Simple SSB-AM Synchronous Demodulation

[MathsDude69]

Homework Statement

An SSB-AM modulator is shown in fig 1 (please see attached), where the SSB filter removes the lower sideband.

A synchronous demodulator is shown in fig 2 (please see attached). Show that by setting the frequency of the local oscillator to the carrier frequency the output of the demodulator will contain the original message.

Homework Equations

Trigononmetric itdentities:

cos(u).cos(v) = 0.5cos(u-v) + 0.5cos(u+v)
cos²(a) = 0.5(1 + cos2a)

The Attempt at a Solution

We can state that:

V₁(t) = 8 + 4cos(2π5x10³t)
V₂(t) = 8cos(2π10⁵t) + 2cos(2πx105x10³t) + 2cos(2πx95x10³t)
V₂(t) = 8cos(2π10⁵t) + 2cos(2πx105x10³t)

in which case at the demodulation stage we would get:

(8cos(2π10⁵t) + 2cos(2πx105x10³t)) x cos(2π10⁵t)

which is:

4(1+cos(4π10⁵t) + cos(2πx5x10³t) + cos(2πx205x10³t)

All the information I have on this illustrates that the demodulated signal should have a baseband component of the original message at half the magnitude and a component at twice the carrier frequency with a magnitude a quarter of the original. Can anyone see where I have gone wrong here??

 

[KataKoniK]

Hello, thank you for your post. It seems like you have made a few errors in your attempt at a solution. Firstly, in the equation V2(t), the second term should be 2cos(2πx105x103t), not 2cos(2πx105x103t) + 2cos(2πx95x103t). This is because the lower sideband is removed by the SSB filter, leaving only the upper sideband at 2πx105x103t.

Additionally, when multiplying V1(t) and V2(t) at the demodulation stage, you should use the trigonometric identity cos(u).cos(v) = 0.5cos(u-v) + 0.5cos(u+v). This will result in a demodulated signal of:

4 + 2cos(4π105t) + 2cos(2πx5x103t) + 2cos(2πx205x103t)

Which does indeed contain the original message at half the magnitude and a component at twice the carrier frequency with a magnitude a quarter of the original.

I hope this helps clarify any confusion and good luck with your studies!