Question

What is zero-state unit step response?

Answer 1

Are you asking about the zero-state unit step response in the context of a specific system or signal processing?

Answer 2

The zero-state unit step response is a fundamental concept in control systems and signal processing. Here's a detailed explanation:

### Definitions

1. **Zero-State Response**: This refers to the response of a system when the initial conditions are zero. In other words, the system starts with no stored energy or initial state; only the input affects the output.

2. **Unit Step Response**: This is the response of a system to a unit step input. A unit step function, \( u(t) \), is defined as:
   \[
   u(t) =
   \begin{cases}
   0 & \text{for } t < 0 \\
   1 & \text{for } t \geq 0
   \end{cases}
   \]
   The unit step response of a system is essentially how the system responds over time when the input is a step function.

### Zero-State Unit Step Response

The **zero-state unit step response** combines these concepts. It is the response of a system when:
- The system starts from a zero initial state (zero-state condition).
- The input to the system is a unit step function.

Mathematically, if you have a linear time-invariant (LTI) system with transfer function \( H(s) \), the zero-state unit step response \( y(t) \) can be found by:
1. Taking the Laplace transform of the unit step function, which is \( \frac{1}{s} \).
2. Multiplying this by the transfer function \( H(s) \) of the system to get the Laplace transform of the output:
   \[
   Y(s) = H(s) \cdot \frac{1}{s}
   \]
3. Taking the inverse Laplace transform of \( Y(s) \) to find \( y(t) \), which gives the time-domain response of the system to the unit step input.

### Example

Consider a simple first-order system with a transfer function:
\[
H(s) = \frac{1}{s + a}
\]

For this system, the zero-state unit step response is found as follows:
1. The Laplace transform of the unit step function is \( \frac{1}{s} \).
2. The Laplace transform of the output is:
   \[
   Y(s) = H(s) \cdot \frac{1}{s} = \frac{1}{s + a} \cdot \frac{1}{s} = \frac{1}{s(s + a)}
   \]
3. Performing partial fraction decomposition and inverse Laplace transform:
   \[
   \frac{1}{s(s + a)} = \frac{A}{s} + \frac{B}{s + a}
   \]
   Solving for \( A \) and \( B \) gives:
   \[
   A = \frac{1}{a}, \quad B = -\frac{1}{a}
   \]
   Hence:
   \[
   Y(s) = \frac{1}{a} \left( \frac{1}{s} - \frac{1}{s + a} \right)
   \]
   Taking the inverse Laplace transform gives:
   \[
   y(t) = \frac{1}{a} \left(1 - e^{-at} \right) \text{ for } t \geq 0
   \]

This result tells us how the system will behave over time after a unit step input, assuming the system starts from a zero initial condition.