1. EENG 2610: Circuit Analysis Class 13: Second-Order Circuits
Oluwayomi Adamo
Department of Electrical Engineering
College of Engineering, University of North Texas

2. The Basic Circuit Equation
Single Node-pair: Use KCL
Single Loop: Use KVL
Differentiating
Differentiating

3. The Response Equations
We need solution for the second-order differential equation:
The solution is:
The complementary solution satisfies the homogeneous equation:

4. Now we focus on homogeneous equation:
Rewrite the equation:
The solution of this equation has the form:
Substitute this solution into the homogeneous equation, we can obtain the characteristic equation:
Solution of the characteristic equation (or natural frequencies):

5. Homogeneous equation:
characteristic equation:
Case 1: Overdamped,
Natural response is the sum of two decaying exponentials:
Case 2: Underdamped,
Natural response is an exponentially damped oscillatory response:
Case 3: Critically damped,

6. Critically damped
Underdamped
(ringing)
Overdamped
Envelope
Figure: Comparison of overdamped, critically damped, and underdamped responses

7. Second-Order Transient Aanalysis
Step 1: Write differential equations that describes the circuit.
Step 2: Derive the characteristic equation, which can be written in the form s2+2ζω0s+ω02=0, where ζ is the damping ratio, ω0 is the undamped natural frequency.
Step 3: The two roots of the characteristic equation will determine the type of response:
If the roots are real and unequal (i.e., ζ>1), the network response is overdamped.
If the roots are real and equal (i.e., ζ=1), the network response is critically damped.
If the roots are complex (i.e., ζ<1), the network response is underdamped.

8. Step 4: The damping condition and corresponding response for the aforementioned three cases outlined are as follows:
Overdamped (ζ>1):
Critically damped (ζ=1):
Underdamped (ζ<1):
Step 5: Two initial conditions, either given or derived, are required to obtain the two unknown coefficients in the response equation.

9. To determine the constants we need
Example 7.7
To determine the constants we need

10. LEARNING EXAMPLE
NO SWITCHING OR
DISCONTINUITY AT t=0.
USE t=0 OR t=0+