Use Partial Fraction Expansion and Laplace Transform tables to solve for V. (t). Include which entries that you utilized.
Transfer function
A transfer function (also known as system function or network function) of a system, subsystem, or component is a mathematical function that modifies the output of a system in each possible input. They are widely used in electronics and control systems.
Convolution Integral
Among all the electrical engineering students, this topic of convolution integral is very confusing. It is a mathematical operation of two functions f and g that produce another third type of function (f * g) , and this expresses how the shape of one is modified with the help of the other one. The process of computing it and the result function is known as convolution. After one is reversed and shifted, it is defined as the integral of the product of two functions. After producing the convolution function, the integral is evaluated for all the values of shift. The convolution integral has some similar features with the cross-correlation. The continuous or discrete variables for real-valued functions differ from cross-correlation (f * g) only by either of the two f(x) or g(x) is reflected about the y-axis or not. Therefore, it is a cross-correlation of f(x) and g(-x) or f(-x) and g(x), the cross-correlation operator is the adjoint of the operator of the convolution for complex-valued piecewise functions.
I have solved parts 1, and 2 I need help solving part 3
![### Circuit Analysis in the S-domain
#### Given Network Details:
- **Components:**
- Resistor: \(2 \Omega\)
- Inductor: \(1 \, \text{H}\)
- Capacitor: \(\frac{1}{2} \, \text{F}\)
- Voltage source: \( e^{-tu(t)} \)
#### Task 1: Redraw and Label the Circuit in S-domain
- **Circuit Conversion:**
- The resistor remains \(2 \Omega\).
- The inductor becomes \(s \, \text{H} = 1 s\).
- The capacitor becomes \(\frac{2}{s} \, \text{F}\).
- **Diagram Description:**
- A series configuration is depicted with elements in the order: 2Ω resistor, s (representing the inductor), and \( \frac{2}{s} \) (representing the capacitor in parallel, labeled \(V_o(s)\)).
#### Task 2: Use Voltage Division to Find \(V_o(s)\)
- **Voltage Division Steps:**
1. **Current Calculation:**
- \( I(s) = \frac{1}{s+1} \cdot \frac{1}{2s + \frac{2}{s}}\)
- Simplifies to:
- \( \frac{1}{(s+1)(2s^2 + s)} \rightarrow \frac{s}{(s+1)(2s^2 + 2s)} \)
2. **Voltage Calculation:**
- \( V_o(s) = \frac{2}{s} \cdot I(s) \)
- Simplifies to:
- \( \frac{s}{s^2 + \frac{5}{2}s + 2} \)
- \( \Rightarrow \frac{2}{(s^2 + 2)(s+1)} \)
### Conclusion:
- The final expression for the output voltage \( V_o(s) \) is:
\[ V_o(s) = \frac{2}{(s^2 + 2)(s+1)} \]
This representation in the S-domain provides a clear framework for analyzing the circuit's behavior under Laplace transform conditions, assuming zero initial conditions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff975724c-f6e6-4f11-80b9-86a7fa975ad8%2F3b4c834b-d3fb-4dc7-a568-8694d4c3ecf6%2F4ouoju_processed.png&w=3840&q=75)
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