# Difference between single tone and dual tone signals?

Editted

Single tone signal has 1 frequency: $$y = \sin(2\cdot \pi \cdot f_1 \cdot t)$$

Dual tone was created by using 2 frequencies:

$$\sin(2\cdot \pi \cdot f_1 \cdot t) + \sin(2\cdot \pi \cdot f_2 \cdot t)$$

How can I define on the plot this difference, whether it is a single tone signal or not?

• well, now that you've changed the formulas, hasn't the question answered itself? Jun 24 at 13:16

It's either, depending how we look at it:

$$\sin(f_1t) + \sin(f_2t) = 2 \cos(.5(f_1 - f_2)t)\sin(.5(f_1 + f_2)t) \tag{1}$$

A common empirical rule is that if $$f_1 \ll f_2$$ (e.g. $$f_2 / f_1 > 5$$), consider them separate - else, a single amplitude-modulated tone, where $$(f_1+f_2)$$ is the "carrier" and $$(f_1-f_2)$$ the "modulator".

This can be observed in a time-frequency representation:

But if we really wanted, we could make STFT represent these as separate, by increasing the window's frequency resolution:

### Code

import numpy as np
from scipy.signal import windows
from ssqueezepy import stft
from ssqueezepy.visuals import imshow, plot

def cosines(freqs, N):
t = np.linspace(0, 1, N)
return np.sum([np.cos(2*np.pi * f * t) for f in freqs], axis=0)

N = 2048
t = np.linspace(0, 1, N)
x1 = cosines([50, 250], N)
x2 = cosines([50, 60],  N)

Sx1, Sx2 = stft(x1), stft(x2)
stft_freqs = np.linspace(0, .5, len(Sx1)) * N

plot(t, x1, title="f1, f2 = 50, 250", show=1)
plot(t, x2, title="f1, f2 = 50, 60",  show=1)

kw = dict(abs=1, xticks=t, yticks=stft_freqs,
xlabel="time [sec]", ylabel="frequency [Hz]")
imshow(Sx1, **kw, title="abs(STFT) | f1, f2 = 50, 250")
imshow(Sx2, **kw, title="abs(STFT) | f1, f2 = 50, 60")

Sx2_2 = stft(x2, windows.dpss(N, 4), n_fft=2048)
stft_freqs = np.linspace(0, .5, len(Sx2_2)) * N
kw['yticks'] = stft_freqs
imshow(Sx2_2, **kw, title="abs(STFT), frequency-localized | f1, f2 = 50, 60")
kw['yticks'] = stft_freqs[:128]
imshow(Sx2_2[:128], **kw, title="frequency-zoomed (same STFT)")


Single tone signal has 1 frequency: $$\sin(2\cdot \pi \cdot f_1)$$

That's not a single tone signal. That is a constant. This is missing the time-dependency!!

You mean something like

$$s_1(t) = \sin(2\pi f_1 t).$$

In a real-valued context, that would be a single tone; in the context of complex signals (and you've been talking about OFDM before, so this is probably the context in which you want to work), this is a signal composed of two complex tones: one at $$-f_1$$, and one at $$+f_1$$.

Dual tone was created by using 2 frequencies: $$\sin(2\cdot \pi \cdot (f_1 + f_2))$$

No. simply no. Even when writing this as

$$s_2(t) = \sin(2\pi(f_1+f_2)t),$$

this is just a the same as your first signal, but it's at frequency $$f_s=f_1+f_2$$ instead.

So, I think you really might have some of your math basics wrong; I'd love to help you, but I think you simply have a wrong definition in your head, nothing I can fix :(