Tài liệu

First order circuits

Science and Technology

INTRODUCTION

Now that we have considered the three passive elements (resistors, capacitors, and inductors) and one active element (the op amp) individually, we are prepared to consider circuits that contain various combinations of two or three of passive elements. In this chapter, we shall examine two types of simple circuits: a circuit comprising a resistor and capacitor and a circuit comprising a resistor and an inductor. These are called RC and RL circuits, respectively. As simple as these circuits are, they find continual applications in electronics, communications, and control system, as we shall see.

We carry out the analysis of RC and RL circuits by applying Kirchhoff’s laws, as we did for resistive circuits. The only difference is that applying Kirchhoff’s laws to purely resistive circuits, results in algebraic differential equations, which are more difficult to solve than algebraic equations. The differential equations resulting from analyzing RC and RL circuits are of the first order. Hence, the circuits are collectively known as first-order circuits.

A first-order circuit is characterized by a first-order diferential equation.

In addition to there being two types of first-order circuits (RC and RL), there are two ways to excite the circuits. The first way is by called source-free circuits, we assume that energy is initially stored in the capacitive or inductive element. The energy circuit and is gradually dissipated in the resistors. Although source-free circuits are by definition free of independent sources, they may have dependent sources. The second way of exiting first-order circuits is by independent sources. In this chapter, the independent sources we will consider are dc sources. The two types of first-order circuits and the two ways of exiting them and add up to the four possible situations we will study in this chapter.

Finally, we consider four typical applications of RC and RL circuits delay and relay circuits, a photoflash unit, and an automobile ignition circuit.

THE SOURCE-FREE RC CIRCUIT

A source-free RC circuit occurs when its dc source is suddenly disconnected. The energy already stored in the capacitor is released to the resistors.

Consider a series combination of a resistor and an initially charged capacitor, as shown in [link]. (The resistor and capacitor may be the equivalent resistance and equivalent capacitance of combination of resistors and capacitors.) Our objective is to determine the circuit response, which, for pedagogic reasons, we assume to be the voltage v(t) across the capacitor. Since the capacitor is initially charged, we can assume that at time t = 0, the initial voltage is

v ( 0 ) = V 0 size 12{v \( 0 \) =V rSub { size 8{0} } } {}

With the corresponding value of the energy stored as

w ( 0 ) = 1 2 CV 0 2 size 12{w \( 0 \) = { {1} over {2} } ital "CV" rSub { size 8{0} } rSup { size 8{2} } } {}

Applying KCL at the top note of the circuit in [link],

i C + i R = 0 size 12{i rSub { size 8{C} } +i rSub { size 8{R} } =0} {}

By definition, iC=Cdv/dt size 12{i rSub { size 8{C} } =C { ital "dv"} slash { ital "dt"} } {} and iR=v/R size 12{i rSub { size 8{R} } = {v} slash {R} } {}. Thus,

C dv dt + v R = 0 size 12{C { { ital "dv"} over { ital "dt"} } + { {v} over {R} } =0} {}

or

dv dt + v RC = 0 size 12{ { { ital "dv"} over { ital "dt"} } + { {v} over { ital "RC"} } =0} {}

This is a first order differential equation, since only the first derivative of v is involved. To solve it, we arrange the terms as

dv v = 1 RC dt size 12{ { { ital "dv"} over {v} } = - { {1} over { ital "RC"} } ital "dt"} {}

integrating both sides, we get

ln v = t RC + ln A size 12{"ln"v= - { {t} over { ital "RC"} } +"ln"A} {}

where ln A is the integration constant. Thus,

ln v A = t RC size 12{"ln" { {v} over {A} } = - { {t} over { ital "RC"} } } {}

Taking powers of e produces

v ( t ) = Ae t / RC size 12{v \( t \) = ital "Ae" rSup { size 8{ - t/ ital "RC"} } } {}

But from the initial conditions, v(0)=A=V0 size 12{v \( 0 \) =A=V rSub { size 8{0} } } {}. Hence,

v ( t ) = V 0 e t / RC size 12{v \( t \) =V rSub { size 8{0} } e rSup { size 8{ - t/ ital "RC"} } } {}

This shows that the voltage response of the RC circuit is an exponential decay of the initial voltage. Since the response is due to the initial energy stored and the physical characteristics of the circuit and not due to some external voltage or current source, it is called the natural response of the circuit.

The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the circuit itself, with no external sources of excitation.

A source free RC circuit.

The natural response is illustrated graphically in [link]. Note that at t = 0, we have the correct initial condition as in [link]. as t increases, the voltage decreases toward zero. The rapidity with which the voltage decreases is expressed in terms of the time constant, denoted by τ size 12{τ} {}, the lower case greek letter tau.

The time constant of a circuit is the time required for the response to decay by a factor of 1/e or 36.6 percent of its initial value.

This implies that at t=τ size 12{t=τ} {}, [link] becomes

V 0 e τ / RC = V 0 e 1 = 0 . 368 V 0 size 12{V rSub { size 8{0} } e rSup { size 8{ { - τ} slash { ital "RC"} } } =V rSub { size 8{0} } e rSup { size 8{ - 1} } =0 "." "368"V rSub { size 8{0} } } {}

or

τ = RC size 12{τ= ital "RC"} {}

In terms of the time constant, [link] can be written as

v ( t ) = V 0 e τ / T size 12{v \( t \) =V rSub { size 8{0} } e rSup { size 8{ { - τ} slash {T} } } } {}

With a calculator it is easy to show that the value of v(t)/V0 size 12{ {v \( t \) } slash {V rSub { size 8{0} } } } {} is as shown in [link]. It is evident from [link] that the voltage v(t) is less than 1 percent of V0 size 12{V rSub { size 8{0} } } {} after 5 τ size 12{τ} {} (five time constants). Thus, it is customary to assume that the capacitor is fully discharged (or charged) after five time constants. In other words, it takes 5 τ size 12{τ} {} for the circuit to reach its final for every time interval of τ size 12{τ} {}, the voltage is reduced by 36.8 percent of its previous value, v(t+τ)=v(t)/e=0.368v(t) size 12{v \( t+τ \) = {v \( t \) } slash {e=0 "." "368"v \( t \) } } {}, regardless of the value of t.

Values of v(t)/V0=eτ/T size 12{ {v \( t \) } slash {V rSub { size 8{0} } =e rSup { size 8{ { - τ} slash {T} } } } } {}

t v ( t ) / V 0 size 12{ {v \( t \) } slash {V rSub { size 8{0} } } } {}
τ size 12{τ} {} 0.36788
2 τ size 12{τ} {} 0.13534
3 τ size 12{τ} {} 0.04979
4 τ size 12{τ} {} 0.01832
5 τ size 12{τ} {} 0.00674
The voltage responce of the RC circuit.

Observe from [link] that the smaller the time constant, the more rapidly the voltage decreases, that is, the faster the response. This is illustrated in [link]. A circuit with a small time constant gives a fast response in that it reaches the steady state (or final state) quickly due to quick dissipation of energy stored, whereas a circuit with a large time constant gives a slow response because it takes longer to reach steady state. At any rate, whether the time constant is small or large, the circuit reaches steady state in five time constants.

Plot of v\Vo = exp(-t/t) for various of the time constant.

With the voltage v(t) in [link], we can find the current iR(t) size 12{i rSub { size 8{R} } \( t \) } {},

i R ( t ) = v ( t ) R = V 0 R e t / τ size 12{i rSub { size 8{R} } \( t \) = { {v \( t \) } over {R} } = { {V rSub { size 8{0} } } over {R} } e rSup { size 8{ - t/τ} } } {}

The power dissipated in the resistor is

p ( t ) = v ( t ) i R ( t ) = V 0 2 R e 2t / τ size 12{p \( t \) =v \( t \) i rSub { size 8{R} } \( t \) = { {V rSub { size 8{0} } rSup { size 8{2} } } over {R} } e rSup { size 8{ - 2t/τ} } } {}

The energy absorbed by the resistor up to time t is

w R ( t ) = 0 t pdt = 0 t V 0 2 R e 2t / τ dt = τV 0 2 2R e 2t / τ 0 t = 1 2 CV 0 2 ( 1 e 2t / τ ) , τ = RC size 12{w rSub { size 8{R} } \( t \) = Int cSub { size 8{0} } cSup { size 8{t} } { ital "pdt"} = Int cSub { size 8{0} } cSup { size 8{t} } { { {V rSub { size 8{0} } rSup { size 8{2} } } over {R} } e rSup { size 8{ - 2t/τ} } ital "dt"} = - { {τV rSub { size 8{0} } rSup { size 8{2} } } over {2R} } e rSup { size 8{ - 2t/τ} } \rline rSub { size 8{0} } rSup { size 8{t} } = { {1} over {2} } ital "CV" rSub { size 8{0} } rSup { size 8{2} } \( 1 - e rSup { size 8{ { - 2t} slash {τ} } } \) ,τ= ital "RC"} {}

Note that as t size 12{t rightarrow infinity } {}, wR()12CV02 size 12{w rSub { size 8{R} } \( infinity \) rightarrow { {1} over {2} } ital "CV" rSub { size 8{0} } rSup { size 8{2} } } {}, which is the same as wC(0) size 12{w rSub { size 8{C} } \( 0 \) } {}, the energy initially stored in the capacitor. The energy that was initially stored in the capacitor is eventually dissipated in the resistor.

In summary:

The key to working with a source-free RC circuit is finding:

  1. The initial voltage v(0)=V0 size 12{v \( 0 \) =V rSub { size 8{0} } } {} across the capacitor.

The time constant τ size 12{τ} {}.

With these two items, we obtain the response as the capacitor voltage vC(t)=v(t)=v(0)et/τ size 12{v rSub { size 8{C} } \( t \) =v \( t \) =v \( 0 \) e rSup { size 8{ { - t} slash {τ} } } } {}. Once the capacitor voltage is first obtained, other variables (capacitor current iC, resistor voltage vR size 12{v rSub { size 8{R} } } {}, and resistor current iR size 12{i rSub { size 8{R} } } {}) can be determined. In finding the time constant τ=RC size 12{τ= ital "RC"} {}, R is often the Thevenin equivalent resistance at the terminals of the capacitor; that is, we take out the capacitor C and find R=RTh size 12{R=R rSub { size 8{ ital "Th"} } } {} at its terminals.

THE SOURCE-FREE RL CIRCUIT

Consider the series connection of a resistor and inductor, as shown in [link]. Our goal is to determine the circuit response, which we will assume to be the current i(t) through the inductor. We select the inductor current as the response in order to take advantage of the idea that the inductor current can not change instantaneously. At t = 0, we assume that the inductor has an initial current I0, or

i ( 0 ) = I 0 size 12{i \( 0 \) =I rSub { size 8{0} } } {}

With the corresponding energy stored in the inductor as

w ( 0 ) = 1 2 LI 0 2 size 12{w \( 0 \) = { {1} over {2} } ital "LI" rSub { size 8{0} } rSup { size 8{2} } } {}

Applying KVL around loop in [link],

v L + v R = 0 size 12{v rSub { size 8{L} } +v rSub { size 8{R} } =0} {}

But vL=Ldi/dt size 12{v rSub { size 8{L} } =L { ital "di"} slash { ital "dt"} } {} and vR=iR size 12{v rSub { size 8{R} } = ital "iR"} {}. Thus,

L di dt + Ri = 0 size 12{L { { ital "di"} over { ital "dt"} } + ital "Ri"=0} {}

Or

di dt + R L i = 0 size 12{ { { ital "di"} over { ital "dt"} } + { {R} over {L} } i=0} {}

Rearranging terms and integrating gives

I 0 i ( t ) di i = 0 t R L dt size 12{ Int cSub { size 8{I rSub { size 6{0} } } } cSup {i \( t \) } { { { ital "di"} over {i} } } size 12{ {}= - Int cSub {0} cSup {t} { { {R} over {L} } ital "dt"} }} {}

ln i / I 0 i ( t ) = Rt L / 0 t -> ln i ( t ) ln I 0 = Rt L + 0 size 12{"ln"i \rline rSub { size 8{I rSub { size 6{0} } } } rSup {i \( t \) } size 12{ {}= - { { ital "Rt"} over {L} } \rline rSub {0} rSup {t} } size 12{ drarrow "ln"i \( t \) - "ln"I rSub {0} } size 12{ {}= - { { ital "Rt"} over {L} } +0}} {}

or

ln i ( t ) I 0 = Rt L size 12{"ln" { {i \( t \) } over {I rSub { size 8{0} } } } = - { { ital "Rt"} over {L} } } {}

Taking the powers of e, we have

i ( t ) = I 0 e Rt / L size 12{i \( t \) =I rSub { size 8{0} } e rSup { size 8{ - ital "Rt"/L} } } {}
The current response of the RL circuit.

This shows that the natural response of the RL circuit is an exponential decay of the initial current. The current response is shown in [link]. It is evident from [link] that the time constant for the RL circuit is

τ = L R size 12{τ= { {L} over {R} } } {}

with τ size 12{τ} {} again having the unit of seconds. Thus, [link] may be written as

i ( t ) = I 0 e t / τ size 12{i \( t \) =I rSub { size 8{0} } e rSup { size 8{ - t/τ} } } {}

With the current in [link], we can find the voltage across the resistor as

v R ( t ) = iR = I 0 Re τ / T size 12{v rSub { size 8{R} } \( t \) = ital "iR"=I rSub { size 8{0} } "Re" rSup { size 8{ { - τ} slash {T} } } } {}

The power dissipated in the resistor is

p = v R i = I 0 2 Re 2t / τ size 12{p=v rSub { size 8{R} } i=I rSub { size 8{0} } rSup { size 8{2} } "Re" rSup { size 8{ - 2t/τ} } } {}
The current response of the RL circuit.

The energy absorbed by the resistor is

w R ( t ) = 0 t pdt = 0 t I 0 2 Re 2t / τ dt = 1 2 τI 0 2 Re 2t / τ / 0 t , τ = L R size 12{w rSub { size 8{R} } \( t \) = Int cSub { size 8{0} } cSup { size 8{t} } { ital "pdt"} = Int cSub { size 8{0} } cSup { size 8{t} } {I rSub { size 8{0} } rSup { size 8{2} } "Re" rSup { size 8{ - 2t/τ} } ital "dt"= - { {1} over {2} } } τI rSub { size 8{0} } rSup { size 8{2} } "Re" rSup { size 8{ - 2t/τ} } \rline rSub { size 8{0} } rSup { size 8{t} } ,τ= { {L} over {R} } } {}

or

w R ( t ) = 1 2 LI 0 2 ( 1 e 2t / τ ) size 12{w rSub { size 8{R} } \( t \) = { {1} over {2} } ital "LI" rSub { size 8{0} } rSup { size 8{2} } \( 1 - e rSup { size 8{ { - 2t} slash {τ} } } \) } {}

Note that as t size 12{t rightarrow infinity } {}, wR()12LI02 size 12{w rSub { size 8{R} } \( infinity \) rightarrow { {1} over {2} } ital "LI" rSub { size 8{0} } rSup { size 8{2} } } {}, which is the same as wL(0) size 12{w rSub { size 8{L} } \( 0 \) } {}, the initial energy stored in the inductor as in [link]. Again, the energy initially stored in the inductor is eventually dissipated in the resistor.

In summary:

The key to working with a source-free RL circuit is to find:

  1. The initial current i ( 0 ) = I 0 size 12{i \( 0 \) =I rSub { size 8{0} } } {} through the inductor.

The time constant τ size 12{τ} {} of the circuit.

With the two terms, we obtain the response as the inductor current iL(t)=i(t)=i(0)eτ/T size 12{i rSub { size 8{L} } \( t \) =i \( t \) =i \( 0 \) e rSup { size 8{ { - τ} slash {T} } } } {}. Once we determine the inductor current iL size 12{i rSub { size 8{L} } } {}, other variables (inductor voltage vL size 12{v rSub { size 8{L} } } {}, resistor voltage vR size 12{v rSub { size 8{R} } } {}, and resistor current iR size 12{i rSub { size 8{R} } } {}) can be obtained. Note that in general, R in [link] is the Thevenin resistance at the terminals of the inductor.

SINGULARITY FUNCTIONS

Before going on with the second half of this chapter, we need to digress and consider some mathematical concepts that will aid our understanding of transient analysis. A basic understanding of singularity functions will help us make sense of the response of first-order circuits to a sudden application of an independent dc voltage or current source.

Singularity functions (also called switching functions) are very useful in circuit analysis. They serve as good approximations to the switching signals that arise in circuits with switching operations. They are helpful in the neat, compact description of some circuit phenomena, especially the step response of RC or RL, circuits to be discussed in the next sections. By definition,

Singularity functions are functions that either are discontinuous or have discontinuous derivatives.

The three most widely used singularity functions in circuit analysis are the unit step, the unit impulse, and the unit ramp functions.

The unit step function u(t) is 0 for negative values of t and 1 for positive values of t.

The unit step function is undefined at t = 0, where it changes abruptly from 0 to 1. It is dimensionless, like other mathematical functions, such as sine and cosine. [link] depicts the unit step function.

It is the same as saying that u(t) is delayed by t0 size 12{t rSub { size 8{0} } } {} seconds, as shown in [link], we simply replace every t by tt0 size 12{t - t rSub { size 8{0} } } {}.

u(t) is advanced by t0 size 12{t rSub { size 8{0} } } {} seconds, as shown in [link]b.

The unit step function.
a) The unit step function delayed by to, b) The unit step advanced by to.

We use the step function to represent an abrupt change in voltage or current, like the changes that occur in the circuits of control systems and digital computers.

We may express in terms of the unit step function as

v ( t ) = V 0 u ( t t 0 ) size 12{v \( t \) =V rSub { size 8{0} } u \( t - t rSub { size 8{0} } \) } {}

If we let t0=0 size 12{t rSub { size 8{0} } =0} {}, then v(t) is simply the step voltage V0u(t) size 12{V rSub { size 8{0} } u \( t \) } {}. A voltage source of V0u(t) size 12{V rSub { size 8{0} } u \( t \) } {} is shown in [link](a); its equivalent circuit is shown in [link](b). It is evident in [link](b) that terminals a-b are short circuited (v = 0) for t < 0 and that v=V0 size 12{v=V rSub { size 8{0} } } {} appears at the terminals for t > 0. Similarly, a current source of I0u(t) size 12{I rSub { size 8{0} } u \( t \) } {} is shown in [link](a), while its equivalent circuit is in [link](b). Notice that for t < 0, there is an open circuit (i = 0), and that i=I0 size 12{i=I rSub { size 8{0} } } {} flows for t > 0.

a) Voltage source of VoU(t), b) Its equivalent circuit.
a) Current source of IoU(t), b) Its equivalent circuit.

The derivative of the unit step function u(t) is the unit impulse function δ(t) size 12{δ \( t \) } {}.

The unit impulse function - also known as the delta function - is shown in [link].

The unit impulse function δ(t) size 12{δ \( t \) } {} is zero everywhere except at t = 0, where it is undefined.

Impulsive currents and voltages occur in electric circuits as a result of switching operations or impulsive sources. Although the unit impulsive function is not physically realizable (just like ideal sources, ideal resistors etc.), it is a very useful mathematical tool.

The unit impulse function.

The unit impulse may be regarded as an applied or resulting shock. It may be visualized as a very short duration pulse of unit area. This may be expressed mathematically as

0 0 + δ ( t ) dt = 1 size 12{ Int cSub { size 8{0 rSup { size 6{ - {}} } } } cSup {0 rSup { size 6{+{}} } } {δ \( t \) ital "dt"} size 12{ {}=1}} {}

Where t=0 size 12{t=0 rSup { size 8{ - {}} } } {} denotes the time just before t = 0 and t=0+ size 12{t=0 rSup { size 8{+{}} } } {} is the time just after t = 0. For this reason, it is customary to write 1 (denoting unit area) beside the arrow that is used to symbolize the unit impulse function. When an impulse function has a strength other than unity, the area of the impulse is equal to its strength. For example, an impulse function 10 δ(t) size 12{δ \( t \) } {} has an area of 10. [link] shows the impulse functions (t+2) size 12{5δ \( t+2 \) } {}, 10 δ(t) size 12{δ \( t \) } {}, and (t3) size 12{ - 4δ \( t - 3 \) } {}.

Three impluse functions.

To illustrate how the impulse function effects to other functions, let us evaluate the integral

a b f ( t ) δ ( t t 0 ) dt size 12{ Int cSub { size 8{a} } cSup { size 8{b} } {f \( t \) δ \( t - t rSub { size 8{0} } \) ital "dt"} } {}

where a<t0<b size 12{a<t rSub { size 8{0} } <b} {}. Since δ(tt0)=0 size 12{δ \( t - t rSub { size 8{0} } \) =0} {} except t=t0 size 12{t=t rSub { size 8{0} } } {}, the integrand is zero except at t0 size 12{t rSub { size 8{0} } } {}. Thus,

a b f ( t ) δ ( t t 0 ) dt = a b f ( t 0 ) δ ( t t 0 ) dt f ( t 0 ) a b f ( t ) δ ( t t 0 ) dt = f ( t 0 ) alignl { stack { size 12{ Int cSub { size 8{a} } cSup { size 8{b} } {f \( t \) } δ \( t - t rSub { size 8{0} } \) ital "dt"= Int cSub { size 8{a} } cSup { size 8{b} } {f \( t rSub { size 8{0} } \) } δ \( t - t rSub { size 8{0} } \) ital "dt"} {} # =f \( t rSub { size 8{0} } \) Int cSub { size 8{a} } cSup { size 8{b} } {f \( t \) δ \( t - t rSub { size 8{0} } \) ital "dt"} =f \( t rSub { size 8{0} } \) {} } } {}

Or

a b f ( t ) δ ( t t 0 ) dt = f ( t 0 ) size 12{ Int cSub { size 8{a} } cSup { size 8{b} } {f \( t \) } δ \( t - t rSub { size 8{0} } \) ital "dt"=f \( t rSub { size 8{0} } \) } {}

This shows that when a function is integrated with the impulse function, we obtain the value of the function at the point where the impulse occurs. This is a highly useful property of the impulse function known as the sampling or sifting property. The special case of [link] is for t0=0 size 12{t rSub { size 8{0} } =0} {}. Then [link] becomes

r ( t ) = t u ( t ) dt = tu ( t ) size 12{r \( t \) = Int cSub { size 8{ - infinity } } cSup { size 8{t} } {u \( t \) ital "dt"} = ital "tu" \( t \) } {}

Integrating the unit step function u(t) results in the unit ramp function r(t); we write

r ( t ) = t u ( t ) dt = tu ( t ) size 12{r \( t \) = Int cSub { size 8{ - infinity } } cSup { size 8{t} } {u \( t \) ital "dt"} = ital "tu" \( t \) } {}

The unit ramp function is zero for negative values of t and has a unit slope for positive values of t.

[link] shows the unit ramp function. In general, a ramp is a function that changes at a constant rate.

The unit ramp function may be delayed or advanced as shown in [link].

The unit ramp function.
The unit ramp function: a) delayed by to, b) advanced by to.

We should keep in mind that the three singularity functions (impulse, step, and ramp) are related by differentiation as

δ ( t ) = du ( t ) dt , u ( t ) = dr ( t ) dt size 12{δ \( t \) = { { ital "du" \( t \) } over { ital "dt"} } ,u \( t \) = { { ital "dr" \( t \) } over { ital "dt"} } } {}

or by integration as

u ( t ) = t δ ( t ) dt , r ( t ) = t u ( t ) dt size 12{u \( t \) = Int cSub { size 8{ - infinity } } cSup { size 8{t} } {δ \( t \) ital "dt"} ,r \( t \) = Int cSub { size 8{ - infinity } } cSup { size 8{t} } {u \( t \) ital "dt"} } {}

although there are many more singularity functions, we are only interested in these three (the impulse function, the unit step function, and the ramp function) at this point.

STEP RESPONSE OF AN RC CIRCUIT

When the dc source of an RC circuit is suddenly applied, the voltage or current source can be modeled as a step function, and the response is known as a step response.

The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source.

The step response is the response of the circuit due to a sudden application of a dc voltage or current source.

An RC circuit with voltage step input.

Consider RC circuit in [link](a) which can be replaced by the circuit in [link](b), where VS size 12{V rSub { size 8{S} } } {}is a constant, dc voltage source. Again, we select the capacitor voltage as the circuit response to be determined. We assume an initial voltage V0 size 12{V rSub { size 8{0} } } {} on the capacitor, although this is not necessary for the step response. Since the voltage of a capacitor cannot change instantaneously,

v ( 0 ) = v ( 0 + ) = V 0 size 12{v \( 0 rSup { size 8{ - {}} } \) =v \( 0 rSup { size 8{+{}} } \) =V rSub { size 8{0} } } {}

where v(0) size 12{v \( 0 rSup { size 8{ - {}} } \) } {} is voltage across the capacitor just before switching and v(0+) size 12{v \( 0 rSup { size 8{+{}} } \) } {} is its voltage immediately after switching. Applying KCL, we have

C dv dt + v V s u ( t ) R = 0 size 12{C { { ital "dv"} over { ital "dt"} } + { {v - V rSub { size 8{s} } u \( t \) } over {R} } =0} {}

or

dv dt + v RC = V s RC u ( t ) size 12{ { { ital "dv"} over { ital "dt"} } + { {v} over { ital "RC"} } = { {V rSub { size 8{s} } } over { ital "RC"} } u \( t \) } {}

where v is the voltage across the capacitor. For t > 0, [link] becomes

dv dt + v RC = V s RC size 12{ { { ital "dv"} over { ital "dt"} } + { {v} over { ital "RC"} } = { {V rSub { size 8{s} } } over { ital "RC"} } } {}

Rearranging terms gives

dv dt = v V s RC size 12{ { { ital "dv"} over { ital "dt"} } = { {v - V rSub { size 8{s} } } over { ital "RC"} } } {}

or

dv v V s = dt RC size 12{ { { ital "dv"} over {v - V rSub { size 8{s} } } } = - { { ital "dt"} over { ital "RC"} } } {}

Integrating both sides and introducing the initial conditions,

ln ( v V s ) / v 0 v t = t RC / 0 t size 12{"ln" \( v - V rSub { size 8{s} } \) \rline rSub { size 8{v rSub { size 6{0} } } } rSup {v left (t right )} size 12{ {}= - { {t} over { ital "RC"} } \rline rSub {0} rSup {t} }} {}

ln ( v ( t ) V s ) ln ( V 0 V s ) = t RC + 0 size 12{"ln" \( v \( t \) - V rSub { size 8{s} } \) - "ln" \( V rSub { size 8{0} } - V rSub { size 8{s} } \) = - { {t} over { ital "RC"} } +0} {}

Or

ln v V s V 0 V s = t RC size 12{"ln" { {v - V rSub { size 8{s} } } over {V rSub { size 8{0} } - V rSub { size 8{s} } } } = - { {t} over { ital "RC"} } } {}

Taking the exponential of both sides

v V s V 0 V s = e t / τ , τ = RC size 12{ { {v - V rSub { size 8{s} } } over {V rSub { size 8{0} } - V rSub { size 8{s} } } } =e rSup { size 8{ - t/τ} } ,τ= ital "RC"} {}

v V s = ( V 0 V s ) e t / τ size 12{v - V rSub { size 8{s} } = \( V rSub { size 8{0} } - V rSub { size 8{s} } \) e rSup { size 8{ - t/τ} } } {}

or

v ( t ) = V s + ( V 0 V s ) e t / τ , t > 0 size 12{v \( t \) =V rSub { size 8{s} } + \( V rSub { size 8{0} } - V"" lSub { size 8{s} } \) e rSup { size 8{ - t/τ} } ,t>0} {}

This is known as the complete response (or total response) of the RC circuit to a sudden application of a dc voltage source, assuming the capacitor is initially charged. The reason for the term “complete” will become evident a little later. Assuming that VS>V0 size 12{V rSub { size 8{S} } >V rSub { size 8{0} } } {}, a plot of v(t0) size 12{v \( t rSub { size 8{0} } \) } {} is shown in [link].

Response of an RC circuit with initially charge capacitor.

If we assume that the capacitor is uncharged initially, we set V0=0 size 12{V rSub { size 8{0} } =0} {} in [link].

We can write alternatively as

v ( t ) = V s ( 1 e t / τ ) u ( t ) size 12{v \( t \) =V rSub { size 8{s} } \( 1 - e rSup { size 8{ - t/τ} } \) u \( t \) } {}

this is the complete step response of the RC circuit when the capacitor is initially uncharged. The current through the capacitor is obtained from [link] using i(t)=Cdv/dt size 12{i \( t \) =C { ital "dv"} slash { ital "dt"} } {}. We get

i ( t ) = C dv dt = C τ V s e t / τ , τ = RC , t > 0 size 12{i \( t \) =C { { ital "dv"} over { ital "dt"} } = { {C} over {τ} } V rSub { size 8{s} } e rSup { size 8{ - t/τ} } ,τ= ital "RC",t>0} {}

or

i ( t ) = V s R e t / τ u ( t ) size 12{i \( t \) = { {V rSub { size 8{s} } } over {R} } e"" lSup { size 8{ - t/τ} } u \( t \) } {}

[link] shows the plot of capacitor voltage v(t) and capacitor current i(t).

Step response of an RC circuit with initially un charged capicitor: a) voltage response, b) current response.

Rather than going through the derivations above, there is a systematic approach – or rather, a short-cut method – for finding the step response of RC or RL circuit. Let us reexamine [link], which is more general than [link]. It is evident that v(t) has two components.

Classically there are two ways of decomposing this into two components. The first is to break it into a “natural response and a forced response” and the second is to break it into a “transient response and a steady-state response”. Starting with natural response and forced response, we write the total or complete response as

Complete response = Natural response (stored energy) + Forced response (independent source)

or

v = v v + v f size 12{v=v rSub { size 8{v} } +v rSub { size 8{f} } } {}

Where

v n = v 0 e τ / T size 12{v rSub { size 8{n} } =v rSub { size 8{0} } e rSup { size 8{ { - τ} slash {T} } } } {}

and

v f = V 0 ( 1 e τ / T ) size 12{v rSub { size 8{f} } =V rSub { size 8{0} } \( 1 - e rSup { size 8{ { - τ} slash {T} } } \) } {}

we are familiar with the natural response vn size 12{v rSub { size 8{n} } } {} of the circuit, as discussed in section 2, vf size 12{v rSub { size 8{f} } } {} is known as forced response because it is produced by the circuit when an external “force” (a voltage source in this case) is applied. It represents what the circuit is forced to do by the input excitation. The natural response eventually dies out along with the transient component of the forced response, leaving only the steady-state component of the forced response.

Another way of looking at the complete response is to break into two components – one temporary and the other permanent, i.e.

Complete response = Transient response (temporary part) + Steady –state response (permanent part)

or

v = v τ + v ss size 12{v=v rSub { size 8{τ} } +v rSub { size 8{ ital "ss"} } } {}

where

v τ = ( V 0 V S ) e τ / T size 12{v rSub { size 8{τ} } = \( V rSub { size 8{0} } - V rSub { size 8{S} } \) e rSup { size 8{ { - τ} slash {T} } } } {}

and

v SS = V S size 12{v rSub { size 8{ ital "SS"} } =V rSub { size 8{S} } } {}

The transient response vτ size 12{v rSub { size 8{τ} } } {} is temporary; it is the portion of the complete response that decays to zero as time approaches infinity. Thus,

The transient response is the circuit’s temporary response that will die out with time.

The steady-state response vSS size 12{v rSub { size 8{ ital "SS"} } } {} is portion of the complete response that remains after the transient response has died out . Thus,

The steady-state response is the behavior of circuit a long time after an external excitation is applied.

The first decomposition of the complete response is in terms of the source of the responses, while the second decomposition is in term of the permanency of the responses. Under certain conditions, the natural response and transient response are the same. The same can be said about the forced response and the steady-state response.

Whichever way we look at it, the complete response in [link] may be written as

v ( t ) = v ( ) + ( v ( 0 ) v ( ) ) e τ / fT size 12{v \( t \) =v \( infinity \) + lline v \( 0 \) - v \( infinity \) rline e rSup { size 8{ { - τ} slash { ital "fT"} } } } {}

where v(0) is the initial voltage at t=0+ size 12{t=0 rSup { size 8{+{}} } } {} and v() size 12{v \( infinity \) } {} is the final steady-state value. Thus, to find the step response of an RC circuit requires three things:

  1. The initial capacitor voltage v(0),
  2. The final capacitor voltage v() size 12{v \( infinity \) } {},
  3. The time constant τ size 12{τ} {},

We obtain item 1 from the given circuit for t < 0 and items 2 and 3 from the circuit for t > 0. Once these items are determined, we obtain the response using [link]. This technique equally applies to RL circuits, as we shall see in the next section.

Note that if the switch changes position at time t=t0 size 12{t=t rSub { size 8{0} } } {} instead of at t = 0, there is a time delay in the response so that becomes

v ( t ) = v ( ) + { v ( t 0 ) v ( ) } e ( τ τ 0 ) / fT size 12{v \( t \) =v \( infinity \) + lline v \( t rSub { size 8{0} } \) - v \( infinity \) rline e rSup { size 8{ { - \( τ - τ rSub { size 6{0} } \) } slash { ital "fT"} } } } {}

where v(t0) size 12{v \( t rSub { size 8{0} } \) } {} is initial value at t=t0+ size 12{t=t rSub { size 8{0} } rSup { size 8{+{}} } } {}. Keep in mind that [link] or [link] applies only to step response, that is, when the input excitation is constant.

STEP RESPONSE OF AN RL CIRCUIT

Consider the RL circuit in [link](a), which may be replaced by the circuit in [link](b). Again, our goal is to find the inductor current i as the circuit response. Rather than apply Kirchhoff’s laws, we will use the simple technique in [link] through [link]. Let the response be the sum of the natural current and the forced current,

i = i n + i f size 12{i=i rSub { size 8{n} } +i rSub { size 8{f} } } {}

We know that the natural response is always a decaying exponential, that is,

in=Aeτ/T size 12{i rSub { size 8{n} } = ital "Ae" rSup { size 8{ { - τ} slash {T} } } } {}

and

τ=LR size 12{τ= { {L} over {R} } } {}

where A is a constant to be determined.

The forced response is the value of the current a long time after the switch in [link](a) is closed. We know that the natural response essentially dies out after five time constants. At that time, the inductor becomes a short circuit, and the voltage across it is zero. The entire source voltage V, appears across R. Thus, the forced response is

i f = V s R size 12{i rSub { size 8{f} } = { {V rSub { size 8{s} } } over {R} } } {}

Substituting [link] and [link] into [link] gives

i = Ae t / τ + V s R size 12{i= ital "Ae" rSup { size 8{ - t/τ} } + { {V rSub { size 8{s} } } over {R} } } {}
An RL circuit with a step input voltage.

We now determine the constant A from the initial value of i. Let I0 size 12{I rSub { size 8{0} } } {} be the initial current through the inductor, which may come from a source other than VS size 12{V rSub { size 8{S} } } {}. since the current through the inductor cannot change instantaneously,

i ( 0 + ) = i ( 0 ) = I 0 size 12{i \( 0 rSup { size 8{+{}} } \) =i \( 0 rSup { size 8{ - {}} } \) =I rSub { size 8{0} } } {}

Thus at t = 0, [link] becomes

I 0 = A + V s R size 12{I rSub { size 8{0} } =A+ { {V rSub { size 8{s} } } over {R} } } {}

From this, we obtain A as

A = I 0 V s R size 12{A=I rSub { size 8{0} } - { {V rSub { size 8{s} } } over {R} } } {}

Substituting for A in [link], we get

i ( t ) = V s R + ( I 0 V s R ) e t / τ size 12{i \( t \) = { {V rSub { size 8{s} } } over {R} } + \( I rSub { size 8{0} } - { {V rSub { size 8{s} } } over {R} } \) e rSup { size 8{ - t/τ} } } {}

This is complete response of the RL circuit. It is illustrated in [link]. The response in [link] may be written as

i ( t ) = i ( ) + [ i ( 0 ) i ( ) ] e τ / T size 12{i \( t \) =i \( infinity \) + lline i \( 0 \) - i \( infinity \) rline e rSup { size 8{ { - τ} slash {T} } } } {}

Where i(0) and i() size 12{i \( infinity \) } {} are the initial and final values of i. Thus, to find the step response of an RL, circuit requires three things:

  1. The initial the inductor current i(0) at t=0+ size 12{t=0 rSup { size 8{+{}} } } {},
  2. The final inductor current i() size 12{i \( infinity \) } {}
  3. The time constant τ size 12{τ} {}

We obtain 1 from the given circuit for t < 0 and items 2 and 3 from the circuit for t > 0. Once these items are determined, we obtain the response using [link]. Keep in mind that this technique applies only for step responses.

Total response of the RL circuit with initial inductor current Io.

Again, if the switching takes place at time t=t0 size 12{t=t rSub { size 8{0} } } {}, instead of t = 0, [link] becomes

i ( t ) = i ( ) + [ i ( t 0 ) i ( ) ] e ( τ t 0 ) / T size 12{i \( t \) =i \( infinity \) + lline i \( t rSub { size 8{0} } \) - i \( infinity \) rline e rSup { size 8{ { - \( τ - t rSub { size 6{0} } \) } slash {T} } } } {}

If I0=0 size 12{I rSub { size 8{0} } =0} {}, then

i ( t ) = V s R ( 1 e t / τ ) u ( t ) size 12{i \( t \) = { {V rSub { size 8{s} } } over {R} } \( 1 - e rSup { size 8{ - t/τ} } \) u \( t \) } {}

This is the step response of the RL circuit with no initial inductor current. The voltage across the inductor is obtained from [link] using v=Ldi/dt size 12{v=L { ital "di"} slash { ital "dt"} } {}. We get

v ( t ) = L di dt = V s L τR e t / τ , τ = L R , t > 0 size 12{v \( t \) =L { { ital "di"} over { ital "dt"} } =V rSub { size 8{s} } { {L} over {τR} } e rSup { size 8{ - t/τ} } ,τ= { {L} over {R} } ,t>0} {}

or

v ( t ) = V s e t / τ u ( t ) size 12{v \( t \) =V rSub { size 8{s} } e rSup { size 8{ - t/τ} } u \( t \) } {}

[link] shows the step response in [link] and [link].

Step response of an RL circuit with no initial inductor current : a)current response, b) voltage response.

SUMMARY

  1. The analysis in this chapter is applicable to any circuit that can be reduced to an equivalent circuit comprising a resistor and a single energy storage element (inductor or capacitor). Such a circuit is the first-order because its behavior is described by a first-order differential equation. When analyzing RC and RL circuits, one must always keep in mind that the capacitor is an open circuit to steady-state dc conditions while the inductor is a short circuit to steady-state dc conditions.
  2. The natural response is obtained when no independent source is present. It has the general form

x ( t ) = x 0 e t / τ size 12{x \( t \) =x left (0 right )e rSup { size 8{ - t/τ} } } {}

where x represents current through (or voltage across) a resistor, a capacitor, or an inductor, and x(0) is the initial value of x. because most practical always have losses, the natural response is a transient response, i.e. it dies out with time.

  1. The time constant τ is the time required for a response to decay to 1/e of its initial value. For RC circuits, τ = RC and for RL circuits, τ=L/R size 12{τ= {L} slash {R} } {}.
  2. The singularity functions include the unit step, the unit ramp function, and the unit impulse functions.

The unit impulse function is

  1. The steady-state response is the behavior of the circuit after an independent source has been applied for a long time. The transient response is the component of the complete response that dies out with time.
  2. The total or complete response consists of the steady-state response and the transient response.
  3. The step response is the response of the circuit to a sudden application of a dc current or voltage. Finding the step response of a first-order circuit requires the initial value x(0+) size 12{x \( 0 rSup { size 8{+{}} } \) } {}, the final value x() size 12{x \( infinity \) } {}, and the time constant τ size 12{τ} {} . with these three items, we obtain the step response as

x ( t ) = x ( ) + ( x ( 0 + ) x ( ) e τ / T ) size 12{x \( t \) =x \( infinity \) + lline x \( 0 rSup { size 8{+{}} } \) - x \( infinity \) e rSup { size 8{ { - τ} slash {T} } } rline } {}

A more general form of this equation is

x ( t ) = x ( ) + [ x ( t o + ) x ( ) e ( τ τ 0 ) / T ] size 12{x \( t \) =x \( infinity \) + lline x \( t rSub { size 8{o} } rSup { size 8{+{}} } \) - x \( infinity \) e rSup { size 8{ { - \( τ - τ rSub { size 6{0} } \) } slash {T} } } rline } {}

Instantaneous value = final + {initial – final} ett0/τ size 12{e rSup { size 8{ - left (t - t rSub { size 6{0} } right )/τ} } } {}.

Đánh giá:
0 dựa trên 0 đánh giá

Tuyển tập sử dụng module này

Nội dung cùng tác giả
 
Nội dung tương tự