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izen
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Homework Statement
How did [itex]\frac{1}{2}x[/itex] come from at k=1?
Homework Equations
The Attempt at a Solution
because k=1 will make the first term at denominator 2(k-1) = [itex]\frac{0}{0}[/itex]
Yes. The [itex]\ \frac{1}{2}x\ [/itex] comes from the fact that the first term has the form 0/0 as k → 1.izen said:Homework Statement
How did [itex]\frac{1}{2}x[/itex] come from at k=1?
Homework Equations
The Attempt at a Solution
because k=1 will make the first term at denominator 2(k-1) = [itex]\frac{0}{0}[/itex]
SammyS said:Yes. The [itex]\ \frac{1}{2}x\ [/itex] comes from the fact that the first term has the form 0/0 as k → 1.
What is [itex]\displaystyle \ \lim_{t\to0}\frac{\sin(t)}{t}\ ?[/itex]
The Fourier transform of a half-wave rectified signal is a combination of the original signal's Fourier transform and its mirror image's Fourier transform. This is due to the fact that the rectification process introduces odd harmonics to the signal.
The amplitude of the Fourier transform of a half-wave rectified signal decreases with increasing frequency. This is because the higher frequency components are filtered out by the rectification process, resulting in a decrease in amplitude for those frequencies.
No, a half-wave rectified signal cannot have negative frequency components in its Fourier transform. This is because the rectification process only allows positive frequency components to pass through, resulting in a purely real Fourier transform.
The power spectral density of a half-wave rectified signal is equal to the square of its Fourier transform. This means that the power spectral density will have a similar shape as the Fourier transform, but with a different scale on the y-axis.
The Fourier series of a half-wave rectified signal only contains odd harmonics, while the Fourier transform includes both odd and even harmonics. This is due to the fact that the Fourier series represents a periodic signal, while the Fourier transform represents a non-periodic signal.