Transfer Function: Determine Vo for Circuit in Fig. 5

[Fronzbot]

Homework Statement

Determine the transfer function for the circuit shown in figure 5 if the output voltage is defined as the voltage drop across the 20$$\Omega$$ resistor. Note there are two sources in this circuit.

Homework Equations

H(s) = $$\frac{Vo}{Vi}$$

The Attempt at a Solution

From the previous question (I needed to find the Voltage and Current across the Inductor and the voltage across the Capacitor all in the 'S' domain) I know that, assuming V_bis the node voltage where the cap, inductor and resistor meet, V_b=$$\frac{120(5s^2+20s+11)}{s(s^2+5s+6)}$$ and, assuming V_a is the node voltage at the current source, V_a=$$\frac{60(11s^2+51s+22)}{s(s^2+5s+6)}$$.

Now what I'm having trouble with is determining what I should use for V_i. Every problem in my book or in class has defined an input voltage somewhere and all of them have only ever had one source. My idea is that I could use superposition to find the transfer function so I determine V_o when the current source is opened and V_i=$$\frac{60}{s}$$V and then add that to the transfer function when the voltage source is shorted and V_i= V_a. Is this a valid approach?

 

[KataKoniK]

Dear student,

Thank you for your forum post. It seems like you are on the right track with using superposition to find the transfer function for this circuit. As you mentioned, there are two sources in this circuit, so it is important to consider each one separately when determining the transfer function.

First, let's define our output voltage as the voltage drop across the 20Ω resistor, which we can call Vout. We can also define our input voltage, Vi, as the voltage at the node where the two sources are connected. Now, using superposition, we can determine the transfer function H(s) as follows:

1. When the current source is opened, the voltage source is still connected, so we can consider the voltage source as our input and use the formula H(s) = Vout/Vi. In this case, Vi will be equal to the voltage at the node where the voltage source is connected, which we can call V1. Using the voltage divider formula, V1 = 120/(1+sRC), where RC is the parallel combination of the 5Ω resistor and the 1/(sC) impedance. Therefore, our transfer function in this case will be:

H1(s) = Vout/V1 = (20Ω)/(5Ω + 1/(sC)) = 4sC/(1+sRC)

2. When the voltage source is shorted, the current source is still connected, so we can consider the current source as our input and use the formula H(s) = Vout/Vi. In this case, Vi will be equal to the voltage at the node where the current source is connected, which we can call V2. Using KCL at this node, we can write the following equation:

(60V - V2)/10Ω + (V2 - Vout)/20Ω = 0

Solving for V2, we get V2 = 3Vout. Therefore, our transfer function in this case will be:

H2(s) = Vout/V2 = (20Ω)/(10Ω + 1/(sC)) = 2sC/(1+sRC)

3. Now, we can add the two transfer functions (H1 and H2) together to get the overall transfer function:

H(s) = H1(s) + H2(s) = 4sC/(1+sRC) +