuit in Fig. 19, vC(0) = 1 V, and dvC/dt evaluated at t = 0 is 2 V/s. Obtain vC(t) , iC(t) and iL(t) for t ≥ 0.
Solution:
For the parallel connected RLC circuit, the differential equation obtained earlier is presented below. Based on the relationship between the capacitor voltage and the rate of change of inductor current, the differential equation using the capacitor voltage can be obtained. From the given initial conditions, an expression for the capacitor voltage can be obtained first. Then an expression for the inductor current is obtained.
Once the capacitor voltage is known, the resistor current and the inductor current can be obtained as follows. From the expression for the capacitor voltage, obtain the resistor current and the capacitor current. Then an expression for the inductor current can be obtained .
It is possible to get the solution using the inductor current as the variable of the differential equation. This technique is slightly longer and it takes more work to solve for the inductor current.
From the value of vC(0), obtain the initial value of rate of change of inductor current.
From the initial value of rate of change of capacitor voltage, obtain the initial value of second derivative of inductor current, as shown below.
Once the inductor current is known, the capacitor voltage can be determined, as shown above. We can use the state-variable approach as follows.
From the differential equation, obtain the A matrix, as shown above. Using the Laplace transforms, the solution can be obtained as shown below.
An RLC circuit is presented in Fig. 2. Given the component values and the initial values as stated in equation (1.16), the time-domain expressions for the capacitor voltage and the current through the circuit can be obtained as follows.
The solution can be obtained as shown below.
It is possible to get the solution in a different way, as shown below.
Next the solution is obtained using the state-variable approach.
After obtaining the inverse, an expression for i(t) can be obtained.
For the circuit in Fig. 20, find obtain v1(t) for the specified values of components.
Solution:
For t > 0, the circuit has no input and the response is due to initial conditions at t = 0. From the circuit in Fig. 20, we get the following equations for t > 0.
The initial current can be determined from the circuit in Fig. 20. For t < 0, the circuit is assumed to be in steady-state and hence the source current flows through R1 and the inductor. Hence at t =0, the capacitor voltage is 1 V and the inductor current is 1 A.
For the circuit in Fig. 21, we get that
Using the Laplace transforms, the solution can be obtained as follows.
The solution obtained using the state-variable approach is presented below.
It can be seen that the use of the state-variable is relatively easy.
For the circuit in Fig. 22, determine iL(t) and vC(t) for t >= 0
Solution:
The initial conditions are obtained to be:
iL(0) = 1 A, vC(0) = 1 V.
From the circuit in Fig. 22, we get the following equations.
We get the solution using the state-variable approach as follows.
Form the differential equation for the circuit in Fig. 23 in terms of R1, R2, C1 and C2 for t > 0. Then for the specified values of components, determine v2(t).
Solution:
For the circuit in Fig. 23, we can form the following equations.
We can form the differential equation, using the specified values of components.
The solution is obtained as follows.
Form the differential equation for the circuit in Fig. 24 in terms of R1, R2, L1 and L2 for t > 0. Then for the specified values of components, determine i1(t) and i2(t).
Solution:
From the circuit in Fig. 24, the following equations can be formed.
The values of components are specified below.
Using the state-variable approach, the solution can be obtained as follows.
This page has explained how to obtain the zero-input response of second-order circuits. Many examples have been presented to explain the process of obtaining the zero-input response in quite a few ways. The next page explains how to obtain the zero-state response of second-order circuits, when excited by an external source.