91Ó°ÊÓ

Consider a system whose input \(x(t)\) and output \(y(t)\) satisfy the first-order differential equation $$\frac{d y(t)}{d t}+2 y(t)=x(t)$$ The system also satisfies the condition of initial rest. (a) (i) Determine the system output \(y_{1}(t)\) when the input is \(x_{1}(t)=e^{3 t} u(t)\) (ii) Determine the system output \(y_{2}(t)\) when the input is \(x_{2}(t)=e^{2 t} u(t)\) (iii) Determine the system output \(y_{3}(t)\) when the input is \(x_{3}(t)=\alpha e^{3 t} u(t)+\) \(\beta e^{2 t} u(t),\) where \(\alpha\) and \(\beta\) are real numbers. Show that \(y_{3}(t)=\alpha y_{1}(t)+\) \(\beta y_{2}(t)\) (iv) Now let \(x_{1}(f)\) and \(x_{2}(f)\) be arbitrary signals such that $$\begin{aligned} &x_{1}(t)=0, \text { for } t

Step by step solution

01

Solve the Differential Equation for Input Function

Given the differential equation \(\frac{dy(t)}{dt} + 2y(t) = x(t)\), substitute \(x_1(t) = e^{3t}u(t)\). To solve, assume a solution of the form \(y(t) = Ae^{3t}\). Substitute into the equation, solving for \(A\) under the condition of initial rest.

02

Determine Output for Input \(x_1(t) = e^{3t}u(t)\)

With \(x_1(t) = e^{3t}u(t)\), solve the differential equation: \(\frac{dy_1(t)}{dt} + 2y_{1}(t) = e^{3t}\). Setting \(y_1(0) = 0\), solve this to obtain \(y_{1}(t)\). The result is \(y_1(t) = \frac{1}{5}e^{3t} - \frac{1}{5}e^{-2t}\).

03

Determine Output for Input \(x_2(t) = e^{2t}u(t)\)

Substitute \(x_2(t) = e^{2t}u(t)\) into the differential equation: \(\frac{dy_{2}(t)}{dt} + 2y_{2}(t) = e^{2t}\). By solving and applying initial conditions, derive \(y_2(t) = \frac{1}{2}e^{2t} - \frac{1}{2}e^{-2t}\).

04

Determine Output for Combined Input \(x_3(t)\)

For \(x_3(t) = \alpha e^{3t} + \beta e^{2t}\), use superposition principle: substitute into the differential equation and solve to find \(y_3(t) = \alpha y_1(t) + \beta y_2(t)\). Thus, the system demonstrates linearity.

05

Show Time Invariance for Input \(x_1(t) = Ke^{2t}u(t)\)

With \(x_1(t) = Ke^{2t}u(t)\), solve the same differential equation, yielding \(y_1(t) = \frac{K}{2}e^{2t} - \frac{K}{2}e^{-2t}\).

06

Show Time Invariance for Shifted Input \(x_2(t) = Ke^{2(t-T)}u(t-T)\)

Shift \(y_1(t)\) found in Step 5 by \(T\): \(y_2(t) = y_1(t-T) = \frac{K}{2}e^{2(t-T)} - \frac{K}{2}e^{-2(t-T)}\). This shows time invariance.

07

Conclusion on System Linearity and Time Invariance

The system is linear (Step 4) and time-invariant (Step 6), thus it is a Linear Time-Invariant (LTI) system. Initial rest confirms causality.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations

To understand the behavior of Linear Time-Invariant (LTI) systems, we often rely on differential equations. A differential equation relates the input and output through derivatives. In this case, the differential equation is \( \frac{dy(t)}{dt} + 2y(t) = x(t) \). Here, \( y(t) \) is the output, and \( x(t) \) is the input.
The process involves applying techniques to solve this equation, typically assuming a form for \( y(t) \), and using any initial conditions known, such as initial rest. This means at the beginning \( y(0) = 0 \). Solutions tell us how the system responds over time.

System Linearity

One of the characteristics of LTI systems is linearity. A system is linear if it satisfies the principles of superposition, which comprises two main conditions: homogeneity and additivity.
- **Homogeneity**: If the input is scaled, the corresponding output should also scale by the same factor.- **Additivity**: If two inputs are applied simultaneously, the output should be the sum of the individual outputs from each input alone.
Demonstrating linearity involves showing that \( y_3(t) = \alpha y_1(t) + \beta y_2(t) \) when \( x_3(t) = \alpha x_1(t) + \beta x_2(t) \). Thus, linearity verifies that the superposition principle is satisfied.

Causality

Causality in systems means that the output at any time \( t \) depends only on input values at the present time and in the past, not future inputs. For our system, causality is integral, and it's considered under the condition of initial rest.
This means that the system output \( y(t) \) is only influenced by \( x(t) \) for \( t \geq 0 \). This makes intuitive sense as it aligns with the cause and effect principle observed in the real world. For an LTI system, causality implies that knowing the input up to any time \( t \) allows us to determine the output up to that point.

Time Invariance

Time invariance signifies that the system's behavior does not change over time. If an input signal \( x(t) \) leads to output \( y(t) \), then a time-shifted input signal \( x(t-T) \) should yield a time-shifted output \( y(t-T) \).
In our exercise, shifting the input \( x_1(t) \) in time led to a corresponding shift in the output by the same amount \( T \), confirmed by checking \( y_2(t) = y_1(t-T) \).
This property is crucial as it ensures the consistency of the system's response over time, making systems more predictable and easier to analyze.