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A modulating signal $m(t)= \sin(2 \pi \nu_m t)$ is transmitted via a carrier of frequency $\nu_c$ using upper sideband (USB) modulation. Show that the USB modulated signal can be demodulated using a reconstructed carrier signal and a low pass filter.

Attempt:

USB (SSB signals in general) can be demodulated by multiplying them with a reconstructed carrier $\cos(2\pi \nu_c t).$ So I first tried to write an expression for the transmitted USB modulated signal:

$$f(t)=A \Big[ m(t) \cos(2 \pi \nu_c t) - \hat{m}(t) \sin(2\pi \nu_c t) \Big]$$

$$f(t)=A \Big[ \sin(2 \pi \nu_m t) \cos(2 \pi \nu_c t) - \hat{m}(t) \sin(2\pi \nu_c t) \Big] \tag{1}$$

Where $A$ is some amplitude, and $\nu_c >> \nu_m$. So to proceed further, I think we need to calculate the Hilbert transform $\hat{m}(t).$ It is possible to do this by finding the imaginary part of the analytic signal:

$$m_a (t) = m(t) + j \hat{m} (t)= 2 \int^\infty_0 M(\nu) \exp(j 2 \pi \nu t) d\nu \tag{2}$$

But the Fourier transform of $m(t)$ is tedious, so I am wondering if there is a simpler way of doing this?

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Apart from scale factors, your USB signal is

$$f(t)=m(t)\cos(2\pi\nu_ct)-\hat{m}(t)\sin(2\pi\nu_ct)\tag{1}$$

(as you've correctly stated in your question). Now if you multiply with a (coherent) carrier $2\cos(2\pi\nu_ct)$, you get

$$\begin{align}g(t)&=2\big[m(t)\cos(2\pi\nu_ct)-\hat{m}(t)\sin(2\pi\nu_ct)\big]\cos(2\pi\nu_ct)\\ &=2m(t)\cos^2(2\pi\nu_ct)-2\hat{m}(t)\sin(2\pi\nu_ct)\cos(2\pi\nu_ct)\\&=m(t) (1+\cos(4\pi\nu_ct))-\hat{m}(t)\sin(4\pi\nu_ct)\tag{2}\end{align}$$

Applying a low pass filter to $(2)$ will filter out all terms at $2\nu_c$, leaving you with the message signal $m(t)$.

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