Let's address each of your questions one by one:

1. Convolution Integral for a Causal LTI System

For a causal LTI system, the output $y(t)$ is computed as the convolution of the input $x(t)$ with the impulse response $h(t)$. A system is causal if its output depends only on the current and past values of the input. Therefore, for a causal system, the impulse response $h(t) = 0$ for $t < 0$.

The convolution integral for a causal system is given by:

$y(t) = (x * h)(t) = \int_{0}^{t} x(\tau) h(t - \tau) \, d\tau$

This integral shows that the output $y(t)$ at time $t$ depends only on the values of the input $x(\tau)$ up to time $t$.

All Possible Integral Combinations for a Causal System:

• For $t < 0$:
  $y(t) = 0$ (since a causal system does not respond to inputs before $t = 0$).

• For $t \geq 0$: $y(t) = \int_{0}^{t} x(\tau) h(t - \tau) \, d\tau$

2. Relationship Between $x(t)$, $h(t)$, and $y(t)$ for an LTI System

For a linear time-invariant (LTI) system:

• Input $x(t)$: The signal that is fed into the system.
• Impulse Response $h(t)$: The system's response to a unit impulse input, $\delta(t)$. This characterizes the entire system.
• Output $y(t)$: The result of the convolution of $x(t)$ with $h(t)$, which represents the system's response to the input.

The relationship is given by the convolution operation:

$y(t) = (x * h)(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) \, d\tau$

This integral signifies that the output is a weighted sum of the impulse response, where the weights are determined by the input signal $x(t)$.

3. Conditions for Stability of an LTI System Based on Impulse Response

An LTI system is BIBO (Bounded Input, Bounded Output) stable if and only if its impulse response $h(t)$ is absolutely integrable. Mathematically, this condition is:

$\int_{-\infty}^{\infty} |h(t)| \, dt < \infty$

This means that the total area under the absolute value of the impulse response must be finite. If this condition is satisfied, then for any bounded input $x(t)$, the output $y(t)$ will also be bounded.

4. Relation Between Impulse Response and Step Response

The impulse response $h(t)$ and the step response $s(t)$ of an LTI system are related as follows:

• The step response $s(t)$ is the response of the system to a unit step input $u(t)$.
• The impulse response $h(t)$ is the derivative of the step response $s(t)$:

$h(t) = \frac{d}{dt} s(t)$

Conversely, the step response can be obtained by integrating the impulse response:

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

For a causal system, this simplifies to:

$s(t) = \int_{0}^{t} h(\tau) \, d\tau$

Would you like further details or explanations on any of these points?

Related Questions:

1. What is the significance of causality in LTI systems?
2. How does the Fourier transform relate to the convolution operation in LTI systems?
3. What is the frequency response of an LTI system, and how is it computed?
4. How does one determine if an LTI system is time-invariant?
5. Can you explain how to find the step response from the impulse response using Laplace transforms?

Tip:

To verify if an LTI system is stable, always check the integrability of its impulse response over the entire time domain.