---
title: "Magnitude and phase spectra of a sum of sinusoids"
subject: "Data Communication"
chapter: 3
key: "semester_5/data_communication:mag-phase-5-sin12t"
source: https://www.bctengineer.com/q/semester_5/data_communication/mag-phase-5-sin12t
site: BCT Engineering
---

# Magnitude and phase spectra of a sum of sinusoids

Data Communication · Chapter 3: Signal Analysis

- Asked in: 2076 Asoj, 4 marks
- Video: [The Spectrum: Representing Signals as a Function of Frequency](https://www.youtube.com/watch?v=xh5pLY3f6H8) — Barry Van Veen
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/mag-phase-5-sin12t
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-3

## Question

Plot the magnitude and phase spectra of X(t) = 5 + sin(12t + 20°) - cos(16t - 60°) + cos(20t + 40°).

## Short answer

Convert: sin(12t+20°)=cos(12t-70°), -cos(16t-60°)=cos(16t+120°). Signal =5+cos(12t-70°)+cos(16t+120°)+cos(20t+40°). Magnitudes: DC=5, amplitude 1 at ω=12,16,20 rad/s; phases -70°, 120°, 40°. Plot vertical lines at those ω. Two-sided gives lines of 0.5 at ±12, ±16, ±20 with phases odd-symmetric.

## Answer

X(t) = 5 + sin(12t + 20°) - cos(16t - 60°) + cos(20t + 40°)

#### Step 1: convert to positive cosines
- sin(12t + 20°) = cos(12t + 20° - 90°) = cos(12t - 70°)
- -cos(16t - 60°) = cos(16t - 60° + 180°) = cos(16t + 120°)
- cos(20t + 40°) is already in the right form.

#### Step 2: table (ω in rad/s; f = ω/2π)
| ω (rad/s) | f (Hz) | Amplitude | Phase |
|---|---|---|---|
| 0 (DC) | 0 | 5 | 0° |
| 12 | 1.91 | 1 | -70° |
| 16 | 2.55 | 1 | +120° |
| 20 | 3.18 | 1 | +40° |

#### Single-sided plots
```text
Magnitude                    Phase (deg)
5 |*                      120 |            *
  |                        40 |                 *
1 |   |    |    |           0 +--+---+----+----+--> w
  +---+----+----+--> w          0  12   16   20
  0  12   16   20            -70|   *
```

#### Two-sided plots
Lines of height 0.5 at ω = ±12, ±16, ±20 and 5 at ω = 0, with phases odd-symmetric: -70°, +120°, +40° at positive ω and +70°, -120°, -40° at negative ω.

The magnitude spectrum is even and the phase spectrum is odd for a real signal.

## Check yourself

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

1. sin(12t+20°) becomes
   A. cos(12t + 110°)
   B. cos(12t + 20°)
   C. cos(12t - 70°)
   D. -cos(12t - 70°)
   Answer: C. Subtract 90°. (Key point: sin conversion)

2. -cos(16t - 60°) becomes
   A. cos(16t - 120°)
   B. cos(16t - 60°)
   C. cos(16t + 60°)
   D. cos(16t + 120°)
   Answer: D. Add 180°. (Key point: Negative cosine)

3. 12 rad/s corresponds to
   A. 1.91 Hz
   B. 12 Hz
   C. 75.4 Hz
   D. 0.5 Hz
   Answer: A. 12/2π. (Key point: Frequency)

4. Two-sided magnitude of each unit cosine is
   A. 1
   B. 0.5
   C. 2
   D. 5
   Answer: B. A/2. (Key point: Two-sided)

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