First order systems are, by definition, systems whose input-output relationship is a first order differential equation. A first order differential equation contains a first order derivative but no derivative higher than first order – the order of a differential equation is the order of the highest order derivative present in the equation.

First order systems contain a single energy storage element. In general, the order of the input-output differential equation will be the same as the number of independent energy storage elements in the system. Independent energy storage cannot be combined with other energy storage elements to form a single equivalent energy storage element. For example, we previously learned that two capacitors in parallel can be modeled as a single equivalent capacitor – therefore, a parallel combination of two capacitors forms a single independent energy storage element.

First order systems are an extremely important class of systems. Many practical systems are first order; for example, the mass-damper system and the mass heating examples from section 6.1 are both first order systems. Higher order systems can often be approximated as first order systems to a reasonable degree of accuracy if they have a dominant first order mode. (System modes will be discussed later in this text.) Understanding first order systems and their responses is an important aspect to design and analysis of systems in general.

First order electrical circuits are a special class of first order systems – they are first order systems which are composed of electrical components. Since the electrical components which store energy are capacitors and inductors, first order circuits will contain either one (equivalent) capacitor or one (equivalent) inductor.

In this textbook, we are really interested only in the analysis of electrical circuits, so the bulk of this chapter, of course, deals with analysis of first order electrical circuits. However, at this stage of your studies, you probably do not have an intuitive grasp of the mechanisms of energy storage in electrical circuits and the response of electrical circuits with energy storage elements. Therefore, this chapter begins in Section 7.1 with a general discussion of the response of first order systems, using a sliding mass as an example within which to frame the basic concepts. This provides a relatively gentle introduction to the nomenclature and mathematics which will be used throughout this chapter, in the context of an example for which the student should have some physical insight. Section 7.1 can be omitted, however, without loss of clarity of the remaining sections. Sections 7.2 and 7.3 present the natural response of RC and RL circuits, respectively (RC circuits have a capacitor as an energy storage element, while RL circuits contain an inductor). The natural response of a system corresponds to the system response to some initial condition, with no forcing function provided to the system. In section 7.4, we present the force response of first order circuits, and in section 7.5 we examine the response of first order circuits to a specific forcing function – a step input.

• Write the general form of the differential equation governing a first order system
• State, in physical terms, the significance of a differential equation’s homogeneous and particular solutions
• Define, from memory, the relationships between a system’s unforced response, zero-input response, natural response, and the homogeneous solution to the differential equation governing the system
• Define, from memory, the relationships between a system’s forced response, zero-state response, and the particular solution to the differential equation governing the system
• Determine the time constant of a first order system from the differential equation governing the system
• Write mathematical expressions from memory, giving the form of the natural and step responses of a first order system
• Sketch the natural response of a first order system from the differential equation governing the system and the system’s initial condition
• Sketch the step response of a first order system from the differential equation governing the system and the amplitude of the input step function
• Write the differential equation governing RC and RL circuits
• Determine the time constant of RC and RL circuits from their governing differential equations
• Determine the time constant of RC and RL circuits directly from the circuits themselves
• Determine initial conditions on arbitrary RC and RL circuits
• Write from memory the form of the natural responses of RC and RL circuits
• Determine the natural response of RC and RL circuits, given the governing differential equation and initial conditions
• Write the form of the differential equations governing forced first order electrical circuits
• Determine the time constant of a forced electrical circuit from the governing differential equation
• Write differential equations governing passive and active first order circuits
• Determine the differential equation governing the step response of a first order electrical circuit
• Write the form of the particular solution of a first order differential equation, to a step input
• Write the form of the step response of a first order electrical circuit
• Determine the final conditions (steady-state response) of a first order electrical circuit, to a step input
• Define DC gain for a circuit and relate it to the steady-state response to a step input
• Determine the step response of a first order electrical circuit from the governing differential equation, the initial conditions, and the final conditions

In this section, we introduce some basic nomenclature relative to first order system responses and illustrate these terms in the context of an example for which the reader may have an intuitive understanding: a mass sliding on a surface. This example, though not directly relevant to the study of electrical circuits, is intended to allow the reader to develop some physical insight into the terminology and concepts relative to the solution of first order differential equations. The concepts and results obtained with this example are then generalized to apply to any arbitrary first order system. These results are used in later sections to provide insight in the analysis of electrical circuits, for which the reader may not yet have an intuitive understanding.

Before discussing first-order electrical systems specifically, we will introduce the response of general first order systems. A general first order system is governed by a differential equation of the form:

$$a_1 \frac{dy(t)}{dt} + a_0 y(t) = f(t), t>t_0 (Eq. 7.1)$$

Where $f(t)$ is the (known) input to the system and $y(t)$ is the response of the system. $a_1$ and $a_0$ are constants specific to the system being analyzed. We assume in equation (7.1) that the input function is applied only for times $t>t_0$. Thus, from equation (7.1), we can only determine the response of the system for times $t>t_0$.

In order to find the solution to equation (7.1), we require knowledge of the system’s initial condition:

$$y(t=t_0) = y_0 (Eq. 7.2)$$

The initial condition, $y_0$, defines the state of the system at time $t=t_0$. Since equation (7.1) describes a system which stores energy, the effect of the initial condition is to provide information as to the amount of energy stored in the system at time $t=t_0$.

The system described by equations (7.1) and (7.2) can be illustrated in block diagram form as shown in Figure 7.1. The output of the system depends upon the initial condition, $y_0$, and the input function $f(t)$. The initial condition provides information relative to the energy stored in the system prior to application of the input function. The input function provides information relative to the energy being applied to the system from external sources. The input-output equation describes how the system transfers the energy initially present in the system and the energy added to the system to the system output.

The solution to equation (7.1) consists of two parts – the homogeneous solution, $y_h (t)$, and the particular solution, $y_p (t)$, as shown below:

$$y(t) = y_h(t) + y_p(t) (Eq. 7.3)$$

The homogeneous solution is due to the properties of the system and the initial conditions applied to the system; it describes the response of the system if no input is applied to the system, so $f(t) = 0$. The homogeneous solution is sometimes called the systems natural response, the unforced response, or the zero input response. Since all physical systems dissipate energy (according to the second law of Thermodynamics) the homogeneous solution must die out with time; thus, $y_h(t) \rightarrow 0$ as $t \rightarrow \infty$.

The particular solution describes the systems response to the particular forcing function applied to the system; the form of the particular solution is dictated by the form of the forcing function applied to the system. The particular solution is also called the forced response or the zero state response.

Since we are concerned only with linear systems, superposition principles are applicable, and the overall system response is the sum of the homogeneous and particular solutions. Thus, equation (7.3) provides the system’s overall response to both initial conditions and the particular forcing function being applied to the system.

The previous concepts are rather abstract, so we provide below an example of the application of the above concepts to a system for whose response the students should have some intuitive expectations. This example is intended to provide some physical insight into the concepts presented above prior to applying these concepts to electrical systems.

As an example of a system which includes energy storage elements we revisit the mass-damper system of section 6.1. The system under consideration is shown in Figure 7.2. The applied force $F(t)$ pushes the mass to the right. The mass’s velocity is $v(t)$. The mass slides on a surface with sliding coefficient of friction $b$, which induces a force which opposes the mass’s motion. The mass will have some initial velocity:

$$v(t=0)=v_0 (Eq. 7.4)$$

Consistent with section 6.1, we consider the applied force to be the input to our system and the mass’s velocity to be the output. Figure 7.3 illustrates the system input-output relationship and initial conditions in block diagram form.

The governing equation for the system was determined in section 4.1 to be the first order differential equation:

$$m \frac{dv(t)}{dt} + bv(t) = F(t) (Eq. 7.5)$$

We consider two cases of specific forcing functions in the following cases. In the first case, the forcing function is zero, and we determine the system’s natural response or the homogeneous solution to equation (7.5) above. In the second case, the forcing function is a constant nonzero force applied to the mass with zero initial velocity.

Let us consider first the case in which the mass has some initial velocity but no external force is applied to the mass. Intuitively, we expect that the velocity of the mass will decrease until the mass comes to rest. In this example, we will determine the solution of the differential equation (7.5) and compare this solution with our expectations.

With no applied forcing function, the differential equation governing the system is:

$$m\frac{dv(t)}{dt} + bv(t) = 0 (Eq. 7.6)$$

The initial condition is given by equation (7.4) above, repeated here for convenience:

$$v(t=0) = v_0$$

Equation (7.6) is a homogeneous differential equation, since there is no forcing function applied to the system. Thus, the particular solution in this case is $y_p(t) = 0$ and our overall solution is simply the homogeneous solution, $y(t) = y_h(t)$.

To solve the above differential equation, we rearrange equation (7.6) to give:

$$\frac{m}{b} \cdot \frac{dv(t)}{dt} = -v(t) (Eq. 7.7)$$

Separating variables in equation (7.7) results in:

$$\frac{dv(t)}{v} = - \frac{b}{m}dt (Eq. 7.8)$$

Incorporating dummy variables of integration and integrating both sides of (7.8) gives:

$$\int_{v_0}^{v(t)} \frac{d\xi(t)}{\xi} = - \frac{b}{m} \int_{0}^{t} d\zeta$$

Which evaluates to:

$$ln(v) |_{v_0}^{v(t)} = - \frac{b}{m} t \Rightarrow ln [v(t)] - ln [v_0] = - \frac{b}{m} t \Rightarrow ln \left[ \frac{v(t)}{v_0} \right] = - \frac{b}{m}t$$

Taking the exponent of both sides of the above provides our final result:

$$v(t) = v_0e^\frac{-bt}{m} (Eq. 7.9)$$

A plot of the response given in equation (7.9) is shown in Figure 7.4. This plot matches our previous expectations, the velocity of the mass at time $t = 0$ is $v_0$ and the velocity decreases exponentially until the mass is (essentially) at rest. Referring to section 6.2, we see that the response of equation (7.9) can be written in terms of a time constant as:

$$v(t) = v_0e^\frac{-t}{\tau}$$

Where the time constant $\tau = \frac{m}{b}$ . This result also agrees with our intuition: as the friction coefficient decreases, the time constant increases and the mass comes to rest more slowly. Likewise, increasing the mass causes the time constant to increase – a larger mass will tend to “coast” for a longer time.

Note The velocity of the mass tells us how much kinetic energy is being stored by the system. The initial condition provides the energy initially stored in the system. The calculated response describes how this energy is dissipated through the sliding friction. No energy is added to the system in this case, since the external applied force is zero.

We will now consider the case in which the mass is initially at rest, and a constant force is applied to the mass at time $t = 0$. Intuitively, we expect the velocity of the mass to increase to some final value; the final velocity of the mass corresponds to the condition in which the frictional force is equal and opposite to the applied force (recall that in our model, the frictional force is proportional to velocity – as the velocity increases, the frictional force opposing the motion also increases). We now solve the governing differential equation for this system and compare the results to our expectations.

The differential equation governing the system, valid for $t > 0$, and initial condition, providing the energy in the system at $t = 0$, are:

$$m \frac{dv(t)}{dt} +bv(t) = F \\ v(t=0)=0 \\ (Eq. 7.10)$$

Where F is the magnitude of the (constant) applied force. Note that since F is constant, and only applied for times $t > 0$, we have a step input with magnitude F. We want to solve the above differential equation for $t > 0$; since the input forcing function can be represented as a step function, this resulting solution is called the step response of the system.

For this case, we have both a nonzero forcing function and an initial condition to consider. Thus, we must determine both the homogeneous and particular solutions and superimpose the result per equation (7.3) above.

The homogeneous solution is determined from:

$$m \frac{dv_h(t)}{dt} + bv_h(t) = 0 (Eq. 7.11)$$

Where $v_h(t)$ is the homogeneous solution. This equation has been solved as case i; the form of the solution is:

$$v_h(t) = K_1e \frac{-bt}{m} (Eq. 7.12)$$

Where $K_1# is (in this case) an unknown constant which will be determined from our initial conditions.

The particular solution is determined from:

$$m \frac{dv_p(t)}{dt} + bv_p(t) = F (Eq. 7.13)$$

Where $v_p(t)$ is the particular solution to the differential equation in equation (7.10). Since the right-hand side of equation (7.13) is constant for $t > 0$, the left-hand side of the equation must also be constant for $t > 0$ and $v_p(t)$ must be constant for $t > 0$. If $v_p(t)$ is constant, $\frac{dv_p(t)}{dt}$ is zero and equation (7.13) simplifies to:

$$bv_p(t)=F$$

So that:

$$v_p(t)=\frac{F}{b} (Eq. 7.14)$$

Superimposing equations (7.12) and (7.14), per the principle expressed in equation (7.3) results in:

$$v(t) =v_k(t) + v_p(t) = K_1e^\frac{-bt}{m} + \frac{F}{b} (Eq. 7.15)$$

We can now use our initial condition, $v(t=0)=0$, to determine the constant $K_1$. Evaluating equation (7.15) at $t = 0$ and applying the initial condition results in:

$$v(t=0) = 0 = K_1e^\frac{-b \cdot 0}{m} + \frac{F}{b} (Eq. 7.16)$$

Since $e^{-b \cdot 0}{m} = 1$, equation (7.16) results in:

$$K_1 = - \frac{F}{b} (Eq. 7.17)$$

Substituting equation (7.17) into equation (7.15) results in the overall solution:

$$v(t) = - \frac{F}{b}e^\frac{-bt}{m} + \frac{F}{b} = \frac{F}{b} \left( 1 - e^\frac{-bt}{m} \right) (Eq. 7.18)$$

If, as in case i, we define the time constant $\tau = \frac{m}{b}$, equation (7.13) can be expressed as:

$$v(t) = \frac{F}{b} \left( 1-e^\frac{-t}{\tau} \right) (Eq. 7.19)$$

A plot of the system response is shown in Figure 7.5. This plot matches our intuitive expectations: the initial velocity is zero; the applied force causes the mass to move. When the frictional and applied forces balance, the velocity of the mass becomes constant. The time constant is determined by the mass and the frictional coefficient; a larger mass results in a longer time constant – it takes longer to get a large mass to its final velocity than a small mass. The frictional coefficient also affects the system time constant; a smaller friction coefficient results in a longer time constant. This result seems counter-intuitive at first, since a smaller frictional coefficient should allow us to accelerate the mass more rapidly. However, the smaller frictional coefficient also results in a higher final velocity – since the time constant is defined by the time required to reach approximately 63.2% of the final velocity, the higher final velocity causes a longer time constant even though the mass is accelerating more rapidly. (If the damping coefficient is zero, the time constant goes to infinity. However, the final velocity also goes to infinity – it takes an infinite amount of time to get to 63.2% of an infinite velocity!)

Note The velocity of the mass again describes the energy stored by the system; in this case, the initial velocity is zero and the system has no energy before the force is applied. The applied force adds energy to the system by causing the mass to move. When the rate of energy addition by the applied force and energy dissipation by the friction balance, the velocity of the mass becomes constant and the energy stored in the system becomes constant.

We use the results of the above examples to re-state some primary results in more general terms. It is seen above that the natural and step responses of first order systems are strongly influenced by the system time constant, $\tau$. The original, general, differential equation – equation (7.1) above – can be re-written directly in terms of the system time constant. We do this by dividing equation (7.1) by the coefficient $a_1$ . This results in:

$$\frac{dy(t)}{dt} + \frac{a_0}{a_1} y(t) = \frac{1}{a_1}f(t), t>t_0 (Eq. 7.20)$$

Defining $\tau = \frac{a_1}{a_0}$, equation (7.20) becomes:

$$\frac{dy(t)}{dt} + \frac{1}{\tau} y(t) = \frac{1}{a_1} f(t) (Eq. 7.21)$$

The initial condition on equation (7.21) is as before:

$$y(t=t_0) = y_0 (Eq. 7.22)$$

The cases of the system homogeneous response (or natural or unforced response) and step response are now stated more generally, for the system described by equations (7.21) and (7.22).

1. Homogeneous response

For the homogeneous response $f(t) = 0$, and the system response is:

$y(t) = y_0e^\frac{-t}{\tau}$ for $t \geq 0 (Eq. 7.23)$

The response is shown graphically in Figure 7.6.

2. Step Response

For a step input of amplitude A, $f(t) = Au_0(t)$ where $u_0(t)$ is the unit step function defined in section 6.1. Substituting this input function into equation (7.21):

$$\frac{dy(t)}{dt} + \frac{1}{\tau}y(t) = \frac{A}{a_1}, t>0 (Eq. 7.24)$$

Using the approach of case 2. of our previous mass-damper system example, we determine the system response to be:

$$y(t) = \frac{A}{a_0} \left[ 1-e^\frac{-1}{\tau} \right] (Eq. 7.25)$$

This response is shown graphically in Figure 7.7.

• A first order system is described by a first order differential equation. The order of the differential equation describing a system is the same as the number of independent energy storage elements in the system – a first order system has one independent energy storage element. (The number of energy dissipation elements is arbitrary, however.)
• The differential equation governing a first order system is of the form:

$$a_1 \frac{dy(t)}{dt} + a_0y(t) = f(t)$$

Where $y(t)$ is the system output, $f(t)$ is the applied input to the system, and $a_0$ and $a_1$ are constants.

• The differential equation governing a first order system can also be written in the form:

$$\frac{dy(t)}{dt} + \frac{1}{\tau}y(t) = \frac{f(t)}{a_1}$$

Where $y(t)$ is the system output, $f(t)$ is the applied input to the system, and $\tau$ is the system time constant.

• The system time constant is a primary parameter used to describe the response of first order systems.
• In this chapter, we considered two types of forcing functions: a zero-input case, in which $f(t) = 0$ and $y(t=0) = y_0$, and a step input case, in which $f(t) = Au_0(t)$ and $y(t=0) = 0$. For the zero-input case, the response is:

$$y(t) = y_0e^\frac{-t}{\tau}$, t>0$$ For the step input case, the so-called //step response// is: $$y(t) = \frac{A}{a_0} \left[ 1-e^{-t}{\tau} \right], t>0$$ * The system response consists of two parts: a //homogeneous solution// and a //particular solution//. The response can also be considered to consist of a //transient response// and a //steady-state response//. The homogeneous solution and the transient response die out with time; they are due to a combination of the system characteristics and the initial conditions. The particular solution and the steady state response have the same form as the forcing function; they persist as $t \rightarrow \infty$. It can be seen from the above that, for the zero-input case, the steady state response is zero (since the forcing function is zero). The steady state step response is $\frac{A}{a-0}$; it is a constant value and is proportional to the magnitude of the input forcing function.

1. In Example 6.2, we examined a body which was subjected to external heating. The system is shown in the figure below. The mass of the body is m, the body material has a specific heat, $c_P$, and is at some temperature $T_B$. The surroundings are at an ambient temperature $T_0$. A heat input $q_{in}$ is applied to the body, the heat dissipation between the body and its surroundings is $q_{out}$. It is common to assume that the heat dissipation $q_{out}$ is proportional to the difference in temperature between the mass and its surrounds, so that $q_{out} = h(T_B - T_0)$. Incorporating this assumption into the governing equation for the system provided in Example 6.2, results in the following differential equation relating $q_{in}$ and $T_B$:

$$mc_p \frac{d(T_B-T_0)}{dt} + h(T_B-T_0) = q_{in}$$

• If the body has some initial temperature $T_i$, and no heat is applied to the body (e.g. $q_{in} = 0$),
  • What is the final temperature of the body?
  • What is the time constant (in terms of $m$, $c_p$, and $h$)?
  • If the body mass is doubled, what is the effect on the time constant? Does this agree with your expectations based on your intuition?
  • Sketch the response of the body temperature ($T_B$) vs. time. Label the initial condition and time constant on the sketch.
• If the body is initially at a temperature $T_0$ (the same as the ambient surroundings) and a constant heat input $q_{in}$ is applied starting at $t=0$,
  • What is the final temperature of the body (in terms of $m$, $c_p$, $q_{in}$, and $h$)
  • What is the effect of doubling the heat input $q_{in}$ on the final temperature? Does this agree with your intuition?
  • What is the effect of doubling the mass on the final temperature. Does this agree with your intuition?
  • What is the effect of doubling the mass on the time constant? Does this agree with your intuition?
  • Sketch the response $T_B$ vs. time. Label the initial temperature, final temperature, and time constant on the sketch.

In this section, we consider source-free circuits containing only resistors and a single capacitor – commonly referred to as RC circuits. Since these circuits contain only a single energy storage element – the capacitor – the governing equations for the circuits will be first order differential equations. Since the circuits are source-free, no input is applied to the system and the governing differential equation will be homogeneous. Thus, in this section we will be examining the natural response of RC circuits.

To begin our investigation of the natural response of RC circuits, consider the simple resistor-capacitor combination shown in Figure 7.8. We assume that the capacitor is initially charged to some voltage, $V_0$, at time $t = 0$ (so that $v(0) = V_0$). We want to determine the capacitor voltage, $v(t)$, for $t>0$.

Applying Kirchhoff’s current law at the positive terminal of the capacitor, as shown in Figure 7.9, results in:

$$i_C(t) + i_R(t) = 0 (Eq. 7.26)$$

Since $i_R(t) = \frac{v(t)}{R}$ and $i_C(t)=C \frac{dv(t)}{dt}$, equation (7.26) can be written in terms of the capacitor voltage as:

$$C \frac{dv(t)}{dt} + \frac{v(t)}{R}=0$$

Separation of variables results in:

$$\frac{dv}{v} = - \frac{1}{RC}dt$$

The integral of the above is:

$$\int_{V_0}^{v(t)} \frac{dv}{v} = - \frac{1}{RC} \int_0^t dt$$

Which evaluates to:

$$ln \left[ v(t) \right] - lm \left[ V_0 \right] = - \frac{t}{RC}$$

Or

$$ln \left[ \frac{v(t)}{V_0} \right] = - \frac{t}{RC}$$

Since $e^{ln(x)} = x$, the above becomes:

$$v(t) = V_0e^\frac{-t}{RC}$$

Since $\frac{dv(t)}{dt} = - \frac{1}{RC}v(t)$ we see that the form of the voltage signal must not change as a result of differentiation. Thus, assume that the voltage signal is of the form:

$$v(t) = Ke^{-st}$$

Where $K$ and $s$ are unknown constants. If we substitute this into the original differential equation, we obtain:

$$-Kse^{-st} = - \frac{K}{RC}e^{-st}$$

This is satisfied if we choose $s= - \frac{1}{RC}$. Employing our initial condition, $v(0) = V_0$, gives:

$$Ke^\frac{-0}{RC} = K = V_0$$

Results in:

$$v(t) = V_0e^\frac{-t}{RC}$$

As before.

The capacitor voltage response is shown graphically in Figure 7.10. The voltage response is a decaying exponential with a time constant:

$$\tau = RC (Eq. 7.27)$$

Thus, if we increase the resistance without changing capacitance, the circuit’s time constant will increase. Likewise, increasing capacitance while maintaining the resistance constant will also increase the system’s time constant. It is important to note that neither the resistance nor the capacitance alone specify the time constant, it is determined by the product of the two – if we simultaneously double the resistance and halve the capacitance, the system’s time constant is unchanged.

We can also obtain the above result by writing the governing differential equation directly in terms of the time constant. Previously, applying KCL to the circuit of Figure 7.9, we obtained:

$$C \frac{dv(t)}{dt} + \frac{v(t)}{R} = 0$$

Which can be re-written (by dividing the equation by the capacitance, $C$, as:

$$\frac{dv(t)}{dt} + \frac{v(t)}{RC}=0$$

Since the time constant for this circuit is $\tau = RC$, the above can be re-written as:

$$\frac{dv(t)}{dt} + \frac{1}{\tau}v(t) = 0 (Eq. 7.28)$$

With initial condition:

$$v(0) = V_0 (Eq. 7.29)$$

The solution to equation (7.28), subject to the initial condition of equation (7.29) is (per our results above):

$v(t) = V_0e^{-t}{\tau}$, for $t \geq 0 (Eq. 7.30)$

Where the time constant is defined per equation (7.27). The above approach matches our previous result. We will use the problem description provided by equations (7.28) and (7.29) and the solution of the form (7.30) most commonly in the subsequent examples. It should be emphasized, however, that the results are not dependent upon the solution approach – either of the other two approaches presented above yield the same conclusions.

The resistance in the time constant of equation (7.27) can be more generally defined as the equivalent overall resistance of the circuit as seen by the capacitor. Thus, if we remove the capacitor from the circuit and create a Thevenin equivalent resistance as seen by the capacitor, the time constant will be the product of the capacitance and this equivalent resistance. We illustrate this point with an example.

Determine the voltage $v(t)$ for the circuit below if $v(0) = 5V$.

We will first solve the problem by writing the first order differential equation governing the system. To aid this process we re-draw the circuit as shown below, labeling the current through the capacitor, defining node “A”, and labeling the voltage at node “A”. For simplicity, we label the capacitor as having capacitance “C” in the figure below.

The voltage at node A, $V_A = v + (3 \Omega) C \frac{dv}{dt}$ is obtained by applying KVL around the outer loop of the circuit.

Applying KCL at node A results in:

$$C \frac{dv}{dt} + \frac{V_A}{2\Omega} + \frac{V_A}{2\Omega} = C \frac{dv}{dt} + V_A = 0$$

Which, when substituting $V_A = v+ 3C \frac{dv}{dt}$ results in:

$$4C \frac{dv}{dt} + V_A = 0$$

This can be placed in the “standard form”, $\frac{dv(t)}{dt} + \frac{1}{\tau}v(t) = 0$, by dividing through by 4C:

$$\frac{dv}{dt} + \frac{1}{4C}V_A = 0$$

Thus, the time constant $\tau = 4C = 8 x 10^{-6}$ seconds.

Removing the capacitor and using circuit reduction to determine an equivalent resistance results in the circuit shown to the left below. The parallel combination of two, $2\Omega$ resistors results in an equivalent $1\Omega$ resistor, as shown in the figure to the right below. The resulting series combinations simplifies to a single $4\Omega$ resistance, so $R_{eq} = 4\Omega$.

The circuit time constant can now be determined from $R_{eq}$ and $C$. Thus,

$$\tau = R_{eq}C = (4\Omega)(2x10^{-6}F) = 8x10^{-6} seconds$$

Since the initial capacitor voltage is given to be 5V, $v(t) = 5e^\frac{-t}{8x10^{-6}}=5e^{1.25x10^5t}V$.

The circuits we have considered so far in this chapter contain no sources, the circuits’ initial conditions are given. In general, we will need to determine the initial conditions from a given source and/or switching operation. For example, a conceptual circuit showing how the initial condition for the circuit of Figure 7.8 can be created is shown in Figure 7.11(a). The switch in Figure 7.11(a) has been closed for a long time; thus, just before the switch opens, the voltage across the capacitor is $V_0$. Opening the switch removes the source from the circuit of interest. Since the voltage across a capacitor cannot change suddenly, the capacitor still has voltage $V_0$ immediately after the switch opens. (Mathematically, we say that $v(t=0)=v(t=0^+)=V_0$. Where time $t=0$ is an infinitesimal time before the switch opens, and the time $t=0^+$ is an infinitesimal time after the circuit opens.) Thus, for times $t>0$, the shaded portion of the circuit of Figure 7.11(a) is identical to the circuit of Figure 7.11(b) from the viewpoint of the capacitor voltage. An example of the analysis of an RC circuit with an included source is provided below; note that the circuit being analyzed is still unforced - the source only provides an initial condition.

Consider the circuit shown below. The switch is originally at position A; at time $t = 0$ seconds, the switch moves to position B in the circuit. We wish to determine the capacitor voltage, $v(t)$ for $t > 0$.

Before time $t = 0$, we assume that the switch has been at position A for a long time – all transients have died off, and any voltages and currents in the circuit have become constant. Since the capacitor voltage-current relationship is $i=C\frac{dv}{dt}$, if all parameters are constant the capacitor current is zero and the capacitor looks like an open circuit. Replacing the capacitor with an open circuit, as shown in the figure below, allows us to determine the voltage across the capacitor before the switch moved to position B. It is fairly easy to see that the capacitor voltage is the same as the voltage across the $12\Omega$ resistor. Since no current flows through the $6\Omega$ resistor in the circuit, the voltage across the $12\Omega$ resistor can be determined by a voltage division between the $12\Omega$ resistor and the $4\Omega$ resistor.

$$v(0^-) = 12V \left[ \frac{12\Omega}{12\Omega + 4\Omega} \right] = 9V$$

Since a capacitor cannot change its voltage instantaneously, the capacitor voltage just after the switch moves to position B is $v(0^+) = v(0^=)=9V$, which gives us our initial condition on the capacitor voltage.

The system time constant can be determined from the capacitance and the equivalent resistance seen by the capacitor. The equivalent resistance can be determined by looking into the capacitor terminals with the switch at position B (recall that we are solving the differential equation for $t>0$). The appropriate circuit is shown below. The equivalent resistance consists of a parallel combination of the $12\Omega$ and $6\Omega$ resistors (note that the $4\Omega$ resistor and the voltage source are no longer relevant to the problem – they are isolated from the capacitor after the switch changes to position B). The equivalent resistance is thus:

$$R_{eq} = \frac{(6\Omega)(12\Omega)}{6\Omega+12\Omega}=4\Omega$$

The system time constant is, therefore, $\tau = R_{eq}C = (4\Omega)(25x10^{-3}F) = 100x10^{-3}$ seconds or 0.1 seconds. Appropriate substitution of the initial condition and time constant into equation (7.30) gives:

$$v(t) = 9e^\frac{-t}{0.1} = 9e^{-10t}V$$

• The natural response of an RC circuit describes the capacitor voltage in a circuit consisting only of resistors and a single equivalent capacitance. The circuit is source-free; the response is entirely due to energy initially stored in the capacitor.
• The differential equation for an unforced RC circuit is of the form:

$$\frac{dv(t)}{dt} + \frac{1}{R_{eq}C}v(t) = 0$$

Where $R_{eq}$ is the equivalent resistance “seen” by the capacitor.

• The RC circuit natural response is of the form:

$$v(t)=V_0e^\frac{-t}{\tau}, for t \geq 0$$

Where $V_0$ is the initial voltage across the capacitor and $\tau$ is the circuit time constant.

• The time constant for any first order system can be determined from the differential equation governing the system. If the governing differential equation is written in the form:

$$\frac{dv(t)}{dt} + \frac{1}{\tau}v(t) = 0$$

The time constant $\tau$ can be determined by inspection. Thus, by comparison with the above differential equation for RC circuits, $\tau = R_{eq}C$.

• Alternatively, for the special case of an RC circuit, the time constant can also be determined by:

$$\tau = R_{eq}C$$

Where $R_{eq}$ is the equivalent resistance “seen” by the capacitor and C is the capacitance in the circuit. (Notice, that to use this relation, we must accurately identify the circuit as a first order RC circuit before proceeding.)

• The capacitor properties can be useful in determining initial conditions for an RC circuit:
  • Capacitors behave like open-circuits when all circuit parameters are constant, and
  • Capacitor voltages cannot change instantaneously

1. The switch in the circuit below moves from position A to position B at time $t=0$.
   1. Write the differential equation governing $v(t), t>0$.
   2. Determine the time constant of the circuit from the differential equation of part 1.
   3. Use the capacitance and the equivalent resistance seen by the capacitor to check your answer to part 2.
   4. Determine the initial condition on the capacitor voltage, $v(t=0^+)$
   5. Determine $v(t), t>0$.

In this section, we consider source-free circuits containing only resistors and a single inductor – commonly referred to as RL circuits. Like RC circuits, these circuits contain only a single energy storage element – the inductor – and the governing equations for the circuits will be first order differential equations. Since the circuits are source-free, no input is applied to the system and the governing differential equation will be homogeneous; the response of the circuit is due entirely to any energy initially stored in the inductor. We will thus be concerned with the natural response of RL circuits.

We will base our discussion of RL circuit natural responses on the series resistor-inductor circuit shown in Figure 7.12. We assume that the inductor has some initial current, $I_0$, flowing through it at time $t = 0$ (so that $i(0) = I_0$). We will determine the inductor current, $i(t)$, for $t>0$.