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A comprehensive overview of control systems and electromagnetics, covering fundamental concepts, types, and applications. It includes detailed explanations, examples, and exercises to enhance understanding. Suitable for students studying electrical engineering or related fields.
Typology: Exercises
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POWERLINE REVIEW CENTER
M EZ Z A N IN E FLO O R , D O N A A M P A R O BUILDING ' C O R N E R E S P A N A G. M. TOLEN TIN O S T R E E T S S A M P A L O C , M A N ILA TEL. NOS. 7 3 5 -7 3 -0 2 Sc 73 3-2 1 -1 8
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Control System - is an arrangement of physical components connected or related in such a manner as to Commend, direct or regulate itself or another system. Purpose of control system The purpose of control system usually identifies or defines the “output” and “input”. If the output and input are given, it is possible to identify or define the nature of system's components.
Three Basic Types of Control Systems:
Two General Categories of Control Systems:
The Control Systems Engineering Problem:
The Representation of the Problem: The Model Three basic representations (models) of physical components and systems are extensively employed in the study of control systems.
The Solution of Linear Constant Coefficient Ordinary Differential Eouation: Consider the class of Differential Equation of the form: A ' _dy_* A d lx > a. - 4 - = > — P d tl to d tl Where: “ t ” is time, the coefficients a t & bt are constant,
X = x{t)
Voltage-force analogy Current-force analogy Force, F Velocity, dx/dt Damping, b Mass, m Spring constant, k
Voltage, u Current, / Resistance, R Inductance, L Elastance, 1/C
Force, F Velocity, dx/dt Damping, b Mass, m Spring constant, k
Current, / Voltage, o Conductance, 6 Capacitance, C Reciprocal of L (ML)
In general, in laying out equivalent electric circuits for mechanical systems, the following rule is observed If the forces act in series in the mechanical system, the electrical elements representing these forces are put in parallel. Forces in parallel are represented by elements in series in electric circuits.
Coulombs Law / Electrostatics: It states that the force between two very small objects separated in a vacuum or free space by a distance which is large compared to their size is proportional to the charge on each and inversely proportional to the square of the distance between them.
Mathematically. F - - — — ■- a R Newtons 4 kS 'R 2
Where: Q x, Q2 = respective charges in Coulomb, R = separation distance in meters £ = permittivity of the medium = S0 • Sr ( for air 8r = 1.0 )
£0 — permittivity constant = 8.854 xlO 12 F/m « ----- *10 9 F/m 36 k a R = unit vector in the direction of R
Electric Field Intensity: It is a vector force on a unit positive test charge. Electric field intensity must be measured by the unit- Newton per Coulomb or Volts per meter.
4 ns-R m Coul Standard Charge Conf/auartion: '
„ Q Volts Newton ,. f ^ ■■ a r ------- o r -------- —
.j^ = -----^ p P :—^ a r ------- Volts^ o r ------- — Newton^ Cylindrical coordinates> , 2ns * r m Coul
P s Volts Newton k = — a R ------- o r -----------
Where: if/ — electric flux in Coulomb (a scalar)
D = electric flux density in Coulomb/m2 (a vector)
dS = differential surface in m2 (a vector) & j> = integral of closed surface
Note: D = £ 0£ rE in Coulomb/m
Gauss Divergence Theorem:
The surface integral of normal component D over the surface is equal to the volume integral of divergence D over the volume integral.
! » • dS ~ \ p vdV = f (V • D )dV in Coulomb J vl 1 V
But p v = div D = V • D from Maxwell’s equation
Where V = ”del” or “nabla” operator = — a v 4-— H-----^ a y
Electric Potential Between Two Points: The potential of point UA” with respect to point “B57 is defined as the work done in moving a unit positive charge <2u from “B” to “A”
Relation ofV toE is E = — V V 9 where V V = gradient of V. Current and Current Density: The electric charges in motion constitute a current, Current density is a vector J.
Poisson’s Equation & Laplace's Equation:
From V • D = p , also D = S * E and E = —V V
Consider, V • 6'(—V F ) — p thus, V • V V = —— or V 2V = —— 8 £ Where: V • VV — divergence of divergence of V = Laplacian of V
Ampere's Law & Magnetic Field:
. The field varies inversely with the distance squared, is independent of the surrounding medium, and
■ I < # x a » H =
in --------
Current Density. J & Magnetic Field. H:
V x H = J = (curl H ) = crE Where: <7 “ conductivity of the material in mho/meter
Curl H =
a , a y (^) a* d d d dx (^) d> H y H z
dHz 8H y A dy dz
a v +
( d n y an dx a>’ +
'dHy dH, dx dy
F I j x Density & Stokes Theorem: (Ampere's Circuital Law)
B = j A * d( = £ b • dS ~ / i0j>J • dS
Where: A = magnetic potential vector. Electromagnetic Waves:
Wave Equations: V x H = (cr + jo)e)E also V x E = ~jcojuH V • E = 0 & V « H = 0 Propagatfon Constant : y - a + jj
Where: a = attenuation factor in neper/m = co
The Ultimate E.E. Review Center
infinity? TA.; steady state V1! ' C. signal B. transient D. force
' 53 + 4 s 2 +85 + 1 2 - 0 A. No, it is unstable C. indeterminate B. Yes, it is stable D. undefined
Gl + G 2
d 2y , Ady dt_ A 0,5, 2 B. 0.5, 4 C. 0.25, 1 D. 2, 4
second order system: 2 — ~ + 4 — + 8>> = 8x
d 2y , dy the second order system:
A. 0.4, 0.
m
- o
\ w IP- (^) K (^) ■---------------^ 1 s (s + 2) c % )
form of a plane disk of radius 4m with a density p s
sin <j)
A. 2/71 Coul B. tc/2 Coul
2 r C. n Coul D. 2n Coul
Coul/m?
s3 + 8 s 2 +145 + 24 = 0 A. stable system C. undefined Bf unstable system D. indeterminate
B K< 10, K >Os D. K<60, K>
A. e~‘ - e - 2' B. - e " + e '2'C. e~'+e~2' D. - e " - e ~ 2‘
following second order: - r f! > + 4 0 ^^ +^ s^ = 8 „ dt2 dt A. 2,1.5 C. 3,0. B. V2 , 1.25 D. V 3. 1
d 2y , ,d y second order: + 5— + 9^ y^ = 9^ u dt2 dt A. 2/3, 2 rad/sec B. 4/5, 2.5 rad/sec C. 13 Find the transfer function C (.s)//?(s) for the system shown.
C. 5/6, 3 rad/sec D. 1/3, 3.5 rad/sec
g l + ^ \ - G xH , - G 2//,
A. l / 4 [ 8 e " ' - 2 r - l j C. 1 / 4 ^ ^ + 7e"2( J
B. 1/4[2/ - 1] D. 1/4 [8e~' - 7e~2 1 j
R(sy
K
s (s + 2) (s 4 10)* f^ —^ ^ A -.
A. 10, 10 Hz B. 240, 0.71 Hz
C. 100, 1.2 Hz D. 60, 2 Hz
1 1 >3 / — (^) ~~ ✓ IVJLkLXu