Question

a) Assume the voltage impulse response of a circuit is $$h(t)=\left\\{\begin{array}{ll}0, & t<0 \\\10 e^{-4 t}, & t \geq 0\end{array}\right.$$ Use the convolution integral to find the output voltage if the input signal is \(10 u(t)\) V. b) Repeat (a) if the voltage impulse response is $$h(t)=\left\\{\begin{array}{ll}0, & t<0 \\\10(1-2 t), & 0 \leq t \leq 0.5 \mathrm{s} \\\0, & t \geq 0.5 \mathrm{s}\end{array}\right.$$ c) Plot the output voltage versus time for (a) and (b) for \(0 \leq t \leq 1\) s.

Short answer

Part (a): \(y(t) = 25(1-e^{-4t})\). Part (b): \(y(t) = 100(t-t^2)\) for \(0 \leq t \leq 0.5\), \(y(t) = 25\) for \(t > 0.5\).\nPlots show respective functions.

Step-by-step solution

Understand the Convolution Integral

The convolution integral is used to find the output of an LTI (Linear Time-Invariant) system when the input and impulse response are given. It is defined as: \( y(t) = \int_{-\infty}^{\infty} h(\tau) x(t - \tau) d\tau \) where \( h(t) \) is the impulse response and \( x(t) \) is the input signal.

Identify Functions for Part (a)

For part (a), the impulse response is \( h(t) = 10e^{-4t} \) for \( t \geq 0 \), and the input signal is \( x(t) = 10u(t) \). The unit step function \( u(t) \) is 1 for \( t \geq 0 \) and 0 otherwise.

Set Up Convolution Integral for Part (a)

Substitute the given functions into the convolution integral: \[ y(t) = \int_{0}^{t} 10e^{-4\tau} \, 10 \, d\tau \] since for \( t < 0 \) the output is zero because \( x(t) = 0 \).

Solve the Integral for Part (a)

Evaluate the integral: \[ y(t) = 100 \int_{0}^{t} e^{-4\tau} \, d\tau = 100 \left[ -\frac{1}{4} e^{-4\tau} \right]_{0}^{t} = 25 \left( 1 - e^{-4t} \right) \] for \( t \geq 0 \).

Identify Functions for Part (b)

For part (b), the impulse response is \( h(t) = 10(1-2t) \) for \( 0 \leq t \leq 0.5 \) and 0 otherwise, and \( x(t) = 10u(t) \) as before.

Set Up Convolution Integral for Part (b)

Substitute the given functions for the integral: \[ y(t) = \int_{0}^{t} h(\tau) \, 10 \, d\tau = 100 \int_{0}^{min(t,0.5)} (1-2\tau) \, d\tau \] for \( t \geq 0 \).

Solve the Integral for Part (b)

Evaluate the integral separately for \( t \leq 0.5 \) and \( t > 0.5 \):For \( 0 \leq t \leq 0.5 \), \[ y(t) = 100 \left[ \tau - \tau^2 \right]_{0}^{t} = 100(t - t^2) \].For \( t > 0.5 \), \[ y(t) = 100 \left[ \tau - \tau^2 \right]_{0}^{0.5} = 100 \times 0.25 = 25 \].

Prepare Plots for Output Voltage

For part (a), plot \( y(t) = 25(1-e^{-4t}) \) over \( 0 \leq t \leq 1 \). For part (b), plot \( y(t) = 100(t - t^2) \) for \( 0 \leq t \leq 0.5 \), and \( y(t) = 25 \) for \( 0.5 < t \leq 1 \). Both plots should show the time on the x-axis and voltage on the y-axis.

Key concepts

LTI Systems

Linear Time-Invariant (LTI) systems are a fundamental concept in signal processing and systems engineering. LTI systems have two main properties: they are linear and time-invariant. These systems adhere to the principles of superposition and homogeneity, meaning the output for a weighted sum of inputs is the same as the sum of the outputs for each input separately.

• **Linearity:** The system's response to a weighted sum of inputs is the sum of the responses to each input weighted individually.
• **Time-Invariance:** The system's characteristics do not change over time. If an input is shifted in time, the output is also shifted by the same amount, without altering its shape or magnitude.

LTI systems are prevalent because they are mathematically manageable and have predictable behaviors, making them ideal for various applications like filtering, control systems, and communications.

Impulse Response

The impulse response of an LTI system is the output when the input is an impulse signal, typically represented by the delta function, \( \delta(t) \). This response characterizes the system's dynamics.

• The impulse response, denoted as \( h(t) \), acts as a "fingerprint" of the system, encapsulating all the information about system's characteristics.
• It allows us to predict the system's behavior with any arbitrary input, using the convolution integral.

The impulse response is crucial because, when combined with the convolution integral, it provides the complete solution for calculating the output of an LTI system for any given input.

Unit Step Function

The unit step function, denoted as \( u(t) \), is a fundamental function in signal processing that serves as a simple model for switch-like signals. It is defined as:

• \( u(t) = 0 \) for \( t < 0 \)
• \( u(t) = 1 \) for \( t \geq 0 \)

The unit step function is often used to "turn on" signals at a specific point and is integral in defining causal systems. In the context of LTI systems, it can be used to represent inputs that start at a specific time, such as the input in our example, where \( x(t) = 10u(t) \). This helps in understanding and computing the system's response using convolution.

Input Signal

The input signal to a system represents the external driving force applied to a system, which produces an output. In signal processing, analyzing the input signal is crucial for predicting system behavior.

• In our example, the input signal is \( x(t) = 10u(t) \), representing a constant voltage that begins at time \( t=0 \).
• This type of signal can often simplify the analysis of systems by allowing us to focus on the response characteristics determined by the impulse response.

Understanding the role of the input signal in the context of an LTI system allows us to apply the convolution integral effectively, calculating the final output based on how it interacts with the system's impulse response. This combination showcases the beauty of linear system theory and its practical value in engineering disciplines.