We also expect the inductor voltage to approach zero as t increases, because the current in the circuit is approaching the constant value of VJR.. 7.51, note that the initial voltage ac
Trang 1constant after the switch has been closed, the current will have reached approximately 63% of its final value, or
i(T) vt V
R R
s - i 0.6321
If the current were to continue to increase at its initial rate, it would reach its final value at / = T; that is, because
di
dt
VJ-l
R = -lV'A L (7.38)
the initial rate at which i{t) increases is
>"£ (7.39)
If the current were to continue to increase at this rate, the expression for i
would be
from which, at t = T,
K
vs
R
V
0.632 - ^
0
0
/l / -, , V,
_ JY '"> = it
V s
1
4T
1
5T
Figure 7.17 A The step response of the RL circuit
shown in Fig 7.16 when / ( ) = 0
0.368 V,
Equations 7.36 and 7.40 are plotted in Fig 7.17 The values given by Eqs 7.37 and 7.41 are also shown in this figure
The voltage across an inductor is Ldi/dt, so from Eq 7.35, for t > 0+,
v = L -R
L /o R f l - W - (V s - I Q R)e-W L K (7.42)
The voltage across the inductor is zero before the switch is closed
Equation 7.42 indicates that the inductor voltage jumps to V s - l () R at
the instant the switch is closed and then decays exponentially to zero
Does the value of v at t = 0+ make sense? Because the initial current
is /() and the inductor prevents an instantaneous change in current, the current is /0 in the instant after the switch has been closed The voltage
drop across the resistor is I Q R, and the voltage impressed across the
induc-tor is the source voltage minus the voltage drop, that is, V s — I { )R
When the initial inductor current is zero, Eq 7.42 simplifies to
v = V s e- (R/L)t (7.43)
If the initial current is zero, the voltage across the inductor jumps to V s We
also expect the inductor voltage to approach zero as t increases, because the current in the circuit is approaching the constant value of VJR Figure 7.18
shows the plot of Eq 7.43 and the relationship between the time constant
Trang 2If there is an initial current in the inductor, Eq 7.35 gives the solution
for it The algebraic sign of /0 is positive if the initial current is in the same
direction as /'; otherwise, /0 carries a negative sign Example 7.5 illustrates
the application of Eq 7.35 to a specific circuit
| Determining the Step Response of an RL Circuit
The switch in the circuit shown in Fig 7.19 has been
in position a for a long time At t = 0, the switch
moves from position a to position b The switch is a
make-before-break type; that is, the connection at
position b is established before the connection at
position a is broken, so there is no interruption of
current through the inductor
a) Find the expression for /(/•) for t ^ 0
b) What is the initial voltage across the inductor just
after the switch has been moved to position b?
c) How many milliseconds after the switch has been
moved does the inductor voltage equal 24 V?
d) Does this initial voltage make sense in terms of
circuit behavior?
e) Plot both i(t) and v(t) versus t
t = 0
Figure 7.19 A The circuit for Example 7.5
Solution
a) The switch has been in position a for a long time,
so the 200 mH inductor is a short circuit across
the 8 A current source Therefore, the inductor
carries an initial current of 8 A This current is
oriented opposite to the reference direction for i;
thus /0 is - 8 A When the switch is in position b,
the final value of i will be 24/2, or 12 A The time
constant of the circuit is 200/2, or 100 ms
Substituting these values into Eq 7.35 gives
i= 12 + ( - 8 - 12)e~'/01
= 12 - 20e~ A, t > 0
b) The voltage across the inductor is
r di
v = L —
dt
= 0.2(200<T1(")
= 40<Tll)'V, t > 0+ The initial inductor voltage is
v{0 + ) = 40 V
c) Yes; in the instant after the switch has been moved to position b, the inductor sustains a cur-rent of 8 A counterclockwise around the newly formed closed path This current causes a 16 V
drop across the 2 fl resistor This voltage drop
adds to the drop across the source, producing a
40 V drop across the inductor
d) We find the time at which the inductor voltage equals 24 V by solving the expression
24 = A0e~ m
for/:
1 , 40
t=w]nT4
= 51.08 X 10"3
= 51.08 ms
e) Figure 7.20 shows the graphs of i(t) and v{t) versus
t Note that the instant of time when the current
equals zero corresponds to the instant of time when the inductor voltage equals the source volt-age of 24 V, as predicted by Kirchhoff s voltvolt-age law
v(y)i(A)
40
32
16
12
8
4
-4
—
l \
I X
f 100 -8
1
200
1—(-300
_ t
400
i
\
500 t (ms)
Figure 7.20 • The current and voltage waveforms for
Example 7.5
Trang 3I / A S S E S S M E N T PROBLE
Objective 2—Be able to determine the step response of both RL and RC circuits
7.5 Assume that the switch in the circuit shown in
Fig 7.19 has been in position b for a long time,
and at t = 0 it moves to position a Find
(a) /-(0+); (b) v(0+); (c) T ,t > 0; (d) i(t), t > 0;
and (e) v(t), t > 0+
NOTE: Also try Chapter Problems 7.35-737
Answer: (a) 12 A;
(b) - 2 0 0 V;
(c) 20 ms;
(d) - 8 + 20<T50' A, t > 0;
(e) -200<T50f V, t > 0+
We can also describe the voltage v(t) across the inductor in Fig 7.16
directly, not just in terms of the circuit current We begin by noting that the voltage across the resistor is the difference between the source voltage and the inductor voltage We write
Yrt V * V{t)
l{t) = R-lf>
(7.44)
where V s is a constant Differentiating both sides with respect to time yields
(7.45)
di
dt
]_dv
R dt'
Then, if we multiply each side of Eq 7.45 by the inductance L, we get an
expression for the voltage across the inductor on the left-hand side, or
v —
-L dv
Putting Eq 7.46 into standard form yields
dv R
— + — v ~ 0
You should verify (in Problem 7.38) that the solution to Eq 7.47 is identi-cal to that given in Eq 7.42
At this point, a general observation about the step response of an
RL circuit is pertinent (This observation will prove helpful later.) When
we derived the differential equation for the inductor current, we obtained
Eq 7.29 We now rewrite Eq 7.29 as
— + — / = —
Observe that Eqs 7.47 and 7.48 have the same form Specifically, each equates the sum of the first derivative of the variable and a constant times the variable to a constant value In Eq 7.47, the constant on the right-hand side happens to be zero; hence this equation takes on the same form as the natural response equations in Section 7.1 In both Eq 7.47 and Eq 7.48, the constant multiplying the dependent variable is the reciprocal of the
time constant, that is, R/L = 1/r We encounter a similar situation in the derivations for the step response of an RC circuit In Section 7.4, we will
use these observations to develop a general approach to finding the
Trang 4natu-The Step Response of an RC Circuit
We can find the step response of a first-order RC circuit by analyzing the
circuit shown in Fig 7.21 For mathematical convenience, we choose the
Norton equivalent of the network connected to the equivalent capacitor
Summing the currents away from the top node in Fig 7.21 generates the
differential equation
C—- + — = I
(7.49)
Division of Eq 7.49 by C gives
dv c v c _ I s
Figure 7.21 A A circuit used to illustrate the step
response of a first-order RC circuit
Comparing Eq 7.50 with Eq 7.48 reveals that the form of the solution for
Vc is the same as that for the current in the inductive circuit, namely,
Eq 7.35 Therefore, by simply substituting the appropriate variables and
coefficients, we can write the solution for i?c directly The translation
requires that l s replace Vg, C replace L, 1/JR replace R, and V Q replace /()
We get
v c = I.R + (V 0 - I s R)e-'/ RC , t > 0 (7.51) -^ Step response of an RC circuit
A similar derivation for the current in the capacitor yields the differential
equation
di 1
(7.52)
Equation 7.52 has the same form as Eq 7.47, hence the solution for i is
obtained by using the same translations used for the solution of
Eq 7.50 Thus
where V () is the initial value of v c , the voltage across the capacitor
We obtained Eqs 7.51 and 7.53 by using a mathematical analogy to
the solution for the step response of the inductive circuit Let's see
whether these solutions for the RC circuit make sense in terms of
known circuit behavior From Eq 7.51, note that the initial voltage
across the capacitor is V {h the final voltage across the capacitor is I S R,
and the time constant of the circuit is RC Also note that the solution
for Vc is valid for t 2: 0 These observations are consistent with the
behavior of a capacitor in parallel with a resistor when driven by a
con-stant current source
Equation 7.53 predicts that the current in the capacitor at t = 0+ is
h ~ K)/^- This prediction makes sense because the capacitor voltage
can-not change instantaneously, and therefore the initial current in the resistor
is VQ/R The capacitor branch current changes instantaneously from zero
at t = 0~ to I s - V()/R at t = 0+ The capacitor current is zero at t = oo
Also note that the final value of v = I S R
Example 7.6 illustrates how to use Eqs 7.51 and 7.53 to find the step
response of a first-order RC circuit
Trang 5Example 7.6 Determining the Step Response of an RC Circuit
The switch in the circuit shown in Fig 7.22 has been
in position 1 for a long time At t = 0, the switch
moves to position 2 Find
a) v 0 (t) for t > 0 and
b) i a {t) for t > 0+
The value of the Norton current source is the ratio of the open-circuit voltage to the Thevenin resistance, or -60/(40 X 103) = - 1 5 mA The resulting Norton equivalent circuit is shown in
Fig 7.23 From Fig 7.23, 1 K R = - 6 0 V and
RC = \0ms We have already noted that
v o (0) = 30 V, so the solution for v 0 is
-t Uv t = 0
/4^-VvV—•-40 V %60kfi
0.25 jaFPp: »„
X
160kO ( ^ )75 V
Figure 7.22 • The circuit for Example 7.6
v 0 = - 6 0 + [30 - (-60)]e~ mu
= - 6 0 + 90e~1(,(" V, t > 0
b) We write the solution for i„ directly from
Eq 7.53 by noting that 7V= - 1 5 mA and
VJR = (30/40) X 10"3, or 0.75 mA:
i a = -2.25e", u u' mA, t > 0+
Solution
a) The switch has been in position 1 for a long time,
so the initial value of v 0 is 40(60/80), or 30 V To
take advantage of Eqs 7.51 and 7.53, we find the
Norton equivalent with respect to the terminals
of the capacitor for t > 0 To do this, we begin by
computing the open-circuit voltage, which is
given by the - 7 5 V source divided across the
40 kfl and 160 ktt resistors:
10-1
(40 + 160) X 10 j (-75) = - 6 0 V
We check the consistency of the solutions for v ()
and i a by noting that
i a = C 1 ^- = (0.25 X 10-6)(-9000fri 0 00
at
Because dv o (0 )/dt = 0, the expression for i Q
clearly is valid only for t > 0+
Next, we calculate the Thevenin resistance, as
seen to the right of the capacitor, by shorting the
- 7 5 V source and making series and parallel
combinations of the resistors:
ttn, = 8000 + 40,000 || 160,000 = 40 101 Figure 7.23 • The equivalent circuit for t > 0 for the circuit shown in Fig 7.22
Trang 6^ A S S E S S M E N T P R O B L E M
Objective 2—Be able to determine the step response of both RL and RC circuits
7.6 a) Find the expression for the voltage across
the 160 kfi resistor in the circuit shown in
Fig 7.22 Let this voltage be denoted v A , and
assume that the reference polarity for the
voltage is positive at the upper terminal of
the 160 kH resistor
NOTE: Also try Chapter Problems 7.51 and 7.53
b) Specify the interval of time for which the expression obtained in (a) is valid
Answer: (a) - 6 0 + 12e~ m)t V;
(b) t > 0*
7.4 A General Solution for Step
and Natural Responses
The general approach to finding either the natural response or the step
response of the first-order RL and RC circuits shown in Fig 7.24 is based
on their differential equations having the same form (compare Eq 7.48
and Eq 7.50) To generalize the solution of these four possible circuits, we
let x(t) represent the unknown quantity, giving x(t) four possible values It
can represent the current or voltage at the terminals of an inductor or the
current or voltage at the terminals of a capacitor From Eqs 7.47, 7.48,
7.50, and 7.52, we know that the differential equation describing any one
of the four circuits in Fig 7.24 takes the form
£ + £ = K
-dt T
(7.54)
where the value of the constant K can be zero Because the sources in the
circuit are constant voltages and/or currents, the final value of x will be
constant; that is, the final value must satisfy Eq 7.54, and, when x reaches
its final value, the derivative dxjdt must be zero Hence
where xy represents the final value of the variable
We solve Eq 7.54 by separating the variables, beginning by solving for
the first derivative:
(x - Kr) "(x - X f )
(7.56)
L<n
Figure 7.24 • Four possible first-order circuits
(a) An inductor connected to a Thevenin equivalent (b) An inductor connected to a Norton equivalent
(c) A capacitor connected to a Thevenin equivalent (d) A capacitor connected to a Norton equivalent
Trang 7General solution for natural and step
responses of RL and RC circuits •
In writing Eq 7.56, we used Eq 7.55 to substitute Xf for Kr We now mul-tiply both sides of Eq 7.56 by dt and divide by x - Xt to obtain
dx - 1
= — d t
X — Xf T
(7.57)
Next, we integrate Eq 7.57 To obtain as general a solution as possible, we
use time t Q as the lower limit and t as the upper limit Time t {) corresponds
to the time of the switching or other change Previously we assumed that
t 0 = 0, but this change allows the switching to take place at any time
Using u and v as symbols of integration, we get
m du
= — / dv
x(t a ) » - Xf
Carrying out the integration called for in Eq 7.58 gives
x(t) = x f + [x(t 0 ) - x f ]e- {t -^\
(7.58)
(7.59)
The importance of Eq 7.59 becomes apparent if we write it out in words:
the unknown variable as a function of time
the final value of the variable
+
the initial the final value of the — value of the variable variable
\y „ -[(-(time of switching)!
e (time constant) (7.60)
Calculating the natural or step response of
RL or RC circuits •
In many cases, the time of switching—that is, r0—is zero
When computing the step and natural responses of circuits, it may help to follow these steps:
1 Identify the variable of interest for the circuit For RC circuits, it is most convenient to choose the capacitive voltage; for RL circuits,
it is best to choose the inductive current
2 Determine the initial value of the variable, which is its value at t Q
Note that if you choose capacitive voltage or inductive current as your variable of interest, it is not necessary to distinguish between
t = ?o and t = tQ 2 This is because they both are continuous vari-ables If you choose another variable, you need to remember that
its initial value is defined at t — $
3 Calculate the final value of the variable, which is its value as t —> oo
4 Calculate the time constant for the circuit
With these quantities, you can use Eq 7.60 to produce an equation describing the variable of interest as a function of time You can then find equations for other circuit variables using the circuit analysis techniques introduced in Chapters 3 and 4 or by repeating the preceding steps for the other variables
Examples 7.7-7.9 illustrate how to use Eq 7.60 to find the step
response of an RC or RL circuit
The expressions / and $ are analogous to 0 and 0 ' Thus x(t ) is the limit of x(t) as t
Trang 8Using the General Solution Method to Find an RC Circuit's Step Response
The switch in the circuit shown in Fig 7.25 has been
in position a for a long time At t = 0 the switch is
moved to position b
a) What is the initial value of v c l
b) What is the final value of v c l
c) What is the time constant of the circuit when the
switch is in position b?
d) What is the expression for v c (t) when t a 0?
e) What is the expression for i(t) when t > 0+?
f) How long after the switch is in position b does
the capacitor voltage equal zero?
g) Plot v c (t) and i(t) versus t
Solution
a) The switch has been in position a for a long time,
so the capacitor looks like an open circuit
Therefore the voltage across the capacitor is the
voltage across the 60 H resistor From the
voltage-divider rule, the voltage across the 60 fl resistor
is 40 X [60/(60 + 20)], or 30 V As the
refer-ence for VQ is positive at the upper terminal of
the capacitor, we have v c (0) — —30 V
b) After the switch has been in position b for a long
time, the capacitor will look like an open circuit
in terms of the 90 V source Thus the final value
of the capacitor voltage is + 90 V
c) The time constant is
r = RC
= (400 X 103)(0.5 X 10-6)
= 0.2 s
d) Substituting the appropriate values for vp v(0),
and t into Eq 7.60 yields
v c {t) = 90 + ( - 3 0 - 90)e- 5'
= 90 - 120e_5'V, t > 0
e) Here the value for r doesn't change Thus we
need to find only the initial and final values for
the current in the capacitor When obtaining the
initial value, we must get the value of l(0+),
because the current in the capacitor can change
instantaneously This current is equal to the
cur-rent in the resistor, which from Ohm's law is
[90 - (-30)]/(400 X 103) = 300 fxA Note that
when applying Ohm's law we recognized that the
Figure 7.25 • The circuit for Example 7.7
capacitor voltage cannot change instantaneously
The final value of i(t) = 0, so
'(0 0 + (300 - 0)e- 5' 300e~5VA, t > 0+
We could have obtained this solution by dif-ferentiating the solution in (d) and multiplying by the capacitance You may want to do so for your-self Note that this alternative approach to finding
i{t) also predicts the discontinuity at t = 0
f) To find how long the switch must be in position b before the capacitor voltage becomes zero, we solve the equation derived in (d) for the time
when v c {t) = 0:
120e" 90 or e 5l = 120
9 0 '
so
' = Mf
= 57.54 ms
Note that when v c = 0, i = 225 JXA and the
voltage drop across the 400 kft resistor is 90 V
g) Figure 7.26 shows the graphs of v c (t) and i(t)
versus t
i(/*A) v c ( V )
300 120
250 100
200 80
150 60
100 40
50 20
0 -20
_ry'
~i \ y
-1 / \
-1 /
y '
/ 200
r -30
i
400
v C
600 800
t (ms)
Figure 7.26 A The current and voltage waveforms for Example 7.7,
Trang 9Example 7.8 Using the General Solution Method with Zero Initial Conditions
The switch in the circuit shown in Fig 7.27 has been
open for a long time The initial charge on the
capacitor is zero At t = 0, the switch is closed Find
the expression for
a) i(t) for t > 0+ and
b) v{t) when f > 0+
0.1 /xF
-A^Ht
:3()kn
Figure 7.27 A The circuit for Example 7.8
Solution
a) Because the initial voltage on the capacitor is
zero, at the instant when the switch is closed the
current in the 30 kO branch will be
/(0 + ) = (7.5)(20)
50
= 3 mA
The final value of the capacitor current will be
zero because the capacitor eventually will
appear as an open circuit in terms of dc current
Thus if = 0 The time constant of the circuit will
equal the product of the Thevenin resistance (as
seen from the capacitor) and the capacitance
Therefore T = (20 + 30)1(^(0.1) X 10- 6 = 5 ms Substituting these values into Eq 7.60 generates the expression
/(f) = 0 -f (3 - 0 ) e 'f / 5 x l ( r 3
= 3e-2(,0fmA, t > 0+
b) To find v(t), we note from the circuit that it
equals the sum of the voltage across the capaci-tor and the voltage across the 30 kf! resiscapaci-tor To find the capacitor voltage (which is a drop in the direction of the current), we note that its initial value is zero and its final value is (7.5)(20), or
150 V The time constant is the same as before, or
5 ms Therefore we use Eq 7.60 to write
v c (t) = 150 + ( 0 - I50)e" mi
= (150 - 150<r2()(") V, t > 0
Hence the expression for the voltage v(t) is
v(t) = 150 - 150^2 0 0' + (30)(3K2,)("
= (150 - 60<T20()f)V, f > 0+
As one check on this expression, note that it pre-dicts the initial value of the voltage across the
20 LI resistor as 150 - 60, or 90 V The instant
the switch is closed, the current in the 20 kO resistor is (7.5)(30/50), or 4.5 mA This current
produces a 90 V drop across the 20 kil resistor,
confirming the value predicted by the solution
The switch in the circuit shown in Fig 7.28 has been
open for a long time At t = 0 the switch is closed
Find the expression for
a) v{t) when t > 0+ and
b) i(i) when t > 0
Solution
a) The switch has been open for a long time, so the initial current in the inductor is 5 A, oriented from top to bottom Immediately after the switch closes, the current still is 5 A, and therefore the initial voltage across the inductor becomes
20 - 5(1), or 15 V The final value of the induc-tor voltage is 0 V With the switch closed, the time constant is 80/1, or 80 ms We use Eq 7.60 to
write the expression for v(t):
v(t) = 0 + (15 - 0)e-'/80x10"s
= 15<T12-5' V, t > 0+ b) We have already noted that the initial value of
Trang 10been closed for a long time, the inductor current
reaches 20/1, or 20 A The circuit time constant is
80 ms, so the expression for /(/) is
/(/) = 20 + ( 5 - 20)<r12J5f
= (20 - 15e_l2-5')A, t > 0
We determine that the solutions for v(t) and i(t)
agree by noting that
V{t) = LJt
= 80 X 10-3[15(12.5K12-5/]
= 15fT12-5'V, / > 0+
NOTE: Assess your understanding of the general solution method by trying Chapter Problems 7.55 and 7.56
Example 7.10 shows that Eq 7.60 can even be used to find the step
response of some circuits containing magnetically coupled coils
Example 7.10 Determining Step Response of a Circuit with Magnetically Coupled Coils
There is no energy stored in the circuit in Fig 7.29
at the time the switch is closed
a) Find the solutions for i m v t) , i u and /2
b) Show that the solutions obtained in (a) make
sense in terms of known circuit behavior
Solution
a) For the circuit in Fig 7.29, the magnetically
cou-pled coils can be replaced by a single inductor
having an inductance of
L X L 2 - M 2 45 - 36
18 - 12 1.5 H
(See Problem 6.41.) It follows that the circuit in
Fig 7.29 can be simplified as shown in Fig 7.30
By hypothesis the initial value of /(, is zero
From Fig 7.30 we see that the final value of i a
will be 120/7.5 or 16 A The time constant of the
circuit is 1.5/7.5 or 0.2 s It follows directly from
Eq 7.60 that
i 0 = 16 - 16e~5'A, / > 0
The voltage v 0 follows from Kirchhoff s
voltage law Thus,
v a - 120 — 7.5/(,
= 120«""* V, t > 0+
To find i[ and /2 we first note from
Fig 7.29 that
3 - ^ + 6 - ^ = 6-r + 1 5 -1
dt dt dt dt
or
dt dt '
7.5 fl
1 = 0
120 V
• I ^ 6 H ^ «
" p 3 H i * 15 H^
Figure 7.29 A The circuit for Example 7.10
120 V
7.5 ft
Hr
>'„ ^1.5 H
Figure 7.30 • The circuit in Fig 7.29 with the magnetically coupled coils replaced by an equivalent coil
It also follows from Fig 7.29 that because
to = ' I + «2»
dt dt dt '
Therefore
a * * = - 2 ^
dt
Because /2(0) is zero we have
Jo
= - 8 + 8e~ 5t A, / > 0
Using Kirchhoffs current law we get
it = 24 - 24e"5/ A, / > 0
b) First we observe that i o (0), /*i(0), and /2(0) are all zero, which is consistent with the statement that