Second-Order Circuit

Finding Initial and Final Values

• Remember that capacitor voltage and inductor current are continuous
• Analyze the values right before and after the switching event

Example 08.2.

(Example 8.1 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 339])

Example 08.3.

Practice Problem 8.1 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 340]

Example 08.4.

(Example 8.2 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 341])

Example 08.5.

(Practice Problem 8.2 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 343])

Natural Response of Series RLC Circuit

• Use KVL to obtain the characteristic equation

• Use initial conditions to solve for $A_{1}$ and $A_{2}$

• Trying to solve for the shared current, $i(t)$

• Characteristic equation:

• Roots (natural frequencies): $s_{1, 2} = -\alpha \pm \sqrt{ a^2 - \omega_{0}^2}$ where $\alpha = \frac{1}{2} \frac{R}{L}$ and $\omega_{0} = \frac{1}{\sqrt{ LC }}$

Damping

• The gradual loss of initial stored energy, evidenced by the decrease in the amplitude of the response
• Dependent on the value of $R$; if there is no $R$, $\alpha = 0$

Overdamped $(\alpha > \omega_{0})$

• Both roots are negative and real
• Decays the slowest

Critically Damped $(\alpha = \omega_{0})$

• Decays the quickest without any ringing

Underdamped $(\alpha < \omega_{0})$

• Two imaginary roots
• Oscillatory response (ringing) due to the storage devices transferring energy to each other
• This means $s_{1, 2} = -\alpha \pm \sqrt{-(\omega_{0}^2 - \alpha^2)} = -\alpha \pm j \omega_{d}$

Example 08.6.

(Example 8.3 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 348])

Example 08.7.

(Practice Problem 8.3 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 348])

Example 08.8.

(Example 8.4 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 348])

Example 08.9.

(Practice Problem 8.4 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 350])

Natural Response of Parallel RLC Circuit

• Use KCL to obtain the characteristic equation

• Use initial conditions to solve for $A_{1}$ and $A_{2}$

• Dampening cases remain the same

• Trying to solve for the shared voltage, $v(t)$

• Characteristic equation:

• Natural frequencies: $s_{1, 2} = -\alpha \pm \sqrt{ a^2 - \omega_{0}^2}$ where $\alpha = \frac{1}{2} \frac{1}{RC}$ and $\omega_{0} = \frac{1}{\sqrt{ LC }}$

Example 08.10.

(Example 8.5 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 352])

Example 08.11.

(Practice Problem 8.5 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 354])

Example 08.12.

(Example 8.6 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 354])

Example 08.13.

(Practice Problem 8.6 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 355])

Step Response of Series RLC Circuit

• Superposition of the natural response (transient response) and steady-state response $v(\infty)$
• However, we are now evaluating the voltage response of the capacitor and not the shared current

Example 08.14.

(Example 8.7 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 357])

Example 08.15.

(Practice Problem 8.7 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 360])

Step Response of Parallel RLC Circuit

• Current response is equal to the transient response and the steady-state response $i(\infty)$

Example 08.16.

(Example 8.8 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 361])

Example 08.17.

(Practice Problem 8.8 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 363])

General Second-Order Circuits

• Our approach for analyzing series/parallel RLC circuits may be extended for any second-order circuit:
  • Evaluate the initial conditions $(x(0), \frac{\mathrm{d}x}{\mathrm{d}t}(0))$ and steady-state response $(x(\infty))$
  • Turn off all independent sources and obtain the transient response using KCL/KVL
  • Determine the characteristic roots of the resulting differential equation
  • The entire response is the sum of the transient response and steady-state response

Example 08.18.

(Example 8.9 [Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., p. 363])