Find the transfer function Recircuit

a) Find the transfer function Recircuit shown in Fig. P13.48(0)
b) Find the transfer function H(s) V/V for the circuit shown in Fig. P13.48(b).
c) Create two different circuits that have the transfer function H(s) V/V 1000/(s+1000). Use components selected from Appendix H and Figs. P13.48(a) and (b).

The Correct Answer and Explanation is :

To address the given problems, we need to analyze the circuits depicted in Figures P13.48(a) and P13.48(b) and derive their transfer functions. Additionally, we’ll design two distinct circuits that realize the specified transfer function using standard components.

a) Transfer Function for the Circuit in Fig. P13.48(a):

Assuming Figure P13.48(a) represents an RC low-pass filter, the circuit consists of a resistor (R) in series with a capacitor (C), and the output voltage (V₀) is taken across the capacitor.

The transfer function H(s) is defined as the ratio of the output voltage to the input voltage (V₀/Vᵢ) in the Laplace domain.

For an RC low-pass filter, the transfer function is:

H(s) = V₀(s) / Vᵢ(s) = 1 / (1 + sRC)

Where:

• s = jω (complex frequency)
• R = resistance
• C = capacitance

b) Transfer Function for the Circuit in Fig. P13.48(b):

Assuming Figure P13.48(b) depicts an RL high-pass filter, the circuit comprises a resistor (R) in series with an inductor (L), and the output voltage (V₀) is taken across the resistor.

The transfer function H(s) for an RL high-pass filter is:

H(s) = V₀(s) / Vᵢ(s) = sL / (R + sL)

c) Designing Two Circuits with Transfer Function H(s) = 1000 / (s + 1000):

The given transfer function resembles a standard low-pass filter with a cutoff frequency of 1000 rad/s. To achieve this transfer function, we can design:

1. RC Low-Pass Filter: For a low-pass filter, the transfer function is: H(s) = 1 / (1 + sRC) To match the given H(s), we set 1/RC = 1000, which implies RC = 1 ms. Using standard component values from Appendix H, we can select:

• R = 1 kΩ
• C = 1 μF This combination yields RC = (1 kΩ)(1 μF) = 1 ms, satisfying the required transfer function.

1. RL Low-Pass Filter: An RL circuit can also realize a similar transfer function. The standard form is: H(s) = R / (R + sL) To match H(s) = 1000 / (s + 1000), we set R = 1000 Ω and L/R = 1 ms, leading to L = 1 H. Using standard components:

• R = 1 kΩ
• L = 1 H This configuration provides the desired transfer function.

Explanation:

The transfer function H(s) = 1000 / (s + 1000) indicates a system with a pole at s = -1000, corresponding to a cutoff frequency of 1000 rad/s. This is characteristic of a first-order low-pass filter, which allows frequencies below the cutoff to pass while attenuating higher frequencies.

In the RC low-pass filter, the resistor and capacitor create a frequency-dependent voltage divider. At low frequencies, the capacitor’s impedance is high, allowing most of the input voltage to appear across it. As frequency increases, the capacitor’s impedance decreases, shunting more current away from the output, thus reducing the output voltage.

For the RL circuit, while it’s typically a high-pass filter, by taking the output across the resistor, the configuration can mimic a low-pass response. At low frequencies, the inductor’s impedance is low, and the voltage drop across the resistor is significant. As frequency increases, the inductor’s impedance rises, reducing the voltage across the resistor.

Selecting standard component values ensures practical implementation. Standard resistors and capacitors are available in preferred values, facilitating the construction of the desired filter with readily available components.

In summary, by analyzing the desired transfer function and understanding the frequency-dependent behavior of resistors, capacitors, and inductors, we can design circuits that meet specific filtering requirements using standard components.

Circuit Diagrams:

1. RC Low-Pass Filter:

   Vi ────R───────┬──── V₀
                  │
                  C
                  │
                 GND

• R = 1 kΩ
• C = 1 μF

1. RL Low-Pass Filter:

   Vi ────L───────┬──── V₀
                  │
                  R
                  │
                 GND

• L = 1 H
• R = 1 kΩ

These diagrams illustrate the configurations of the RC and RL circuits that achieve the specified transfer function.