How Hilbert Cancells SSB Sideband

Start with a simple baseband tone

Take

$$m(t)=\cos(\omega_m t)$$

and carrier

$$\cos(\omega_c t).$$

If we simply multiply them:

$$s(t)=m(t)\cos(\omega_ct)$$

then

$$s(t)=\cos(\omega_mt)\cos(\omega_ct).$$

Using

$$\cos A\cos B =\frac12[\cos(A+B)+\cos(A-B)],$$

we get

$$\boxed{s(t)=\frac12\cos((\omega_c+\omega_m)t)+\frac12\cos((\omega_c-\omega_m)t)}$$

So we get two frequencies:

$$\boxed{\omega_c+\omega_m}$$

and

$$\boxed{\omega_c-\omega_m}.$$

These are the upper sideband (USB) and lower sideband (LSB).

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We want only one

Suppose we want the upper sideband:

$$\omega_c+\omega_m.$$

We need another signal that produces the same two frequencies but with the unwanted one having the opposite sign.

This is where the Hilbert transform comes in.

For

$$m(t)=\cos(\omega_mt),$$

the Hilbert transform is

$$\boxed{\hat m(t)=\sin(\omega_mt)}.$$

Now multiply this by the sine carrier:

$$\hat m(t)\sin(\omega_ct)=\sin(\omega_mt)\sin(\omega_ct).$$

Using

$$\sin A\sin B=\frac12[\cos(A-B)-\cos(A+B)],$$

we get

$$\boxed{\hat m(t)\sin(\omega_ct)=\frac12\cos((\omega_c-\omega_m)t)-\frac12\cos((\omega_c+\omega_m)t)}$$

Look carefully.

First branch

$$m\cos(\omega_ct)$$

gives

$$+\frac12\cos(\omega_c+\omega_m)t+\frac12\cos(\omega_c-\omega_m)t.$$

Second branch

$$\hat m\sin(\omega_ct)$$

gives

$$-\frac12\cos(\omega_c+\omega_m)t+\frac12\cos(\omega_c-\omega_m)t.$$

Now we have exactly what we need.

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Add/subtract the branches

For the upper sideband, use

$$\boxed{s_{\text{USB}}=m(t)\cos(\omega_ct)-\hat m(t)\sin(\omega_ct)}$$

Substitute the two expressions:

$$s_{\text{USB}}=\left[\frac12\cos(\omega_c+\omega_m)t+\frac12\cos(\omega_c-\omega_m)t\right]$$ $$-\left[\frac12\cos(\omega_c-\omega_m)t-\frac12\cos(\omega_c+\omega_m)t\right].$$

Now combine terms.

For the lower sideband:

$$+\frac12-\frac12=0.$$

So it cancels.

For the upper sideband:

$$+\frac12+\frac12=1.$$

So it remains.

Therefore:

$$\boxed{s_{\text{USB}}=\cos((\omega_c+\omega_m)t)}$$

Only the upper sideband remains.