This page describes how to obtain the zero-state response of a series RLC circuit. The previous page has described how the homogeneous response or the zero-input response can be obtained. Hence the focus in this page is on obtaining the zero-state response. To obtain the zero-state response, it is necessary to form the differential equation first and then obtain the solution.
The next page shows how we can form the differential equation and the Laplace transform functions for some circuits. Once the differential equation or the Laplace transform function is set up, the solution process is similar to what is outlined in this page.
After the next page, some interactive pages are presented, and then a page containing some worked examples is presented. The interactive pages contain one interactive page for each type of input, and a few other pages.
For the circuit in Fig. 25, we get the differential equation as follows.
It is possible to form the differential equation, with the current as the variable, as shown below.
Out of the two differential equations (1.4) and (1.5), the differential equation expressed by equation (1.4) is usually the preferred equation, since the differential equation formed with the capacitor voltage as the variable contians just the input voltage. On the other hand, the differential equation formed with the current as the variable is expressed in terms of the derivative of the input voltage.
The Laplace transform function can be obtained either from the differential equation or from the transformed circuit. If the transformed circuit is used to derive the Laplace transform function for the capacitor voltage, the effort needed to form the differntial equation can be avoided.
For the circuit in Fig. 26, we can get the Laplace transform function for the capacitor voltage. We need to use the voltage division rule and the transform impedances of the elements.
We can also form the state equation for solution using state-transition matrix.
Equation (1.8) presents the general form of the state equation. In equation (1.8), A is the system matrix. For a second-order system, A is a 2 × 2 square matrix and B is a a column matrix with 2 rows. It is preferable to select variables that represent energy stored in a system as the state variables. For the circuit in Fig. 25, you can choose the capacitor voltage and the inductor current as the state variables. For the circuit in Fig. 25, we can form the state equation as follows.
From equation (1.12), matrices A and B can be identified.
On solving equation (1.9), we can obtain y(t). It is illustrated subsequently how we can get y(t).
We can use the Laplace transforms for the state-variable approach.
We can obtain Y(s), as shown by equation (1.14). The inverse of Y(s) yields y(t), as shown by equation (11.6). The state-transition matrix can be obtained, as shown by equation (11.6).
When the initial conditions are of zero-value, y(t) can be obtained, as shown by equation (1.18). From the A matrix, (sI - A) matrix can be obtained. The Lapalce transform of the state-transition matrix can be obtained, as shown by equation (1.18).
We obtain the Laplace transform functions for the capacitor voltage and the inductor current, as shown by equations (1.19) and (1.20). Since the functions obatined are the same as before, there is no need to illustrate how to obtain solution using the state-variable approach from the Lapalce transforms perspective.
Depending on the values of R, L, and C, the circuit can be over-damped, critically damped or under-damped. The input can be one of the following:
We obtain the zero-state response for each type of circuit for each of the inputs listed above. We start with the over-damped circuit. The solution in each of the cases can be obtained using the differential equations, the Laplace transforms or the state-variable approach. By using two techniques, we can verify that the solutions obtained are correct. You can also understand the similarities between the two approaches. The Laplce transforms approach is preferable for the unit-step, the impulse and the ramp input. The differential equations is much simpler for the the sinusoidal and the complex sinusoid input.
It is somewhat laborious to obtain the solution using the state-tansition matrix and hence this apporach is illustrated for getting the step and impulse response only. The use of state equations is the preferred method for obtaining a solution using a program-based numerical method. This technique is also illustrated. The advantage of the state-variable approach is that an n-th system can be represented by n first-order equations. Solution of n first-order equations tends to be easier than solving a single n-th order equation. Moreover, obtaining formulating the state equation is easier than forming an n-th order differential equation. However, in this problem we make use of the differential equation to get the state equation.
The following prefixes to sub-section titles are used in this page.
The values of components used for the over-damped circuit are presented below. The initial conditions are set to be of zero value.
From the specified values of the components and the nature of input, we get equation (2.3) from equation (2.2).
The solution is illustrated above. Form the auxiliary equation, as shown by equation (2.4). Get the nature of the homogeneous response, as shown by equation (2.5). These steps are common for obtaining to any other type of input and they will not be repeated. Then obtain the particular solution. Since the rate of change in input is zero for t > 0, the particular solution does not vary and its value can be obtained from equation (2.3) by setting the first derivative and the second derivative of capacitor voltage to be zero. Equation (2.6) expresses the particular solution. The zero-state response or the unit-step response is the sum of the particular solution and the complementary solution and we get equation (2.7). The values of A and B are evaluated form the initial conditions, as shown by equation (2.8). The unit-step response is expressed by equation (2.9). This explanation is not repeated for the other solutions. The current through the circuit is expressed by equation (2.10).
The impulse response is the derivative of the unit-step response. It is possible to assume that the impulse response has the same the form as the complementary solution, expressed by equation (2.5). We can obtain its first and second derivatives, then subsitute them unto equation (2.12) and obtain the solution, but this approach is tedious. For the same values of components, the unit-step response has already been obtained and it is presented earlier by equation (2.9). Let the unit-step response of capacitor voltage be called α(t). Then equation (2.9) can be re-presented as shown below.
The impulse response of capacitor voltage is expressed by equation (2.14). It is the practice to refer to the impulse response by h(t). The current through the circuit is expressed by equation (2.15). It has been assumed that the initial current through the circuit, expressed by i(0-), is zero. From equation (2.15), it is seen that i(0+), is 1 Amp.When an impulse voltage is applied, there is a step change in current. We can find the step change as shown below.
When an impulse voltage is applied, the rate of change of current at t = 0 is expressed by equation (2.13). Since the value of L = 1 H, we obtain that i(0+), is 1 Amp.
From the specified values of the components and the nature of input, we get equation (2.16) from equation (2.15). The auxilairy equation and the complementary solution have been presented earlier, by equations (2.4) and (2.5).
The particular solution is presented by equation (2.17). The ramp response of the capacitor voltage is obtained, as shown by equations (2.18), (2.19) and (2.20). The current through the cicrcuit is obtained, as shown by equation (2.21).
The ramp input is the integral of the unit-step input. Hence the ramp response can be obatined as the integral of the unit-step response. The constant of integration is evaluated from the initial condition. Equations (2.13), (2.22) and (2.23) illustrate the process.
The response of the over-damped circuit to an exponential input is obtained now. The pole of the exponential function is distinct from the the poles of the circuit.
We can obtain the particular solution and the zero-state response as follows.
The response of the over-damped circuit to an exponential input is obtained now. The pole of the exponential function coincides with one of the the poles of the circuit.
We can obtain the particular solution and the zero-state response as follows.
The response of the over-damped circuit to a sinusoidal input is obtained now.
The input is specified by equation (2.37). The cosine function is the real part of an exponential function. Hence we first obtain the response due to the corresponding exponential function and then take its real part as the particular solution.
We can verify the solution using the jω notation, as shown below.
The response to the sinusoidal input is then obtained, as shown below.
It takes some work to get the solution expressed by equation (2.48).
The derivative can be replaced by the pole of the input. The particular solution is obtained, as shown below.
The zero-state response can then be obtained.
We obtain the solution to the same circuit, using the Laplace transforms. When the circuit is over-damped, we can express equation (1.6) differently, as shown below.
Equation (3.2) is obtained from equation (1.6). We can express the denominator as the product of two factors and equation (3.2) can be expanded into sum of partial fractions, as shown by equation (3.3). Equation (3.4) presents the corresponding time-domain expression.
The valaues of components and the source voltage are specified by equation (3.5).
Equation (3.6) expresses the partial fractions and the capacitor voltage is expressed by equation (3.7).
It is better to use the Laplace transforms to obtain the impulse response. This technique is easier than using the differential equations approach.
The current through the circuit can be obtained, as shown above.
Expressions for the capacitor voltage and the current through the circuit can be obtained, as shown above.
Expressions for the capacitor voltage and the current through the circuit can be obtained, as shown above.
Expressions for the capacitor voltage and the current through the circuit can be obtained, as shown above.
Using the Laplace transforms approach is more difficult than using the differential equations approach. Expanding the Laplace transform function into partial fractions is laborious. It may seem that equation (3.28) is different from equation (2.48). We can verify that they are the same, as shown by equation (3.29).
Using the Laplace transforms approach is more difficult than using the differential equations approach. Expanding the Laplace transform function into partial fractions is laborious.
The component values can be substituted into state-equation (11.2).
Once we have the A matrix, we can obtain the state-transition matrix as follows.
After forming the characteristic equation, the eigenvalues are determined, as shown by equation (4.3). By use of Cayley-Hamilton theorem, we get equation (4.4) and the expressions for α0 and α1 are obtained, and equation (4.5) presents their values.
The state-transition that is obtained is displayed by equation (4.6). Equations (4.7) and (4.8) show the steps involved in evaluating the integral in equation (1.9).
Equation (4.9) presents the integral. As shown by equation (4.10), the state vector equals the product of the state-transition matrix and the integral, when the initial conditions are of zero value. The state vector can be obtained, as shown by equations (4.11) and (4.12). It can be seen that we obtain the same results, as obtained earlier.
Matlab script for obtaining the response is presented below.
% Solution for series RLC Over-damped circuit: % Zero-State Response for unti step input R=5; % Value in Ohms L=1; % Value in Henry C=1/6; % Value in Farad y = [0; 0]; % Initial capacitor voltage,Initial current step=0.01; % Step size is 0.01 sec, Period = 2.5; % Response calaculated for 2.5 seconds
A = [0 1/C; -1/L -R/L]; % A square, matrix B = [0; 1/L]; % B column matrix
incR = [0;0]; % increments initialized
for n= 1: Period/step; zeit(n) = (n-1)/100; % Time computed stored in an array vcap(n) = y(1); cur(n) = y(2); incr = (A*y + B)*step; % From equation (4.1) y = y + incr; % increments added to Previous values end;
subplot(211)
plot(zeit,vcap)
title('Capacitor Voltage')
ylabel('Volt')
axis([0 2.5 0 1])
grid
subplot(212)
plot(zeit,cur)
title('Current')
ylabel('Amp')
xlabel('Time in second')
axis([0 2.5 0 0.25])
grid
The plots obtained are presented below.
It can be seen that the script is relatively simple and self-explanatory. It is seen that a single line of script calculates the increments. We add the increments to the previous values and the values of capacitor voltage and current can be stored in arrays to be used later for plotting later.
It can be seen that we obtain the same results, as obtained earlier. The use of state-variable technique is not difficult when the impulse response is to be obtained.
The script used for obtaining the unit-step response can be modified easily to get the response to the complex sinusoid input. The script for obtaining the the response to the complex sinusoid input is presented below.
% Solution for series RLC Over-damped circuit: % Zero-State Response for Complex Sinusoid input R=5; % Value in Ohm L=1; % Value in Henry C=1/6; % Value in Farad y = [0; 0]; % Initial capacitor voltage,Initial current step=0.002; % Step size is 0.002 sec, Period = 2.0; % Response calaculated for 2 seconds
A = [0 1/C; -1/L -R/L]; % A square, matrix B = [0; 1/L]; % B column matrix
incR = [0;0]; % increments initialized
for n= 1: Period/step; zeit(n) = (n-1)*step; % Time computed stored in an array vcap(n) = y(1); cur(n) = y(2); vs(n) = exp(-4.0*(n-1)*step)*cos(3.0*(n-1)*step); incr = (A*y + B*vs(n))*step; y = y + incr; % increments added to Previous values end;
subplot(211)
plot(zeit,vcap)
title('Capacitor Voltage')
ylabel('Volt')
axis([0 2 0 0.2])
grid
subplot(212)
plot(zeit,cur)
title('Current')
ylabel('Amp')
xlabel('Time in second')
axis([0 2 -0.1 0.1])
grid
The changes made to the earlier script are only a few. The step size and the duration for simulation have been changed. A line is added to define the input at each instant and the increments are calculated, taking into account the input. The dimensions for the plot have been changed. The plots obtained are shown below.
For the circuit in Fig. 24, the components can be selected such that it is critically-damped, and the solution can be obtained using the following techniques.
The impulse response is the derivative of the unit-step response.
The input is specified by equation (5.33). The cosine function is the real part of an exponential function. Hence we first obtain the response due to the corresponding exponential function and then take its real part as the particular solution.
We can verify the solution using the jω notation, as shown below.
The response to the sinusoidal input is then obtained, as shown below.
The particular solution is obtained as follows.
The zero-state response to the complex sinusoid input is then obtained, as shown below.
For the same component values, the solutions are obtained using the Laplace transforms.
Equation (6.19) shows that the solution obtained using the Laplace transforms approach is the same as that obtained earlier using the differential equations approach.
Equation (7.1) presents the stae equation for the critically-damped system. The A and B matrices are presented by equation (7.2).
From the characteristic equation, we get the eigenvalues. Once the eigenvalues are known, the coefficients α0 and α1 can be determined.
Equation (7.7) presents the state-transition matrix. The integral is evaluated, as shown by equations (7.8) and (7.9).
The capacitor voltage is determined as shown by equations (7.10) and (7.11).
The state-transition matrix has been already obtained, since the same component values are used.
The capacitor voltage is determined as shown by equations (7.13) and (7.14). The same expression has been obtained earlier.
For the circuit in Fig. 24, the components can be selected such that it is critically-damped, and the solution can be obtained using the following techniques.
Given the component values and the zero initial values as stated in equation (8.1), the time-domain expressions for the capacitor voltage and the current through the circuit can be obtained as follows.
The capacitor voltage can be determined as shown below.
The complementary solution for the under-damped circuit is not repeated for other inputs. The current through the circuit can be determined as shown below.
The impulse response is the derivative. For the same values of components, the unit-step response has already been obtained and it is presented earlier by equation (8.9). Let the unit-step response of capacitor voltage be called α(t). Then equation (8.9) can be re-presented as shown below.
The impulse response of the capacitor voltage obtained is expressed by equation (8.15).
Given the component values and the zero initial values as stated in equation (8.1), the time-domain expressions for the capacitor voltage and the current through the circuit can be obtained as follows.
The capacitor voltage can be determined as shown below.
The capacitor voltage can be determined as shown below.
Find the response to a complex exponential function and take its real part as the particular solution.
We can verify the above solution, using the jω notation.
The response to sinusoidal input is obtained as shown below.
Define Q(D) and evaluate it as shown below.
The particular solution can then be obtained.
The zero-state response is obtained next.
When the exciation pole coincides with one of the poles of the system, the multiplicity of one of the poles is two. In this case, the solution is obtained as shown below.
The occurrence of two poles at the same location gives rise to t being present in the solution. The response to complex sinusoid is obtained as shown below.
When the pole of the complex sinusoid input coincides with one of the poles of the under-damped system, it is preferable to use the differential equations approach to ge the solution. The use of Laplace transforms is somewaht tricky for this problem.
From the circuit in Fig. 25, we get an expression for the capacitor voltage, as shown below.
We can get an expression for the current, as shown below.
With an impulse input, we get the following equations.
The impulse response of the capacitor voltage and the current through the circuit are expressed by equations (9.7) and (9.9).
We can get an expression for the capacitor voltage, as shown below.
An expression for the current through the circuit is now obtained.
An expression for the current through the circuit is now obtained.
It is necessary to resort to the complex differentiation theorem, as illustrated by equations (9.28) and (9.29). It is seen that the solution obtained is the same, as that obtained using the differential equations approach.
Next the solution is obtained using the state-variable approach. This approach is used only for the unit-step response and the impulse response.
Equation (1.12) presents the state equation. Find the eigenvalues next.
Next the coefficients are obtained.
Then the state transition matrix is obtained.
Then the integral is evaluated.
Finally we can obtain the capacitor voltage and the current through the circuit.
We can make use of equation (10.10) to get the impulse response.
We can obtain the capacitor voltage and the current through the circuit.
The values of components used for the undamped circuit are presented below. The value of inductance is 1 H and the value of capacitance is 0.25 F. The initial conditions are set to be of zero value.
The capacitor voltage can be determined as shown below.
The current through the circuit can be determined as shown below.
For the same component values, as specified by equation (11.1), the impulse response of capacitor voltage can be obtained as shown below.
For the same component values, as specified by equation (11.1), the ramp response of capacitor voltage can be obtained as shown below.
Use the assumed complementary solution, obtained earlier for the unit-step response.
The current through the circuit can be determined as shown below.
For the same component values, as specified by equation (11.1), the response of capacitor voltage to an exponential input can be obtained as shown below.
Use the assumed complementary solution, obtained earlier for the unit-step response.
For the same component values, as specified by equation (11.1), the response of capacitor voltage to a sinusoidal input can be obtained as shown below. Use the assumed complementary solution, obtained earlier for the unit-step response.
For the same component values, as specified by equation (11.1), the response of capacitor voltage to a sinusoidal input can be obtained as shown below. In this case, the excitation pole is at the same location as one of the poles of the circuit.
The particular solution is now obtained. Use the assumed complementary solution, obtained earlier for the unit-step response.
If the input is a cosine function, the response can be obtained as shown below.
For the same component values, as specified by equation (11.1), the response of capacitor voltage to a complex sinusoid input can be obtained as shown below.
Use the assumed complementary solution, obtained earlier for the unit-step response.
From the circuit in Fig. 25, we can get an expression for the capacitor voltage, as shown below.
The current through the circuit can be determined as shown below.
For the specified component values of the undamped circuit, the capacitor voltage and the current through the circuit are obtained, as sown below.
From the circuit in Fig. 25, we get the following equations.
We can get an expression for the capacitor voltage, as shown below.
From the circuit in Fig. 25, we get the following equations.
We can get an expression for the capacitor voltage, as shown below.
From the circuit in Fig. 25, we can get an expression for the capacitor voltage, as shown below.
From the circuit in Fig. 25, we can get an expression for the capacitor voltage, as shown below. In this case, the excitation pole is at the same location as one of the poles of the circuit.
Since the denominator factor is raised to the power of 2, it indicates the occurrence of two roots at the same location. Since the exciation function is a sine function, we can expect the particular solution to contain the cosine function. The next step is shown below.
We make use of the complex differentiation theorem, as shown by equation (12.20). Expand the function for the capacitor voltage into partial fractions and then obtain an expression for the capacitor voltage in time domain.
From the circuit in Fig. 25, we can get an expression for the capacitor voltage, as shown below.
The illustration fo solution using the state variables approach is restricted to obtaining the unit-step response and the impulse response. This technique is laborious, but it can be used to obtain the zero-state response to other inputs, as well as obtaining the total solution.
Now the solution is obtained using the state-variable approach, for the unit-step input.
It is necessary to represent differential equation (13.1) by a state equation, as shown by equation (13.2). Equations (13.3) and (13.4) defines the two states, needed for a second-order system. From the above equations, we get the following equations.
Next the coefficients are obtained.
Then the state transition matrix is obtained.
Then the integral is evaluated.
Finally we can obtain the capacitor voltage and the current through the circuit.
It can be seen that the process is long, but on the other hand it is a well-defined process.
The state transition matrix, obtained earlier for this undamped system, is presented below.
Then the integral is evaluated. It is easy, since the input is an impulse function.
Next we can obtain the capacitor voltage and the current through the circuit.
This page has explained how to obtain the response of a series RLC circuit to commonly used excitation functions. In the case of other circuits, it is necessary to form the circuit equation. Once the differential equation or the Laplace transform function is obtained, the solution is similar to what has been described in this page. The next page illustrates how we can form the circuit equations for some circuits.