# Spectral effects of multiplying a signal by itself?

This has to do with creating the synchronization circuit for MSK in a DSP as described in "Minimum Shift Keying" by Subbarayan Pasupathy published in IEEE Communications Magazine.

At the beginning of the circuit he shows signal passing through a "squarer". The explanation reads:

"Although the MSK signal $s(t)$ has no discrete components which can be used for synchronization, it produces string discrete spectral components at $2f_+$ and $2f_-$ when passed through the squarer. (The squarer, in effect, doubles the modulation index and produces an FSK signal with $\delta_f = 1/T$, known as Sunde's FSK. Sunde's FSK has 50 percent of its total power in the line components at the two transmitter frequencies.)"

$s(t)$ consists of a signal containing components at $\pm 300 \mathrm{Hz}$ around the carrier frequency.

I'm confused by this because when I multiply the signal by itself it doesn't seem to produce a signal with components at twice the offset frequencies. Is squaring a signal different than multiplying it by itself (sample by sample, not convolution with itself)?

Thank you

• Are you using a high-enough sampling frequency? In your simulation, you'll need a sampling frequency that is more than double the maximum frequency that can ever appear in the system. – MBaz Oct 9 '14 at 21:05

Is squaring a signal different than multiplying it by itself

In the frequency domain this is equivalent to convolving the original frequency spectrum with itself. The convolution "spreads" the signal out. If the signal is at $\pm300$ Hz around a carrier frequency $f_c$, then you should see energy at $\pm(2f_c + 600)$ Hz, $\pm2f_c$ Hz, $\pm(2f_c - 600)$ Hz, $0$ Hz, and $\pm600$ Hz.