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2. The input signal x(t) shown below is a step function with an amplitude = 5. A step function has a value of zero before t=0, but has a value equal to its amplitude for all time after t=0. The x(t) is applied to a LTI system described by the impulse response, h(t) shown below. Find the output y(t) of the LTI system (Assume y(t) = x(t) * h(t)). Show all work. h(t) 5 x(t) 2 time (t, sec) 0 1 -2 -1 3 0 time (t, sec) 1 2

2. The input signal x(t) shown below is a step function with an amplitude = 5. A step function has a value of zero before t=0, but has a value equal to its amplitude for all time after t=0. The x(t) is applied to a LTI system described by the impulse response, h(t) shown below. Find the output y(t) of the LTI system (Assume y(t) = x(t) * h(t)). Show all work. h(t) 5 x(t) 2 time (t, sec) 0 1 -2 -1 3 0 time (t, sec) 1 2

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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**Problem Statement:** 2. The input signal \( x(t) \) shown below is a step function with an amplitude of 5. A step function has a value of zero before \( t = 0 \), but has a value equal to its amplitude for all time after \( t = 0 \). The \( x(t) \) is applied to an LTI (Linear Time-Invariant) system described by the impulse response \( h(t) \) shown below. Find the output \( y(t) \) of the LTI system (Assume \( y(t) = x(t) * h(t) \)). Show all work. **Diagrams Explanation:** **Step Function \( x(t) \) Graph:** - **x-axis (Time, t in seconds):** - The time axis ranges from 0 to 2 seconds. - **y-axis (Amplitude):** - The amplitude axis ranges from 0 to 5. - **Function Behavior:** - From \( t < 0 \), \( x(t) = 0 \). - From \( t = 0 \) onwards, \( x(t) = 5 \). This maintains a constant value (step) until \( t = 2 \). **Impulse Response \( h(t) \) Graph:** - **x-axis (Time, t in seconds):** - The time axis ranges from -2 to 2 seconds. - **y-axis (Amplitude):** - The amplitude axis ranges from 0 to 3. - **Function Behavior:** - The impulse response forms a triangle centered at \( t = 0 \). - At \( t = -1 \), \( h(t) \) rises linearly to a peak of 3 at \( t = 0 \). - From \( t = 0 \), it decreases linearly back to 0 at \( t = 1 \). - \( h(t) = 0 \) outside the interval \([-1, 1]\). **Solution Approach:** To find the output \( y(t) \) of the system, perform the convolution of \( x(t) \) and \( h(t) \). The convolution formula is given by: \[ y(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) d\t
Transcribed Image Text:**Problem Statement:** 2. The input signal \( x(t) \) shown below is a step function with an amplitude of 5. A step function has a value of zero before \( t = 0 \), but has a value equal to its amplitude for all time after \( t = 0 \). The \( x(t) \) is applied to an LTI (Linear Time-Invariant) system described by the impulse response \( h(t) \) shown below. Find the output \( y(t) \) of the LTI system (Assume \( y(t) = x(t) * h(t) \)). Show all work. **Diagrams Explanation:** **Step Function \( x(t) \) Graph:** - **x-axis (Time, t in seconds):** - The time axis ranges from 0 to 2 seconds. - **y-axis (Amplitude):** - The amplitude axis ranges from 0 to 5. - **Function Behavior:** - From \( t < 0 \), \( x(t) = 0 \). - From \( t = 0 \) onwards, \( x(t) = 5 \). This maintains a constant value (step) until \( t = 2 \). **Impulse Response \( h(t) \) Graph:** - **x-axis (Time, t in seconds):** - The time axis ranges from -2 to 2 seconds. - **y-axis (Amplitude):** - The amplitude axis ranges from 0 to 3. - **Function Behavior:** - The impulse response forms a triangle centered at \( t = 0 \). - At \( t = -1 \), \( h(t) \) rises linearly to a peak of 3 at \( t = 0 \). - From \( t = 0 \), it decreases linearly back to 0 at \( t = 1 \). - \( h(t) = 0 \) outside the interval \([-1, 1]\). **Solution Approach:** To find the output \( y(t) \) of the system, perform the convolution of \( x(t) \) and \( h(t) \). The convolution formula is given by: \[ y(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) d\t
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