Math Problem Statement

Find the transfer function Vo(s)/Vi(s) for the given circuit involving resistors, inductors, and capacitors in the Laplace domain.

Solution

This circuit looks like a combination of resistors, inductors, and capacitors, and you're tasked with finding the transfer function Vo(s)Vi(s)\frac{V_o(s)}{V_i(s)} in the Laplace domain.

Here's a step-by-step outline of how we can solve this:

1. Impedances in the Laplace Domain:

  • For the resistor, the impedance is R=6ΩR = 6 \, \Omega.
  • For the inductor, the impedance is ZL=sLZ_L = sL, where LL is the inductance. So, for the two inductors:
    • The impedance of the first inductor is s×1=ss \times 1 = s.
    • The impedance of the second inductor is 2s2s.
  • For the capacitor, the impedance is ZC=1sCZ_C = \frac{1}{sC}. The capacitor's impedance is given as 1s\frac{1}{s}, meaning C=1FC = 1 \, \text{F}.

2. Equivalent Impedance of the Circuit:

We need to analyze how these components are connected.

  • The voltage source is on the left with Vi(s)V_i(s).
  • There is a series combination of 3Ω3 \, \Omega resistance, sΩs \, \Omega inductance, and 2sΩ2s \, \Omega inductance.
  • The output voltage Vo(s)V_o(s) is across the parallel combination of:
    • A 6 Ω\Omega resistor.
    • A 1sΩ\frac{1}{s} \, \Omega capacitor.

Let's first find the equivalent impedance of the parallel section: Zparallel=(16+s1)1=61+6sZ_{parallel} = \left( \frac{1}{6} + \frac{s}{1} \right)^{-1} = \frac{6}{1 + 6s}

3. Overall Impedance:

Now, the total impedance ZtotalZ_{total} in series with the voltage source will be the sum of the series components and the parallel impedance: Ztotal=3+s+2s+61+6s=3+3s+61+6sZ_{total} = 3 + s + 2s + \frac{6}{1 + 6s} = 3 + 3s + \frac{6}{1 + 6s}

4. Transfer Function:

The transfer function Vo(s)Vi(s)\frac{V_o(s)}{V_i(s)} is given by the voltage divider formula: Vo(s)Vi(s)=ZparallelZtotal\frac{V_o(s)}{V_i(s)} = \frac{Z_{parallel}}{Z_{total}} Substituting the values: Vo(s)Vi(s)=61+6s3+3s+61+6s\frac{V_o(s)}{V_i(s)} = \frac{\frac{6}{1 + 6s}}{3 + 3s + \frac{6}{1 + 6s}}

Now, we can simplify this expression to get the transfer function in a more usable form. I'll proceed with this simplification.The transfer function Vo(s)Vi(s)\frac{V_o(s)}{V_i(s)} simplifies to:

Vo(s)Vi(s)=2(s+1)(6s+1)+2\frac{V_o(s)}{V_i(s)} = \frac{2}{(s + 1)(6s + 1) + 2}

This is the expression for the ratio of the output voltage to the input voltage in the Laplace domain.

Would you like further details on this, or any specific questions about the steps? Here are five questions you might explore:

  1. How would you analyze the behavior of this transfer function as s0s \to 0 (DC analysis)?
  2. What is the steady-state gain of this system?
  3. How would this circuit behave for high-frequency signals (as ss \to \infty)?
  4. How can you convert this transfer function back to the time domain?
  5. What is the physical interpretation of the poles and zeros of this transfer function?

Tip: Understanding the poles and zeros of a transfer function helps in predicting the stability and frequency response of a circuit.

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Math Problem Analysis

Mathematical Concepts

Circuit Analysis
Laplace Transform
Impedance
Transfer Function

Formulas

Impedance of Resistor: Z_R = R
Impedance of Inductor: Z_L = sL
Impedance of Capacitor: Z_C = 1/(sC)
Transfer Function: Vo(s)/Vi(s) = Z_parallel / Z_total

Theorems

Voltage Divider Rule
Laplace Transform of Circuit Elements

Suitable Grade Level

Undergraduate Electrical Engineering