In this page, we find out how we can obtain the zero-input response of first-order circuits. In some cases, the initial condition is specified directly, whereas it may have to be determined in some cases. The solution can be obtained either by forming a differential equation and solving it or by drawing the transform network and obtaining the inverse Laplace transform expression. Both techniques are illustrated.
In this section, the zero-input response of an RC circuit is obtained. The response of capacitor voltage can be called the homogeneous response. It can also be called the natural response or the zero-input response. It is also the transient response and in this case, the steady-state response has zero value.
The circuit is presented in Fig. 3. Assume that the capacitor holds charge for t < 0 and its voltage at t = 0 is E Volts . That is, vC(0) = E. At t = 0, the switch is closed and the voltage across the capacitor for t > 0 is obtained as follows. Once the switch is closed, the capacitor voltage equals the voltage across the resistor. Hence we get equation ( 3.1). . It is seen that the direction of current is such that the capacitor gets discharged. If current flows into the positive terminal of the capacitor leading to increase in the charge held by capacitor, then both the current and the rate of change of capacitor voltage have the same sign. In this case, the current flows out of the positive terminal of the capacitor leading to decrease in the charge held by capacitor. Hence the capacitor current and the rate of change of capacitor voltage have opposite signs, as shown by equation (3.2). The direction for capacitor current is shown in Fig. 3. It means that if this current i(t) is positive, then the rate of change of capacitor voltage is negative. Then the negative of the rate of capacitor voltage is positive, equaling the current. The voltage across the resistor is the product of resistance and the resistor current, as expressed by equation (3.3). Since the capacitor voltage is equal to the resistor voltage, we get equation (3.4). Equation (3.4) is the first-order homogeneous differential equation for the RC circuit in Fig. 3.
Equation (3.5) is obtained from equation from (3.4). We get equation (3.6) from equation (3.5) by re-arranging the terms. The solution of the homogeneous differential equation is presented by equation (3.7).
Let us try getting the solution using Laplace transforms. The Laplace transform for the derivative of capacitor voltage is expressed by equation (3.8). Then equation (3.5) can be presented in terms of Laplace transforms, as shown by equation (3.9). The Laplace transform for the capacitor voltage is presented by equation (3.10) and the capacitor voltage in time domain is expressed by equation (3.11).
The technique illustrated above is not the recommended technique. The purpose of using Laplace transform is to avoid forming the differential equation. We should draw the transformed network and then obtain the solution, as illustrated below.
Using transform representation of a capacitor, the capacitor in circuit in Fig. 3 can be represented as shown in Fig. 4. The capacitor is replaced by a voltage source and the transform reactance of the capacitor. The value of voltage source is the initial value of capacitor voltage, applied as a step function. Using voltage division among impedances, an expression for the capacitor voltage VC(s) can be obtained as shown below. Note that the capacitor voltage is equal to the voltage across the resistor.
Equation (3.12) presents the Laplace transform function representing the capacitor voltage and its inverse transform is presented by equation (3.13). It can be seen that the use of the transformed network makes it easier to obtain the solution. Obtaining the solution based on the transformed network is the preferred approach.
The product RC of the RC circuit is called its time constant and is named as t in Fig. 5 shown below. The unit for time constant is second. It is seen that
The response of an RL circuit due to initial condition is now obtained.
An RL circuit is shown in Fig. 6a. The switch remains at position A before t < 0 for a sufficiently long time. Let switch be thrown over to position B at t = 0. Then an expression for inductor current for t ≥ 0 is obtained.
Given that the switch is in position A for t < 0 for a sufficiently long time implies that the rate of change of inductor current is zero and its current is a steady value. When the rate of change of inductor current is zero, the inductor voltage is also zero, and hence all the source current flow through the inductor for t < 0. We can state that the circuit is in steady-state for t < 0. If the circuit is at steady-state for t < 0, the rates of change in the circuit is zero. Equation (3.15) presents the condition that exists for t < 0 and equation (3.16) specifies the initial value of inductor current at t = 0, when the switch is thrown over from position A to position B.
The circuit that needs to be considered for t ≥ 0 is shown in Fig. 6b. It is seen that
The magnitude of voltage across the resistor is equal to that of the inductor voltage, but the sign of resistor voltage is opposite to that of the inductor voltage. Hence we get equation (3.17). We can re-arrange the terms and obtain the homogeneous differential equation for the circuit in Fig. 6b. The homogeneous differential equation is displayed by equation (3.18). The solution to equation (3.18) is presented below by equation (3.19).
The ratio of inductance to resistance is the time constant of the circuit and its unit is second, as shown by equation (3.20).
The solution displayed by equation (3.19) can be obtained using Laplace transform.
The equation in terms of Laplace transforms is presented by equation (3.21). You need to know how to express the derivative of a time function by its equivalent Laplace transform expression. Equation (3.21) can be re-arranged, and we get equation (3.22). The solution is also presented by equation (3.22).
The circuit in Fig. 6c shows the transformed network. The inductor in time domain is replaced by its transformed reactance and an impulse voltage source, expressed in terms of Laplace transform.
Applying Kirchoff's voltage law to the loop, we get equation (3.23), where the impulse voltage is expressed as the sum of the drop across the inductive reactance and the resistor. The solution is expressed by equation (3.24). It is seen that it is easy to obtain the solution by using the transformed network.
For the circuit in Fig. 7a, obtain vo(t) and vC1(t) for t ≥ 0.
SOLUTION:
It is preferable that the variable of the differential equation be such that it represents energy in a system. Then the solution of the variable is usually continuous, as long as there is no impulse input function. The capacitor has stored energy and it is a function of its voltage. Hence it is preferable to form the differential equation for a capacitor voltage. But for the circuit in Fig. 7, it is difficult to implement this recommendation. If the variable of the differential equation does not represent energy in a system, the solution for that variable may have discontinuity, even if there is no impulse input function or no impulse signal created by the initial condition. To illustrate what is meant by continuity and discontinuity, the values of capacitor voltages and the resistor current can be determined for two instants, one just before the switch is closed, referred to as t = 0-, and one just after the switch is closed, referred to as t = 0+.
Equation (3.25a) states that the voltage across C1 does not change from t = 0- to t = 0+. It stays at E Volts. Equation (3.25b) states that the voltage across C2 does not change from t = 0- to t = 0+. It stays at zero Volt. Both these voltages are continuous and there is no sudden change at t = 0. On the other hand, the resistor current at t = 0- is zero, and it is E/R at t = 0+. There is a step change at t = 0, as expressed by equation (3.25c). The current through the resistor has a discontinuity at t = 0, since its derivative is infinity.
From the circuit in Fig. 7a, we get that
Equation (3.26a) states that the voltage across capacitor C1 equals the voltage drop across the resistor and the voltage across capacitor C2. The voltage across capacitor C2 is the output voltage, vo(t). We get equation (3.26b) by differentiating equation (3.26a) with respect to time. The same current flows through both capacitors. Since current is flowing out of the positive terminal of C1, the voltage across C1 falls. On the other hand, the current charges C2 and we get equation (3.27). From equations (3.26b) and (3.27), we get equation (3.28), by replacing the rate of change of capacitor voltage by a suitable expression.
Equation (3.28) can be expressed as shown by equation (3.29), where t is the time constant of the circuit. The assumed homogeneous solution is also presented by equation (3.29). The initial value of current can be obtained, as shown by equation (3.30), and the natural response of the current in the circuit is expressed by equation (3.31). Note that the initial value of resistor current is specified at t = 0+, indicating that it is the value obtained just after closing the switch. As stated earlier, the current is not continuous at t = 0. In such a case, it is preferable to refer to time as either as t = 0- or t = 0+, so that it can be inferred that there is a step change at t = 0. By integrating the expression for current, we obtain an expression for the output voltage, as expressed by equation (3.32).
Next we get an expression for the voltage across capacitor C1.
The voltage across C1 is the sum of the output voltage and the drop across the resistor, as expressed by equation (3.33). On re-grouping terms, we get equation (3.34) from equation (3.33). On simplifying equation (3.34), we get equation (3.35). Equation (3.36) verifies that the solution is correct.
The same problem can be solved using the Laplace transforms. From Fig. 7b,
The Laplace expression transform for the output voltage is obtained using the voltage division rule, as shown by equation (3.36). The inverse transform of output voltage is expressed by equation (3.37).
It can be seen from the plots shown above that both the capacitor voltages are equal in steady-state and the steady-state current is zero. It can also be seen that the current rises suddenly at t = 0 and then it decays. The step change in current does not lead to any discontinuity in the capacitor voltages. It takes an impulse current to cause a step change in the voltage across a capacitor.
We can try a variation of the circuit in Fig. 7a, by letting R = 0 W. The resulting circuit is presented below.
For the circuit in Fig. 8a, obtain vo(t) for t ≥ 0.
SOLUTION:
Let us get the solution using the differential equation. For the circuit in Fig. 8a, we can form the following equations.
Equation (3.38a) presents the equation that links the capacitor voltage and capacitor current. When the capacitor current flows into the positive terminal of a capacitor, its voltage rises and its rate of voltage is positive. In this case, the loop current flows through capacitors C1 and C2, connected in series. The voltage applied to the circuit is the initial voltage of capacitor C1, as stated by equation (3.38b). Hence we get equation (3.39). Then the voltage across capacitor C2 is obtained as shown by equation (3.40).
We can an expression for the voltage across C1 as shown above. The current flows into the negative terminal of C1 and hence the change in the voltage of C1 due to current is negative. By subtracting the change from the initial value, we get the voltage across C1 as shown by equation (3.41). For t > 0, the voltage across C1 is the same as the voltage across capacitor C2, as shown by equation (3.42). It has to be so since both the capacitors are connected in parallel. But there is a step change in the voltage of C1, just before t = 0 and just after t = 0. Equation (3.43) presents voltage of C1 at t = (0-) and at t = (0+). An expression for the impulse current supplied by C1 is expressed by equation (3.44). It is the same expression, presented by equation (3.39). Since an impulse current flows out of C1 into C2, there is a step change in voltages of C1 and C2. The voltage across C1 falls suddenly and voltage across C2 rises suddenly. The process is illustrated by the sketch shown below. It can also be verified that the energy stored in C1 at t = (0-) is greater than the energy stored in both capacitors at t = (0+). The sudden change in energy stored has occurred because of the impulse current. The impulse current is due to the difference in voltages of capacitors at t = 0.
The voltage across C2 at t = (0+) can be determined, using the concept of charge balance. From the circuit in Fig. 8b, we get that from t = (0-) to t = (0+), C1 loses charge and C2 gains charge, where the loss of charge equals the gain in charge, The charge lost by C1, as well as the charge gained by C2 are presented by equation (3.45a). On solving equation (3.45b), we get equation (3.45c), which yields the output voltage vo(0+). An expression for charge lost is presented by equation (3.45d), whereas an expression for charge gained is presented by equation (3.45e). It is seen that the charge lost by C1 equals the charge gained by C2.
We can make use of Laplace transforms and get the solution.
It is seen that the use of Laplace transforms makes it easier to get the solution. Equation (3.45f) presents an expression for the output voltage. This expression is obtained from the inverse transform of Vo(s).
We can get the solution also from the results we have obtained for the circuit in Fig. 7a. When R = 0, the circuits in Fig. 7a and 8a are the same.
Equation (3.32) presents the output voltage obtained for the circuit in Fig. 3.7a. When the value of resistance is zero, the time constant becomes zero and the exponential term is of zero value, as stated by equation (3.46g). Then we get equation (3.47), which is the result obtained earlier.
For the circuit in Fig. 9, obtain iR(t), i2(t) and i1(t) for t ≥ 0.
SOLUTION:
The solution using differential equation is presented first. For this solution, the circuit in Fig. 9 is re-drawn for t ≥ 0 and the circuit is presented in Fig. 10. For this circuit, the variable of the differential equation does not represent the energy stored in the inductors, since it is inconvenient. On the other hand, the differential equation is formed for the voltage across the resistor. Since all the three elements are connected in parallel, it is also the voltage across the inductors. This voltage is discontinuous at t = 0, whereas the currents through the inductors are continuous. The energy stored in an inductor is a function of its current. Even though there is no impulse function present in the circuit, the voltage across the resistor is discontinuous at t = 0. On the other hand, there can be discontinuity in inductor current, only if there is an impulse input function or if an impulse voltage is caused by the initial condition. As expressed by equation (3.48a), current i1(t) is continuous. Equation (3.48b) expresses that current i2(t) is continuous. On the other hand, voltage vR(t) is discontinuous at t = 0, as expressed by equation (3.48c).
As stated by equation (3.49a), the current through inductor L1 is the sum of current through inductor L2 and the current through the resistor. Note the direction for currents, marked in Fig. 10. The initial value of current through L1 is IS, and the initial value of current through L2 is zero. After the switch is closed, the current through the resistor at t = 0+ is IS. This is because the current through an inductor cannot change suddenly. Since the current through L2 cannot change suddenly from its value, the entire current through L1 flows through the resistor. Hence we get equation (3.49b). Differentiate equation (3.49a) and we get equation (3.50).
Since all the three elements are in parallel, the voltage across all of them is the same, as shown by equation (3.51).. This voltage is assigned to be the resistor voltage. Equation (3.52) expresses how the current in each element is related to its voltage. There is a negative sign for L1, since its current is leaving its positive terminal.
Equation (3.50) can now be presented, as shown by equation (3.53). Re-arrange the terms and get equation (3.54). The solution of the differential equation is presented by equation (3.55). Once the resistor voltage is known, the resistor current can be expressed, as shown by equation (3.56).
From the expression for the resistor voltage, an expression for current through inductor L2 can be obtained, as shown by equation (3.57). We get equation (3.58) from equation (3.57). Since the current through inductor L1 is the sum of current through inductor L2 and the current through the resistor, we get equation (3.59). We can verify the solution by checking what the initial current through inductor L1 is. It can be seen that it is equal to IS, the value derived earlier.
Next the solution using Laplace transforms is presented.
The transformed network is presented in Fig. 11a. Here the inductor L1 is represented by its transform reactance and a current source. The two inductors in parallel can be combined and the resultant circuit is shown in Fig. 11b. using the current division rule, an expression for the resistor current can be obtained, as shown by equation (3.60). The time-domain expression for the resistor current is presented by equation (3.61). We have obtained earlier the same expression.
Now an expression for current through inductor L2 can be obtained. From Fig. 11a and equation (3.60), we get equation (3.62). The time-domain expression for current through inductor L2 is presented by equation (3.63). We have obtained earlier the same expression.
The plots of currents are presented above. It can be seen that both i1(t) and i2(t) are continuous, whereas iR(t) is discontinuous. Because the voltage that appear across the inductors at t = 0 is finite, the inductor currents are continuous at t = 0. A step change in resistor current leads to a step change in the resistor voltage. This step voltage is applied to the inductors also, but a step voltage causes no sudden change in inductor currents. An impulse voltage applied to an inductor causes a step change in inductor current.
We can replace the load resistor in Fig. 9 by an open circuit, and the resultant circuit is presented in Fig. 12a.
An expression for inductor current i2(t) can be obtained, as shown below. From the circuit in Fig. 12b, we find that the voltage across the two inductors is the same voltage. Because of the direction assigned to the inductor currents, we get equation (3.64). On integrating both sides, we get equation (3.65). Equation (3.66) specifies the values of i1(0-) and i2(0-). Hence we can obtain the value of i2(0+), as shown by equation (3.67).
We can obtain the same result using the Laplace transforms. The transformed network is shown in Fig. 12c. Based on the current division rule, we get equation (3.68) and inductor current i2(t) can be expressed by equation (3.69)
It is possible to obtain an expression for i2(0+), based on the volt-seconds associated with the two inductors. Equation (3.70) states that the volt-seconds lost by L1 equals the volt-seconds gained by L2. The volt-seconds lost and the volt-seconds gained can be expressed in terms of the inductances and their currents, as shown by equation (3.71). An expression for i2(0+) is then obtained, as shown by equation (3.72).
The step changes in inductor currents and the impulse inductor voltage can be illustrated by means of a sketch, as shown below. The impulse voltage across the inductors at t = 0 is due to the step current input, as shown in Fig. 12c. It can also be verified that the energy stored in L1 at t = (0-) is greater than the sum of energy stored in both inductors at t = (0+). The sudden change in energy stored has occurred because of the impulse voltage and the redistribution of current in inductors. The impulse voltage across the inductor is due to the initial current in L1 at t = (0-) and the magnitude of the impulse voltage can be obtained form the inverse transform of VL(s) marked in Fig. 12c.
The expression for for i2(0+), presented by equation (3.69) can also be derived from the expression for for i2(0+), presented by equation (3.58). When the resistor, shown in the circuit in Fig. 10 is replaced by an open-circuit, the resultant circuit is presented by Fig. 12b. Then the time constant present as a part of the exponential term in equation (3.58) is infinite and the exponential term has zero. It can be seen that both equations (3.58) and (3.69) are the same when the time constant of the circuit is infinite.
In this section, many worked examples are presented. In most of the examples, the aim is to obtain the zero-input response. Some examples illustrate the relationship between voltage and current in elements. Strictly, they should have been presented in the chapter on elements. But this section is more suited for such examples, since we have studied differential equations and Laplace transforms since then. In some of the examples, the zero-state response is obtained. These examples are presented in this section, because the response in these cases is obtained without forming a differential equation.
Example 1:
The circuit shown in Fig. 13 is at rest for t < 0. Find vy(0+).
Solution:
From the circuit in Fig. 13, we can obtain the circuit shown below.
The circuit circuit shown in Fig. 13 is at rest for t < 0. Hence the initial voltage across the 1 F capacitor is 12 V, and the initial voltage across the 2 F capacitor is 8 V. Hence equations (3.73) and (3.74) can be formed, as shown below.
We can draw the transformed circuit, as shown in Fig. 14 and the voltage across the 3 F capacitor can be obtained using the voltage division rule. The answer is the same, as obtained earlier.
Example 2:
Obtain vC(t) for t ≥ 0 for the circuit in Fig. 15. Switch A opens at t = 0 and switch B closes at t = 10 s. For t < 0 s, switch A is closed and switch B is open for t < 10 s. The circuit is at rest for t < 0.
Solution:
From the circuit in Fig. 15, we can obtain the circuit shown below.
The circuit that remains connected for t < 0 is shown in Fig. 16a. Since the circuit is at rest for t < 0, it means that the responses in the circuit tend to be dc values, as the source is a dc signal. That is, the rate of change of capacitor voltage is zero, implying that the capacitor current is zero. When the capacitor current is zero, the capacitor voltage is obtained as shown by equation (3.76). The voltage across the capacitor is the same as that across the 15 W resistor. The voltage across the 15 W resistor can be obtained using the voltage division rule. The voltage across the capacitor for the period, defined by 0 < t < 10 s, can be obtained from the circuit in Fig. 16b. The time constant of this circuit is 30 s. The expression for the capacitor voltage is presented by equation (3.78). The capacitor voltage at t = 10 s is obtained, as shown by equation (3.79).
For t > 10 s, the circuit to be considered is shown in Fig. 16c. The time constant of this circuit is 12 s. Then the solution obtained is as shown by equations (.380) and (3.81) presented below.
The solution using Laplace transforms can be obtained based on the transformed network presented in Fig. 17. For the circuit in Fig. 17a, we get that
For t > 10 s, the circuit to be considered connected is shown in Fig. 17b. From this circuit we get the following equations.
You need to know how a time-shifted function is represented in Laplace transforms.
Example 3:
Obtain iL(t) for t ≥ 0 for the circuit in Fig. 8. Switch A opens at t = 0 second and switch B closes at t = 1 s. For t < 0 s, switch A is closed and switch B is open for t < 1 s.
Solution:
From the circuit in Fig. 18, we can obtain the circuit shown below.
The circuit that remains connected for t < 0 is shown in Fig. 19a. It is assumed that the circuit is at rest for t < 0. Then the responses in the circuit tend to be dc values, since the source is a dc signal. That is, the rate of change of inductor current is zero, implying that inductor voltage is zero. When the inductor voltage is zero, the inductor current is obtained as shown by equation (3.87). The inductor acts as a short-circuit for t < 0, and hence the initial inductor current at for t = 0 second is 4 A. The time constant of the circuit is one second, as stated by equation (3.88). Then the inductor current for 0 < t < 1 second can be expressed, as shown by equation (3.89).
For t > 1 s, the circuit that is to be considered is shown in Fig. 19c. The time constant t of this circuit is 2.5 s. Then we can get equation (3.90) and (3.91).
The transformed circuits for 0 < t < 1 second and for t > 1 second are shown in Fig. 20. We can obtain the inductor current for 0 < t < 1 second as shown below.
We can obtain the inductor current for t > 1 second as shown below.
Example 4:
For the circuit in Fig. 21, obtain the following values.
Solution:
From the circuit in Fig. 21, we can obtain the equations shown below.
Because of the inductor, the current through the inductor cannot change suddenly at t = 0, since no impulse voltage is applied across the inductor in this circuit. However, the inductor voltage can suddenly at t = 0. It means the inductor voltage at t = 0+ can be different from the the inductor voltage at t = 0-, as shown by equation (4.5).
Example 5:
Given that
find the following:
Solution:
From the circuit in Fig. 22 and equations (4.6) and (4.7, we can obtain the equations shown below.
The total energy discharged by the capacitor should equal its initial energy, as seen from equations (4.10) and (4.11).
Example 6:
For the circuit in Fig. 23, the source current is specified by the side of the circuit. Find expressions for vR(t), vL(t) and vS(t) and sketch these functions.
Solution:
Given the current through the resistor, its voltage can be obtained, as shown by equation (4.12).
The inductor voltage is the product of its inductance and the rate of change of current. Equation (4.13) expresses the inductor voltage.
The source voltage is the sum of the resistor voltage and the inductor voltage, as expressed by equation (4.14). The plots of vR(t), vL(t) and vS(t) are presented below.
Example 7:
For the circuit in Fig. 25, the inductor current is specified by the side of the circuit. Find expressions for iR(t), iS(t) and vS(t) and sketch these functions.
Solution:
These functions can also be expressed, as shown below.
Example 8:
For the circuit shown in Fig. 27, the source voltage is specified by equation (4.21). Sketch vS(t). Find expressions for iR(t) and iC(t), and sketch iC(t).
Solution:
Equation (4.22) expresses the source voltage in terms of step and ramp functions. Then the resistor current and the capacitor current can be obtained, as shown by equations (4.23) and (4.24).
Example 9:
For the circuit shown in Fig. 27, the source current is specified by the side of the circuit. Sketch iS(t). Find expressions for vR(t), vL(t) and vS(t). Sketch vL(t) and vS(t).
Solution:
Example 10:
Find an expression for vO(t) for each of the source functions, specified by equations (4.28) and (4.29).
Solution:
For the input specified by equation (4.28), we get the following equations.
For the input specified by equation (4.29), we get the following equations.
Example 11:
Find an expression for vO(t) for each of the source functions, specified by equations (4.28) and (4.29).
Solution:
For the input specified by equation (4.28), we get the following equations.
For the input specified by equation (4.29), we get the following equations.
Example 12:
Find an expression for vO(t) for each of the source functions, specified by equations (4.42) and (4.43).
Solution:
We can express the output in terms of Laplace transforms, as shown below.
For the input specified by equation (4.42), we get the following equations.
Equations (4.45) and (4.46) show how an expression for vO(t) can be obtained using Laplace transforms. We can obtain the same answers without recourse to Laplace transforms and equations (4.47) and 94.48) illustrate this process. For the input specified by equation (4.43), we can get the solution using both techniques.
In this page, it is shown how we can obtain the zero-input response of a first-order circuit. Worked examples have been presented how the solution in response to inputs can be obtained, when there is no need to form a differential equation. But for many first-order circuits, it is necessary to form a differential equation before a solution can be obtained. The next page takes up the topic of obtaining the zero-state response and the total response of first-order circuits.