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Line spectra of 12 + 6sin(140πt+30°) - 9cos(80πt-70°)

Short answer

Rewrite: 6sin(140πt+30°)=6cos(140πt-60°), -9cos(80πt-70°)=9cos(80πt+110°). Single-sided: DC 12, 40 Hz amplitude 9 phase 110°, 70 Hz amplitude 6 phase -60°. Two-sided: at ±40 Hz, amplitude 4.5 phases ±110°; at ±70 Hz, amplitude 3 phases ∓60°.

Plot the line spectrums of X(t) = 12 + 6sin(140πt + 30°) - 9cos(80πt - 70°).

X(t) = 12 + 6 sin(140πt + 30°) - 9 cos(80πt - 70°)

Step 1: write every term as a positive cosine

  • 6 sin(140πt + 30°) = 6 cos(140πt + 30° - 90°) = 6 cos(140πt - 60°)
  • -9 cos(80πt - 70°) = 9 cos(80πt - 70° + 180°) = 9 cos(80πt + 110°)

Step 2: frequencies

  • 140π = 2π·70, so f = 70 Hz
  • 80π = 2π·40, so f = 40 Hz

Step 3: table (single-sided)

f (Hz)AmplitudePhase
0 (DC)120°
409+110°
706-60°
Amplitude spectrum           Phase spectrum
12 |*                    110 |        *
 9 |  |                    0 +--+-----+------+--> f
 6 |  |     |                 0 40     70
   |  |     |             -60|               *
   +--+-----+---> f
   0  40    70

Two-sided spectrum

Each cosine of amplitude A gives lines of A/2 at ±f with phases ±φ.

f (Hz)-70-4004070
Magnitude34.5124.53
Phase+60°-110°0+110°-60°

The fundamental frequency is the gcd of 40 and 70, which is 10 Hz, so the period is 0.1 s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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