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3

This is a consequence of what was discussed in your previous question, that any real-valued function like $\sin(\omega t)$ consists of both positive and negative frequencies. Multiplying by a complex exponential $e^{i\omega t}$ simply shifts frequency by $\omega$. There are many ways to show this, one way is to simply look it up in a table of Fourier ...

1

A real-valued multiply is the same as an AM modulation, where the result includes both upper and lower sidebands. (Similar to the spectrum of an AM radio station on an SDR waterfall.) A complex-valued multiply is more like an SSB modulation (plus carrier, depending on modulator depth and offset). So you only get one sideband, upper sideband if the ...

4

That's just the math behind it – everything is alright with these results! You need to write down the formula of the real-valued $\sin(t)$ in terms of $e^{j2\pi t}$ and $e^{-j2\pi t}$, and you'll see that, as shown in your plots, the real-valued harmonic oscillations have a positive and a negative frequency component – so multiplying two of these yields four ...

1

If I understand your question correctly, the issue is that you have one graph producing a complex output of an FFT and the other producing a floating point output from an FFT. When you produce an FFT from complex data using a floating point output, you will find that the signal is reflected across the y axis at zero. This is because the floating point ...

0

The output of the FFT you used is complex. When you input a strictly real signal to a FFT that produces a complex result, half of the complex result is the complex conjugate of the other half (this gets rid of all the unwanted imaginary components, when everything is summed up), mirrored. When you display only the magnitudes in an FFT plot, a complex ...

6

Because mathematically, a function like $\sin(\omega t)$ has an angular frequency of $\omega$ and $-\omega$. Consider: $$e^{i\omega t} + e^{-i\omega t}$$ By Euler's formula this can be expanded to: $$\cos(\omega t)+i\sin(\omega t) + \cos(-\omega t)+i\sin(-\omega t)$$ By the trig identity $\sin(x) + \sin(-x) = 0$ this simplifies to:  \cos(\omega t)+ \...

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