When talking about input and output impedances, what are we comparing them to?

Question

Let's consider a theoretical highpass filter:

^{simulate this circuit – Schematic created using CircuitLab}

I'm trying to show that the worst-case input and output impedance is R but the whole concept of input / output impedance isn't really clicking for me.

Resistors have impedance. Capacitors have impedance. How can an input or output node have impedance? I think that they must be comparing the node to something but I'm not exactly sure.

Initially I was thinking that input impedance must be the amount of impedance that would be met as the current traveled through my RC filter. If the frequency was high, C1 would look like a short and the current would flow through R1 to ground. If the frequency was low, C1 would look like an open circuit and impedance would be infinite. Having a low input impedance is the worst-case scenario and I just showed that the lowest it could be is R1. So that seems to work.

Following that train of thought, I figured that output impedance must be the amount of impedance that would be met as the current traveled through ..... the rest of my RC filter? This train of thought suggests that C1 is irrelevant because the only place left for the current to go is through R1. That is the right answer but it seems like I got there incorrectly.

Is this right? If not, can you please explain it?

Answer 1

§ The impedance of your circuit seen from the input is the impedance of the capacitor (depends on frequency) plus the impedance of the resistor (just R), seen from the output the two impedances are parallel so the impedance is the impedance of the capacitor parallel with the impedance of the resistor. Zc1//R1 calculated as Ztotal^(-1)=Zc1^(-1)+R1^(-1). and Zc1 can be calculated as 1/(2 * pi * f * C) where f is the frequency and C is the capacitance of C1 in farads

§ Look up thevenin equivalent, it will help you understand how an output/ input can have an impedance. the impedance is basically just the slope between the voltage and current, if an increase in voltage at an input or output of 1v leads to an increase in current of 0.1A then the input/output has an impedance of 10ohm

§ It is important to note that the way the impedance of a capacitor or an inductor differs from the resistance of a resistor is in that they don't dissapate energy due to the current*voltage in them, that is because there is a 90deg. phase difference between voltage and current in capacitors and inductors.

Answer 2

The simple view, for the DC case, is that the voltage source has some internal resistance, and your circuit has some resistance to ground. These form a resistor divider, so as soon as your sink is connected, the voltage on the input node is a fraction of the source voltage.

^{simulate this circuit – Schematic created using CircuitLab}

In both cases, the source has an output impedance of \$100\Omega\$, but the input impedance of the sink determines what is measured on the voltmeter. In the same way, the measured value varies with source impedance if the sink impedance is constant.

Voltmeters ideally have a very high input impedance, which makes the measured value fairly independent from the source output impedance (as long as it is a few orders of magnitude smaller than the voltmeter's).

Now extending that to AC, we have capacitors and inductors that introduce complex-valued resistance. Your filter has close to infinite input impedance at DC, and close to R1 at "high" frequencies, with some roll-off in between. Nothing is known about the source, but unless it is very stiff (i.e. low output impedance), the voltage measured at your input node will be affected by the filter, i.e. your filter feeds back into the source.

If there are multiple sinks connected (e.g. a passive crossover network on a speaker), this means that the voltage on the filter output is different if all paths are connected than when just one is present.

If you go even higher in frequency, you'd reduce C1 to shift the cut-off frequency upwards. As C1 nears the capacitance of the cable, you end up with a frequency-dependent voltage-divider between the cable and your filter.

In addition, if cables get longer than a fraction of the wavelength, the time it takes for the signal to travel further affects impedance over frequency: for a signal with a frequency \$f\$ and a coxial cable of length \$0.66\frac{c}{4f}\$, if the filter fully blocks the signal and presents infinite input impedance at this frequency, the signal is reflected back along the cable, and the peak of the wave arrives at the source when the source sends out the negative peak, so these cancel out, and from the source's point of view, the output is shorted to ground (i.e. the cable has a frequency dependent input impedance on the source side and output impedance on the sink side).

The latter is only relevant for frequencies above a few tens of MHz, because \$c\$ is fairly large.

Answer 3

Your question appears to be about the difficulties of finding the impedances of a high pass filter. Consider the possibility that many do not even mention impedance when analyzing filters. As is the case in this Wikepedia article about high pass filters.

Impedance is usually used with reference to the output or input of an working circuit such as a transmitter or antenna. It is useful to talk about a working circuit's impedance as this relates to maximum power transfer. A worthy goal in electronic circuit design.

A high pass filter, such as the one in the question, is usually not talked about in such a way. Instead a filter may be talked about in terms what it could do if added to a working circuit. Such as orders of magnitude of attenuation and cut off frequencies.

Using the schematic of the High Pass Filter in the question, the impedance from Vin to ground is the sum of the impedances of the 1uF capacitor and the 100 ohm resistor. It is assumed this is what the OP calls the "input impedance". The impedance of a capacitor is dependent on frequency and is given by this equation:

The impedance of the resister is simply its value, 100 ohms.

So the total impedance from Vin to ground is:

impedance = capacitor impedance + resistor impedance
impedance = (1/(2*Pi*frequency*0.000001)) + 100

(Note: We need to convert micro-Farads to Farads so 1uF becomes 0.000001F.)

A similar problem is discussed here in greater detail.

The impedance from Vout to ground is only the 100 ohm resistor. The capacitor's far node is unconnected and therefore does not contribute. It is assumed this is what the OP calls the "output impedance". The impedance of a resistor is simply its resistance.

So the impedance from Vout to ground is:

impedance = 100