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Consider a signal x(t) = 4 rect (t/6) and its Fourier - transform is x(ω). Determine the area under the curve in the ω-domain.
- a)∞
- b)8π
- c)24
- d)4
Correct answer is option 'C'. Can you explain this answer?
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Consider a signal x(t) = 4 rect (t/6)and its Fourier - transform is x(...
Thus, the correct answer is: 24
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Consider a signal x(t) = 4 rect (t/6)and its Fourier - transform is x(...
The Fourier transform of x(t) = 4 rect(t/6) can be calculated as follows:
The rectangular function rect(t/6) can be expressed as the difference of two step functions:
rect(t/6) = u(t/6) - u(t/-6),
where u(t) is the unit step function.
The Fourier transform of a step function u(t) is given by:
F{u(t)} = 1/(jw) + πδ(w),
where F{ } denotes the Fourier transform, j is the imaginary unit, w is the frequency variable, and δ(w) is the Dirac delta function.
Using linearity and time scaling properties of the Fourier transform, we can calculate the Fourier transform of rect(t/6) as:
F{rect(t/6)} = F{u(t/6) - u(t/-6)}
= F{u(t/6)} - F{u(t/-6)}
= (1/(jw) + πδ(w)) - (1/(jw) + πδ(w))
= 0.
Therefore, the Fourier transform of x(t) = 4 rect(t/6) is 0.
The rectangular function rect(t/6) can be expressed as the difference of two step functions:
rect(t/6) = u(t/6) - u(t/-6),
where u(t) is the unit step function.
The Fourier transform of a step function u(t) is given by:
F{u(t)} = 1/(jw) + πδ(w),
where F{ } denotes the Fourier transform, j is the imaginary unit, w is the frequency variable, and δ(w) is the Dirac delta function.
Using linearity and time scaling properties of the Fourier transform, we can calculate the Fourier transform of rect(t/6) as:
F{rect(t/6)} = F{u(t/6) - u(t/-6)}
= F{u(t/6)} - F{u(t/-6)}
= (1/(jw) + πδ(w)) - (1/(jw) + πδ(w))
= 0.
Therefore, the Fourier transform of x(t) = 4 rect(t/6) is 0.
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