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To complement the other quite appropriate answers: Say you build a real signal $s(t)$ with bandwidth $B$ that transmits $R$ bits per second. Its spectrum $S(f)$ extends from $-B$ to $B$. The modulation property of the Fourier Transform tells you that the spectrum of $$s(t)\cos(2\pi f_c t)$$ is $$\frac{1}{2} \large[ S(f+f_c)-S(f-f_c) \large].$$ Note that ...


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Because bit rate depends on the bandwidth and not on the carrier frequency. Of course at higher frequencies you have more bandwidth, and thus you can transmit more data. But 1 MHz in lower frequencies and 1 MHz at higher frequencies have no difference on the data rate. Other effects may need to be taken into consideration though at higher frequencies. For ...


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If you are a bass and sing A 220 Hz and B 247 Hz at 120 beats per minute, you can sing data at a certain rate. If you are a soprano and sing two quarter notes A 880 Hz and B 988 Hz at the same BPM, your data rate isn't any higher. If you sing more notes (higher bandwidth) you could communicate a more complex score (e.g. a higher data rate). So bandwidth (...


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