In previous chapter we considered circuits with a single storage element (a capacitor or an inductor). Such circuits are first-order because the differential equations describing them are first-order. In this chapter we will consider circuits containing two storage elements. These are known as second order circuits because their responses are described by differential equations that contain second derivatives.

Typical examples of second-order circuits are RLC circuits, in which the three kinds of passive elements are present. Examples of such circuits are shown in Figure 1(a) and Figure 1(b). Other examples are RC and RL circuits, as shown in Figure 1(c) and Figure 1(d). It is apparent from Figure 1 that a second-order circuit may have two storage elements of different type or the same type (provided elements of the same type cannot be represented by an equivalent single element). An op amp circuit with two storage elements may also be a second-order circuit. As with first-order circuits, a second-order circuit may contain several resistors and dependent and independent sources.

Our analysis of second-order circuits will be similar to that used for first-order. We will first consider circuits that are excited by the initial conditions of the storage elements. Although these circuits may contain dependent sources, they are free of independent sources. These source-free circuits will give natural responses as expected. Later we will consider circuits that are excited by independent sources. These circuits will give both the transient response and the steady-state response. We consider only dc independent sources in this chapter.

We begin by learning how to obtain the initial conditions for the circuit variables and their derivatives, as this is crucial to analyzing second-order circuits. Then we consider series and parallel RLC circuits such as shown in Fig 1 for the two cases of excitation: by initial conditions of the energy storage elements and by step inputs. Later we examine other types of second-order circuits.

An understanding of the natural response of the series RLC circuit is a necessary background for future studies in filter design and communications networks.

Consider the series RLC circuit shown in Figure 2. The circuit is being excited by the energy initially stored in the capacitor and inductor. The energy is represented by the initial capacitor voltage V0 and initial inductor current I0. Thus, at t = 0,

Applying KVL around the loop in Figure 1,

To eliminate the integral, we differentiate with respect to t and rearrange terms. We get

This is a second-order differential equation and is the reason for calling the RLC circuits in this chapter second-order circuits. Our goal is to solve Equation 4. To solve such a second-order differential equation requires that we have two initial conditions, such as the initial value of i and its first derivative or initial values of some i and v. The initial value of i is given in Equation 2. We get the initial value of the derivative of I from Equation 1 and Equation 3; that is,

Ri ( 0 ) + L di ( 0 ) dt + V 0 = 0 Ri ( 0 ) + L di ( 0 ) dt + V 0 = 0 size 12{ ital "Ri" \( 0 \) +L { { ital "di" \( 0 \) } over { ital "dt"} } +V rSub { size 8{0} } =0} {}

Or

with the two initial conditions in Equation 2 and Equation 5, we can now solve Equation 4. Our experience in the preceding chapter on first-order circuits suggests that the solution is of exponential form. So we let

where A and s are constants to be determined. Substituting Equation 7 into Equation 5 and carrying out the necessary differentiations, we obtain

As 2 e st + AR L se st + A LC e st = 0 As 2 e st + AR L se st + A LC e st = 0 size 12{ ital "As" rSup { size 8{2} } e rSup { size 8{ ital "st"} } + { { ital "AR"} over {L} } ital "se" rSup { size 8{ ital "st"} } + { {A} over { ital "LC"} } e rSup { size 8{ ital "st"} } =0} {}

Or

Since i=Aesti=Aest size 12{i= ital "Ae" rSup { size 8{ ital "st"} } } {} is assumed solution we are trying to find, only the expression in parentheses can be zero:

This quadratic equation is known as the characteristic equation of the differential Equation 4, since the roots of the equation dictate the character of i. The two roots of Equation 8 are

A more compact way of expressing the roots is

Where

The roots s1 and s2 are called natural frequencies, measured in nepers per second (Np/s), because they are associated with the natural response of the circuit; ω0 is known as the resonant frequency or strictly as the undamped natural frequency or the damping factor, expressed in nepers per second. In terms of α and ω0, Equation 8 can be written as

The variables s and ω0ω0 size 12{ω rSub { size 8{0} } } {} are important quantities we will be discussing throughout the rest of the text.

The two values of s in Equation 11 indicate that there are two possible solutions for I, each of which is of the form of the assumed solution in Equation 7; that is,

and

i 2 = A 2 e s 2 t i 2 = A 2 e s 2 t size 12{i rSub { size 8{2} } =A"" lSub { size 8{2} } e rSup { size 8{s rSub { size 6{2} } t} } } {}

Since Equation 4 is a linear equation, any linear combination of the two distinct solutions i1 and i2 is also a solution of Equation 4. A complete or total solution of Equation 4 would therefore require a linear combination of i1i1 size 12{i rSub { size 8{1} } } {} and i2i2 size 12{i rSub { size 8{2} } } {}. Thus, the natural response of the series RLC circuit is

Where the constants A1 and A2 are determined from the initial values i(0) and di(0)/dt in Equation 2 and Equation 5.

From Equation 11, we can infer that there are three types of solutions:

1. If α>ω0α>ω0 size 12{α>ω rSub { size 8{0} } } {}, we have the overdamped case.

2. If α=ω0α=ω0 size 12{α=ω rSub { size 8{0} } } {}, we have the critically damped case.

3. If α<ω0α<ω0 size 12{α<ω rSub { size 8{0} } } {}, we have the underdamped case.

We will consider each of these cases separately.

From Equation 9 and Equation 11, α>ω0α>ω0 size 12{α>ω rSub { size 8{0} } } {} implies C>4L/R2C>4L/R2 size 12{C> {4L} slash {R rSup { size 8{2} } } } {}. When this happens, both roots s1s1 size 12{s rSub { size 8{1} } } {} and s2s2 size 12{s rSub { size 8{2} } } {} are negative and real. The response is

which decays and approaches zero as t increases. Figure 3(a) illustrates a typical overdamped response.

When α=ω0α=ω0 size 12{α=ω rSub { size 8{0} } } {}, C=4L/R2C=4L/R2 size 12{C= {4L} slash {R rSup { size 8{2} } } } {} and

For this case, Equation 15 yields

i ( t ) = A 1 e − αt + A 2 e − αt = A 3 e − αt i ( t ) = A 1 e − αt + A 2 e − αt = A 3 e − αt size 12{i \( t \) =A rSub { size 8{1} } e rSup { size 8{ - αt} } +A rSub { size 8{2} } e rSup { size 8{ - αt} } =A rSub { size 8{3} } e rSup { size 8{ - αt} } } {}

Where A3=A1+A2A3=A1+A2 size 12{A rSub { size 8{3} } =A rSub { size 8{1} } +A rSub { size 8{2} } } {}. This cannot be the solution, because the two initial conditions cannot be satisfied with single constant. What then could be wrong? Our assumption of an exponential solution is incorrect for the special case of critical damping. Let us go back to Equation 4. When α=ω0=R/2Lα=ω0=R/2L size 12{α=ω rSub { size 8{0} } = {R} slash {2L} } {}, Equation 4 becomes

d 2 i dt 2 + 2α di dt + α 2 i = 0 d 2 i dt 2 + 2α di dt + α 2 i = 0 size 12{ { {d rSup { size 8{2} } i} over { ital "dt" rSup { size 8{2} } } } +2α { { ital "di"} over { ital "dt"} } +α rSup { size 8{2} } i=0} {}

or

If we let

then Equation 18 becomes

df dt + αf = 0 df dt + αf = 0 size 12{ { { ital "df"} over { ital "dt"} } +αf=0} {}

which is a first-order differential equation with solution f=A1e−αtf=A1e−αt size 12{f=A rSub { size 8{1} } e rSup { size 8{ - αt} } } {}, where A1A1 size 12{A rSub { size 8{1} } } {} is a constant. Equation 19 then becomes

e αt di dt + e αt αi = A 1 e αt di dt + e αt αi = A 1 size 12{e rSup { size 8{αt} } { { ital "di"} over { ital "dt"} } +e rSup { size 8{αt} } αi=A"" lSub { size 8{1} } } {}

or