24/05/2018, 20:20

Second order circuits

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 ...

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: a) series RLC circuit, b) parallel RLC circuit, c) Rl circuit, d) RC circuit.

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 [link](a) and [link](b). Other examples are RC and RL circuits, as shown in [link](c) and [link](d). It is apparent from [link] 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.

A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements.

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.

A source-free series RLC circuit.

Consider the series RLC circuit shown in [link]. 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,

v ( 0 ) = 1 C ∫ − ∞ 0 idt = V 0 size 12{v ( 0 ) = { {1} over {C} } Int cSub { size 8{ - infinity } } cSup { size 8{0} } { ital "idt"} =V rSub { size 8{0} } } {}
i ( 0 ) = I 0 size 12{i ( 0 ) =I rSub { size 8{0} } } {}

Applying KVL around the loop in [link],

Ri + L di dt + 1 C ∫ − ∞ t idt = 0 size 12{ ital "Ri"+L { { ital "di"} over { ital "dt"} } + { {1} over {C} } Int cSub { size 8{ - infinity } } cSup { size 8{t} } { ital "idt"} =0} {}

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

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

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 [link]. 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 [link]. We get the initial value of the derivative of I from [link] and [link]; that is,

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

di ( 0 ) dt = − 1 L ( RI 0 + V 0 ) size 12{ { { ital "di" ( 0 ) } over { ital "dt"} } = - { {1} over {L} } ( ital "RI" rSub { size 8{0} } +V rSub { size 8{0} } ) } {}

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

i = Ae st size 12{i= ital "Ae" rSup { size 8{ ital "st"} } } {}

where A and s are constants to be determined. Substituting [link] into [link] and carrying out the necessary differentiations, we obtain

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

Ae st ( s 2 + R L s + 1 LC ) = 0 size 12{ ital "Ae" rSup { size 8{ ital "st"} } ( s rSup { size 8{2} } + { {R} over {L} } s+ { {1} over { ital "LC"} } ) =0} {}

Since i=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:

s 2 + R L s + 1 LC = 0 size 12{s rSup { size 8{2} } + { {R} over {L} } s+ { {1} over { ital "LC"} } =0} {}

This quadratic equation is known as the characteristic equation of the differential [link], since the roots of the equation dictate the character of i. The two roots of [link] are

s 1 = − R 2L + ( R 2L ) 2 − 1 LC size 12{s rSub { size 8{1} } = - { {R} over {2L} } + sqrt { ( { {R} over {2L} } ) rSup { size 8{2} } - { {1} over { ital "LC"} } } } {}
s 2 = − R 2L − ( R 2L ) 2 − 1 LC size 12{s rSub { size 8{2} } = - { {R} over {2L} } - sqrt { ( { {R} over {2L} } ) rSup { size 8{2} } - { {1} over { ital "LC"} } } } {}

A more compact way of expressing the roots is

s 1 = − α + α 2 − ω 0 2 , s 2 = − α − α 2 − ω 0 2 size 12{s rSub { size 8{1} } = - α+ sqrt {α rSup { size 8{2} } - ω rSub { size 8{0} } rSup { size 8{2} } } ,s rSub { size 8{2} } = - α - sqrt {α rSup { size 8{2} } - ω rSub { size 8{0} } rSup { size 8{2} } } } {}

Where

α = R 2L , ω 0 = 1 LC size 12{α= { {R} over {2L} } ,ω rSub { size 8{0} } = { {1} over { sqrt { ital "LC"} } } } {}

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, [link] can be written as

s 2 + 2αs + ω 0 2 = 0 size 12{s rSup { size 8{2} } +2αs+ω rSub { size 8{0} } rSup { size 8{2} } =0} {}

The variables s and ω0 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 [link] indicate that there are two possible solutions for I, each of which is of the form of the assumed solution in [link]; that is,

i 1 = A 1 e s 1 t , size 12{i rSub { size 8{1} } =A rSub { size 8{1} } e rSup { size 8{s rSub { size 6{1} } t} } ,} {}

and

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

i ( t ) = A 1 e s 1 t + A 2 e s 2 t size 12{i ( t ) =A rSub { size 8{1} } e rSup { size 8{s rSub { size 6{1} } t} } +A"" lSub {2} size 12{e rSup {s rSub { size 6{2} } t} }} {}

Where the constants A1 and A2 are determined from the initial values i(0) and di(0)/dt in [link] and [link].

From [link], we can infer that there are three types of solutions:

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

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

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

We will consider each of these cases separately.

Overdamped case ( α>ω0 size 12{α>ω rSub { size 8{0} } } {})

From [link] and [link], α>ω0 size 12{α>ω rSub { size 8{0} } } {} implies C>4L/R2 size 12{C> {4L} slash {R rSup { size 8{2} } } } {}. When this happens, both roots s1 size 12{s rSub { size 8{1} } } {} and s2 size 12{s rSub { size 8{2} } } {} are negative and real. The response is

i t = A 1 e s 1 t + A 2 e s 2 t size 12{i left (t right )=A rSub { size 8{1} } e rSup { size 8{s rSub { size 6{1} } t} } +A rSub {2} size 12{e rSup {s rSub { size 6{2} } t} }} {}

which decays and approaches zero as t increases. [link](a) illustrates a typical overdamped response.

a) Overdamped, b) critically damped response, c) underdamped response.

Critically damped case ( α=ω0 size 12{α=ω rSub { size 8{0} } } {})

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

s 1 = s 2 = − α = − R 2L size 12{s rSub { size 8{1} } =s rSub { size 8{2} } = - α= - { {R} over {2L} } } {}

For this case, [link] yields

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+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 [link]. When α=ω0=R/2L size 12{α=ω rSub { size 8{0} } = {R} slash {2L} } {}, [link] becomes

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

d dt ( di dt + αi ) + α ( di dt + αi ) = 0 size 12{ { {d} over { ital "dt"} } ( { { ital "di"} over { ital "dt"} } +αi ) +α ( { { ital "di"} over { ital "dt"} } +αi ) =0} {}

If we let

f = ( di dt + αi ) size 12{f= ( { { ital "di"} over { ital "dt"} } +αi ) } {}

then [link] becomes

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

which is a first-order differential equation with solution f=A1e−αt size 12{f=A rSub { size 8{1} } e rSup { size 8{ - αt} } } {}, where A1 size 12{A rSub { size 8{1} } } {} is a constant. [link] then becomes

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

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} } } {}

This can be written as

d dt ( e αt i ) = A 1 size 12{ { {d} over { ital "dt"} } ( e rSup { size 8{αt} } i ) =A rSub { size 8{1} } } {}

Integrating both sides yields

e αt i = A 1 t + A 2 size 12{e rSup { size 8{αt} } i=A rSub { size 8{1} } t+A rSub { size 8{2} } } {}

or

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

where A2 size 12{A rSub { size 8{2} } } {} is another constant. Hence, the natural response of the critically damped circuit is a sum of two terms: a negative exponential and a negative exponential multiplied by a linear term, or

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

A typical critically damped response is shown in [link](b). In fact, [link](b) is a sketch of i(t)=te−αt size 12{i ( t ) = ital "te" rSup { size 8{ - αt} } } {}, which reaches a maximum value of e−1/α size 12{ {e rSup { size 8{ - 1} } } slash {α} } {} at t=1/α size 12{t= {1} slash {α} } {}, one time constant, and the decays all the way to zero.

Underdamped case ( α<ω0 size 12{α<ω rSub { size 8{0} } } {})

For α<ω0 size 12{α<ω rSub { size 8{0} } } {}, C<4L/R2 size 12{C< {4L} slash {R rSup { size 8{2} } } } {}. The roots may be written as

s 1 = − α + − ( ω 0 2 − α 2 ) = − α + jω d size 12{s rSub { size 8{1} } = - α+ sqrt { - ( ω rSub { size 8{0} } rSup { size 8{2} } - α rSup { size 8{2} } ) } = - α+jω rSub { size 8{d} } } {}
s 2 = − α − − ( ω 0 2 − α 2 ) = − α − jω d size 12{s rSub { size 8{2} } = - α - sqrt { - ( ω rSub { size 8{0} } rSup { size 8{2} } - α rSup { size 8{2} } ) } = - α - jω rSub { size 8{d} } } {}

Where j=−1 size 12{j= sqrt { - 1} } {}and ωd=ω02−α2 size 12{ω rSub { size 8{d} } = sqrt {ω rSub { size 8{0} } rSup { size 8{2} } - α rSup { size 8{2} } } } {}, which is called the damping frequency. Both ω0 size 12{ω rSub { size 8{0} } } {} and ωd size 12{ω rSub { size 8{d} } } {} are natural frequencies because they help determine the natural response; while

ω0 size 12{ω rSub { size 8{0} } } {} is often called the undamped natural frequency, ωd size 12{ω rSub { size 8{d} } } {} is called the damped natural frequency. The natural response is

i ( t ) = A 1 e − ( α − jω d ) t + A 2 e − ( α + jω d ) t = e − αt ( A 1 e jω d t + A 2 e − jω d t ) alignl { stack { size 12{i ( t ) =A rSub { size 8{1} } e rSup { size 8{ - ( α - jω rSub { size 6{d} } ) t} } +A rSub {2} size 12{e rSup { - ( α+jω rSub { size 6{d} } ) t} }} {} # size 12{ {}=e rSup { size 8{ - αt} } ( A rSub { size 8{1} } e rSup { size 8{jω rSub { size 6{d} } t} } +A rSub {2} size 12{e rSup { - jω rSub { size 6{d} } t} } size 12{ ) }} {} } } {}

Using Euler’s identities,

e jθ = cos θ + j sin θ , e − jθ = cos θ − j sin θ size 12{e rSup { size 8{jθ} } ="cos"θ+j"sin"θ,e rSup { size 8{ - jθ} } ="cos"θ - j"sin"θ} {}

We get

i ( t ) = e − αt [ A 1 ( cos ω d t + j sin ω d t ) + A 2 ( cos ω d t − j sin ω d t ) e − αt [ ( A 1 + A 2 ) cos ω d t + j ( A 1 − A 2 ) sin ω d t ] alignl { stack { size 12{i ( t ) =e rSup { size 8{ - αt} } [ A rSub { size 8{1} } ( "cos"ω rSub { size 8{d} } t+j"sin"ω rSub { size 8{d} } t ) +A rSub { size 8{2} } ( "cos"ω rSub { size 8{d} } t - j"sin"ω rSub { size 8{d} } t ) } {} # =e rSup { size 8{ - αt} } [ ( A rSub { size 8{1} } +A rSub { size 8{2} } ) "cos"ω rSub { size 8{d} } t+j ( A rSub { size 8{1} } - A rSub { size 8{2} } ) "sin"ω rSub { size 8{d} } t ] {} } } {}

Replacing constants (A1+A2) size 12{ ( A rSub { size 8{1} } +A rSub { size 8{2} } ) } {} and j(A1−A2) size 12{j ( A rSub { size 8{1} } - A rSub { size 8{2} } ) } {} with constants B1 size 12{B rSub { size 8{1} } } {} and B2 size 12{B rSub { size 8{2} } } {}, we write

i ( t ) = e − αt [ ( B 1 cos ω d t + j ( B 2 ) sin ω d t ] size 12{i ( t ) =e rSup { size 8{ - αt} } [ ( B rSub { size 8{1} } "cos"ω rSub { size 8{d} } t+j ( B rSub { size 8{2} } ) "sin"ω rSub { size 8{d} } t ] } {}

With the presence of sine and cosine functions, it is clear that the natural response for this case is exponentially damped and oscillatory in nature. The response has a time constant of 1/α size 12{ {1} slash {α} } {} and a period of T=2π/ωd size 12{T= {2π} slash {ω rSub { size 8{d} } } } {}. [link](c) depicts a typical underdamped response. [link] assumes for each case that i(0) = 0.

Once the inductor current i(t) is found for RLC series circuits as shown above, other circuit quantities such as individual element voltages can easily be found. For example, the resistor voltage is vR=Ri size 12{v rSub { size 8{R} } = ital "Ri"} {}, and the inductor voltage is vL=Ldi/dt size 12{v rSub { size 8{L} } =L { ital "di"} slash { ital "dt"} } {}. The inductor current i(t) is selected as the key variable to be determined first in order to take advantage of [link].

We conclude this section by noting the following interesting, peculiar properties of an RLC network:

1. The behavior of such a network is captured by the idea of damping, which is the gradual loss of the initial stored energy, as evidenced by the continuous decrease in the amplitude of the response. The damping effect is due to the presence of the resistance R. The damping factor  determines the rate at which the response is damped. If R = 0, then α=0 size 12{α=0} {}, and we have an LC circuit with 1/LC size 12{ {1} slash { sqrt { ital "LC"} } } {}as the undamped natural frequency. Since α<ω0 size 12{α<ω rSub { size 8{0} } } {} in this case, the response is not only undamped but also oscillatory. The circuit is said to be lossless, because dissipating or damping element (R) is absent. By adjusting the value of R, the response may be made undamped, overdamped, critically damped, or underdamped.

2. Oscillatory response is possible due to the presence of the two types of storage elements. Having both L and C allows the flow of energy back and forth between the two. The damped oscillation exhibited by the underdamped response is known as ringing. It stems from the ability of the storage elements L and C to transfer energy back and forth between them.

3. Observe from [link] that the waveforms of the responses differ. In general, it is difficult to tell from the waveforms the difference between the overdamped and critically damped responses. The critically damped case is the borderline between the underdamped and overdamped cases and it decays the fastest. With the same initial conditions, the overdamped case has the longest settling time, because it takes the longest time to dissipate the initial stored energy. If we desire the the fastest response without oscillation or ringing, the critically damped circuit is right choice.

Parallel RLC circuits find many practical applications, notably in communications networks and filter designs.

Consider the parallel RLC circuit shown in [link]. Assume initial inductor current I0 and initial capacitor voltage V0,

i ( 0 ) = I 0 = 1 L ∫ ∞ 0 v ( t ) dt size 12{i ( 0 ) =I rSub { size 8{0} } = { {1} over {L} } Int cSub { size 8{ infinity } } cSup { size 8{0} } {v ( t ) ital "dt"} } {}
v ( 0 ) = V 0 size 12{v ( 0 ) =V rSub { size 8{0} } } {}

Since the three elements are in parallel, they have the same voltage v across them. According to passive sign convention, the current is entering each element; that is, the current through each element is leaving the top node. Thus, applying KCL at the top node gives

v R + 1 L ∫ − ∞ t vdt + C dv dt = 0 size 12{ { {v} over {R} } + { {1} over {L} } Int cSub { size 8{ - infinity } } cSup { size 8{t} } { ital "vdt"} +C { { ital "dv"} over { ital "dt"} } =0} {}
A source-free parallel RLC.

Taking the derivative with respect to t and dividing by C results in

d 2 v dt 2 + 1 RC dv dt + 1 LC v = 0 size 12{ { {d rSup { size 8{2} } v} over { ital "dt" rSup { size 8{2} } } } + { {1} over { ital "RC"} } { { ital "dv"} over { ital "dt"} } + { {1} over { ital "LC"} } v=0} {}

We obtain the characteristic equation by replacing the first derivative by s and the second derivative by s 2 size 12{s rSup { size 8{2} } + { {1} over { ital "RC"} } s+ { {1} over { ital "LC"} } =0} {} . By following the same reasoning used in establishing [link] through [link], the characteristic equation is obtained as

s 2 + 1 RC s + 1 LC = 0 size 12{s rSup { size 8{2} } + { {1} over { ital "RC"} } s+ { {1} over { ital "LC"} } =0} {}

The roots of the characteristic equations are

s 1,2 = − 1 2 RC ± ( 1 2 RC ) 2 − 1 LC size 12{s rSub { size 8{1,2} } = - { {1} over {2 ital "RC"} } +- sqrt { ( { {1} over {2 ital "RC"} } ) rSup { size 8{2} } - { {1} over { ital "LC"} } } } {}

Or

s 1,2 = − α ± α 2 − ω 0 2 size 12{s rSub { size 8{1,2} } = - α +- sqrt {α rSup { size 8{2} } - ω rSub { size 8{0} } rSup { size 8{2} } } } {}

Where

α = 1 2 RC , ω 0 = 1 LC size 12{α= { {1} over {2 ital "RC"} } ,ω rSub { size 8{0} } = { {1} over { sqrt { ital "LC"} } } } {}

The names of these terms remain the same as in the preceding section, as they play the same role in the solution. Again, there are three possible solutions, depending on whether α>ω0 size 12{α>ω rSub { size 8{0} } } {}, α=ω0 size 12{α=ω rSub { size 8{0} } } {}, or α<ω0 size 12{α<ω rSub { size 8{0} } } {}. Let us consider these cases separately.

Overdamped case ( α>ω0 size 12{α>ω rSub { size 8{0} } } {})

From [link], α>ω0 size 12{α>ω rSub { size 8{0} } } {} when L>4R2C size 12{L>4R rSup { size 8{2} } C} {}. The roots of characteristic equation are real and negative. The response is

v ( t ) = A 1 e s 1 t + A 2 e s 2 t size 12{v ( t ) =A rSub { size 8{1} } e rSup { size 8{s"" lSub { size 6{1} } t} } +A rSub {2} size 12{e rSup {s rSub { size 6{2} } t} }} {}

Critically damped case ( α=ω0 size 12{α=ω rSub { size 8{0} } } {})

For α=ω0 size 12{α=ω rSub { size 8{0} } } {}, L=4R2C size 12{L=4R rSup { size 8{2} } C} {}. The roots are real and equal so that the response is

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

Underdamped case ( α<ω0 size 12{α<ω rSub { size 8{0} } } {})

When α<ω0 size 12{α<ω rSub { size 8{0} } } {}, L<4R2C size 12{L<4R rSup { size 8{2} } C} {}. In this case the roots are complex and may be expressed as

s 1,2 = − α ± jω d size 12{s rSub { size 8{1,2} } = - α +- sqrt {α rSup { size 8{2} } - ω rSub { size 8{0} } rSup { size 8{2} } } } {}

Where

ω d = ω 0 2 − α 2 size 12{ω"" lSub { size 8{d} } = sqrt {ω rSub { size 8{0} } rSup { size 8{2} } - α rSup { size 8{2} } } } {}

The response is

v ( t ) = e
0