logistic_guy
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Find the Fourier transform of the half-cosine pulse shown.
half cosine pulse - 2
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logistic_guy
Senior Member
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- Apr 17, 2024
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This signal is the same as the previous one except it is shifted to the right by \(\displaystyle \frac{T}{2}\).
Let us call the previous signal \(\displaystyle h(t)\).
\(\displaystyle h(t) = A\cos \frac{\pi t}{T} \ \text{rect}\frac{t}{T}\)
Then, the signal in the op is:
\(\displaystyle g(t) = h\left(t - \frac{T}{2}\right) = A\cos \frac{\pi [t - \frac{T}{2}]}{T} \ \text{rect}\frac{t - \frac{T}{2}}{T}\)
We can find the Fourier transform of this signal from scratch, but since we already know the Fourier transform of \(\displaystyle h(t)\), we can use it along with the time shifting property to find the Fourier transform of \(\displaystyle h\left(t - \frac{T}{2}\right)\).
\(\displaystyle \textcolor{red}{\text{Time shifting property.}}\)
If
\(\displaystyle \mathcal{F}\{h(t)\} = H(f)\)
Then,
\(\displaystyle \mathcal{F}\{h(t - t_0)\} = H(f) e^{-j2\pi f t_0}\)
\(\displaystyle H(f) = \frac{A}{2\pi}\left(\frac{\cos \pi fT}{f + \frac{1}{2T}} - \frac{\cos \pi fT}{f - \frac{1}{2T}}\right)\)
Then,
\(\displaystyle G(f) = \mathcal{F}\{g(t)\} = \mathcal{F}\left\{h\left(t - \frac{T}{2}\right)\right\} = H(f)e^{-j2\pi fT/2} = \frac{A}{2\pi}\left(\frac{\cos \pi fT}{f + \frac{1}{2T}} - \frac{\cos \pi fT}{f - \frac{1}{2T}}\right) e^{-j\pi fT}\)
Let us call the previous signal \(\displaystyle h(t)\).
\(\displaystyle h(t) = A\cos \frac{\pi t}{T} \ \text{rect}\frac{t}{T}\)
Then, the signal in the op is:
\(\displaystyle g(t) = h\left(t - \frac{T}{2}\right) = A\cos \frac{\pi [t - \frac{T}{2}]}{T} \ \text{rect}\frac{t - \frac{T}{2}}{T}\)
We can find the Fourier transform of this signal from scratch, but since we already know the Fourier transform of \(\displaystyle h(t)\), we can use it along with the time shifting property to find the Fourier transform of \(\displaystyle h\left(t - \frac{T}{2}\right)\).
\(\displaystyle \textcolor{red}{\text{Time shifting property.}}\)
If
\(\displaystyle \mathcal{F}\{h(t)\} = H(f)\)
Then,
\(\displaystyle \mathcal{F}\{h(t - t_0)\} = H(f) e^{-j2\pi f t_0}\)
We are given:
\(\displaystyle H(f) = \frac{A}{2\pi}\left(\frac{\cos \pi fT}{f + \frac{1}{2T}} - \frac{\cos \pi fT}{f - \frac{1}{2T}}\right)\)
Then,
\(\displaystyle G(f) = \mathcal{F}\{g(t)\} = \mathcal{F}\left\{h\left(t - \frac{T}{2}\right)\right\} = H(f)e^{-j2\pi fT/2} = \frac{A}{2\pi}\left(\frac{\cos \pi fT}{f + \frac{1}{2T}} - \frac{\cos \pi fT}{f - \frac{1}{2T}}\right) e^{-j\pi fT}\)