System Impulse Response and Step Response Definition

In the realm of continuous time systems, two fundamental characteristics that describe a system's behavior are the impulse response and the step response. The impulse response, typically denoted as $\(h(t)\)$, represents the output of a system when the input is a Dirac delta function, $\(\delta(t)\)$. It is a crucial characteristic for Linear Time-Invariant (LTI) systems because it completely characterizes the system. The step response, denoted as $\(s(t)\)$, represents the output of a system when the input is a unit step function, $\(u(t)\)$.

Relationship between Step Response and Impulse Response

For any continuous time Linear Time-Invariant (LTI) system, there exists a direct and important relationship between its impulse response $\(h(t)\)$ and its step response $\(s(t)\)$. The step response is obtained by integrating the impulse response over time. This means that if you know the system's impulse response, you can find its step response by performing integration.

The mathematical relationship is given by:

$$$s(t) = \int_{-\infty}^{t} h(\tau) d\tau$$$

Conversely, the impulse response is the derivative of the step response:

$$$h(t) = \frac{ds(t)}{dt}$$$

In this problem, we are given the impulse response $\(h(t) = \delta(t - 1) + \delta(t - 3)\)$ for a continuous time system, and we need to find the value of its step response at a specific time, $\(t = 2\)$. To do this, we will use the integration formula.

Calculating the Step Response from Impulse Response

We are given the impulse response $\(h(t)\)$ as:

$$$h(t) = \delta(t - 1) + \delta(t - 3)$$$

To find the step response $\(s(t)\)$, we integrate $\(h(t)\)$ with respect to $\(\tau\)$ from $\(-\infty\)$ to $\(t\)$:

$$$s(t) = \int_{-\infty}^{t} (\delta(\tau - 1) + \delta(\tau - 3)) d\tau$$$

Due to the linearity property of integration, we can split the integral into two separate integrals:

$$$s(t) = \int_{-\infty}^{t} \delta(\tau - 1) d\tau + \int_{-\infty}^{t} \delta(\tau - 3) d\tau$$$

A key property of the Dirac delta function is that its integral is the unit step function. Specifically, the integral of $\(\delta(x - a)\)$ from $\(-\infty\)$ to $\(x\)$ is $\(u(x - a)\)$, where $\(u(x)\)$ is the unit step function.

$$$\int_{-\infty}^{x} \delta(\xi - a) d\xi = u(x - a)$$$

Applying this property to each term in our expression for $\(s(t)\)$:

• The integral $\(\int_{-\infty}^{t} \delta(\tau - 1) d\tau\)$ becomes $\(u(t - 1)\)$.
• The integral $\(\int_{-\infty}^{t} \delta(\tau - 3) d\tau\)$ becomes $\(u(t - 3)\)$.

Therefore, the step response $\(s(t)\)$ of the system is:

$$$s(t) = u(t - 1) + u(t - 3)$$$

Recall the definition of the unit step function $\(u(x)\)$:

Evaluating Step Response at t = 2

We need to find the value of the step response $\(s(t)\)$ specifically at $\(t = 2\)$. Substitute $\(t = 2\)$ into the derived expression for $\(s(t)\)$:

$$$s(2) = u(2 - 1) + u(2 - 3)$$$

Simplify the terms inside the unit step functions:

$$$s(2) = u(1) + u(-1)$$$

Now, let's evaluate each unit step function using its definition:

• For $\(u(1)\)$: Since the argument $\(1\)$ is greater than or equal to $\(0\)$ ($\(1 \ge 0\)$), the value of $\(u(1)\)$ is $\(1\)$.
• For $\(u(-1)\)$: Since the argument $\(-1\)$ is less than $\(0\)$ ($\(-1 < 0\)$), the value of $\(u(-1)\)$ is $\(0\)$.

Substitute these values back into the equation for $\(s(2)\)$:

$$$s(2) = 1 + 0$$$

$$$s(2) = 1$$$

Therefore, the value of the step response of the continuous time system at $\(t = 2\)$ is $\(1\)$.