How to solve Second Order Differential Circuits?

1. Second Order Differential Circuits
A PRESENTATION BY:
ABHIJEET GUPTA (140110111001)
DARSHAK PADSALA (140110111008)
HIREN PATEL (140110111035)
PRERAK TRIVEDI (140110111045)

2. What is Second Order Differential Circuits?
• Second order systems are, by definition, systems whose input-
output relationship is a second order differential equation.
• A second order differential equation contains a second order
derivative but no derivative higher than second order.
• Second order systems contain two independent energy
storage elements, per our previous comments pertaining to
the relationship between the number of energy storage
elements in a system and the system order.

3. 2nd-Order Differential Equation
• A 2nd-order differential equation has the form:
• Solution of a 2nd-order differential equation requires two
initial conditions: x(0) and x’(0)

4. • Since 2nd-order circuits have two energy-storage types, the circuits
can have the following forms:
1) Two capacitors
2) Two inductors
3) One capacitor and one inductor
A) Series RLC circuit
B) Parallel RLC circuit

5. Form Of The Solution To Differential Equations
• A 2nd-order differential equation has the form: order differential
equation has the form:
where x(t) is a voltage v(t) or a current i(t)
The general solution to differential equation has two parts:
• x(t) = xh + xp = homogeneous solution + particular solution
• where xh is due to the initial conditions in the circuit and xp is due to the
forcing

6. The Forced Response
• The forced response is due to the independent sources in the circuit
for t > 0.
• Since the natural response will die out once the circuit reaches
steady-state (under dc conditions), the forced response can be
found by analyzing the circuit at t = ∞.
• In particular,
xf = x(∞)

7. The Natural Response
• To find the natural response, set the forcing function f(t) (the right-
hand side of the DE) to zero.
• Substituting the general form of the solution Aest yields the
characteristic equation:
• Finding the roots of this quadratic (called the characteristic roots or
natural frequencies) yields:

8. Possible Responses
• The roots of the quadratic equation above may be real and distinct,
repeated, or complex. Thus, the natural response to a 2nd-order
circuit has 3 possible forms:
1) Overdamped response
2) Critically damped response
3) Underdamped response

9. Overdamped Response
• Roots are real and distinct [
(a1)2 > 4ao ]
• Here both A1 and A2 are real,
distinct and negative. The
general solution is given by
• The motion (current) is not
oscillatory, and the vibration
returns to equilibrium.
A1 + A2

10. Critically damped response
• Here the roots are negative, real
and equal,
• i.e. so A1 = A2 = s = -a1/2
• The general solution is given by
• The vibration (current) returns to
equilibrium in the minimum time
and there is just enough damping
to prevent oscillation.

11. Underdamped response
• Roots are complex [(a1)2 < 4ao ]
so s1 , s2 = α ± jβ
• Show that the solution has the
form:
• In this case, the motion
(current) is oscillatory and the
amplitude decreases
exponentially.`

12. All Possible Response at glance

13. Series and Parallel RLC Circuits
Two common second-order circuits are now considered:
• Series RLC circuits
• Parallel RLC circuits.
Relationships for these circuits can be easily developed such that the
characteristic equation can be determined directly from component
values without writing a differential equation for each example.
• A general 2nd-order characteristic equation has the form:
s2 + 2αs + wo
2 = 0
• Where α = damping coefficient, wo = resonant frequency

14. Suppose instead of a sinusoidal source we had a
slowly varying square waveform or a sharp turn on
of voltage. How would a LRC circuit behave?
We can start by using Kirchoff’s laws again.
R
Vin
C
L
Vout
C
Q
R
dt
dQ
dt
Qd
L
C
Q
dt
dI
LIRVVVV CLR


2
2
0
This is a second order differential equation that can be solve for the general and
particular solutions.
Transients in a Series RLC circuit

15. The solutions to the quadratic above determine the
form of the solutions. We will just state the
solutions for different value of R, L and C.
R
Vin
C
L
Vout
Overdamped:
21
LCRC






dampedCritically:
21
LCRC






DampedUnder:
21
LCRC






0
112

LC
x
RC
x

16. Since the voltage across the circuit is common to all
three circuit elements, the current through each
branch can be found using Kirchoff’s Current Law,
(KCL) which is given as:
This is a second order differential equation that can be solve for the general and
particular solutions.
Transients in a Parallel RLC circuit

17. Procedure for analyzing 2nd-order circuits
1) Find the characteristic equation and the natural response
[A] Determine if the circuit is a series RLC or parallel RLC (for t > 0
with independent sources shorted). If the circuit is not series RLC
or parallel RLC determine the describing equation of capacitor
voltage or inductor current.
[B] Obtain the characteristic equation. Use the standard formulas for
α and wo for a series RLC circuit or a parallel RLC circuit. Use these
values of α and wo in the characteristic equation as: s2 + 2αs + wo
2.

18. [C] Find the roots of the characteristic equation (characteristic roots).
[D] Determine the form of the natural response based on the type of
characteristic roots :
Real distinct Roots
Complex Roots
Repeated Roots

19. Procedure for analyzing 2nd-order circuits
2) Find the forced response – Analyze the circuit at t = ∞ to find xf = x(∞).
3) Find the initial conditions, x(0) and x’(0).
A) Find x(0) by analyzing the circuit at t = 0- find all capacitor voltages and
inductor currents)
B) Analyze the circuit at t = 0+ (use the values for vC and iL found at t = 0- in the
circuit) and find dvC(0+)/dt = iC(0+)/C or diL(0+)/dt = vL(0+)/L.
4) Find the complete response
A) Find the total response, x(t) = xh + xp
B) Use the two initial conditions to solve for the two unknowns in the total
response

20. THANK YOU