Math Problem Statement

An analog LTI system with impulse response function h(t) = u(2 − t) + u(t − 2) is presented with an input x(t) = t[u(t − 2) − u(t − 4)]. a) Determine and plot the system output y(t) = x(t) ∗ h(t)

Solution

To solve this problem, we need to compute the output y(t)y(t), which is the convolution of the input x(t)x(t) with the impulse response h(t)h(t). The output can be expressed as:

y(t)=x(t)h(t)=x(τ)h(tτ)dτy(t) = x(t) \ast h(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) d\tau

Step 1: Analyze the input and impulse response functions

We start by analyzing both the input x(t)x(t) and the impulse response h(t)h(t):

  • The impulse response is given as: h(t)=u(2t)+u(t2)h(t) = u(2 - t) + u(t - 2) This is a combination of step functions u(t)u(t). Specifically:

    • u(2t)u(2 - t) is a step function that equals 1 for t2t \leq 2 and 0 for t>2t > 2.
    • u(t2)u(t - 2) is a step function that equals 0 for t<2t < 2 and 1 for t2t \geq 2.

    So, h(t)h(t) is 1 everywhere, as the sum of these two step functions results in: h(t)=1for allt.h(t) = 1 \quad \text{for all} \, t.

  • The input x(t)x(t) is given as: x(t)=t[u(t2)u(t4)]x(t) = t[u(t - 2) - u(t - 4)] This is a piecewise function, nonzero only between t=2t = 2 and t=4t = 4. It can be expressed as:

    \begin{cases} t, & 2 \leq t \leq 4, \\ 0, & \text{otherwise}. \end{cases}$$

Step 2: Set up the convolution integral

We now compute the convolution y(t)=x(t)h(t)y(t) = x(t) \ast h(t), which simplifies due to h(t)=1h(t) = 1 for all tt. The convolution becomes: y(t)=x(τ)dτy(t) = \int_{-\infty}^{\infty} x(\tau) d\tau

Since x(τ)x(\tau) is nonzero only in the interval [2,4][2, 4], the integral reduces to: y(t)=24τdτy(t) = \int_2^4 \tau \, d\tau

Step 3: Perform the integration

Let's now compute this integral: y(t)=24τdτ=[τ22]24y(t) = \int_2^4 \tau \, d\tau = \left[\frac{\tau^2}{2}\right]_2^4 Substituting the limits: y(t)=422222=16242=82=6y(t) = \frac{4^2}{2} - \frac{2^2}{2} = \frac{16}{2} - \frac{4}{2} = 8 - 2 = 6

Step 4: Final Output

The output y(t)y(t) will be a constant value y(t)=6y(t) = 6 for the time interval where x(t)x(t) and h(t)h(t) overlap. Since x(t)x(t) is nonzero only between t=2t = 2 and t=4t = 4, the convolution will be nonzero in this region.

Plotting y(t)y(t)

The output y(t)y(t) is a piecewise function:

\begin{cases} 6, & 4 \leq t \leq 6, \\ 0, & \text{otherwise}. \end{cases}$$ I can generate a plot for you as well if you'd like! Would you like me to provide more details or help you with a plot? --- Here are some related questions you might explore: 1. How does the convolution change if the impulse response is not constant? 2. What would happen if the input function $$x(t)$$ was shifted by a different amount? 3. How does the unit step function impact the behavior of linear time-invariant (LTI) systems? 4. What are the properties of the convolution integral? 5. How would the output change if the limits of the input $$x(t)$$ were altered? **Tip:** In convolution problems, identifying where the functions are nonzero simplifies the integration significantly.

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Math Problem Analysis

Mathematical Concepts

Convolution
LTI Systems
Impulse Response
Unit Step Function

Formulas

y(t) = x(t) ∗ h(t) = ∫_{−∞}^{∞} x(τ) h(t − τ) dτ
h(t) = u(2 − t) + u(t − 2)
x(t) = t[u(t − 2) − u(t − 4)]

Theorems

Convolution Theorem
Properties of Unit Step Function

Suitable Grade Level

Undergraduate (Engineering/Mathematics)