Final answer:
The Fourier transform of g₁(t) = cos(ωt)u(t) is π(δ(ω - ω₀) + δ(ω + ω₀)) * 1/(jω) + π²δ(ω).
Step-by-step explanation:
To find the Fourier transform of the function g₁(t) = cos(ωt)u(t), let's break it down. The Fourier transform of cos(ωt) is given by:
F{cos(ωt)} = π(δ(ω - ω₀) + δ(ω + ω₀))
where δ represents the Dirac delta function, and ω₀ is the angular frequency of the cosine function. The Fourier transform of u(t) is:
F{u(t)} = 1/(jω) + πδ(ω)
where j is the imaginary unit. Now, to calculate the Fourier transform of g₁(t), we can multiply the transforms of cos(ωt) and u(t):
F{g₁(t)} = F{cos(ωt)} * F{u(t)}
F{g₁(t)} = π(δ(ω - ω₀) + δ(ω + ω₀)) * (1/(jω) + πδ(ω))
After simplifying and using the property that δ(-x) = δ(x), we obtain:
F{g₁(t)} = π(δ(ω - ω₀) + δ(ω + ω₀)) * 1/(jω) + π²δ(ω)
This is the Fourier transform of the function g₁(t) = cos(ωt)u(t).