EGR 2201 Unit 9First-Order Circuits

Read Alexander & Sadiku, Chapter 7. Homework #9 and Lab #9 due next

week. Quiz next week.

Review: DC Conditions in a Circuit with Inductors or Capacitors

Recall that when power is first applied to a dc circuit with inductors or capacitors, voltages and currents change briefly as the inductors and capacitors become energized.

But once they are fully energized (i.e., “under dc conditions”), all voltages and currents in the circuit have constant values.

To analyze a circuit under dc conditions, replace all capacitors with open circuits and replace all inductors with short circuits.

What About the Time Before DC Conditions?

We also want to be able to analyze such circuits during the time while the voltages and currents are changing, before dc conditions have been reached.

This is sometimes called transient analysis, because the behavior that we’re looking at is short-lived. It’s the focus of Chapters 7 and 8 in the textbook.

The circuits we’ll study in this unit are called first-order circuits because they are described mathematically by first-order differential equations.

We’ll study four kinds of first-order circuits: Source-free RC circuits

Source-free RL circuits

RC circuits with sources

RL circuits with sources

Four Kinds of First-Order Circuits

We use the term natural response to refer to the behavior of source-free circuits.

And we use the term step response to refer to the behavior of circuits in which a source is applied at some time.

So our goal in this unit is to understand the natural response of source-free RC and RL circuits, and to understand the step response of RC and RL circuits with sources.

Natural Response and Step Response

Consider the circuit shown. Assume that at time t=0, the capacitor is charged and has an initial voltage, V0.

As time passes, the initial charge on the capacitor will flow through the resistor,gradually dischargingthe capacitor.

This results in changingvoltage v(t) and currents iC(t) and iR(t), which we wish to calculate.

Natural Response of Source-Free RC Circuit (1 of 2)

Applying KCL,

Therefore

Therefore

Natural Response of Source-Free RC Circuit (2 of 2)

This equation is an example of a first-order differential equation. How do we solve it for v(t)?

Math Detour: Differential Equations Differential equations arise frequently in

science and engineering. Some examples:

The equations above are all called linear ordinary differential equations with constant coefficients.

82 vdt

dv

tvdt

dv

dt

vd365

2

2

tdt

dv

dt

vd

dt

vdsin83

3

3

4

4

A first-order diff. eq.

A second-order diff. eq.

A fourth-order diff. eq.

Solving Differential Equations The differential equations on the previous

slide are quite easy to solve. The ones shown below are more difficult.

In a later math course you’ll learn many techniques for solving such equations.

tvdt

dvsin)7(2 3

2

272 tx

v

t

v

072 vdt

dvt Non-constant coefficient

Non-linear

A partial differential equation

A Closer Look at Our Differential Equation

Note that our equation, , contains two constants, R and C.

It also contains two variables, v and t. Also, t is the independent variable, while v

is the dependent variable. We sometimes indicate this by writing v(t) instead of just v.

Our goal is to write down an equation that expresses v(t) in terms of t, such as:

(But neither of those is right!)

Solving Our Differential Equation To solve our equation, , use a technique

called separation of variables. First, separate the variables v and t:

Then integrate both sides:

Then raise e to both sides:where A is a constant

Apply the Initial Condition At this point we have:

The last step is to note that if we set t equal to 0, we get:

But we assumed earlier that the initial voltage is some value that we called V0. So A must be equal to V0, and therefore:

The Bottom Line I won’t expect you to be able to reproduce

the derivation on the previous slides. The important point is

to realize that whenever we have a circuit like this

the solution for v(t) is:

Graph of Voltage Versus Time

Here’s a graph of

This curve is called a decaying exponential curve.

Note that at first the voltage falls steeply from its initial value (V0). But as time passes, the descent becomes less steep.

The Time Constant The values of R and

C determine how rapidly the voltage descends.

The product RC is given a special name (the time constant) and symbol ():

The greater is, the more slowly the voltage descends.

Don’t Confuse t and Since

and = RC, we can write

Remember that t is a variable (time). But is a constant.

Rules of Thumb After one time

constant (i.e., when t = ), the voltage has fallen to about 36.8% of its initial value.

After five time constants (i.e., when t = 5), the voltage has fallen to about 0.7% of its initial value. For most practical purposes we say that the capacitor is completely discharged and v = 0 after five time constants.

Comparing Different Values of

The greater is, the more slowly the voltage descends, as shown below for a few values of .

Finding Values ofOther Quantities

We’ve seen that From this equation we can use our prior

knowledge to find equations for other quantities, such as current, power, and energy. For example, using Ohm’s law we find that

The Keys to Working witha Source-Free RC Circuit

1. Find the initial voltage across the capacitor.

2. Find the time constant

3. Once you know these two items, voltage as a function of time is:

4. Once you know the voltage, solve for any other circuit variables of interest.

More Complicated Cases A circuit that looks more complicated at

first might be reducible to a simple source-free RC circuit by combining resistors. Example: Here we can combine the three

resistors into a single equivalent resistor, as seen from the capacitor’s terminals.

Where Did V0 Come From?

In previous examples you’ve simply been given the capacitor’s initial voltage, V0.

More realistically, you have to find V0 by considering what happened before t = 0. Example: Suppose you’re told that the switch in

this circuit has been closed for a long time before it’s opened at t = 0. Can you find the capacitor’svoltage V0 at time t = 0?

Consider the circuit shown. Assume that at time t=0, the inductor is energized and has an initial current, I0.

As time passes, the inductor’s energy will gradually dissipateas current flowsthrough the resistor.

This results in changingcurrent i(t) and voltages vL(t) and vR(t), which we wish to calculate.

Natural Response of Source-Free RL Circuit (1 of 2)

Applying KVL,

Therefore

Therefore

Natural Response of Source-Free RL Circuit (2 of 2)

This first-order differential equation is similar to the equation we had for source-free RC circuits.

Solving Our Differential Equation To solve our equation, first separate the

variables i and t:

Then integrate both sides:

Then raise e to both sides:

Apply the Initial Condition At this point we have:

The last step is to note that if we set t equal to 0, we get:

But we assumed earlier that the initial current is some value that we called I0. So A must be equal to I0, and therefore:

The Bottom Line I won’t expect you to be able to reproduce

the derivation on the previous slides. The important point is

to realize that whenever we have a circuit like this

the solution for i(t) is:

The Time Constant The values of R and L determine how

rapidly the current decreases from its initial value to 0.

Recall that for RC circuits we defined the time constant as

For RL circuits, we define it as

The greater is, the more slowly the current decreases from its initial value.

Don’t Confuse t and Since

and = L/R, we can write

Remember that t is a variable (time). But is a constant.

Graph of Current Versus Time

Here’s a graph of

It’s a decaying exponential curve, with the current falling steeply from its initial value (I0). But as time passes, the descent becomes less steep.

Rules of Thumb After one time

constant (i.e., when t = ), the currenthas fallen to about 36.8% of its initial value.

After five time constants (i.e., when t = 5), the current has fallen to about 0.7% of its initial value. For most practical purposes we say that the inductor is completely de-energized and i = 0 after five time constants.

Finding Values ofOther Quantities

We’ve seen that From this equation we can use our prior

knowledge to find equations for other quantities, such as voltage, power, and energy. For example, using Ohm’s law we find that

The Keys to Working witha Source-Free RL Circuit

1. Find the initial current through the inductor.

2. Find the time constant

3. Once you know these two items, current as a function of time is:

4. Once you know the current, solve for any other circuit variables of interest.

More Complicated Cases A circuit that looks more complicated at

first might be reducible to a simple source-free RL circuit by combining resistors. Example: Here we can combine the resistors

into a single equivalent resistor, as seen from the inductor’s terminals.

Where Did I0 Come From?

In previous examples you’ve simply been given the inductor’s initial current, I0.

More realistically, you have to find I0 by considering what happened before t = 0. Example: Suppose you’re told that the switch in

this circuit has been closed for a long time before it’s opened at t = 0. Can you find the inductor’scurrent I0 at time t = 0?

Where We Are We’ve looked at:

Source-free RC circuits Source-free RL circuits

We still need to look at: RC circuits with sources RL circuits with sources

Before doing this, we’ll look at some mathematical functions called singularity functions (or switching functions), which are widely used to model electrical signals that arise during switching operations.

Three Singularity Functions The three singularity functions that

we’ll study are:

The unit step function

The unit impulse function

The unit ramp function

The Unit Step Function The unit step function u(t) is 0 for

negative values of t and 1 for positive values of t:

𝑢 (𝑡 )={0 , 𝑡<01 ,𝑡>0

We can obtain other step functions by shifting the unit step function to the left or right…

…or by multiplying the unit step function by a scaling constant:

Shifting and Scaling the Unit Step Function

𝑢 (𝑡−2 )={0 , 𝑡<2 s1 , 𝑡>2 s

3𝑢 (𝑡 )={0 , 𝑡<03 , 𝑡>0

As with any mathematical function, we can “flip” the unit step function horizontally by replacing t with t:

Flipping the Unit Step Function Horizontally

𝑢 (−𝑡 )={1 , 𝑡<00 , 𝑡>0

By adding two or more step functions we can obtain more complex “step-like” functions, such as the one shown below from the book’s Practice Problem 7.6.

Adding Step Functions

Step functions are useful for modeling sources that are switched on (or off) at some time:

Using Step Functions to Model Switched Sources

The Unit Impulse Function The unit step function’s derivative is the

unit impulse function (t), also called the delta function.

The unit impulse function is 0 everywhere except at t =0, where it is undefined:

It’s useful for modeling “spikes” that can occur during switching operations.

𝛿 (𝑡 )={ 0 , 𝑡<0Undefined , 𝑡=0

0 , 𝑡>0

We can obtain other impulse functions by shifting the unit impulse function to the left or right, or by multiplying the unit impulse function by a scaling constant:

We won’t often use impulse functions in this course.

Shifting and Scaling the Unit Impulse Function

The Unit Ramp Function The unit step function’s integral is the

unit ramp function r(t). The unit ramp function is 0 for negative

values of t and has a slope of 1 for positive values of t:

𝑟 (𝑡 )={0 , 𝑡<0𝑡 , 𝑡≥0

We can obtain other ramp functions by shifting the unit ramp function to the left or right…

…or by multiplying the unit ramp function by a scaling constant:

Shifting and Scaling the Unit Ramp Function

𝑟 (𝑡−0.2 )={0 ,𝑡<0.2 s𝑡 , 𝑡≥0.2 s

4𝑟 (𝑡 )={ 0 ,𝑡<04 𝑡 ,𝑡≥ 0

By adding two or more step functions or ramp functions we can obtain more complex functions, such as the one shown below from the book’s Example 7.7.

Adding Step and Ramp Functions

A circuit’s step response is the circuit’s behavior due to a sudden application of a dc voltage or current source.

We can use a step function to model this sudden application.

Step Response of a Circuit

We distinguish the following two times: t = 0 (the instant just

before the switch closes) t = 0+ (the instant just

after the switch closes)

Since a capacitor’s voltage cannot change abruptly, we know that v(0) = v(0+) in this circuit.

But on the other hand, i(0) i(0+) in this circuit.

t = 0 versus t = 0+

Assume that in the circuit shown, the capacitor’s initial voltage is V0 (which may equal 0 V).

As time passes after the switch closes, the capacitor’s voltagewill gradually approachthe source voltage VS.

This results in changingvoltage v(t) and current i(t), which we wish to calculate.

Step Response of RC Circuit (1 of 2)

Applying KCL, for t>0,

Therefore

Separating variables,

Step Response of RC Circuit (2 of 2)

Integrating both sides,

Solution of Our Differential Equation

Raising e to both sides and rearranging:

Applying the initial condition and letting = RC:

Finally,

𝑣 (𝑡 )={ 𝑉 0 , 𝑡<0

𝑉 𝑆+(𝑉 0−𝑉 𝑆)𝑒− 𝑡𝜏 ,𝑡≥ 0

The Bottom Line I won’t expect you to be able to reproduce

the derivation on the previous slides. The important point is

to realize that for a circuit like this:

the solution for v(t) is:

𝑣 (𝑡 )={ 𝑉 0 , 𝑡<0

𝑉 𝑆+(𝑉 0−𝑉 𝑆)𝑒− 𝑡𝜏 ,𝑡≥ 0

Graph of Voltage Versus Time

Here’s a graph of

(assuming V0< VS). This is a saturating

exponential curve.

Note that the voltage at first rises steeply from its initial value (V0), and then gradually approaches its final value (VS).

𝑣 (𝑡 )={ 𝑉 0 , 𝑡<0

𝑉 𝑆+(𝑉 0−𝑉 𝑆)𝑒− 𝑡𝜏 ,𝑡≥ 0

Rules of Thumb, etc. We can repeat many of the same remarks

as for source-free circuits, such as:The greater is, the more

slowly v(t) approaches its final value.

For most practical purposes, v(t) reaches its final value after 5.

Knowing v(t), we can use Ohm’s law to find current, and we can use other familiar formulas to find power, energy, etc.

Another Way of Looking At It We can rewrite the complete response

as:

Here v(0) is the initial value and v() is the final, or steady-state, value.

This same equation works for source-free RC circuits too, since setting v() to 0 gives .

The Keys to Finding anRC Circuit’s Step Response

1. Find the capacitor’s initial voltage .

2. Find the capacitor’s final voltage .

3. Find the time constant

4. Once you know these items, voltage is:

5. Once you know the voltage, solve for any other circuit variables of interest.

More Complicated Cases A circuit that looks more complicated at

first might be reducible to a simple RC circuit by combining resistors. Example: Here, for t > 0 we can combine the

resistors into a single equivalent resistor, as seen from the capacitor’s terminals.

Two Ways of Breaking It Down

There are two useful ways of looking at the response

1. As the sum of a natural response and a forced response.

2. As the sum of a transient response and a steady-state response. This way is of more interest to us.

Natural Response Versus Forced Response

We can think of the complete response

as being the sum of: 1. A natural response

that depends the capacitor’s initial charge.

2. Plus a forced response that depends on the voltage source.

Transient Response Versus Steady-State Response

We can think of the complete response

as being the sum of: 1. A transient response that dies away as time

passes.

2. Plus a steady-state response that remains after the transient response has died away.

“Under DC Conditions” = Steady-State

Recall that earlier we used the term “under dc conditions” to refer to the time after an RC or RL circuit’s currents and voltages have “settled down” to their final values.

This is just another way of referring to what we’re now calling steady-state values. So way can say that in the steady state,

capacitors look like open circuits and inductors look like short circuits.

Assume that in the circuit shown, the inductor’s initial current is I0 (which may equal 0 A).

As time passes, the inductor’s currentwill gradually approacha steady-state value.

This results in changingcurrent i(t) and voltage v(t), which we wish to calculate.

Step Response of RL Circuit (1 of 2)

Using the same sort of math we used previously for RC circuits, we find

where = L/R, just as for source-free RL circuits.

Step Response of RL Circuit (2 of 2)

𝑖 (𝑡 )={ 𝐼 0 ,𝑡<0𝑉 𝑆

𝑅+(𝐼 0−𝑉 𝑆

𝑅 )𝑒−𝑡𝜏 , 𝑡>0

Another Way of Looking At It

We can rewrite

as:

Here i(0) is the initial value and i() is the final, or steady-state, value.

This same equation works for source-free RL circuits too, since setting i() to 0 gives .

Graph of Voltage Versus Time

Here’s a graph of

(assuming i(0) > i(∞)).

Rules of Thumb, etc. We can repeat many of the same remarks

as for previous circuits, such as:The greater is, the more

slowly i(t) approaches its final value.

For most practical purposes, i(t) reaches its final value after 5.

Knowing i(t), we can use Ohm’s law to find voltage, and we can use other familiar formulas to find power, energy, etc.

The Keys to Finding anRL Circuit’s Step Response

1. Find the inductor’s initial current .2. Find the capacitor’s final current .

3. Find the time constant

4. Once you know these items, current is:

5. Once you know the current, solve for any other circuit variables of interest.

More Complicated Cases A circuit that looks more complicated at

first might be reducible to a simple RL circuit by combining resistors. Example: Here we can combine the resistors

into a single equivalent resistor, as seen from the inductor’s terminals.

A General Approach for First-Order Circuits (1 of 3)

As noted on page 276 of the textbook:Once we know , , and , almost all

the circuit problems in this chapter can be solved using the formula

What is x here? It could be any current or voltage in a first-order circuit.

A General Approach for First-Order Circuits (2 of 3)

So to find x(t) in a first-order circuit, where x could be any current or voltage:

1. Find the quantity’s initial value .2. Find the quantity’s final value .

3. Find the time constant: for an RC circuit. for an RL circuit.

4. Once you know these items, solution is:

A General Approach for First-Order Circuits (3 of 3)

The equation from the previous slide, always graphs as either:

A decaying exponential curve if the initial value x(0) is greater than the final value x().

Or a saturating exponential curve if the initial value x(0) is less than the final value x().