- #1
JustPeter
- 2
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Hi Guys,
I'm having trouble with the following:
A finite-time signal is the result of a filter G(t) applied to a signal. The filter is simply “on” (1) for t ∈ [0,T] and off (“0”) otherwise. If x(t) is the signal, and x(ω),its Fourier transform, compute the Fourier transform of the filtered signal. Next, take a simple sine for x(t), x(t) = sin(ω0t), and compute the Fourier transform for the finite-time signal. Write the result, it must involve the filter, and integrations should stretch [−∞,∞]
I don't really know what to do exactly, with the first problem.
I can try calculating the Fourier transform of the filter:
G(ω)= ∫0T e-iωtdt = -1/(iω)⋅(e-iωT-1)
The Fourier transform of the signal is: x(ω)
The convolution theorum says that the convolution of two functions is the product of the Fourier-transformed functions. Which makes: G(ω)x(ω).
But I have the idea that this isn't right. Could one of you guys assist me?
Peter
I'm having trouble with the following:
A finite-time signal is the result of a filter G(t) applied to a signal. The filter is simply “on” (1) for t ∈ [0,T] and off (“0”) otherwise. If x(t) is the signal, and x(ω),its Fourier transform, compute the Fourier transform of the filtered signal. Next, take a simple sine for x(t), x(t) = sin(ω0t), and compute the Fourier transform for the finite-time signal. Write the result, it must involve the filter, and integrations should stretch [−∞,∞]
I don't really know what to do exactly, with the first problem.
I can try calculating the Fourier transform of the filter:
G(ω)= ∫0T e-iωtdt = -1/(iω)⋅(e-iωT-1)
The Fourier transform of the signal is: x(ω)
The convolution theorum says that the convolution of two functions is the product of the Fourier-transformed functions. Which makes: G(ω)x(ω).
But I have the idea that this isn't right. Could one of you guys assist me?
Peter