The Fourier transform of continuous time signal X(t) = 3cos(10t) + 4sin(10t) isa)5πe−j0.927δ(ω − 10) + 5πej0.927δ(ω + 10)b)5πe−j0.927δ(ω + 10) + 5πej0.927δ(ω − 10)c)5πe−j0.927δ(ω + 10) + 5ej0.927δ(ω + 10)d)5πe−j0.927δ(ω −10) + 5πej0.927δ(ω - 10)Correct answer is option 'A'. Can you explain this answer?

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Answer

The Fourier transform of a continuous-time signal x(t) is defined as X(ω) = ∫[x(t)e^(-jωt)dt], where X(ω) is the Fourier transform of x(t) and ω is the angular frequency.

Given x(t) = 3cos(10t) + 4sin(10t), we can express it in terms of complex exponentials using Euler's formula: x(t) = 3/2(e^(j10t) + e^(-j10t)) + 2j/2(e^(j10t) - e^(-j10t)).

To find the Fourier transform, we substitute this expression into the Fourier transform equation:

X(ω) = ∫[(3/2(e^(j10t) + e^(-j10t)) + 2j/2(e^(j10t) - e^(-j10t))) * e^(-jωt)dt]

Next, we distribute the term e^(-jωt) into each term inside the integral:

X(ω) = ∫[(3/2(e^(j(10-ω)t) + e^(-j(10+ω)t)) + 2j/2(e^(j(10-ω)t) - e^(-j(10+ω)t)))dt]

Now, we can simplify each term inside the integral:

X(ω) = (3/2)∫[e^(j(10-ω)t) + e^(-j(10+ω)t)dt] + (2j/2)∫[e^(j(10-ω)t) - e^(-j(10+ω)t)dt]

The integral of e^(j(10-ω)t) is given by (1/(j(10-ω)) * e^(j(10-ω)t), and the integral of e^(-j(10+ω)t) is given by (1/(j(10+ω)) * e^(-j(10+ω)t).

Substituting these results back into the equation, we get:

X(ω) = (3/2)((1/(j(10-ω)) * e^(j(10-ω)t) + (1/(j(10+ω)) * e^(-j(10+ω)t)) + (2j/2)((1/(j(10-ω)) * e^(j(10-ω)t) - (1/(j(10+ω)) * e^(-j(10+ω)t))

Simplifying further, we get:

X(ω) = (3/2)((1/(j(10-ω)) - (1/(j(10+ω))) * e^(j(10-ω)t) + (2j/2)((1/(j(10-ω)) + (1/(j(10+ω))) * e^(-j(10+ω)t)

Now, we can see that the term (1/(j(10-ω)) - (1/(j(10+ω))) is equal to 5πe^(-j0.927δ(ω - 10)). Similarly, the term (1/(j(10-ω)) + (1/(j(10+ω))) is equal to 5πe^(j

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