Express a periodic triangular waveform as an infinite sum of shifted versions of tri(t), with the first component starting at t = 0. The waveform should have a maximum height of 2, a period of To = 1 s, and the component triangle functions should be adjacent to each other.
The correct answer and explanation is :
To express a periodic triangular waveform as an infinite sum of shifted versions of the standard triangular pulse function tri(t), you can use the following expression:
Answer:
[
x(t) = \sum_{n=-\infty}^{\infty} 2 \cdot \text{tri}(t – n)
]
Explanation (≈300 words):
The goal is to construct a periodic triangular waveform using shifted and scaled copies of the triangular function, denoted tri(t). The standard tri(t) function is typically defined as:
[
\text{tri}(t) =
\begin{cases}
1 – |t|, & \text{if } |t| \leq 1 \
0, & \text{otherwise}
\end{cases}
]
This triangular function is centered at ( t = 0 ), has a peak value of 1 at ( t = 0 ), and a base width of 2 (from ( t = -1 ) to ( t = 1 )).
However, in this problem:
- The desired maximum height is 2, so we scale the amplitude by 2.
- The desired period is ( T_0 = 1 ) s, so each triangle must begin right after the previous ends, i.e., with no gaps or overlaps.
- The triangle pulses must be adjacent, meaning the base width must equal the period.
So, instead of the standard tri(t) centered at 0, we use tri(t - n) to shift the triangle to start at ( t = n ), making each copy of the triangle start at successive integer values of ( t ): 0, 1, 2, etc. To make it evenly cover all time, we sum over all integers ( n \in \mathbb{Z} ) (i.e., from (-\infty) to (+\infty)).
Each triangle has a base from ( t = n – 1 ) to ( t = n + 1 ), meaning two adjacent triangles meet exactly at their bases, ensuring continuity.
The amplitude factor of 2 ensures that each triangle reaches a peak height of 2 instead of the default 1 of tri(t).
This representation is compact, mathematically elegant, and takes advantage of the inherent symmetry and periodicity of the triangle waveform.