To say "180 degrees out of phase" instead of "inverted" is actually a misnomer: yeah, ok it's sort of equivalent in the case of a regular waveform, but the inverted signal used in balanced transmission is not the product of a phase shift, and this nomenclature is completely wrong when we're talking about audio signals, which are not regular i.e. repetitive, signals. Other than that thanks for the vid, nice clear presentation.
Keep in mind that being balanced does not equal noise rejection. In accordance with V=IR, if A and -A are not the same impedance from driving to receiving end, the *voltage* noise induced in the wires will not be perfectly equal and not be perfectly rejected. It goes much further than that, of course... That's just one example.
For years I've been looking into the differences in the wiring between balanced and unbalanced, and it wasn't until this video I actually understood what was going on. I'm familiar with the theory of differential pairs, but didn't put two and two together. Thanks!
YEEEEEEEEEE This was an amazing and easy to understand explanation of a balanced signal, Thank you! I was expecting to be going into the hole of why some headphones have larger impedance then others again.
No, when subtracting a negative from a positive, this equals a positive. eg, if the signal is 2, then the out of phase signal is -2. Therefore 2- -2 = 4, this is the "difference". Because the noise is originally in phase it is then cancelled out because you are subtracting a positive from a posative. eg if the noise is 3, then 3-3=0 So you end up with a loud signal with no noise...... hope this helps :)
Looking for some good monitors, and i came across some monitors with balances and some with unbalanced. Thanks for the information, it was super helpful!
pffft. chill re: the intro. If you want good intro's search for good intro's, the content here was exactly what I was looking for, concise and technical. Balance!
fuck dude, i was sitting i pain with this shit for like two evenings straight and you blew my teacher out of the water in five minutes, youre fantastic
I have a Scarlett 2i4 with 1 pair of balanced plug outputs and 2 pairs of unbalanced RCA. My monitors have just a pair of RCA inputs. Which outputs shall I use? Thanks!
Thank you for a great explanation. Now I know that in my next DIY unbalanced interconnects,.only positive wire (center RCA '3 screws type' conector pin) carries actually actually the signal in the silver plated copper shielded wire. Thank you!
So, do the headphones have be balanced too, or just the cable ? In other words can I use a balanced cable on any headphones? If they make them ofcourse Thanks 😊.
sometimes it is unavoidable that we have an unballance output that must be connected to amplifier input which need balance input. what is the best way to do this? thanks
Yes and from the engineer's perspective it is, but on the consumer end I have measured rejections of 40dB which does not cut it for my purposes (and I did get the rejection to 80dB with some work)
Very good explanation , it is just to understand what enter to the imput of the operactional amplifier is the difference of the 2 signals, the second (line blue) is the same of the red one , but out of fase in 180° . Filanally the OA does the rest .
Conclusion: "All sound systems need to be balanced". Where the run of cable is short then the noise on the signal wire is negligible. For example a 50 cm interconnect between pre-amp and power-amp would not need to be balanced unless there was an extreme condition. Many amplifiers do not have the option of balanced inputs. Moreover those cables can be shielded even if unbalanced. This could be by way of a co-axial signal cable, or a shield in addition to the signal and return wires.
That's right ! My question is: The + & - signals need to be an equal amplitude from output, or there's need to be different ?!? I thing is need to be equal, but I tested a circuit from acknowledged brand, and the signal from out isn't equal. So I changed the resistor to higher on opamps - signal from 10K to 22K, on + in I leaved the 10K, and now its equal. :)
I think is the best explanation i have seen. But just to be sure i got it right... The noise on the two signals before they arrive at the amplifier are in phase? And in the amplifier the two signals and noise polarity are reversed? So noise cancel each other because now they are out of phase 180 degrees? And the signal (amplitude) is duplicated because the 2 signals are are now in phase so they were added?
see my post above about inverted vs phase nomenclature; the noise signals as measured at the inputs of the differential amp are the same polarity, but the audio signals are opposite polarity. The diff amp produces an amplified version of the difference between the signals at the + and the - input. This is ok for the audio because we've fed the two inputs with opposite polarity signals. It's also great for noise cancelling because the +/- nature of the diff amp means that the noise signals effectively are nulled out in the amp, because the two leads, being in very close proximity to each other, pick up the same amplitude of noise signal. Google balanced lines, or twisted pairs, for better explanation.
Great video. What would happen if say, I connected an unbalanced stereo (single signal line for left and right) such as phono L R in to a balanced amplifier, (2 signal line each for left and right), would that simply not work properly? Or would it be just as effective as having an ordinary unbalanced signal to amp? Try to illustrate: monoL --------> balanced signal A ---------------- balanced amp ?`¦> balanced signal -A ------------ balanced amp monoR --------> balanced signal A ---------------- balanced amp ?`¦> balanced signal -A ------------ balanced amp My reason for asking is, I am making a custom audio switcher that has 4 selectable stereo inputs and balanced stereo outputs (required for balanced active speakers) I suppose it would be easier for me to just leave the unbalanced option and just make a balanced audio switcher, or is there some clever way I can jury rig the unbalanced signals into the balanced ones? Thanks for any reply :3
Is the source signal brought back into phase at the end of the cable itself or by the receiving input at the end? Can using a female to male xlr converting adapter (male/male or female/female) reverse the phase again and thus making the received signal noise only?
Hi, I thought I'd answer the question on Alectrosystem's behalf. The input of the amplifier does the work as explained in the video. i.e. Mackie SRM450 version 1 and version 2. They require a 'balanced' input and the 2 signals run to an operational amplifier (chip) which amplifies the difference in the 2 signals. The difference is the music / audio since the noise will be the same. You can use cables to convert to 'unbalanced' but not the other way round. The noise is on top of the signal so you can't separate just the noise.
I can see the advantage of having one signal the complete polar opposite to the other, but this would still work fine with both signals completely in phase but with one stronger than the other. As long as the interference is exactly the same polarity and amplitude on both wires, then even if the two audio signals were in phase but of different amplitudes, it would still work (although you'd need to turn the gain up more to get the same volume out of the amp).
This is very good. Thank you so much. I liked the simplicity and clarity, and the equation at the end. I will go on and research how the amp substracts signal 1 minus signal 2. I do have a question, on 1:40 you say that the ground cable is the return path for the signal. I don't understand this. Please help. Thanks again.
3:18 video says -> "these wires are connected to a differential amplifier which subtracts the out of phase signal from the in phase signal" I'm confused, if the 'out of phase signal' (music) is subtracted from the 'in phase signal' (noise) doesn't that mean that we’re keeping the noise and subtract the music? I would think that a differential amp (which only amplifies the "difference" in the signal) would subtract the in phase signal from the out of phase signal. Right?
Here is my question (I am a dilettante when it comes to the electrical component of recording): I plug a balanced cable into my guitar. How does that signal become inverted on the -A side?
Are you sure that your guitar use TRS connectors? I thought they use TS, which is unbalanced. The signal will never be balanced before both the source, the cable, and the connection in the other end is balanced.
So... I should use balanced if i'm doing a live show (hypothetical) because the distance between the output and the amplifier is greater? and If i'm just doing something at home and just recording or something I wouldn't need to use balanced?
So instead of saying we are subtracting the +A from -A we can say that we are basically just flipping one of them back to in phase which knocks the Noise out of phase therefor cancelling the noise out. Correct?
Because ALL the wires act like an antenna. This means any external noise the wires pick up will be in phase (i.e. the same noise on each wire). I think he’s talking about external noise, like a CB radio that bleeds into your PA or a vacuum cleaner that causes a hum in your guitar amp. He's NOT talking about internal noise (like noisy single-coil pickups on your strat).
He's saying they're being added because they're being subtracted and they're inverse. Look at the equation at 3:58 and think about it for a second it'll make sense.
Everything makes sense except for the part where he says, "However the noise is in phase". If the two signals are 180 degrees out of phase with each other on a Balanced audio system, then why does the diagram show the noise following A and -A along their wave trajectory? Is their some deeper science behind noise waves which always keeps them in phase with each other? If so are they oriented as + or - or neither when processed by the differential amp.? Besides that the video was succinct!
In my opinion they are in phase because you can assume that both wires works same as antenna and capture the same amount of noise cause they are close to each other and positioned same in space, however I assume that some part of noises are still left on The output, but it's relatively small
It's just a wrong graphical representation but the idea is correct. Think of it this way, if you would plot alone the noise signal in both of the wires, it would look the same. The noise is just mounted on out of phase signals, but the noise itself in phase.
consumer end? surely an engineer is more likely to need better noise reduction for their precision of work, consumer brings to mind somebody just plugging a guitar into an amp. what were your purposes?
