To sketch the signal $x(t) u(l - t)$, we need to break down and understand the components of the signal:

Components:

1. $x(t)$: This is the main signal that varies with time $t$.
2. $u(l - t)$: This is a unit step function, where $u(t)$ is defined as: 1 & \text{if } t \geq 0 \\ 0 & \text{if } t < 0 \end{cases}$$ In this case, the argument is $$(l - t)$$, so the unit step function shifts to trigger at $$t = l$$. The signal is non-zero only when $$t \leq l$$.

Behavior of the signal $x(t) u(l - t)$:

• For $t \leq l$: The unit step function is 1, so the signal is simply $x(t)$.
• For $t > l$: The unit step function is 0, so the signal becomes 0.

Steps to Sketch:

1. Sketch $x(t)$ normally.
2. Apply the unit step function $u(l - t)$, which limits the signal to the interval $(-\infty, l]$.
3. After $t = l$, the signal will be zero.

Do you have more details about the form of $x(t)$ (e.g., is it a sine wave, exponential, etc.)? If it's a general signal, we can sketch it based on this logic.

Would you like further details or clarifications?

Related Questions:

1. How would the graph change if $x(t)$ were a ramp function?
2. What would happen if the unit step was $u(t - l)$ instead?
3. Can you explain the physical meaning of $x(t) u(l - t)$?
4. How would the signal behave if $l$ were negative?
5. How does the Fourier transform of a unit step function look?

Tip: The unit step function is often used to model switching behavior in signals, turning a signal on or off at a specific time.