Sorry to deflate your argument, but it's not correct. I think where you are coming from is that for you the term "impedance" sounds as if it's the difficulty that a signal has in getting into a piece of equipment, a bit like furniture piled up in a doorway. So you say that a low impedance input presents little difficulty in getting the signal in and therefore the signal is not distorted or corrupted in its passage. It's certainly one way of thinking about inputs and outputs of electronic gear, but it's not the conventional one, and it does not conform with how "impedance" is defined.

If you are happy with "resistance", then we can develop that idea first. When talking about audio gear, an input resistance is considered to be connected across the input and ground, that is, in parallel with and not in series with the input. So a signal sees no difficulty in reaching an input amplifier or whatever is behind the input connector, but the amount of current that must be supplied depends on how much of the input voltage flows to ground through the input resistance. So a high resistance causes little current to flow and a low resistance causes more current to flow. The voltage is not affected as long as the source can continue to supply the demanded current.

For the resistance of an output, it's normal to consider this as being a resistance in series with an output voltage source. With no load, the whole of the source voltage appears on the output, but as you begin to take current, some of the output voltage is lost across the output resistance. So we can see that when an output drives an input, some of the original output voltage is developed across the output resistance and the rest is developed across the load's input resistance. When I first started working with studio audio circuits, input and output resistances were always matched, usually at 600 Ohms, and this meant that when connecting the output of one device to the input of another, in all cases you only got half the nominal output voltage passed across to the input device. The equipment was specified accordingly.

The concept works for both a.c. and d.c., providing the circuits are purely resistive and have no inductance and capacitance in their input and output circuits. This is never the case, however, as even a simple interconnecting cable and connector has capacitance and (to a lesser degree) inductance. It's well known that a long coax cable connecting a guitar that has a standard piezo pickup to its amplifier can roll off the high frequencies due to the capacitance of the cable. So we introduce a frequency-dependence to the resistance terms, and this is called impedance. It's the same concept as resistance, but it's a vector and not a linear quantity, and you have to look out for both amplitude and phase changes. The real part of the impedance is the resistance and the imaginary part is the reactance.

To sum up: inputs and outputs have an impedance, which is a vector quantity. In a lot of cases and over most of the audio frequency range, it's the resistance that we need to deal with. A high-impedance input does not demand much current for the voltage you apply to it, and therefore has only a small effect on a signal being driven from a device with a low output impedance.

Yes, it's a tough argument without mentioning the aspect of capacitance, which is where the whole idea came from actually. I should have mentioned that in the original post. Or was I trying to make a subtle point making fun of the manufacturers who only bring into the equation the parallel resistance value when quoting an input impedance ;) hmm... The thought being that having an extremely high input resistance can't always be a better thing, because when you factor in the inevitable line/cable capacitance, you will get much more significant high rolloff when you plug into something with an extremely high input impedance than if you had a lower impedance input relatively speaking.

I guess (and I know that this is where I am wrong) I am referring to the term impedance in place of coloration, sort of to say that an unimpeded AC range shows all frequencies as a natural flat response, while an impeded AC range is distorted and shaped. Circuits which have need for an input capacitor are sometimes prone to low end rolloff when either the capacitor itself, or the series or parallel resistor is too small. Again, I would consider this to be high impedance seeing as how the AC range is affected.

Maybe I've got impedance and reactance backwards... in fact I think that I might.

What you may not be taking account of is the source impedance of the output that is driving the high-capacitance input. At audio frequencies and using normal connecting cable, the capacitances of the output driver, the cable, the connectors and the receiving input all sum together, and form an R-C filter in which the R term is the output resistance of the driving circuit. To a first approximation, the input resistance of the receiving circuit does not come into the equation, as in modern equipment it is usually much higher than the output resistance of the driving circuit.

I read your first post and thought "what?". I then saw Boswell's measured response which I think put's many of your thoughts back on the right road.

What has not been mentioned as yet is that you are relating input impedance, resistance or whatever to tone and colouration. While this may be your current experience, this is certainly not an engineering "feature" and there is absolutely no reason why changing input impedances should change colouration albeit with caveats relating to the signal source impedances and interconnect impedances (which could be the true source of any colouration rather than input impedance per se).

This is much as you found when trying to form a LP filter by adding capacitance to your guitar amp input. In reality this was far more complex than you thought, as a passive guitar's output impedance varies wildly.

In the real world there are many sources that require very high input input impedance (like your passive Guitar). One of these back in the day was "crystal" microphones. The only way to get these to sound anything approaching some sort of fidelity was to load them with 10MOhm or more. This applies to ceramic pickups often used on acoustic guitars too. On the other hand, some sources are better with a low impedance as Boswell mentioned. His comments about "matched" I/O are most critical for noise considerations as this gives maximum power transfer and hence the lowest noise (look up "maximum power transfer theory") which is common in RF design. With other sources the load impedance can be almost irrelevant - almost!

As you say, you will indeed have a tough argument with this one.

Edit: Boswell beat me to the punch on the "source" of colouration there!

I appreciate you both taking the time to set me straight.

Boswell, post: 376686 wrote: What you may not be taking account of is the source impedance of the output that is driving the high-capacitance input. At audio frequencies and using normal connecting cable, the capacitances of the output driver, the cable, the connectors and the receiving input all sum together, and form an R-C filter in which the R term is the output resistance of the driving circuit. To a first approximation, the input resistance of the receiving circuit does not come into the equation, as in modern equipment it is usually much higher than the output resistance of the driving circuit.

Interesting, I never thought of it that way. The output resistance being the series resistor and the cable network acting as a capacitor in parallel. I always thought of it as the cable network capacitance and the input parallel resistor being in parallel, as well as any other parallel resistances/loads in the vicinity, and calculating the rolloff that way. I think your way more accurately describes what actually happens though. What if someone were to wire their guitar's volume pot as a purely variable resistor from signal path to ground instead of as a voltage divider. Something tells me that if this actually worked more people would be doing it.

MrEase, post: 376687 wrote: I read your first post and thought "what?". I then saw Boswell's measured response which I think put's many of your thoughts back on the right road.

What has not been mentioned as yet is that you are relating input impedance, resistance or whatever to tone and colouration. While this may be your current experience, this is certainly not an engineering "feature" and there is absolutely no reason why changing input impedances should change colouration albeit with caveats relating to the signal source impedances and interconnect impedances (which could be the true source of any colouration rather than input impedance per se).

This is much as you found when trying to form a LP filter by adding capacitance to your guitar amp input. In reality this was far more complex than you thought, as a passive guitar's output impedance varies wildly.

In the real world there are many sources that require very high input input impedance (like your passive Guitar). One of these back in the day was "crystal" microphones. The only way to get these to sound anything approaching some sort of fidelity was to load them with 10MOhm or more. This applies to ceramic pickups often used on acoustic guitars too. On the other hand, some sources are better with a low impedance as Boswell mentioned. His comments about "matched" I/O are most critical for noise considerations as this gives maximum power transfer and hence the lowest noise (look up "maximum power transfer theory") which is common in RF design. With other sources the load impedance can be almost irrelevant - almost!

As you say, you will indeed have a tough argument with this one.

Edit: Boswell beat me to the punch on the "source" of colouration there!

To quote the wiki: Electrical impedance extends the concept of [="http://en.wikipedia.org/wiki/Electrical_resistance"]resistance[/]="http://en.wikipedia…"]resistance[/] to AC circuits, describing not only the relative [[url=http://="http://en.wikipedia…"]amplitudes[/]="http://en.wikipedia…"]amplitudes[/] of the [="http://en.wikipedia.org/wiki/Voltage"]voltage[/]="http://en.wikipedia…"]voltage[/] and [[url=http://="http://en.wikipedia…"]current[/]="http://en.wikipedia…"]current[/], but also the relative [="http://en.wikipedia.org/wiki/Phase_(waves)"]phases[/]="http://en.wikipedia…"]phases[/]. When the circuit is driven with [[url=http://="http://en.wikipedia…"]direct current[/]="http://en.wikipedia…"]direct current[/] (DC), there is no distinction between impedance and resistance; the latter can be thought of as impedance with zero phase angle.

This I suppose is misleading because it equates impedance to the AC equivalent of resistance. That's where I was coming from in my initial post. I have some reading to do for sure. In my engineering classes, we are just touching on the topic of voltage dividers... hopefully we will get into more interesting things soon.

Guitarfreak, post: 376693 wrote: ... What if someone were to wire their guitar's volume pot as a purely variable resistor from signal path to ground instead of as a voltage divider. Something tells me that if this actually worked more people would be doing it.

Some guitars I have seen do have their controls wired (or mis-wired) in this way.

It would work if the pickups were a current source. A current source has infinite output impedance and so the voltage developed across the pot (actually wired as a rheostat in this case) would be proportional to the value of the resistance between wiper and ground.

Both the electromagnetic type of pickups for steel string guitars and the piezo type often fitted to nylon-strung acoustic instruments are senstive to the load placed on them. Changing the load not only changes the amplitude of the output but also the frequency response ("tone"). Most performers would want the tone to remain the same as the "volume" knob is operated and be controlled by separate "tone" knobs on the instrument.

Guitarfreak, post: 376693 wrote:
To quote the wiki: Electrical impedance extends the concept of [="http://en.wikipedia.org/wiki/Electrical_resistance"]resistance[/]="http://en.wikipedia…"]resistance[/] to AC circuits, describing not only the relative [[url=http://="http://en.wikipedia…"]amplitudes[/]="http://en.wikipedia…"]amplitudes[/] of the [="http://en.wikipedia.org/wiki/Voltage"]voltage[/]="http://en.wikipedia…"]voltage[/] and [[url=http://="http://en.wikipedia…"]current[/]="http://en.wikipedia…"]current[/], but also the relative [="http://en.wikipedia.org/wiki/Phase_(waves)"]phases[/]="http://en.wikipedia…"]phases[/]. When the circuit is driven with [[url=http://="http://en.wikipedia…"]direct current[/]="http://en.wikipedia…"]direct current[/] (DC), there is no distinction between impedance and resistance; the latter can be thought of as impedance with zero phase angle.

This I suppose is misleading because it equates impedance to the AC equivalent of resistance. That's where I was coming from in my initial post. I have some reading to do for sure. In my engineering classes, we are just touching on the topic of voltage dividers... hopefully we will get into more interesting things soon.

I don't think this is particularly misleading but more that you misunderstood. In AC circuits with a pure resistance, voltage and current are in phase whereas with a pure capacitance, the current will 90 degrees ahead (of in phase) of the voltage and with inductance, current is 90 degrees behind voltage. This is independent of frequency.

Whenever you have a combination of resistance and reactance (either or both of inductance or capacitance) the relationship between phase and voltage becomes variable with frequency - which is the basic reason we can produce filters. With filters beyond simple RC or RL types, the maths rapidly becomes very complex and historically this has been almost prohibitive to design. Fortunately for people like me, Anatol Zverev produced an incredible work entitled "Handbook of Filter Synthesis" which became my bible back in the day. I still find my copy useful but nowadays you would be a) very lucky to find a copy and b) be very blessed to actually follow all the maths it entails.

Personally I applaud your efforts to learn all this but, and I'm sure Boswell will back me up with this, you do have to walk before you can run. It will take a lifetime but you will get there. I started playing with electronic in my pre-teens but rarely understood what I was making (using modified published designs much as you are doing). As I went through senior school and later University all this "experience" (mainly of failures!) held me in good stead and I lost track of the number of times when the penny dropped on something I had tried years earlier. Have I said this before???? ;<)