# Why we do that in AM demodulation?

Discussion in 'Wireless & RF Design' started by idmond, Feb 14, 2013.

1. ### idmond Thread Starter Member

Oct 5, 2010
18
1
[SIZE=-1]Hey all,

[/SIZE][SIZE=-1]"In AM synchronous demodulation, Why we don't divide m(t)coswt by cos(wt)
instead of multiplying [/SIZE][SIZE=-1]by cos(wt)?"[/SIZE]
[SIZE=-1]
When I was taking a class in Analog Communication... I recall that I was watching one of the video lectures on signals and [/SIZE][SIZE=-1]systems from Berkeley University, the Professor was talking [/SIZE][SIZE=-1](in that video recording of the lecture) [/SIZE][SIZE=-1]about[/SIZE][SIZE=-1] amplitude modulation and why when we want to [/SIZE][SIZE=-1]demodulate an AM modulated signal , in [/SIZE][SIZE=-1]synchronous[/SIZE][SIZE=-1] detection, we multiply the AM modulated signal = m(t)cos(wt) by cos(wt)[/SIZE], where w is the carrier frequency, and why we don't just divide by cos(wt) to get m(t) back, since this can be easily implemented by a simple divider circuit.

well, he didn't answer that question in the lecture, may be he just left it as an assignment for his students, I don't know. but it just got me thinking about it, if it's all about thinking mathematically, then it seems like it's more intuitive and a whole lot easier if we just divide by coswt, why we go through all the trouble and multiply then we have to know the trig identity of (cos(a)cos(b)) and then put a BPF centered at w after the output...

[SIZE=-1]what are your thoughts on this?
[/SIZE]

2. ### MrChips Moderator

Oct 2, 2009
19,117
6,146
The straight forward answer is that in a radio receiver demodulation is performed by a mixer circuit which is essentially analog multiplier.

3. ### idmond Thread Starter Member

Oct 5, 2010
18
1
so why not a divider circuit?

4. ### MrChips Moderator

Oct 2, 2009
19,117
6,146
Because a divider circuit is not easy to make.

5. ### idmond Thread Starter Member

Oct 5, 2010
18
1
can you please elaborate more on this?

6. ### Papabravo Expert

Feb 24, 2006
12,284
2,723
Take a look at an analog mixer and you will see a simple circuit with a few components. Multiplication happens naturally in such a circuit. AFAIK there is no similarly simple divider circuit that can be realized with a handful of components. I suppose you could make a divider if you had a reciprocal circuit and a multiplier(mixer).

In any case the other simple answer is that multiplication and division are not the point. The point is that with periodic functions (sinewaves), multiplication produces signal energy at the input frequencies, but also on the sidebands which are the sum and the difference of the two input frequencies.

7. ### idmond Thread Starter Member

Oct 5, 2010
18
1
so what you're saying is that it's possible to implement such a divider circuit that can do the AM demodulation but it's gonna be a lot easier if we did the demodulation using the multiplier scheme because the multiplier circuit is simple and easy to implement. right?

8. ### Tesla23 AAC Fanatic!

May 10, 2009
374
88
Dividing by cos(wt) would result in terrible noise performance. The optimum filter in the presence of AWGN is a matched filter, which can be implemented by multiplying by cos(wt). This has the effect of not weighting the period around the zero crossings very highly - as there is little signal power here, and weights the peaks of the waveform highest, where there is most signal power. If you divide by cos(wt) you will magnify the noise around the zero crossings where there is little signal. Read up on matched filters, and try to work out the S/N at the output of a modulator that divides by cos(wt).

9. ### Papabravo Expert

Feb 24, 2006
12,284
2,723
Zero crossings are problematical in a divider -- aren't they?

10. ### idmond Thread Starter Member

Oct 5, 2010
18
1
Tesla23, let me get this straight:

In the real world, noise is added to the received signal and it is going to look something like this, m(t)coswt+n(t), where n(t) is the added noise When the modulated signal passes through the channel, If we then divide by coswt, we will get, m(t)+n(t)/coswt, and since the cosine function ranges from -1 to 1, thus for values of cosine less than 1 or more than -1, the noise term will be amplified, because now I divide the noise term by a value less than 1, and its effect will be maximum around the zero crossings where the cosine function has a very small value there and thus the noise will be maximum there , so we will get poor SNR. did I get this right?

11. ### Tesla23 AAC Fanatic!

May 10, 2009
374
88
idmond, after a little analysis I'll change my answer. Assuming that the signal plus noise has been bandpass filtered around ω, then the composite signal can be represented by

$S(t) = m(t)cos(\omega t) + n_I(t)cos(\omega t) + n_Q(t)sin(\omega t)$

if you demodulate this by dividing by cos(ωt) then you get (ideally):

$S(t)/cos(\omega t) = m(t) + n_I(t) + n_Q(t)tan(\omega t)$

The key thing to note is that although the division by small numbers does amplify the noise in the quadrature channel, it stays at the carrier frequency (and harmonics) and (ideally) does not mix to baseband where it would add to the demodulated signal.

So if you low-pass filter the output the result is

$m(t) + n_I(t)$

the same outcome as the traditional approach of multiplying by cos(ωt) and low pass filtering.

There is clearly a practical issue of handling the singularities around where cos(ωt)=0, but ideally the resultant tan(ωt) has no low frequency components that would make this demodulator worse than the traditional one.

What you can achieve in practice depends on how you handle the division around the zero crossings. It certainly looks feasible that you could achieve reasonable performance (e.g. with a blanking period), but compared to the traditional approach it looks more complex and with, at best, slightly worse performance.

I still don't like it though - you are multiplying by a clipped version of 1/cos(ωt), which has lots of high frequency components, so you are relying on there being no components at comparable frequencies on the signal. So the implementation will be sensitive to high frequency signals that leak into the signal path. Multiplying by cos(ωt) is much cleaner - only extracting the signal component desired.

If you understand about signal spaces, AM is simply modulating the amplitude of the cos(ωt) basis function, and the simplest way to extract the amplitude of this component is to dot product the received signal vector with cos(ωt). What you are doing when you multiply by 1/cos(ωt) and low pass filter is you are forming the dot product with 1/cos(ωt). Now a clipped version of 1/cos(ωt) is periodic and so has a fourier series (with very poor convergence properties due to the singularities), but importantly it has a cos(ωt) component. This is the only one you need for your demodulator, and it extracts out the cos(ωt) component in the signal. All the other harmonics in the clipped version of 1/cos(ωt) will also extract out components of the signal if they exist - so you need very good filtering to ensure that they don't.

Last edited: Mar 4, 2013
12. ### Tesla23 AAC Fanatic!

May 10, 2009
374
88
Just to expand on what I mean by responding to harmonics, if you feed

$S(t) = m_1(t)cos(\omega t) + m_3(t)cos(3\omega t) + m_5(t)cos(5\omega t) + m_7(t)cos(7\omega t)$

into a demodulator that multiplies by cos(ωt) and LPF you will get:

$m_1(t)$

if you feed it into your demodulator where you divide by cos(ωt) and LPF you will get:

$m_1(t) - m_3(t) + m_5(t) - m_7(t)$

It is (very) sensitive to frequencies other than what you want.

13. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,208
427
A divider can be built to divide ONE input frequency, but not an infinite number of them. A multiplier can multiply an infinite number of frequencies simultaneously...such as real world communications.