Question

(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator together with a carrier at 100 kHz, with only the upper sideband retained. In the coherent detector used to recover the local oscillator supplies a sinusoidal wave of frequency 100.02 kHz. Determine the frequency components of the detector output. (b) Repeat your analysis, assuming that only the lower sideband is transmitted.

Answer 1

Answer:

Explanation:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

$100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz$

Carrier frequency 100kHz

Upper side band

$100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz$

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

$100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz$

At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

$100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz$

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

$100k, 99.9k, 99.8k\ \ and \ \99.6kHz$

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

$100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz$