What does it mean for a signal to have impedance?
What does it mean for signals to have impedance? To me, impedance is kinda like an "extended resistance" that also includes phase shift
$$Z = R + jX $$It makes sense to me that components like capacitors ($Z_C = \frac{1}{j\omega C}$) and inductors ($Z_L = j\omega L$) have impedances because they impact signals, but what does it mean for a signal to have impedance?
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What does it mean for signals to have impedance?
I expect that what it means is that a particular signal source has an effective series resistance (for a voltage source) or, an effective parallel resistance for a current source.
Typically, a signal generator might have an output impedance of 50 Ω (for example).
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Impedance
Yes, impedance is "extended" resistance. Impedance is really a complex value, with resistance it's real part. The imaginary part represents pure inductance or capacitance (with opposite sign). Sometimes we say "impedance" when really only the resistance is relevant. That's OK since resistance is impedance, with the special case of the imaginary part being zero.
The model
It is often useful to model power supplies and signals as being a perfect voltage source in series with some impedance. Equivalently, you can model it as a current source with the same impedance in parallel. This is in fact what the Thevenin and Norton equivalent sources are all about. Look those up if you've not heard of them. Then you can ask questions here if you don't understand something.
I'll use the Thevenin model (perfect voltage source in series with finite impedance) here going forward because it's probably a little easier for you to visualize. We are more used to thinking of voltage signals than current signals (although the latter is perfectly valid).
Power supply impedance example
Let's say we have a 12 V power supply. You leave it unloaded and put your perfect voltmeter across it, and it reads exactly 12 V. Hopefully that makes sense so far. Note that the current the supply is producing is 0 in this case. It's easy to produce the right value when you don't have to work at it.
What does the supply do when you put a load on it, meaning it has to produce current to maintain the specified voltage? Let's say you put a 6 Ω power resistor across the supply. Ideally 2 A would flow. Now you measure the supply voltage and find it's actually only 11.8 V. It dropped 200 mV with a 2 A load. In our model of a perfect 12 V supply with some resistance in series, that resistance is (200 mV)/(2 A) = 100 mΩ. If you then test the supply voltage at 1 A and find 11.9 V, and at 3 A and fine 11.7 V, then the Thevenin model seems to fit well.
Signal impedance
The same thing works for signals too. The signal out of an opamp with feedback might be quite low, because the feedback actively makes the opamp push or pull harder when something is loading the output. However, signals from real world transducers, for example, usually have significant impedance.
Consider a dynamic microphone, which is basically a coil of fine wire. These are usually designed to have about 600 Ω impedance. If the microphone produces a 1 mV signal open-circuit, then that drops to 500 µV when the microphone amplifier loads it with another 600 Ω. The internal impedance of the microphone and whatever load you put on it form a voltage divider, and you only get the output of that voltage divider. If the 600 Ω microphone is loaded with 1 kΩ, then you get (1 kΩ)/(600 Ω + 1 kΩ) = 62.5% of the theoretical ideal voltage source in the microphone. The 1 mV open-circuit signal becomes 625 µV when loaded with 1 kΩ.
Noise pickup
Impedance of signals is more than just how much the level drops when loaded. A big issue is that high impedance signals pick up more noise from the environment than low impedance signals. Let's consider the noise being capacitively coupled from the environment to the signal. There is now a voltage divider formed by that capacitance and the signal impedance. Let's say we have an EKG signal with 1 MΩ impedance, with 10 pF coupling to a nearby 120 V 60 Hz power cable. 10 pF at 60 Hz has about 265 MΩ impedance magnitude. The 120 V from the power wire will be attenuated 266 times onto the EKG signal. That means there will be (120 V)/266 = 450 mV of 60 Hz hum on the EKG signal. Since EKG signal are usually only a few mV, this hum totally swamps the intended signal, making the EKG useless.
The lower the impedance of the signal, the less of the same ambient noise is picked up. For example, if the impedance were only 1 kΩ, then the power line hum would be 1000 times less, or 450 µV. That's still a large fraction of the intended signal, so not acceptable for an EKG. However, consider the same ambient noise coupling to the signal running from a power amplifier to a loudspeaker. Most loudspeakers have 8 Ω impedance, and signal levels are volts, not millivolts. The 60 Hz pickup on a speaker wire is totally irrelevant.
Dealing with noise pickup
So what can you do to mitigate noise pickup? How are EKGs, for example, possible at all? There are several strategies:
- Make the impedance lower. We already saw why and how that works.
- Amplify the signal at the source. With a larger signal, the same amount of noise pickup matters less. 1 mV of hum on a 1 mV microphone signal would be totally unacceptable, but the same 1 mV on a 5 V audio signal is a lot less objectionable (although still far from "HiFi").
- Move away from the noise source. The capacitance between two conductors drops off with distance. Lower capacitance coupling to the same noise results in less noise level on the signal.
- Shield it. If you enclose the signal wires in a grounded conducting tube, then all that nasty noise is shunted to ground and it's nice and quiet inside the tube. This is what shielded cable is all about.
- Use differential signalling. You realize that some amount of noise pickup on the wire is inevitable. The trick is to use two symmetric signals of opposite polarity, then make sure both are coupled the same to the ambient environment. The signal is the difference between the two, but the noise will be the same on both. If the receiver subtracts the two voltages, then in theory it gets the signal with the noise cancelling out.
Twisted pair wire is useful for this. The point of twisting the two wires is so that each ends up having the same coupling to the environment in the long run.
In practice, a combination of these strategies are usually used. Quality microphones, for example, use differential signalling and shielded cable. They can also contain a small battery-powered amplifier directly in the microphone. That both increases the signal level and decreases the signal impedance on the long wire.
EKG signals are high impedance and low level, so you have to use every possible trick to deal with noise. Differential signalling and shielded cable are essential. With an EKG, a neutral point on the body is driven to the average of the two signals. This is often connected to the right leg, since it's furthest from the heart and therefore favors one of the signals least over the other. This right leg signal drives the body to a neutral voltage, actively attempting to cancel the inevitable noise coupled to the body from the environment.
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R is a real resistance that stores energy as heat and temperature rise which depends on the thermodynamic property of thermal resistance which is impacted by an enclosure to get hotter or cooler by forced airflow.
Everything has some real resistance, even insulators with an electrode-insulation interface (called Effective Series Resistance or ESR) with some capacitance. Also, batteries and inductors have some series resistance.
But X is always a stored resistance referred to as "reactance". Inductors have +ve reactance and Capacitors gave -ve reactance and for sine wave currents the polarity refers to how the voltage "reacts".
The ideal L & C components are considered lossless with no resistance but one must be aware of the actual R values from the design and datasheet.
Energy stored in both L and C may be readily computed and be transferred from or to real power with some R value.
E.g. a Battery is a massive capacitor xx,xxx Farads above a minimum chemical voltage can power a lamp which has resistance.
For more details:
AI answer:
Impedance is a measure of the opposition that a circuit presents to the flow of an alternating current (AC) signal. It is represented by a complex number that has both magnitude and phase. Impedance is a combination of resistance (which causes loss or dissipation of energy) and reactance (which does not cause loss but instead alters the phase relationship between the voltage and current in the circuit).
Reactance is a measure of how much a circuit opposes the change in the flow of an AC signal. It is determined by the circuit's inductance and capacitance and is measured in ohms. Inductive reactance (XL) results from an inductor's tendency to resist changes in the current flowing through it, while capacitive reactance (XC) results from a capacitor's tendency to resist changes in the voltage across it. Both XL and XC are proportional to the frequency of the AC signal.
In a circuit with impedance, the loss or dissipation of energy is represented by resistance, which is also measured in ohms. Resistance is a property of a material or component that determines how much current flows through it for a given voltage applied across it. In a circuit with resistance, the voltage and current are in phase, meaning they rise and fall together.
In summary, impedance is a measure of the total opposition to the flow of an AC signal, and it includes both resistance (which causes loss or dissipation of energy) and reactance (which does not cause loss but instead alters the phase relationship between the voltage and current in the circuit). The reactance is further divided into inductive and capacitive reactance, which are proportional to the frequency of the AC signal.
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