TimeDomain Circuit

1
Time-Domain Circuit
This course is concerned with the analysis of
networks (which are composed of circuit elements)
and their responses to time-varying stimuli. So
far weve examined the basic waveforms of which
stimuli and responses are composed, and weve
examined the basic elements of which networks are
composed. Now its time to start putting them
together. In this section, well begin to
consider networks containing more than one type
of circuit element, and well examine the
behavior of such networks in the time-domain
under both DC and transient (switched) stimuli.
2
Time-Domain Circuit
Analysis of a network which is purely resistive
or not purely resistive (i.e., a network which
contains inductors and/or capacitors) is done by
writing a system of mesh or node equations for
the network. When energy-storage elements
(capacitors and inductors) are present, these are
first-order or second-order differential
equations.
3
Network Equations
Consider the single-mesh circuit below, which is
stimulated by the Voltage source e(t). The mesh
current is i(t), and our previous study of the
basic circuit elements tells us that the Voltages
across the resistor and the inductor are
The current-Voltage relationship for the
capacitor is
vR(t)
-


R
vC(t)
C
or
-
e(t)
-

L
vL(t)
-
4
Network Equations
Now, apply KVL around the mesh
This is an integrodifferential equation, because
it contains an integral term and a differential
term. We can turn it into a differential equation
vR(t)
differential equation by differentiating both
sides to get rid of the integral term
-


R
vC(t)
C
-
e(t)
-

L
vL(t)
-
5
Network Equations
Rearranging things a bit,
Which is of the form
vR(t)
-

An equation of this form is an nth order,
ordinary, linear differential equation with
constant coefficients. In our equation, n 2,
so its a second-order equation.

R
vC(t)
C
-
e(t)
-

L
vL(t)
-
6
Network Equations
If the equation or equations describing a network
are linear, then the network is linear, which
means the superpostion theorem applies to it. To
analyze the response of a network to some
stimulus, we must solve the network equations.
This may be done in the time domain (you saw an
example of this in EET152/EET207), or in the
transform domain.
Solving network equations using Laplace transform
techniques has many advantages. Well study
those techniques soon, but first were going to
examine the time-domain approach.
vR(t)
-


R
vC(t)
C
-
e(t)
-

L
vL(t)
-
7
Transient Response
Consider the RC network shown below. We want to
find the response vc(t). First, lets find the
network equation
vR(t)
-

This is a simple first-order differential
equation.
R
C

u(t)
-
vC(t)
-
8
Transient Response
A simple first-order differential equation like
this is solved by making an educated guess as
to the form of the solution, then finding the
constants that make it fit. If u(t) is an ideal
step, then for t gt 0,
vR(t)
-

Our educated guess is
R
C

u(t)
-
vC(t)
-
9
Transient Response
Lets check our guess
vR(t)
-

R
C

u(t)
-
vC(t)
Our guess is correct.
-
10
Transient Response
So
So the solution is correct. Now, lets plot the
solution. Well let R 1W and C 1F, to
simplify calculations. Also, remember that the
time constant t RC.
vR(t)
-

R
C

u(t)
-
vC(t)
-
11
Transient Response
Now, lets plot the solution. Well let R 1W
and C 1F, to simplify calculations. Also,
remember that the time constant t RC. Note
that, before the step is applied at t 0, both
i(t) and vC(t) are stable at 0. This is the
initial condition. Also note that, as t
approaches 5t,
i(t) approaches a stable value of 0 A, and
vC(t) approaches a stable value of 1 V. When
the circuit Voltages and currents approach a
stable condition after the application of a
transient stimulus, they are approaching their
steady-state values.
12
Transient Response
When the step is applied, the current immediately
jumps to 1 A, but then gradually returns to zero
(steady state) Also, vC(t) approaches a
steady-state value of 1V. The portion of the
plot between the stable initial condition and the
steady-state condition is called the transient
response. The steady-state portion of the plot
is the steady-state response.
13
RC Network and Feedback
Now lets look at the response of this circuit in
a slightly different way. The capacitor is
initially uncharged, so vC(t) 0 for t lt 0. The
capacitors voltage will only rise above zero if
current flows into it through the resistor. The
current which may flow into it depends on vR(t) ,
but
When the step occurs at t 0, the voltage on the
capacitor is zero, so the current flowing into it
is large. In fact, the current will never be
larger. As current flows into the capacitor, vC
rises. This means that vR drops and so does
the current! The rate at which vC increases is
proportional to the current, so the closer vC
approaches its steady-state value, the more
slowly it approaches its steady-state value.
vR(t)
-

R
C

u(t)
-
vC(t)
-
14
RC Network and Feedback
This RC circuit could be used to model an
everyday event, such as leaving a stoplight when
it changes from red to green. If the speed limit
is 45 mph and the light turns green when t 0,
one might say the desired speed of the car is a
step function
The car cant accelerate instantly (which would
require infinite torque, more than even a vintage
GTO would have), just as the voltage across a
capacitor cant change instantly (which would
require infinite current).
vR(t)
Further, to avoid encounters with the long arm of
the law, the driver must monitor the speed and
make sure he backs off the gas as the car
approaches 45 mph just as the current is
reduced as the capacitor voltage approaches
steady-state!
-

R
C

u(t)
-
vC(t)
-
15
RC Network and Feedback
The accelerating car and the charging capacitor
can both be modeled as a negative feedback loop.
Consider the system diagram shown below
vC(t)
vR(t)
i(t)

S
u(t)
(1/C) ?idt
1/R
-
For the car, the summing junction is the driver.
The driver compares the speedometer reading to
the speed limit (the steady-state speed he wants
to achieve) and develops an error signal (his
brain saying to himself Im now only 5 mph below
the speed limit, better tell the foot to ease off
the gas). The engine is the 1/R gain block,
converting the error signal to torque. The mass
of the car is the integrator, and the speedometer
is the feedback path.
16
RC Network and Feedback
For the RC circuit, the summing junction
represents the resistor. vR(t) is the voltage
across the resistor, which is the diference
between the input
vC(t)
vR(t)
i(t)

S
u(t)
(1/C) ?idt
1/R
-
voltage u(t) and the output vC(t). The 1/R gain
block represents the resistor converting the
error voltage to the charging current, and the
capacitor is represented by the integrator. The
feedback path is the connection between the
capacitor and the resistor. The closer the
output voltage gets to the input, the smaller the
error voltage, which reduces the charge current,
so the output voltage approaches the input less
rapidly.
17
RC Network and Feedback
Lets analyze this system diagram mathematically
(ohhhh, noooooo!!!!!!)
vC(t)
vR(t)
i(t)

S
u(t)
(1/C) ?idt
1/R
-
Obviously,
18
RC Network and Feedback
We would get the same result if the integrator
were simply an integrator (no 1/C), if the gain
of the gain block were changed to 1/RC
vC(t)
vR(t)
i(t)

S
?idt
u(t)
1/RC
-
The relationship between input and output is
still
19
RC Network and Feedback
Fot the sake of generality, we can replace u(t)
with a general input signal vin(t), and replace
vC(t) with a general output, vout(t)
vout(t)
vR(t)
i(t)

S
?idt
vin(t)
1/RC
-
This diagram and equation can be used to model
any feedback loop in which the output is the
integral of the difference between the input and
the output. There are many such feedback loops
in everyday life, and they all behave like RC
networks.
20
RC Network and Feedback
A few slides back, the concepts of transient
response and steady-state response were
introduced. For a system represented by a
feedback
vout(t)
vR(t)
i(t)

S
?idt
vin(t)
1/RC
-
loop, the transient response is the response
during the interval for which the error signal is
significant. For the RC network, this is the
interval during which the voltage across the
resistor is nonzero, and the capacitor voltage is
significantly different from the input voltage.
21
RC Network and Feedback
For the car and driver, the transient response is
the interval from the time the light turns green
until the car reaches the speed limit.
vout(t)
vR(t)
i(t)

S
?idt
vin(t)
1/RC
-
For the car and driver, the steady-state response
begins when the car reaches the speed limit, and
the driver simply has to regulate the speed to
keep from getting pulled over.
22
RC Network and Feedback
For the RC network, the steady-state response
begins when the resistor voltage has dropped to
very close to zero.
vout(t)
vR(t)
i(t)

S
?idt
vin(t)
1/RC
-
when this happens, the charging current is also
nearly zero, and the output voltage is stable.
23
DC Steady-State Circuit
Lets consider the RC network as itself rather
than as a feedback loop for a moment. For a
unit-step excitation, steady-state is reached
when the output voltage is equal to the input
voltage. In this state, vR 0, so no current
flows throught the capacitor. In effect, the
capacitor is an open circuit. In fact, we could
model the steady-state response by simply taking
the capacitor out of the circuit and replacing it
with an open circuit. All currents and voltages
are equal to their steady-state values.
vR(t) 0
vR(t) 0
-
-


R
R
C
C


u(t)
-
vC(t) 1
-
vC(t) 1
-
-
24
DC Steady-State Circuit
If there are more than one capacitor in the
network, replace them all with open circuits and
solve for the steady-state voltages and currents.
Also, if there are inductors, replace them with
short circuits.
vR(t) 0
vR(t) 0
-
-


R
R
C
C


u(t)
-
vC(t) 1
-
vC(t) 1
-
-
25
Initial Circuit
Lets assume that the capacitor in our circuit is
initially uncharged. That is, for t lt 0, vC(t)
0. The capacitor voltage cannot change
instantaneously, because that would require
infinite current. Therefore, for a very brief
time after the input voltage changes from 0 to 1
(due to the unit step), the capacitor voltage is
still zero. In other words, for that very, very,
very brief moment, the capacitor behaves like a
short circuit, and all of the input voltage
appears across the resistor. The initial
voltages and currents can be found by replacing
the capacitor with a short circuit.
vR(t) 0
vR(t) 0
-
-


R
R
C
C


u(t)
-
vC(t) 1
-
vC(t) 1
-
-
26
Initial Circuit
In fact, if the network includes many capacitors,
replace them all with short circuits to solve for
the initial conditions, and replace all inductors
with open circuits. The initial circuit with
capacitors and inductors replaced by short and
open circuits, respectively, is the reverse
(sort of) of the steady-state circuit in which
capacitor and inductors are replaced by open and
short circuits, respectively.
vR(t) 0
vR(t) 0
-
-


R
R
C
C


u(t)
-
vC(t) 1
-
vC(t) 1
-
-
27
Single Time-Constant Circuits
Stanley calls these single time-constant
circuits, but they are really first-order
circuits circuits which can be modeled using a
single first-order differential equation. This
is the type of circuit weve been considering so
far, now we have two names for it Weve
considered an simple RC circuit in some detail,
lets do the same thing for a simple RL circuit.
Well assume that the switch closes at t 0, but
was open for a very long time prior to t 0.
That means that the current is zero at t 0 (an
initial condition).
When the switch closes, KVL tells us
The first term on the right side is the voltage
across the inductor, the second term is the
resistor voltage.
28
Single Time-Constant Circuits
The general solution of this differential
equation is
The general solution is turned into a specific
solution by using the circuits initial
conditions to find the value of the constant K.
29
Single Time-Constant Circuits
We already know the initial condition i(t) 0
at t 0. The switch closes at precisely t 0,
so one could argue that there are two parts of
the instant t 0. We could call the first part,
just before the switch closes, t 0-. We could
then call the second part, just after the switch
closes, t 0. The inductor current cant
change instantaneously, so i(t0-) i(t0) 0.
We can plug this into the general solution
30
Single Time-Constant Circuits
The inductor current for this simple RL circuit
is given by
If we replaced the unit-step source in our RC
circuit with the same battery and switch, the
battery voltage would be
These functions are almost identical, and clearly
behave in the same way.
31
Single Time-Constant Circuits
We can define the time constant t as t RC for
the RC circuit, and t L/R for the RL circuit.
Then,
If we replaced the unit-step source in our RC
circuit with the same battery and switch, the
battery voltage would be
These functions are almost identical, and clearly
behave in the same way.
32
Initial Conditions
The solutions weve used so far were obtained
with initial conditions of zero. For the RL
circuit, the inductor current was zero at t 0
for the RC circuit, the capacitor voltage was
zero at t 0. What if the initial conditions
are nonzero?
In the last chapter, we developed a model for a
capactior with a nonzero initial voltage. Lets
use it, and find a solution for a circuit
containing a capacitor with an initial condition
of E0 volts
33
Initial Conditions
The solutions weve used so far were obtained
with initial conditions of zero. For the RL
circuit, the inductor current was zero at t 0
for the RC circuit, the capacitor voltage was
zero at t 0. What if the initial conditions
are nonzero?
In the last chapter, we developed a model for a
capactior with a nonzero initial voltage. Lets
use it, and find a solution for a circuit
containing a capacitor with an initial condition
of E0 volts
Clearly,
is a solution.
34
Initial Conditions
Earlier in this chapter, we found that we could
find the initial voltages and currents by
replacing capacitors with short circuits. If we
do that here, we find that for t 0,
Substituting this into
results in