---
title: "DTFS spectrum of {1, −0.5, 0, 0, 0.5}"
subject: "Signals and Systems"
chapter: 2
key: "semester_4/signals_and_systems:dtfs-five-point-periodic"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-five-point-periodic
site: BCT Engineering
---

# DTFS spectrum of {1, −0.5, 0, 0, 0.5}

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2079 Asoj, 6 marks; 2079 Chaitra, 6 marks (asked 2 times)
- Video: [Determine DTFS of the signal and draw the spectrum | Numerical 1 on DTFS | EnggClasses](https://www.youtube.com/watch?v=HlEse_jD3ow) — EnggClasses
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-five-point-periodic
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

Find the frequency domain representation (both magnitude and phase plot) of a periodic signal defined as x[n] = {1, −0.5, 0, 0, 0.5} (x[0] = 1), with period N = 5.

## Short answer

Compute ak = (1/5) sum x[n] e^{-j 2pi kn/5}, giving ak = 0.2 + j0.2 sin(2pi k/5). Magnitudes: |a0|=0.2, |a1|=|a4|≈0.276, |a2|=|a3|≈0.232. Phases: 0°, 43.6°, 30.5°, -30.5°, -43.6°.

## Answer

The signal has period $N = 5$, so $\omega_0 = 2\pi/5$ and

$$
a_k = \frac{1}{5}\sum_{n=0}^{4}x[n]\,e^{-jk(2\pi/5)n}
$$

One period: $x[0] = 1$, $x[1] = -0.5$, $x[2] = 0$, $x[3] = 0$, $x[4] = 0.5$. Because of periodicity $x[4] = x[-1]$, so it is easier to sum over $n = -1, 0, 1$ (note $x[-1] = 0.5$):

$$
\begin{aligned}
a_k &= \frac{1}{5}\left[x[-1]e^{jk2\pi/5} + x[0] + x[1]e^{-jk2\pi/5}\right] \\
&= \frac{1}{5}\left[1 + 0.5e^{jk2\pi/5} - 0.5e^{-jk2\pi/5}\right] \\
&= \frac{1}{5}\left[1 + 0.5\,(2j)\sin\frac{2\pi k}{5}\right] \\
&= \frac{1}{5}\left[1 + j\sin\frac{2\pi k}{5}\right]
\end{aligned}
$$

**Magnitude and phase:**

$$
|a_k| = \frac{1}{5}\sqrt{1 + \sin^2\frac{2\pi k}{5}}, \qquad \angle a_k = \tan^{-1}\left(\sin\frac{2\pi k}{5}\right)
$$

#### Values over one period
| $k$ | $\sin(2\pi k/5)$ | $a_k$ | $\lvert a_k\rvert$ | $\angle a_k$ |
|---|---|---|---|---|
| 0 | 0 | $0.2$ | 0.200 | $0^\circ$ |
| 1 | 0.951 | $0.2 + j0.190$ | 0.276 | $43.56^\circ$ |
| 2 | 0.588 | $0.2 + j0.118$ | 0.232 | $30.45^\circ$ |
| 3 | $-0.588$ | $0.2 - j0.118$ | 0.232 | $-30.45^\circ$ |
| 4 | $-0.951$ | $0.2 - j0.190$ | 0.276 | $-43.56^\circ$ |

(Since $a_{k+5} = a_k$: $a_{-1} = a_4$, $a_{-2} = a_3$.)

#### Magnitude and phase plots (repeat every 5)
```text
 |a_k|  (repeats every 5)
 0.276     |     |
 0.232  |  |     |  |
   0.2  |  |  |  |  |
       -+--+--+--+--+--> k
       -2 -1  0  1  2

 angle a_k (deg)
 43.56           |
 30.45           |  |
       -+--+--o--+--+--> k
-30.45  |  |
-43.56     |
       -2 -1  0  1  2
```

The magnitude spectrum is even and the phase spectrum is odd in $k$, as expected for a real $x[n]$.

**Check (synthesis at $n = 0$):** $\sum_{k=0}^{4}a_k = 5\times 0.2 + j(0) = 1 = x[0]$. Correct.

## Check yourself

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

1. What is w0 for N = 5?
   A. pi/5
   B. 2pi/5
   C. 2pi
   D. 5/2pi
   Answer: B. w0 = 2pi/N = 2pi/5. (Key point: Fundamental frequency)

2. Why sum over n = -1, 0, 1?
   A. x[2] and x[3] are non-zero
   B. x[4] equals x[-1] by periodicity, so the non-zero samples sit at -1, 0, 1
   C. To avoid complex numbers
   D. Because only odd n matter
   Answer: B. Sums can be over any period; x[4] = x[-1] = 0.5 simplifies the work. (Key point: Useful trick)

3. What is a_k?
   A. (1/5)(1 + j sin(2pi k/5))
   B. (1/5)(1 + cos(2pi k/5))
   C. (1/5)(1 - j sin(2pi k/5))
   D. (1/5)(1 + 2j sin(2pi k/5))
   Answer: A. 0.5 e^(jθ) - 0.5 e^(-jθ) = j sinθ, giving (1/5)(1 + j sin θ). (Key point: Result for a_k)

4. What is a_0?
   A. 1
   B. 0
   C. 0.2 (the mean of the samples)
   D. 0.5
   Answer: C. a_0 = (1 - 0.5 + 0 + 0 + 0.5)/5 = 0.2. (Key point: Value of a_0)

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