Download Exam 2 Equation Sheet

Exam 2 Equation Sheet
Capacitor
dv C
dt
=
q Cv
Inductor
di
=
vL L L
dt
=
W
=
iC C
=
W
1
2
Li2
1
2
2
Cv
=
q2
=
2C
1
2
=
iL (t) iL (t o ) +
qv
v C=
(t) v(t o ) +
1
idt
C∫
1
vdt
L∫
First Order Circuits
Natural Response
RC-Circuit:
v C (t)
= VOC + [v c (0 − ) − VOC ]e − t / τ where=
τ R ThCeq and VOC = v Th
RL-Circuit:
τ L eq / R Th and ISC =
iL (t)= ISC + [iL (0 − ) − ISC ]e − t / τ where =
iN
Forced Response
Step 1: Use KVL or KCL to derive the differential equation dx/dt + ax(t) = f(t)
Step 2: The general solution is of the form
x(t) =
x f (t) + xn (t) where x n (t) =
Ae-t/τ
Step 3: Assume the forcing response to be the form
-at
when v s (t), is (t) = Ke-at
B e
x f (t) =  1
(t) K sin( ωt)
B1 sin(ωt) + B2 cos(ωt) when v s (t), is=
Step 4: Substitute assumed solution into Step 1 to find constants B 1 and B 2 .
Step 5: Apply initial conditions [v C (0−) or i L (0−)] to the general solution to obtain the
constant A.
Step 6: Solve the circuit problem and interpret
Second Order Circuits
Step 1: Use KVL, KCL, v L = Ldi L /dt, and i C = Cdv C /dt to obtain two first order response
equation. Be sure that these equations dependent on v C and i L .
Step 2: Set all sources equal to zero to obtain the Natural Response (x n (t)) and solve
the characteristic equation for the roots.
s2 + 2αs + ω02 = 0
⇒ s1,s2 = −α ± α 2 − ω02
Step 3: The natural response of the circuit depends only on the values of R, L, and C
which dictates the types of roots s 1 and s 2 . The 3 possible solutions are
•
Critical Damping ― α = ω0 : x critical (t) = e −αt (A1t + A 2 )
•
Overdamping ― α > ω0 : x over
=
(t) A1es1t + A 2 es2 t
•
Underdamping – α < ω0 : xunder (t) = e−αt (A1 cos ωd t + A 2 sin ωd t), where ωd =
Step 4: The forcing response x f (t) has the same form as the effective source:
ω02 − α 2
Forcing Function
K = constant
Ke− at
K sinat or K cosat
Assumed Solution
x f (t) = B
x f (t) = Be− at
=
x f (t) B1 cosat + B2 sinat
Find the constant B and write out the complete response.
Step 5: Find the initial conditions x 2 (0−) and dx 2 (0−)/dt. This is a two-step process;
•
Find the steady-state values for the capacitor’s voltage v C (0−) and the inductor’s
current i L (0−) before the switch is thrown.
•
To find the derivative dx 2 (0+)/dt after the switch has been thrown, use equation 2 in
Step 1 and solve for it numerically.
Step 6: Apply the initial conditions of Step 5 to find the constants A 1 and A 2 .
•
Setup up two equations involving x 2 (0+) and dx 2 (0+)/dt and solve for A 1 and A 2
•
Write down the Total Response for the circuit: x(t) = x n (t) + x f (t)
Step 7: Solve the circuit problem and interpret