FT of Cosine and Sine

The Fourier transform of cosine and sine functions produces pairs of Dirac delta impulses in the frequency domain, showing that a pure sinusoid contains energy at exactly one frequency. This result is the foundation for analysing bandpass signals and modulation in radio frequency systems.

Core Concept

A cosine or sine signal oscillates forever and is not absolutely integrable. Its Fourier transform therefore cannot be evaluated by direct integration. Instead, we use Euler's formulas to write the sinusoids as sums of complex exponentials, and then use the known FT pair for complex exponentials.

Euler's formulas are: cos(2*pi*f0*t) = (1/2)[e^(j2*pi*f0*t) + e^(-j2*pi*f0*t)] and sin(2*pi*f0*t) = (1/2j)[e^(j2*pi*f0*t) - e^(-j2*pi*f0*t)]. Since the FT of e^(j2*pi*f0*t) is delta(f - f0), each complex exponential contributes one impulse in the spectrum.

The cosine spectrum has two real impulses of equal weight 1/2 at f = +-f0. The sine spectrum has two imaginary impulses, positive at f = -f0 and negative at f = +f0. This imaginary, antisymmetric character reflects the odd nature of the sine function.

Key Formula

FT{cos(2*pi*f0*t)} = (1/2)[delta(f - f0) + delta(f + f0)]. FT{sin(2*pi*f0*t)} = (1/2j)[delta(f - f0) - delta(f + f0)] = -j/2 [delta(f - f0) - delta(f + f0)]. In angular frequency: FT{cos(omega0*t)} = pi[delta(omega - omega0) + delta(omega + omega0)] and FT{sin(omega0*t)} = pi/j [delta(omega - omega0) - delta(omega + omega0)]. Note the pi factor in the omega convention.

Problem: Find FT of x(t) = cos(2*pi*100*t) + sin(2*pi*200*t)

Step 1: Apply FT of cosine
  FT{cos(2*pi*100*t)} = (1/2)*delta(f-100) + (1/2)*delta(f+100)

Step 2: Apply FT of sine
  FT{sin(2*pi*200*t)} = -j/2 * delta(f-200) + j/2 * delta(f+200)

Step 3: Add by linearity
  X(f) = (1/2)*delta(f-100) + (1/2)*delta(f+100)
        -j/2 * delta(f-200) + j/2 * delta(f+200)

Step 4: Magnitude at each spectral line
  |X(100)| = 1/2  (cosine component)
  |X(-100)| = 1/2
  |X(200)| = 1/2  (sine component)
  |X(-200)| = 1/2

Final Answer:
  Four spectral impulses at f = +-100 Hz and +-200 Hz
  Cosine impulses are real; sine impulses are imaginary

Exam Tip: In the omega convention, FT{cos(omega0*t)} = pi[delta(omega-omega0) + delta(omega+omega0)]. The pi factor is often forgotten. In the f convention there is no pi: FT{cos(2*pi*f0*t)} = (1/2)[delta(f-f0) + delta(f+f0)]. Always identify the convention before writing down the answer. Mixing them gives results that are off by a factor of 2*pi.

Properties Summary

• Cosine pair (f): cos(2*pi*f0*t) <-> (1/2)[delta(f-f0) + delta(f+f0)], real symmetric spectrum.
• Sine pair (f): sin(2*pi*f0*t) <-> -j/2 [delta(f-f0) - delta(f+f0)], imaginary antisymmetric spectrum.
• Cosine pair (omega): cos(omega0*t) <-> pi[delta(omega-omega0) + delta(omega+omega0)].
• Sine pair (omega): sin(omega0*t) <-> pi/j [delta(omega-omega0) - delta(omega+omega0)].
• A sinusoid is real, so its spectrum satisfies conjugate symmetry: X(-f) = X*(f).
• Modulation: x(t)*cos(2*pi*f0*t) <-> (1/2)[X(f-f0) + X(f+f0)], shifting the baseband spectrum to +-f0.
• Energy: sinusoids have infinite energy (integral of cos^2 is infinite), so they are power signals, not energy signals.

Quick Revision

• Use Euler formulas: cos = (e^jx + e^-jx)/2, sin = (e^jx - e^-jx)/(2j).
• FT{cos(2*pi*f0*t)} = (1/2)[delta(f-f0) + delta(f+f0)]: two real impulses.
• FT{sin(2*pi*f0*t)} = -j/2 [delta(f-f0) - delta(f+f0)]: two imaginary impulses.
• The f convention has no pi factor; the omega convention carries pi.
• Each impulse in the spectrum corresponds to one complex exponential component.
• Cosine spectrum is real and even; sine spectrum is imaginary and odd.
• Exam trap: writing FT{cos(omega0*t)} = (1/2)[delta(omega-omega0) + delta(omega+omega0)] when the omega convention actually requires a pi coefficient. This error loses marks because the omega-domain pair always carries the pi factor.