Hi

My main question is Q2. Please help me with it.

Q1: In DC current electrons simply follow a 'smooth' path without any fluctuations and that's what direct non-alternating current is.

Now let me tell you what I think of AC current. As you can see in this diagram there are two terminals A and B. e(-) stands for electrons and arrows stand for their direction of motion. When electrons are moving toward left of the page, the current or voltage is +ve and when electrons are moving towards right of the page the current is -ve.

Each terminal, A and B, switches between +ve and -ve. When A is +ve, B is -ve. If the frequency of AC is 50 Hz then each terminal would switch between opposite polarities 100 times during one second. So, it is all about push and pull of electrons.

In a wire electronic current flows more like sound waves rather water waves. Isn't it so? So, would you define an alternating current a longitudinal wave? A longitudinal wave can also be shown as a sine wave as is shown here.

Q2: When two waves meet there is going to be interference. This applet is a good one to see this. For example, in this configuration both waves have same amplitude and are in phase which results in double amplitude. I hope everything is correct so far. Now we turn to the main question. As the figure shows noise wave is 'riding' the main signal wave. I'm having a lot of difficulty to picturize it. When a signal wave is superimposed with noise wave, we are going to get a new third wave, aren't we? So, technically speaking noise wave is not riding the signal wave. I think I will wait for your reply to this and then will move on with the next part of the question. Thank you for the help.

Regards
PG

 

In a wire electronic current flows more like sound waves rather water waves. Isn't it so? So, would you define an alternating current a longitudinal wave?

No I would not, for two reasons. First, electromagnetic waves are typically more like transverse waves than longitudinal waves. Second, you are discussing AC in the context of circuit theory. Circuit theory is an abstraction where there is no spatial dimensions, and the only dimension is time. All spatial variables are lumped into the circuit component values and key assumptions must be met. Hence and AC waves are waves in time only and the classification of longitudinal versus transverse is not even meaningful.

However, when circuit dimensions become large compared to the wavelength, spatial waves exist, either as guided waves along wires, or as free-space waves when radiation occurs. In this case, the waves are much better described as transverse waves.

 

When a signal wave is superimposed with noise wave, we are going to get a new third wave, aren't we? So, technically speaking noise wave is not riding the signal wave.

Rather than calling it a wave, call it a signal. The true signal will be a sum of the main signal and the noise signal. If I understand you correctly, there is this "third" signal which is the sum of the other two. So, saying the noise rides the wave is just a figurative statement. The two signals are superposed or added together.

I think the author is using poor language in saying "superimposed", because the correct word is superposed. Superposition has a special meaning with signals and noise. The word superimpose means to put something on top of something else, leaving both entities still visible. The picture does show the noise superimposed on the signal, and this is incorrect as you point out. All one would see is one signal (this third signal you spoke of) that is the superposition of the noise and signal.

 

Thanks a lot, Steve.

Now I would jump to main question. So, there we have a third signal which is sum of other two. I was studying about rejection of common mode signals in amplifier topic. It is said that an amplifier 'extracts' the main signal and rejects the common mode noise. How can it extract the actual signal from the third signal which is sum of main signal and noise? You can't separate milk from the tea. Please help me with it.

Best wishes
PG

 

Thanks a lot, Steve.

Now I would jump to main question. So, there we have a third signal which is sum of other two. I was studying about rejection of common mode signals in amplifier topic. It is said that an amplifier 'extracts' the main signal and rejects the common mode noise. How can it extract the actual signal from the third signal which is sum of main signal and noise? You can't separate milk from the tea. Please help me with it.

Best wishes
PG

We are free to express signals as sums of other signals. This works for linear systems. One mathematical rule is that any signal pair that drive a system can be decomposed into a common mode signal and a differential mode signal. In this case, the noise happens to be common mode, by its nature, and you have deliberately constructed the main signals to be differential mode only. The method of extraction is based on the idea that your circuit amplifies the differential mode and rejects the common mode.

Let's make a comparison with a hand operated air pump used to pump up an inflatable tire. You and your friend decide to work together and pull and push on the pump. Your output air flow will depend on the differential mode of pulling and pushing and it will not depend the common mode. If you face each other and both pull towards your chest and both push away from your chest at the same time, the pump operates and the tire gets inflated. However, if he pushes while you pull, and vice versa, the tire is not inflated. If somehow you do a combination of both common mode driving and differential mode driving, the output airflow/pressure will only depend on the differential part of the driving.

Another counter example is a two man saw for cutting trees. Two people have to apply force in the same direction and this works with the common mode drive of two people. If the two people operate by applying differential mode forces, then the saw does not move and the tree does not fall down. Again, if you do a combination of both differential and common mode, only the common mode cuts the tree.

So the idea of common mode and differential mode is more common (pardon the pun) than people realize. The key thing is that there needs to be two inputs, and a common ground reference to define the relations between the signals and the mode components. In cases where the system responds differently to the two modes, it makes sense to express the signals and system response in terms of those modes.

 

I thought I'd put together a cheat sheet to help you comprehend the common mode and differential mode definitions and the context for which they are most useful.

Please see attached drawing. You'll see a general system that is to be driven on top. This system is assumed to be referenced to a ground level. On the left, the two input signals V1 and V2 are general signals that could be applied. On the right, the same signals are shown in terms of differential and common mode signals. The conversion formulas are also there.

Hopefully you can see that there is an implied reference for the common mode signal. Even if you drive the inputs differentially with no explicit ground reference, there is an implied ground level for any common noise that is picked up on the terminals.

 

Thanks a lot for the help, Steve.

Please see the attachment and let me know if I have made any progress in understanding the topic. Thank you for your time.

Best wishes
PG

 

I'm not sure I understand your drawing. Are your plots versus time? if so, this is not correct. The signals are not separated in time, but are different in character. I'll try to put together some actual plots of examples signals and add them to my cheat sheet.

 

Hi

No they are not time separated and that's the reason in the description above "Q" I said "and at the same time". The push and pull of electrons (i.e. electric current) is happening at the same time. I hope it's little clearer now. Thanks.

Regards
PG

 

Hi

No they are not time separated and that's the reason in the description above "Q" I said "and at the same time". The push and pull of electrons (i.e. electric current) is happening at the same time. I hope it's little clearer now. Thanks.

Regards
PG

OK, I think I get it now, but your drawing is still confusing because you are showing one source Vin, but two signals. You can't really add those signals into one effective signal. You should draw at least two input signals, like my diagram on the left, and preferably use three signal, like my diagram on the right. Otherwise, the definitions of your signals in the graphs/colors are not representable in the schematic. However, let me try to clarify what i think you are saying.

So compare your diagram to the one I showed with V1 and V2. Forget about the noise signal and focus on the main signal. How do they compare? From what I see, you have a V1 signal, but your V2 signal is -V1. This means that Vdm=2*V1=-2*V2 and Vcm=0 if you use the formulas I provided.

Now consider your noise signal. You have V1=V2 now, hence Vdm=0 and Vcm=V1=V2.

In the end, your noise signal is a true common mode signal, and your main signal is a pure differential signal. With the values of Vdm and Vcm clearly defined in terms of two separate signals (main and noise), you could move forward and analyze the circuit response.

Now to answer your questions.

First, I don't really find your diagram correct because you show one source Vin feeding the circuit, but you have two signals with different characteristics. One is a differential signal and one is a common mode signal. Hence, it would be preferable to use my diagram right side to express these signals in a less confusing way.

Second, as far as what to conclude, I would say that you have just now characterized the nature of the input signals, both main signal and noise. You now know what types of signals they are and can figure out how your circuit will respond to these signals.

Third, as far as how your circuit will respond, you have to decide if you want to analyze assuming an ideal opamp or a real one. An ideal opamp does not care about the common mode signal at all, but a real one does.