---
title: "FS of ±1.5 square wave, ω₀ = π"
subject: "Signals and Systems"
chapter: 2
key: "semester_4/signals_and_systems:square-wave-fs-fundamental-pi"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/square-wave-fs-fundamental-pi
site: BCT Engineering
---

# FS of ±1.5 square wave, ω₀ = π

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2072 Magh, 5 marks
- Video: [Finding Fourier coefficients for square wave](https://www.youtube.com/watch?v=WTD8jg8fXoE) — Khan Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/square-wave-fs-fundamental-pi
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

Find the Fourier Series coefficient of the following continuous time periodic signal x(t) = 1.5 for 0 ≤ t < 1; −1.5 for 1 ≤ t < 2, with fundamental frequency π.

## Short answer

Period T=2, ω₀=π. a0 = 0. For k odd: ak = 3/(jkπ) = -j3/(kπ), magnitude 3/(|k|π) (e.g., |a1|≈0.955). Even harmonics are zero. Phase is -π/2 for positive odd k.

## Answer

The fundamental frequency is $\omega_0 = \pi$, so the period is $T = 2\pi/\omega_0 = 2$. Over one period, $x(t) = 1.5$ on $0 \le t < 1$ and $x(t) = -1.5$ on $1 \le t < 2$.

```text
 x(t)
  1.5  ____      ____      ____
      |    |    |    |    |
 -----+----+----+----+----+----- t
      0    1    2    3    4
           |____|    |____|
 -1.5
```

#### DC coefficient
$$
a_0 = \frac{1}{T}\int_0^2 x(t)\,dt = \frac{1}{2}\left[1.5(1) - 1.5(1)\right] = 0
$$

#### Coefficients for k ≠ 0
$$
a_k = \frac{1}{T}\int_0^T x(t)e^{-jk\omega_0 t}dt = \frac{1}{2}\left[\int_0^1 1.5\,e^{-jk\pi t}dt - \int_1^2 1.5\,e^{-jk\pi t}dt\right]
$$
Evaluate each integral, using $e^{-jk\pi} = (-1)^k$ and $e^{-j2k\pi} = 1$:
$$
\begin{aligned}
\int_0^1 e^{-jk\pi t}dt &= \frac{1 - e^{-jk\pi}}{jk\pi} = \frac{1-(-1)^k}{jk\pi} \\
\int_1^2 e^{-jk\pi t}dt &= \frac{e^{-jk\pi} - e^{-j2k\pi}}{jk\pi} = \frac{(-1)^k - 1}{jk\pi}
\end{aligned}
$$
Substituting:
$$
\begin{aligned}
a_k &= \frac{1.5}{2}\cdot\frac{[1-(-1)^k] - [(-1)^k - 1]}{jk\pi} \\
&= \frac{1.5}{2}\cdot\frac{2[1-(-1)^k]}{jk\pi} = \frac{1.5\,[1-(-1)^k]}{jk\pi}
\end{aligned}
$$
So
$$
a_k = \begin{cases} \dfrac{3}{jk\pi} = -j\dfrac{3}{k\pi}, & k \text{ odd} \\ 0, & k \text{ even} \end{cases}
$$

| k | ±1 | ±2 | ±3 | ±5 |
|---|---|---|---|---|
| $\lvert a_k\rvert$ | $3/\pi$ ≈ 0.955 | 0 | $1/\pi$ ≈ 0.318 | $3/(5\pi)$ ≈ 0.191 |
| $\angle a_k$ | ∓π/2 | — | ∓π/2 | ∓π/2 |

The coefficients are purely imaginary and odd ($a_{-k} = -a_k$). This is expected, because $x(t)$ is real and has odd symmetry about $t = 0$ (half-wave symmetry also removes the even harmonics). In trigonometric form:
$$
x(t) = \frac{6}{\pi}\left[\sin\pi t + \frac{1}{3}\sin 3\pi t + \frac{1}{5}\sin 5\pi t + \dots\right]
$$

**Answer:** $a_0 = 0$, $a_k = -j\,3/(k\pi)$ for odd $k$, and $a_k = 0$ for even $k$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With fundamental frequency w0 = pi, the period is:
   A. T = 1
   B. T = pi
   C. T = 2
   D. T = 2 pi
   Answer: C. T = 2 pi/w0 = 2. (Key point: Period from w0)

2. a_0 for this signal is:
   A. 1.5
   B. 0
   C. 0.75
   D. 3
   Answer: B. The areas of +1.5 and -1.5 each over 1 s cancel, so the average is zero. (Key point: DC term)

3. What is a_k?
   A. 3/(jk pi) for odd k and 0 for even k
   B. 3/(jk pi) for all k
   C. 0 for odd k and 3/(jk pi) for even k
   D. 1.5/(jk pi) for odd k
   Answer: A. a_k = 1.5[1 - (-1)^k]/(jk pi). (Key point: Coefficients)

4. Why are the coefficients purely imaginary and odd?
   A. Because x(t) is real and odd
   B. Because x(t) is even
   C. Because the period is 2
   D. Because a_0 = 0
   Answer: A. A real odd signal has purely imaginary, odd coefficients. (Key point: Nature of spectrum)

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