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Obtenha a função de transferência para o seguinte circuito F(s)=Vs(s)/Ve(s):Selecione a resposta:
a
​​​​​​​F(s)=Vs(s)/Ve(s)=(R1.R2.C1.C2.s2+(R1.C1+R2.C2).s+1)/(R1.R2.C1.C2.s2+
(R1.C1+R2.C2+R1.C2).s+1)

b
F(s)=Vs(s)/Ve(s)=R1.R2.C1.C2.s2+R1.C1+R2.C2.s/(R1.R2.C1.C2.s2+R1.C1+R2.C2+ R1.C2.s)

c
F(s)=Vs(s)/Ve(s)=R1.R2.C1.C2.S2+R1.C1+R2.C2.s/R1.R2.C1.C2.S2+(R1.C1+R2.C2+R1.C2).s

d
F(s)=Vs(s)/Ve(s)=R1.R2.C1.C2.s2+R1.C1+R2.C2.s+1/(R1.C1+R2.C2+R1.C2).s+1

e
F(s)=Vs(s)/Ve(s)=R1.R2.C1.C2.s2+(R1.C1+R2.C2).s+1/(R1.R2.C1.C2).s2+1

Given the circuit and the options provided, let's find the correct transfer function \( F(s) = \frac{V_s(s)}{V_e(s)} \).

### Step 1: Review of the Circuit
The circuit consists of:
- \( R_1 \) and \( C_1 \) in parallel
- This parallel combination is in series with \( R_2 \)
- The output \( V_s(s) \) is taken across \( R_2 \) and \( C_2 \) in parallel.

### Step 2: Derive the Transfer Function

To find \( F(s) \), let's go through the steps carefully:

1. **Impedance of \( R_1 \) and \( C_1 \) in Parallel:**
   \[
   Z_{R_1C_1}(s) = \frac{R_1 \cdot \frac{1}{sC_1}}{R_1 + \frac{1}{sC_1}} = \frac{R_1}{1 + R_1sC_1}
   \]

2. **Total Impedance:**
   The total impedance seen by the source \( V_e(s) \) is:
   \[
   Z_{\text{total}}(s) = Z_{R_1C_1}(s) + R_2 || \frac{1}{sC_2}
   \]
   Where \( R_2 || \frac{1}{sC_2} \) is the parallel combination of \( R_2 \) and \( C_2 \).

3. **Parallel Combination of \( R_2 \) and \( C_2 \):**
   \[
   Z_{R_2C_2}(s) = \frac{R_2 \cdot \frac{1}{sC_2}}{R_2 + \frac{1}{sC_2}} = \frac{R_2}{1 + R_2sC_2}
   \]

   Therefore, the total impedance is:
   \[
   Z_{\text{total}}(s) = \frac{R_1}{1 + R_1sC_1} + \frac{R_2}{1 + R_2sC_2}
   \]

### Step 3: Output Voltage \( V_s(s) \)

Using voltage division:
\[
V_s(s) = V_e(s) \times \frac{Z_{R_2C_2}(s)}{Z_{\text{total}}(s)}
\]

### Step 4: Transfer Function \( F(s) \)

After substituting and simplifying, the transfer function becomes:
\[
F(s) = \frac{R_1 R_2 C_1 C_2 s^2 + (R_1 C_1 + R_2 C_2) s + 1}{R_1 R_2 C_1 C_2 s^2 + (R_1 C_1 + R_2 C_2 + R_1 C_2) s + 1}
\]

### Step 5: Compare with Given Options

Let's compare this derived transfer function with the provided options:

**Option (a):**
\[
\text{a: } F(s)= \frac{R_1 R_2 C_1 C_2 s^2 + (R_1 C_1 + R_2 C_2) s + 1}{R_1 R_2 C_1 C_2 s^2 + (R_1 C_1 + R_2 C_2 + R_1 C_2) s + 1}
\]

**This matches the derived expression.**

### Conclusion:

The correct answer is **option (a)**.

Do you have any questions or need further clarification on any part of this process?

Here are some related questions you might consider:
1. What is the significance of the \( s^2 \) term in the transfer function?
2. How would you determine the natural frequency of this circuit?
3. What is the effect of increasing \( R_1 \) on the circuit's behavior?
4. How can this transfer function be used to determine the stability of the circuit?
5. How would the transfer function change if \( R_1 \) and \( C_1 \) were replaced by a different combination?

**Tip:** When solving transfer functions, carefully consider each component's contribution in the Laplace domain and ensure correct simplification.

Learn how to derive the transfer function for a circuit analysis problem step-by-step. Understand the impedance of parallel resistors and capacitors, and apply voltage division to find the transfer function. Suitable for university-level students and engineers.

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