A time-limited rectangular pulse defined by f(t) = 0 for t < 0, f(t) = E for 0 ≤ t ≤ a, and f(t) = 0 for t > a is best described as which type of standard signal?

Difficulty: Easy

Correct Answer: A pulse (rectangular) function

Explanation:


Introduction / Context:
Standard test signals—step, impulse, ramp, and pulse—are frequently used in systems and signal analysis. Correct classification helps in understanding system responses and Fourier/Laplace transforms. Here, the signal is nonzero only over a finite interval [0, a] with amplitude E, which is a classic rectangular pulse.


Given Data / Assumptions:

  • f(t) = 0 for t < 0.
  • f(t) = E for 0 ≤ t ≤ a.
  • f(t) = 0 for t > a.


Concept / Approach:
A rectangular pulse can be written as the difference of two step functions: f(t) = E[u(t) − u(t − a)]. It is time-limited (finite duration) and has finite energy. It is not a single step (which stays at E forever after t ≥ 0), nor a shifted step at t = a alone (that would be zero until t ≥ a and then E thereafter).


Step-by-Step Solution:

Recognize finite support: f(t) ≠ 0 only on 0 ≤ t ≤ a.Represent with steps: f(t) = E[u(t) − u(t − a)].Therefore, it is a rectangular pulse (amplitude E, width a, starting at t = 0).


Verification / Alternative check:

Graphing shows a horizontal line at E between t = 0 and t = a, zero elsewhere—defining a pulse.


Why Other Options Are Wrong:

Unit step: would remain at E for all t ≥ 0.Shifted step at t = a: zero until t = a, then E forever (not finite-duration).Impulse: infinitely narrow and area-normalized; not the case here.


Common Pitfalls:

Confusing pulse with step or mistaking the finite duration as two impulses at boundaries.


Final Answer:

A pulse (rectangular) function

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