For t > 0, what is the output voltage Vc(t) of the series RLC circuit excited by a unit impulse function δ(t)?

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Understand the Problem

The question is asking for the output voltage Vc(t) of a series RLC circuit that has been excited by a unit impulse function δ(t). It provides several possible expressions for Vc(t) and asks the respondent to identify the correct one for t > 0.

Answer

The output voltage $V_c(t)$ is given by: $$ V_c(t) = \frac{2}{\sqrt{3}} e^{-\frac{1}{2}t} \sin\left(\frac{\sqrt{3}}{2}t\right) $$
Answer for screen readers

The correct expression for the output voltage $V_c(t)$ for $t > 0$ is:

$$ V_c(t) = \frac{2}{\sqrt{3}} e^{-\frac{1}{2}t} \sin\left(\frac{\sqrt{3}}{2}t\right) $$

Steps to Solve

  1. Identify the Circuit Parameters

The given RLC circuit consists of a resistor $R = 1 , \Omega$, an inductor $L = 1 , H$, and a capacitor $C = 1 , F$.

  1. Determine the Characteristic Equation

The characteristic differential equation for a series RLC circuit can be expressed as:

$$ L \frac{d^2 V_c(t)}{dt^2} + R \frac{dV_c(t)}{dt} + \frac{1}{C} V_c(t) = 0 $$

Substituting the circuit values:

$$ \frac{d^2 V_c(t)}{dt^2} + V_c(t) = 0 $$

  1. Find the Damping Factor

The damping factor $\zeta$ and natural frequency $\omega_0$ are given by:

$$ \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1 \cdot 1}} = 1 $$

Given that $R < 2\sqrt{L/C}$, this circuit is underdamped.

  1. Write the Solution for Underdamped Response

The general solution for the underdamped response is:

$$ V_c(t) = e^{-\zeta \omega_0 t} \left( A \cos(\omega_d t) + B \sin(\omega_d t) \right) $$

Where $\omega_d = \omega_0 \sqrt{1 - \zeta^2}$. Since $\zeta = \frac{R}{2\sqrt{L/C}} = \frac{1}{2}$, we find:

$$ \omega_d = 1 \sqrt{1 - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2} $$

  1. Apply Initial Conditions

Since the circuit is excited by a unit impulse, the initial conditions are:

$$ V_c(0) = 0 \quad \text{and} \quad \frac{dV_c(t)}{dt} \bigg|_{t=0} = 1 $$

  1. Solve for Coefficients A and B

We substitute the values into the general solution and apply the initial conditions:

$$ V_c(t) = e^{-\frac{1}{2} t} \left( A \cos\left(\frac{\sqrt{3}}{2}t\right) + B \sin\left(\frac{\sqrt{3}}{2}t\right) \right) $$

  1. Find the Final Expression

After substituting initial conditions, we can derive that the output voltage $V_c(t)$ takes the form. The coefficient values are determined to give us one of the proposed answers.

The correct expression for the output voltage $V_c(t)$ for $t > 0$ is:

$$ V_c(t) = \frac{2}{\sqrt{3}} e^{-\frac{1}{2}t} \sin\left(\frac{\sqrt{3}}{2}t\right) $$

More Information

This expression represents the transient response of the output voltage across the capacitor in an underdamped RLC circuit due to the impulse input. The form includes an exponential decay multiplied by a sinusoidal function, which is typical for underdamped responses.

Tips

  • Forgetting to apply the initial conditions correctly.
  • Confusing between oscillatory and non-oscillatory responses in RLC circuits.
  • Incorrectly calculating the damping ratio and natural frequency.
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