Demodulating signal by passing incoming carrier wave through low pass filter as opposed to multiplying by carrier wave? Demodulation through many resources found online says that the incoming signal is multiplied by the carrier wave and then passed through a low pass filter that removes the carrier wave from the modulated signal. I understand this.
However, we know that the incoming signal is already multiplied by the carrier wave therefore, shouldn't passing the incoming signal through a low pass filter remove the carrier wave anyway?
For example.
The former states :
Modulated Signal * Carrier wave  = incoming signal
Incoming signal * (Wave of same frequency and phase)  ---&gt; low pass filter = modulated signal
The latter states(my question):
Modulated Signal * Carrier wave  = incoming signal
Incoming signal ---&gt; low pass filter (removes the carrier wave)= modulated signal
I would like to know why the former is stated rather than the latter and the benefits of it.
I suppose the incoming signal can include noise that could somewhat alter the signal therefore multiplying by a wave of same frequency can help, but I'm not too sure about this.
What you're missing is that multiplying a signal by a sine wave changes the frequency at which the signal appears.  Running a signal through a linear time-invariant filter (what you mean above when you just say &quot;filter&quot;) can emphasize or de-emphasize the signal content at certain frequencies, but it cannot move the signal in frequency.
The classic explanation for this is to consider a signal that's a single sine wave at some frequency $f_{RF}$.  So, $x(t) = \cos(2 \pi f_{RF} t)$.  Put it through a low-pass filter and it's just gone.  However, if you multiply that signal by a local oscillator, you generate signals at the sum and difference of the signal frequency and the carrier frequency.  Let your local oscillator signal be $x_{lo}(t) = \cos(2 \pi f_{LO} t)$.
Multiply them together, and you get $$x_{IF}(t) = x_{RF}(t) x_{LO}(t) = \cos(2 \pi f_{RF} t) \cdot \cos(2 \pi f_{LO} t)$$
Now it's just trigonometry: $$x_{IF}(t) = \frac{1}{2}\left(\cos \left(2 \pi f_{RF} - f_{LO} \right) t + \cos \left(2 \pi f_{LO} + f_{RF} \right) t \right)$$
Now you have something useful for a radio receiver: the signal at $f_{RF} - f_{LO}$.  The only problem is that you have the other signal, too.  So now you run that through a low-pass filter -- and the result is your desired signal.