# How is it possible to ever have frequency on a DC circuit?

Preface

Please read this carefully. I put a lot of time into it and am genuinely attempting to understand. I've read numerous articles about DC and AC circuits and searched numerous places to get this answer, but I'm not sure I know the correct terms that are involved.

Also, yes I've googled "can a dc circuit ever have frequency?" And the answers you get are that DC circuits have a frequency of 0.

Background

I'm attempting to build a circuit which examines an input frequency and converts it to MIDI. Here's the schematic to the circuit from the official Arduino site (which is incorrect in at least one place):

Error Explained

You can see that for some reason the original designer drew the output from OpAmp2 even though it has no Input.

My Attempt To Test

Since I was attempting to test the code from that article. I just wanted to read some values off of A0. Just to see the thing work so I could begin to reverse engineer the obviously erroneous circuit.

I won't reproduce all of the code but here's the important part: Note: The code is from the article where I obtained the schematic (https://www.arduino.cc/en/Tutorial/AnalogToMidi)

void setup() {
// put your setup code here, to run once:
Serial.begin(115200);
pinMode(11, OUTPUT);
// Set available bandwidth between 75Hz and 600Hz
meter.setBandwidth(75.00, 600.00);
// Intialize A0 at sample rate of 45kHz
meter.begin(A0, 45000);
}

void loop() {
float frequency = meter.getFrequency();
Serial.prinln(frequency);
if (frequency > 0)
{
// Find the index of the corresponding frequency
int noteIndex = searchForNote(frequency);
int note = notePitch[noteIndex];


Basically, all I cared about was reading the frequency I got back from the meter.getFrequency() method and printing it out to the serial monitor.

I whipped up the following circuit:

simulate this circuit – Schematic created using CircuitLab

(Now) Obvious Results

So I fired up the code and opened up serial monitor and started spinning the pot knob, but as you probably already know, Serial monitor only prints out -1.0 (negative 1.0).

There's No Frequency On a DC Circuit

At first, I was very confused. Why wouldn't it ever get any different value even though the voltage input was changing?

Then, it hit me. There's no frequency on a DC circuit.

The Question

How would the code ever read frequency off of the circuit? How could there ever be frequency on a DC circuit? How is that possible?

I understand that I am totally ignoring the work done by the input in the original circuit (mic to op amp), but I also understand that the input will only be a series of currents and voltages which are simply amplified by the opamp anyways.

Pulses Become Frequency?

Is that the point, that they will become pulses and those pulses will have a frequency?

EDIT : Adding GetFrequency Code You can see the source code of the ArduinoFrequencyMeter at: https://github.com/arduino-libraries/AudioFrequencyMeter/blob/master/src/AudioFrequencyMeter.cpp

Here's the source of the getFrequency() method:

float AudioFrequencyMeter::getFrequency()
{
float frequency = -1;

if (checkMaxAmp > amplitudeThreshold) {
frequency = (float)(sampleRate / period);

if ((frequency < minFrequency) || (frequency > maxFrequency)) {
frequency = -1;
}
}

return frequency;
}


It is interesting that if the checkAmp is not greater than amplitudeThreshold then it will return -1. So it is likely that is what is happening in the case where I'm seeing all -1.

But, because of the way the code is, you cannot differentiate that problem from the problem of the frequency falling outside of the range -- since frequency out of range also returns the same value.

I will probably alter the code at some point and have it return -1, -2, -3 for each case.

• Something for you to ponder: do you think it would be possible to have both DC and AC in the same circuit at the same time - one superimposed on the other? – brhans Oct 15 '18 at 12:46
• You never have DC unless your device will be on forever and has been on forever. – PlasmaHH Oct 15 '18 at 12:48
• @brhans That's a good question and only because of my reading about topics like capacitors and (very little reading) about diodes (for protection from AC) I know it is true that you can have both. But I don't understand it either. – raddevus Oct 15 '18 at 12:49
• @PlasmaHH That makes sense to me...as the circuit is powering up and down it is in a state of AC until it reaches it's DC level of stabilization, I guess, right? – raddevus Oct 15 '18 at 12:50
• How does the getFrequency() function work? Does it have a lower limit on the frequency it can measure? – Elliot Alderson Oct 15 '18 at 14:24

"DC" is a lot like saying "frictionless". There are two ways you could define them:

1. DC is zero frequency. Frictionless is zero friction.

2. Any frequency that is too small for us to care about (i.e. too long in time) is DC. Any friction that is too small for us to care about is frictionless.

Definition #1 is a nice theoretical definition. There is no ambiguity about what it means: zero is zero. It's what you will often find in textbooks.

Definition #2 is more practical. No natural cycle has zero frequency, as its period would be an infinite amount of time. Every mechanical system has some friction. But in both cases, these effects may be so small that we don't care about them. So we can treat them as if they are zero.

So what frequency is the cutoff for DC? It varies with each problem, and is sometimes a matter of opinion. Part of being an EE is knowing when you can assume that things are in steady-state.

As for your Arduino question... You give it a signal, and it tries to determine the frequency of it. It's not magic; there is an algorithm involved. The algorithm has a minimum and maximum frequency; you are correct that it returns -1 if the signal seems to be above the maximum or below the minimum frequency.

Here's why. Suppose the software can read frequencies as low as 1 Hz. That means in the worst case, you would have to wait 1 second (*note 1) to find the frequency.

Now suppose you change the minimum frequency to 0.1 Hz. The worst-case waiting time will be 10 seconds. At 0.001 Hz, 1000 seconds. As the frequency approaches the "ideal" zero frequency of DC, the waiting time will become inifite.

You don't want to wait forever to have the algorithm tell you that you are at DC. Therefore, it will return -1 unless it has found a frequency in the correct range.

(*note 1) Yeah, maybe 0.5 second because of the Nyquist rate. It still doesn't change my argument that the waiting time becomes infinite.

• That was a fantastic and lucid explanation of things I'm experiencing but (since I'm not actually an EE - I'm a software dev) couldn't really even understand how to ask the questions. Thanks very much for helping me gain more understanding. – raddevus Oct 15 '18 at 20:26

simulate this circuit – Schematic created using CircuitLab

Figure 1. Potentiometer connection.

Your potentiometer test circuit is wired incorrectly. In Figure 1a if the pot is turned to minimum you will short-circuit the 5 V supply to GND. Instead wire it as Figure 1b so that A0 can be swept from 0 to 5 V. Only 5/10k = 0.5 mA will flow through R2.

How would the code ever read frequency off of the circuit? How could there ever be frequency on a DC circuit? How is that possible?

It is possible and is inherent in some designs and deliberately designed into other circuits.

DC power supply

simulate this circuit

Figure 1. A mains-powered DC power supply gives a DC output but with 100 or 120 Hz ripple. (Taken from my answer to how does the filter capacitor work in a rectifier circuit?).

Figure 2. Simulator results. (Click to enlarge.)

• The top blue trace shows the voltage out of the power supply and presented to the load. The capacitor is charged up on each mains half-cycle but in the gaps where the voltage falls away the voltage droops. The result is a DC voltage with an AC ripple.

Audio amplifiers

simulate this circuit

Audio amplifiers require a DC bias on the audio signal (which is AC) when the amplifier is powered from a single-ended power supply.

Figure 3. The op-amp is powered from a single supply so it can't accept negative input signals or output negative voltages.

The solution for Figure 3 is:

• Add DC blocking capacitor C1 on the input.
• Bias the op-amp to half-supply with R1 and R2.
• Amplify with OA1. (In this case the gain is 1 and it is just a buffer amplifier.)
• Remove the DC component with C2.

Plain-old telephone system (POTS)

Figure 4. A GPO 746 rotary telephone transmits AC superimposed on the 50 V DC power.

Another familiar case is the standard telephone. The phone is powered from 50 V DC from the exchange but audio is superimposed on the DC and transmitted over the line.

• Thanks that is great info. I'll check my actual potentiometer circuit at home again but I believe I just drew it incorrectly on the schematic. I tested it against another very simple analogRead() circuit and read values from 7 - 1023 as I varied the pot, so I think it is right but I'll check again. – raddevus Oct 15 '18 at 20:52