• The response of a second-order circuit is overdamped, underdamped, or critically damped as shown in Table 8.2.. • In determining the natural response of a second-order circuit, we firs
Trang 1296 Natural and Step Responses of RLC Circuits
To find the maximum value of ^ sp , we find the smallest positive value of
time where dv &p /dt is zero and then evaluate v sp at this instant The
expression for t max is
(See Problem 8.67) For the component values in the problem statement,
we have
R 4 X 103
a = —- = — - — = 666.67 rad/s,
2L
and
f109
<°d = A / T T ~ (666.67)2 = 28,859.81 rad/s
1.2 Substituting these values into Eq 8.99 gives
^max = 53.63 flS
Now use Eq 8.98 to find the maximum spark plug voltage, v sp (t metx ):
V U = -25,975.69 V
b) The voltage across the capacitor at r max is obtained from Eq 8.97 as
^ c ( W ) = 262.15 V The dielectric strength of air is approximately 3 x 10 6 V / m , so this result tells us that the switch contacts must be separated by 262.15/3 X 10 6, or 87.38, fj,m to prevent arcing at the points at ?max
In the design and testing of ignition systems, consideration must
be given to nonuniform fuel-air mixtures; the widening of the spark plug gap over time due to the erosion of the plug electrodes; the relationship between available spark plug voltage and engine speed; the time i t takes the primary current to build up to its initial value after the switch
is closed; and the amount of maintenance required to ensure reliable operation
We can use the preceding analysis of a conventional ignition system
to explain why electronic switching has replaced mechanical switching in today's automobiles First, the current emphasis on fuel economy and exhaust emissions requires a spark plug with a wider gap This, in turn, requires a higher available spark plug voltage These higher voltages (up
to 40 kV) cannot be achieved with mechanical switching Electronic switching also permits higher initial currents in the primary winding of the autotransformer This means the initial stored energy in the system is larger, and hence a wider range of fuel-air mixtures and running condi-tions can be accommodated Finally, the electronic switching circuit elim-inates the need for the point contacts This means the deleterious effects
of point contact arcing can be removed from the system
NOTE: Assess your understanding of the Practical Perspective by trying Chapter Problems 8.68 and 8.69
Trang 2Summary
The characteristic equation for both the parallel and
series RLC circuits has the form
s 2 + 2as + O)Q = 0,
where a = 1/2RC for the parallel circuit, a = R/2L for
the series circuit, and col = ^/LC for both the parallel
and series circuits (See pages 267 and 286.)
The roots of the characteristic equation are
*1.2 B -a ± Vo2"
(4-(See page 268.)
• The form of the natural and step responses of series
and parallel RLC circuits depends on the values of a 2
and col; s u c r l responses can be overdamped,
underdamped, or critically damped These terms
describe the impact of the dissipative element (R) on
the response The neper frequency, a, reflects the effect
of R (See pages 268 and 269.)
• The response of a second-order circuit is overdamped,
underdamped, or critically damped as shown in
Table 8.2
• In determining the natural response of a second-order
circuit, we first determine whether it is over-, under-, or
critically damped, and then we solve the appropriate equations as shown in Table 8.3
• In determining the step response of a second-order cir-cuit, we apply the appropriate equations depending on the damping, as shown in Table 8.4
• For each of the three forms of response, the unknown
coefficients (i.e., the As, B s, and Ds) are obtained by
evaluating the circuit to find the initial value of the
response, x(0), and the initial value of the first deriva-tive of the response, dx(Q)/dt
• When two integrating amplifiers with ideal op amps are connected in cascade, the output voltage of the second integrator is related to the input voltage of the first by an ordinary, second-order differential equation Therefore, the techniques developed in this chapter may be used to analyze the behavior of a cascaded integrator (See pages 289 and 290.)
• We can overcome the limitation of a simple integrating amplifier—the saturation of the op amp due to charge accumulating in the feedback capacitor—by placing a resistor in parallel with the capacitor in the feedback path (See page 291.)
TABLE 8.2 The Response of a Second-Order Circuit is Overdamped, Underdamped, or Critically Damped
The Circuit is When Qualitative Nature of the Response
Overdamped a 2 > oil
Underdamped
Critically damped
a" < oj{)
2 2
The voltage or current approaches its final value without oscillation The voltage or current oscillates about its final value
The voltage or current is on the verge of oscillating about its final value
TABLE 8.3 In Determining the Natural Response of a Second-Order Circuit, We First Determine Whether i t is Over-, Under-,
or Critically Damped, and Then We Solve the Appropriate Equations
Damping Natural Response Equations
Overdamped x(t) = A^e* 1 ' + A 2 e S2 '
Underdamped x{t) - (B x cos <a d t + B 2 sin o) d t)e'
Critically damped x(t) = {Dj + D 2 )e~°"
Coefficient Equations JC(0) = Ai + A2 ; dx/dt(0) = A { s { + A 2 s 2
x(0) - Bi ;
dx/dt(0) = -aBi + <o d B 2 , where o> d = VOJQ - a 2
-v(0) = D 2 , dx/dt(0) = D i - aD 2
Trang 3298 Natural and Step Responses of RLC Circuits
TABLE 8.4 In Determining the Step Response of a Second-Order Circuit, We Apply the Appropriate Equations Depending
on the Damping
Overdamped x(t) = X f + A[ eiV + A 2 e* 2 '
Underdamped x(t) = Xf + (B[ cos <o d t + B' 2 sin a> d t)e~
Critically damped x(t) = X f + D[ te~ al + D' 2 e~ al
a where X* is the final value of x(t)
Coefficient Equations
x(0) = X f + A\ + A 2;
dx/dt(0) = A\ s, + A 2 s 2
x(0) - X f + B\ ; dx/dt(0) = -aB\ + <o d B' 2
x(0) = X f + D' 2;
dx/dt(0) = D\ - aD' 2
Problems
Sections 8.1-8.2
8.1 The resistance, inductance, and capacitance in a
parallel RLC circuit are 2000 ft, 250 mH, and
10 nF, respectively
a) Calculate the roots of the characteristic equation
that describe the voltage response of the circuit
b) Will the response be over-, under-, or critically
damped?
c) What value of R will yield a damped frequency
of 12 krad/s?
d) What are the roots of the characteristic equation
for the value of R found in (c)?
e) What value of R will result in a critically damped
response?
8.2 The circuit elements in the circuit in Fig 8.1 are
R = 200 ft, C = 200 nF, and L = 50 mH The
ini-tial inductor current is - 4 5 mA, and the iniini-tial
capacitor voltage is 15 V
a) Calculate the initial current in each branch of
the circuit
b) Find v(t) for t > 0
c) Find i L {t) for t > 0
8.3 The resistance in Problem 8.2 is increased to
PSPICE 312.5 ft Find the expression for v(t) for t > 0
MULTISIM
8.4 The resistance in Problem 8.2 is increased to 250 ft
PSPICE Find the expression for v(t) for t > 0
MULTISIM
8.5 a) Design a parallel RLC circuit (see Fig 8.1) using
component values from Appendix H, with a
res-onant radian frequency of 5000 rad/s Choose a
resistor or create a resistor network so that the
response is critically damped Draw your circuit
PSPICE
MULTISIM
b) Calculate the roots of the characteristic equa-tion for the resistance in part (a)
8.6 a) Change the resistance for the circuit you
designed in Problem 8.5(a) so that the response
is underdamped Continue to use components from Appendix H Calculate the roots of the characteristic equation for this new resistance b) Change the resistance for the circuit you designed
in Problem 8.5(a) so that the response is over-damped Continue to use components from Appendix H Calculate the roots of the character-istic equation for this new resistance
8.7 The natural voltage response of the circuit in Fig 8.1 is
v(t) = 75<r800,)'(cos 6000/ - 4 sin 60000V, t > 0, when the inductor is 400 mH Find (a) C; (b) R;
( c ) V0; ( d ) /0; a n d ( e ) /L( / )
8.8 Suppose the capacitor in the circuit shown in
Fig 8.1 has a value of 0.1 juF and an initial voltage
of 24 V The initial current in the inductor is zero The resulting voltage response for / s 0 is
v(t) = -8e- 250t + 32^1 0 0 0' V
a) Determine the numerical values of R, L, a,
and <w0
b) Calculate i R (t), i L (t), and i c (t) for t > 0+ 8.9 The voltage response for the circuit in Fig 8.1 is known to be
500*
v(f) = Dite'™ + D 2 e -500/ t > 0
Trang 4The initial current in the inductor (/()) is -10 mA,
and the initial voltage on the capacitor (VQ) is 8 V
The inductor has an inductance of 4 H
a) Find the values of R, C, D h and D 2
b) Find i c (t) for t > 0+
8.10 The natural response for the circuit shown in Fig 8.1
is known to be
v(t) = -l\e- im + 20e-400' V, t > 0
If C = 2 /xF and L = 12.5 H, find i L (i) + ) in
milli-amperes
8.11 Tlie initial value of the voltage v in the circuit in
Fig 8.1 is zero, and the initial value of the capacitor
current, /c(0+), is 45 mA The expression for the
capacitor current is known to be
i c (t) = A x e- m)t + A 2 e~ m \ t > 0+,
when R is 250 ft Find
a) the values of a, co {) , L, C, A h and A 2
Hint: di c (0 + ) di L (0 + ) di R (Q + ) -v(0) 1 /c(0+)
b) the expression for v(t), t 2: 0,
c) the expression for i R (t) > 0,
d) the expression for i L {t) £: 0
8.12 Assume the underdamped voltage response of the
circuit in Fig 8.1 is written as
v(t) = (A x + A 2 )e~ al cos (o ( ,t + y'(^i _ A 2 )e~ at sin a> d t
The initial value of the inductor current is /(), and
the initial value of the capacitor voltage is V {) Show
that A 2 is the conjugate of A]_ (Hint: Use the same
process as outlined in the text to find A\ and A 2 )
8.13 Show that the results obtained from Problem 8.12—
that is, the expressions for A x and A 2 —are consistent
with Eqs 8.30 and 8.31 in the text
8.14 In the circuit in Fig 8.1, R = 5 k f t , L = 8 H ,
PSPICE c = 125 np , v0 = 30 V, and /0 = 6 mA
MULTISIM
a) Find v{t) for t > 0
b) Find the first three values of t for which dv/dt is
zero Let these values of t be denoted * j , t 2 ,
and r3
c) Show that t 3 — t\ — T (l
d) Show that t 2 - t x = T d /2
e) Calculate v(t { ), v(t 2 ), and v(t 3 )
f) Sketch v(t) versus t for 0 < t < t 2
8.15 a) Find v(t) for t > 0 in the circuit in Problem 8.14
if the 5 kfi resistor is removed from the circuit
MULTISIM
b) Calculate the frequency of v(t) in hertz
c) Calculate the maximum amplitude of v(t) in volts
8.16 In the circuit shown in Fig 8.1, a 2.5 H inductor is
PSPICE shunted by a 100 nF capacitor, the resistor R is
MULTISIM adjusted for critical damping, V 0 = - 1 5 V, and
/() = —5 mA
a) Calculate the numerical value of R
b) Calculate v(t) for t > 0
c) Find v{t) when i c (t) = 0
d) What percentage of the initially stored energy remains stored in the circuit at the instant /c(r) isO?
8.17 The resistor in the circuit in Example 8.4 is changed
«PICE to 3200 a
MULTISIM
a) Find the numerical expression for v{t) when
t > 0
b) Plot v(t) versus t for the time interval
0 s f < 7 ras Compare this response with
the one in Example 8.4 (R = 20kft) and
Example 8.5 (R = 4 kH) In particular, compare peak values of v(t) and the times when these
peak values occur
8.18 The two switches in the circuit seen in Fig P8.18
PSPICE operate synchronously When switch 1 is in position
a, switch 2 is in position d When switch 1 moves to position b, switch 2 moves to position c Switch 1 has
been in position a for a long time At t = 0, the
switches move to their alternate positions Find
v 0 (t) for t > 0
Figure P8.18
i n
8.19 The resistor in the circuit of Fig P8.18 is increased
PSPICE f rom 100 H to 200 O Find v a (t) for t > 0
MULTISIM
8.20 The resistor in the circuit of Fig P8.18 is increased
"sn« from 100 ft to 125 ft Find vjt) for t s 0
MULTISIM
8.21 The switch in the circuit of Fig P8.21 has been in
PSPICE position a for a long time At J = 0 the switch
moves instantaneously to position b Find v 0 (t) for
/ s 0
t = ()
a \ / b
16 X lO3/*
!4nF ? 15.625 H A
Trang 58.22 T h e inductor in the circuit of Fig P8.21 is decreased
to 10 H Find v a {t) for t > 0
8.23 T h e inductor in t h e circuit of Fig P8.21 is decreased
to 6.4 H Find v 0 (t) for t > 0
Section 8.3
8.24 F or t h e circuit in E x a m p l e 8.6, find, for t > 0,
PSPICE ( a ) v { t ) ( b ) iR{t) a n d ( c ) / c ( 0
MULTISIM
8.25 For t h e circuit in E x a m p l e 8.7, find, for t 3: 0,
«"« (a) v(t) a n d (b) i c (t\
8.26 For t h e circuit in E x a m p l e 8.8, find v(t) for t > 0
8.27 T h e switch in t h e circuit in Fig P8.27 h a s b e e n o p e n
MULTISIM a l ° n § t n n e ^ e f ° r e closing at t = 0 Find
a) v 0 (t) for t > 0+ ,
b) i L {t) for t > 0
Figure P8.27
156.25 a
PSPICE
MULTISIM
8.28 U s e t h e circuit in Fig P8.27
a) Find the total energy delivered to t h e inductor
b) Find the total energy delivered t o t h e equivalent
resistor
c) Find the total energy delivered to t h e capacitor
d) Find t h e total energy delivered by t h e
equiva-lent current source
e) Check t h e results of parts (a) through ( d )
against t h e conservation of energy principle
A s s u m e that at t h e instant t h e 60 m A dc current
source is applied t o t h e circuit in Fig P8.29, the
ini-tial current in t h e 50 m H inductor is - 4 5 m A , a n d
the initial voltage on t h e capacitor is 15 V (positive
at the u p p e r terminal) Find the expression for i L {t)
f o r / > 0 if # equals 200 H
Figure P8.29
8.29
PSPICE
MULTISIM
60 mA '/.(0M5(
8.30 T h e resistance in t h e circuit in Fig P8.29 is changed
PSPICE t o 312.5 a Find i L (t) for t > 0
MULTISIM
8.31 T h e resistance in t h e circuit in Fig P8.29 is changed
PSPICE t o 250 ft F i n d i L {t) for t > 0
MULTISIM
8.32 T h e switch in t h e circuit in Fig P8.32 h a s b e e n
PSPICE o p e n a long time before closing at t = 0 Find i r (t)
Figure P8.32
3 kfl
15 V
8.33 Switches 1 a n d 2 in t h e circuit in Fig P8.33 a r e syn-PSPKE chronized W h e n switch 1 is o p e n e d , switch 2 closes and vice versa Switch 1 has b e e n o p e n a long time
before closing a t / = 0 Find i L (t) for t 2: ()
8.34 T h e switch in the circuit in Fig P8.34 has been open
PSPICE for a long time before closing at t = 0 Find v 0 (t)
mnsiM f o r / > 0 >
Figure P8.34
12 V
400 a
^
/ = 0 1.25/xF
+
M l 251
8.35 a) F o r t h e circuit in Fig P8.34, find i 0 for t > 0
PSPICE b) Show that your solution for L is consistent with
MULTISIM ' ' v
the solution for v 0 in P r o b l e m 8.34
8.36 T h e switch in t h e circuit in Fig P8.36 h a s b e e n
PSPICE o p e n a long time b e f o r e closing at t = 0 A t t h e
t i m e t h e switch closes, t h e capacitor has n o s t o r e d
energy Find v a for t > 0
Figure P8.36
7.5 V
250 a
^
/ = 0
!4H
+
v 0 , 25 fiF
Figure P8.33
5 kfl Switch 1
| ) 6 0 m A
Trang 68.37 There is no energy stored in the circuit in Fig P8.37
PSPICE when the switch is closed at t = 0 Find v() (t)
™ f o r r > 0
Figure P8.37
12V
400 n
^
t = 0
1.25/nF:
+ 1
vAl.25H
8.38 a) For the circuit in Fig P8.37, find i a for t > 0
b) Show that your solution for i a is consistent with
the solution for v„ in Problem 8.37
PSPICE
MULTISIM
Section 8.4
8.39 The initial energy stored in the 31.25 nF capacitor
in the circuit in Fig P8.39 is 9 /xJ The initial energy
stored in the inductor is zero The roots of the
char-acteristic equation that describes the natural
behav-ior of the current i are -4000 s_1 and -16,000 s_1
a) Find the numerical values of R and L
b) Find the numerical values of /(0) and di(0)/dt
immediately after the switch has been closed
c) Find i(t) for
d) How many microseconds after the switch closes
does the current reach its maximum value?
e) What is the maximum value of/ in milliamperes?
f) Find v L {t) for t > 0
Figure P8.39
31.25 nF
8.40 a) Design a series RLC circuit (see Fig 8.3) using
component values from Appendix H, with a
res-onant radian frequency of 20 krad/s Choose a
resistor or create a resistor network so that the
response is critically damped Draw your circuit
b) Calculate the roots of the characteristic
equa-tion for the resistance in part (a)
8.41 a) Change the resistance for the circuit you
designed in Problem 8.40(a) so that the response
is underdamped Continue to use components
from Appendix H Calculate the roots of the
characteristic equation for this new resistance
b) Change the resistance for the circuit you designed in Problem 8.40(a) so that the response
is overdamped Continue to use components from Appendix Ff Calculate the roots of the characteristic equation for this new resistance
8.42 The current in the circuit in Fig 8.3 is known to be
i = 51<r2(,,),)' cos 1500f + B 2 e~ 2mt sin 1500/, t > 0
The capacitor has a value of 80 nF; the initial value
of the current is 7.5 mA; and the initial voltage on
the capacitor is -30 V Find the values of R, L, B h
and B 2
8.43 Find the voltage across the 80 nF capacitor for the
circuit described in Problem 8.42 Assume the refer-ence polarity for the capacitor voltage is positive at the upper terminal
8.44 In the circuit in Fig P8.44, the resistor is adjusted
PSPICE for critical damping The initial capacitor voltage is iiimsm 1 5 y a n cj t n e jn{tial inductor current is 6 mA
a) Find the numerical value of R
b) Find the numerical values of i and di/dt
immedi-ately after the switch is closed
c) Find v c (t) for t a 0
Figure P8.44
+ TA
!320nF
R
-vw-125 mH
8.45 The switch in the circuit shown in Fig P8.45 has
PSPICE been in position a for a long time At t = 0, the
switch is moved instantaneously to position b Find
/(0 for t > 0
Figure P8.45
8012
A M
-10 H
8.46 The switch in the circuit in Fig P8.46 on the next
page has been in position a for a long time At / = 0, the switch moves instantaneously to position b a) What is the initial value of t>fl?
b) What is the initial value of dvjdtl c) What is the numerical expression for vjf) for t > 0?
PSPICE MULTISIM
Trang 7302 Natural and Step Responses of RLC Circuits
75 V
1 kO
A A A
-o.i MF:
/ = 0
2kO
A W
-3 kO
- A A A —
400 mH
8.47 The switch in the circuit shown in Fig P8.47 has
PSPICE been closed for a long time The switch opens at
* ™ r = 0.Find
a) i () (t) for t > 0,
b) v a {t) for t > 0
Figure P8.47
f = 0
300 a
>v
80 V 6 2.5 mH • 500 n |f/ , , ( 0 +
:4() nF
8.48 The switch in the circuit shown in Fig P8.48 has
been closed for a long time The switch opens at
t = 0 Find v t) (t) for t > 0
Figure P8.48
8.49 The circuit shown in Fig P8.49 has been in operation
PSPICE for a long time At t = 0, the source voltage suddenly
MULTI5IM jumps to 250 V Find v () {t) for t > 0
Figure P8.49
8kH
^ A V - 160mH
+
v„[t)
8.50 The initial energy stored in the circuit in Fig P8.50
PSPICE is zero Find v() {t) for t > 0
250 n
60 V
8.51 The capacitor in the circuit shown in Fig P8.50 is
changed to 4 ju.F The initial energy stored is still
zero Find v () (t) for t SE 0
8.52 The capacitor in the circuit shown in Fig P8.50 is changed to 2.56 ^tF The initial energy stored is still
zero Find v a (t) for t > 0
8.53 The switch in the circuit of Fig P8.53 has been in PSPICE position a for a long time At t = 0 the switch moves instantaneously to position b Find
a) v o (0 + )
b) dv () (Q + )/dt
c) v„(t) for t > 0
Figure P8.53
8.54 The switch in the circuit shown in Fig P8.54 has
been closed for a long time before it is opened at
t = 0 Assume that the circuit parameters are such
that the response is underdamped
a) Derive the expression for vjt) as a function of
Vg, a, co d , C, and R for t > 0
b) Derive the expression for the value of t when the magnitude of v 0 is maximum
Figure P8.54
t = 0
Liv a {t)
8.55 The circuit parameters in the circuit of Fig P8.54
PSPICE are R = 4800 ft, L = 64 mH, C = 4 nF, and
«iunsm ^ = _7 2 V
a) Express v a (t) numerically for t & 0
b) How many microseconds after the switch opens
is the inductor voltage maximum?
Trang 8c) What is the maximum value of the inductor
voltage?
d) Repeat (a)-(c) with R reduced to 480 ft
8.56 The two switches in the circuit seen in Fig P8.56
PSPICE operate synchronously When switch 1 is in
position a, switch 2 is closed When switch 1 is
in position b, switch 2 is open Switch 1 has been in
position a for a long time At ( = 0, it moves
instan-taneously to position b Find v c (t) for t > 0
Figure P8.56
w 4 (') £1811
8.57 Assume that the capacitor voltage in the circuit of
Fig 8.15 is underdamped Also assume that no
energy is stored in the circuit elements when the
switch is closed
a) Show that dvc/dt = (a)Q/a) (i )Ve~ lxl s'mw d t
b) Show that dv c /dt = 0 when t = mrfo) th where
n = 0,1,2
c) Let t n = mr/aj, and show that v c (t„)
d) Show that
1 v c { h ) - V
(X = 777" m :—: — ,
T d v c (h) - V
where T d - t 3 - t {
8.58 The voltage across a 100 nF capacitor in the circuit
of Fig 8.15 is described as follows: After the switch
has been closed for several seconds, the voltage is
constant at 100 V The first time the voltage exceeds
100 V, it reaches a peak of 163.84 V This occurs
7r/7 ms after the switch has been closed The second
time the voltage exceeds 100 V, it reaches a peak of
126.02 V This second peak occurs 3-77-/7 after the
switch has been closed At the time when the switch
is closed, there is no energy stored in cither the
capacitor or the inductor Find the numerical values
of R and L (Hint: Work Problem 8.57 first.)
Section 8.5
8.59 Show that, if no energy is stored in the circuit
shown in Fig 8.19 at the instant v s jumps in value,
then dvjdt equals zero at t = 0
8.60 a) Find the equation for v ( ,(t) for 0 < t < rsat in
the circuit shown in Fig 8.19 if u„i(0) = 5 V and
vJQ) = 8 V
b) How long does the circuit take to reach saturation?
8.61 a) Rework Example 8.14 with feedback resistors
Ri and R 2 removed
b) Rework Example 8.14 with v ol (0) = - 2 V a n d
v o (0) = 4 V
8.62 a) Derive the differential equation that relates
the output voltage to the input voltage for the circuit shown in Fig P8.62
b) Compare the result with Eq 8.75 when
Rtd - R 2 C 2 = RC in Fig 8.18
c) What is the advantage of the circuit shown in Fig P8.62?
Figure P8.62
8.63 The voltage signal of Fig P8.63(a) is applied to
PSPICE the cascaded integrating amplifiers shown in
MULT1SIM Fig P8.63(b) There is no energy stored in the capacitors at the instant the signal is applied
a) Derive the numerical expressions for v a (t) and
v a i(t) for the time intervals 0 < t < 0.5 s and
0.5 s < t < t m
b) Compute the value of t S(lt
Figure P8.63 MrrAO
80
0
- 4 0
0.5
1
1
f ( s )
(a)
Trang 93 0 4 Natural and Step Responses of RLC Circuits
8.64 The circuit in Fig P8.63(b) is modified by adding a
PSPICE i M i l resistor in parallel with the 500 nF capacitor
mTISIM and a 5 MH resistor in parallel with the 200 nF
capacitor As in Problem 8.63, there is no energy
stored in the capacitors at the time the signal is
applied Derive the numerical expressions for v n (t)
and v o[ (t) for the time intervals 0 < t < 0.5 s and
t > 0.5 s
8.65 We now wish to illustrate how several op amp
cir-cuits can be interconnected to solve a differential
equation
a) Derive the differential equation for the
spring-mass system shown in Fig P8.65(a) Assume
that the force exerted by the spring is directly
proportional to the spring displacement, that
the mass is constant, and that the frictional force is directly proportional to the velocity of the moving mass
b) Rewrite the differential equation derived in (a)
so that the highest order derivative is expressed
as a function of all the other terms in the
equa-tion Now assume that a voltage equal to d 2 x/dt 2
is available and by successive integrations
gen-erates dx/dt and x We can synthesize the
coeffi-cients in the equations by scaling amplifiers, and
we can combine the terms required to generate
d 2 x/dt 2 by using a summing amplifier With these ideas in mind, analyze the interconnection shown in Fig P8.65(b) In particular, describe the purpose of each shaded area in the circuit and describe the signal at the points labeled B, Figure P8.65
K
M
-*(0 —
(a)
Ri
(b)
Trang 10C, D, E, and F, assuming the signal at A
repre-sents d 2 x/dt 2 Also discuss the parameters R; R],
Ci; R 2 , C2; R$, R4', R5, Re', and R 7, i?8 in terms
of the coefficients in the differential equation
Sections 8.1-8.5
8.66 a) Derive Eq 8.92
m S S m b ) Derive Eq 8.93
c) Derive Eq 8.97
8.67 Derive Eq 8.99
PRACTICAL
8.68 a) Using the same numerical values used in the
Practical Perspective example in the text, find
the instant of time when the voltage across the
capacitor is maximum
PRACTICAL
PERSPECTIVE
b) Find the maximum value of v c
c) Compare the values obtained in (a) and (b) with and vc(rmax)
PRACTICAL p j c r PERSPECTIVE 6
V
8.69 The values of the parameters in the circuit in
8.21 are R = 3 O ; L = 5 mH; C = 0.25 juF;
ic — 12 V; and a = 50 Assume the switch opens
when the primary winding current is 4 A
a) How much energy is stored in the circuit at
t = 0+? b) Assume the spark plug does not fire What is the maximum voltage available at the spark plug? c) What is the voltage across the capacitor when the voltage across the spark plug is at its maxi-mum value?
8.70 Repeat Problem 8.68 using the values given in
Problem 8.69