5
$\begingroup$

The point I'm confused about is the antenna. Depending on the environment, there are tons of different frequencies passing through you (FM, AM, 4G, SAT, etc). The electrons in the wire will move as the radio frequencies pass by. One wave will move the electrons at a certain rate. Another wave will move the same electrons at a different rate. All these waves will interfere with each other. You get some weird movement or none at all on the wire depending on the different frequencies hitting the antenna.

I kind of understand a tuning circuit (LC). How is the tuner picking up anything when the antenna is giving the LC circuit a complex wave? It's like you mixed up paint and the LC tuner somehow unmixes it. For some reason, it's not clicking for me. So what if it resonates at a certain frequency. The wave it's feeding is complete gibberish. Two different frequencies hitting the antenna could cancel out each other so electrons don't move.

I do not understand how a tuner (LC) can pick up a certain frequency.

Once I understand that this only makes sense for a specific frequency. FM radio works on a range of frequencies. How does a tuner "catch" a range of them?

https://physics.stackexchange.com/questions/326727/how-can-an-antenna-pick-up-thousands-of-frequencies-at-the-same-time

https://physics.stackexchange.com/questions/223469/how-does-the-tuner-really-work-in-a-crystal-set

https://physics.stackexchange.com/questions/8310/how-does-a-digital-radio-tuner-work

Edit: I'd like to give everybody a green checkmark. :) I still have a long way to go to be satisfied with my understanding but this is a great start. I've always been interested in radios. I'm going to start by creating my own AM then FM radio. I just needed the theory because anybody can put a kit together. I want to know the WHY in detail. I have so many more questions but I think this is enough to satisfy this post. You guys are the best!

$\endgroup$
  • 1
    $\begingroup$ Hello and welcome to ham.stackexchange.com! $\endgroup$ – rclocher3 Oct 26 at 18:04
2
$\begingroup$

There's something called the "principle of superposition" — in a linear system (which we can consider an antenna and the "front end" of a receiver to be), if the current resulting from signal A is $C_A$, and the current resulting from signal B is $C_B$, then the current resulting from both signals at once is simply $C_{A}+C_{B}$. Even if you have a million signals, they all just add to one another linearly, without being "modified". And as long as each one has its own frequency, we can use things like LC filters to pick out the one we want. In a time-domain graph it might look like "complete gibberish", but all of the original structure is still there (and much more easily seen in a frequency-domain plot).

It is a little bit like mixing paint and separating it out again, but imagine that every different color of paint was made of different-sized particles. When you mix the paints together, it looks like a muddy mess, but all the individual particles are still in there. If you had some very good, very fine mesh filters that could filter particles precisely by their size, you could separate one color back out of the mix! With real paints, that's not practical, but with real radio waves, it is.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Yours is probably the closest answer. I'm just having a mental block. The antenna at the end is coiled up and feeds the inductor which is part of the LC circuit. It looks like a transformer to me. The wave is complete gibberish. you have all kinds of different frequencies hitting the antenna at the same time potentially. The LC circuit only responds to a certain frequency of charging and discharging. How is the LC circuit pulling out that one frequency? I just don't see it. $\endgroup$ – Derpy Oct 27 at 2:53
  • $\begingroup$ I wonder if there's an animation that shows a complex wave feeding an LC circuit. Showing how the LC circuit only charges and discharges from certain parts of the complex wave that match up. What I mean by charging and discharging is how the capacitor and inductor both charge each other at a certain rate. $\endgroup$ – Derpy Oct 27 at 2:57
  • 1
    $\begingroup$ @Derpy how's this for a start? $\endgroup$ – hobbs - KC2G Oct 27 at 6:02
  • $\begingroup$ that really helps! I think I'm gonna have to really study superposition. For example, if you have two different frequencies that come in phase periodically, doesn't that destroy the original amplitude (AM radio)? How does AM radio work when other frequencies can influence the amplitude when they align in phase? $\endgroup$ – Derpy Oct 27 at 14:50
  • 1
    $\begingroup$ @Derpy any other frequency can't stay in phase, and if it's a significantly different frequency, it passes through being in phase quickly enough that the filter doesn't react very much. The closer the frequency is to the desired frequency, the longer it can maintain phase coherence... which is why filters have roll-off. $\endgroup$ – hobbs - KC2G Oct 27 at 15:34
2
$\begingroup$

It's not like mixing paint. That's because it's a just a simple sum or linear mix, rather than a chaotic or non-linear mix with intermodulation products.

It's like a duet with a soprano and a bass singer. You can easily transcribe the low frequency bass voice and semi-ignore the soprano, or vice versa, because the frequency ranges are so different, and your ear's cochlea has a filter mechanism that is a mechanical analog to electronic LC filters.

And tuners have a bandwidth. A filter can be either narrow or wide, depending on the design spec.

| improve this answer | |
$\endgroup$
  • $\begingroup$ That's some really great insight. I'll study how the ear does it. That might help me to see more clearly. $\endgroup$ – Derpy Oct 27 at 15:10
2
$\begingroup$

Consider a swing, like the kind found at a playground. If you sit on it and shift your weight forward and backward at just the right rhythm, you can get the swing to go very high.

It goes high because the combination of the swing and the mass of your body is resonant at a particular frequency. When you shift your weight to "pump" the swing, you add just a little more energy to the swing. And when you pump at the right time, this extra energy is added to the stored energy from all the previous pumps, so with each swing you go a little higher than the last one. But this only works if you pump at the resonant frequency.

If you pump at some other frequency, you just jiggle around a little bit. You don't go higher and higher, because the actions of each pump don't reinforce each other.

Imagine you are swinging along happily, and simultaneously you receive a phone call, and your phone in your pocket is on vibrate. The vibration from your phone is also a shifting of weight, just as you are doing to pump the swing. But it is at a much higher frequency. Does it alter your motion on the swing? Technically yes, but the effect is very small because the vibration is not at the swing's resonant frequency. Imagine any perturbation you like: perhaps another person on the swing with you, but pumping at some other frequency. These actions might alter the swinging motion a little bit, but the swing responds most significantly to its resonant frequency, even if there are other oscillations going on at the same time.

An LC filter is a resonant system, like a swing. The difference is a swing involves an oscillation between gravitational potential energy (at the top of the swing) and kinetic energy (at the bottom of the swing), whereas an LC filter oscillates between energy stored in the electric and magnetic field of the capacitor and inductor respectively. The LC filter will respond strongly to oscillations at its resonant frequency, while other oscillations at other frequencies have only a negligible effect.

All modulations, not just FM, can be considered a "range" of frequencies. The only signal which is exactly just one frequency is an unmodulated carrier, which contains no information and so isn't used for communication. Some modulations use a wider range of frequencies than others, but no practical modulation uses a range of zero width.

That said, how can a filter work when the signal consists of a range of frequencies?

Real filters, even swings, have a resonant frequency where they are most sensitive. As the frequency deviates above or below that resonant frequency, the filter response diminishes, but it does not immediately drop to zero. The objective in designing a filter for a radio is to design a filter which passes the range of frequencies allocated to the signal, but no more. A very simple filter, like one made of a single inductor and capacitor, is "good enough" for some applications. But a radio designed for performance rather than simplicity will have more complicated filters, with more than one inductor and capacitor, to make the filter perform better. There are often multiple stages of filtering. Filtering, whether analog or digital, is a significant part of designing a radio, and a complex topic in itself.

| improve this answer | |
$\endgroup$
2
$\begingroup$

Radio waves are waves, which have many similarities to other waves, which allows us to consider analogies. Here's one: if you put your ear firmly to the end of a long tube, the sound will be different than if you weren't listening through the tube. Instead of hearing the usual mix of high-pitched sounds, low-pitched sounds, and mid-range-pitch sounds, you will hear mostly sounds near the resonant frequency of the tube (and harmonics). What's happening is that sounds of all frequencies are entering the other end of the tube, but sounds of frequencies near the resonant frequency of the tube (and harmonics) propagate through the tube better, because the sound waves reflecting off the far end of the tube (your ear) reinforce the sound waves coming in at that narrow range of frequencies. At other frequencies, the sound waves coming in partially cancel themselves out with reflections of the sound waves. This happens because a tube is a resonator.

Filtering in analog RF circuits is somewhat like that. Analog filters use resonance to reinforce desired frequencies, and cancel (attenuate) undesired ones.

You mention radio waves interfering with each other. "Interfere" might not be the best word for someone learning about the phenomenon, because two people interfering in a narrow hallway can't get by each other, but two waves interfering pass right through each other. They might cancel each other at one point for an instant, but the waves continue on.

You might be confused, because in one paragraph I talk about waves canceling through resonance, and in another I say that waves pass right through each other. It's true that two waves pass through each other without being annihilated. So how can waves cancel? They cancel at a point where the two waves meet with opposite amplitudes. If the two waves have the same frequency and amplitude and meet at a point where they are 180° out of phase, then they will consistently sum to zero at that point, even though the waves themselves continue on.

One situation in which two waves meet with the same frequency is a standing wave, in which a wave meets its reflection. If the reflection is perfect, meaning that it has the same amplitude as the original wave, then the waves will consistently cancel each other at certain points and sum to zero. The points where the sum of wave and its reflection are minimized, ideally to zero, are called nodes. But at other points the waves do not always sum to zero. The animation below, borrowed from the Wikipedia page, illustrates a standing wave well: the blue and green waves are the original wave and its reflection, and the red wave is the sum of the blue and green waves. See how the amplitude of the red wave is always zero at the nodes, which are spaced a half-wavelength apart?

animation of a standing wave

So to sum up, an analog filter circuit selects a narrow range of frequencies because the filter components cleverly use resonance to preferentially transmit those frequencies, similar to the way that a long tube capped on one end preferentially transmits sound waves of a narrow range of frequencies.

| improve this answer | |
$\endgroup$
  • $\begingroup$ The audio tube is a great analogy. It almost parallels with waveguides Although waveguides are only high pass filters, not resonant until the final turn(s). $\endgroup$ – KX4UQ Oct 27 at 23:29
1
$\begingroup$

Propagation of waves in free space is both electronic and magnetic in nature. Both field components combine to induce the movement of electrons in an antenna.

All frequencies are added together to make up the radio spectrum (a single entity per polarization, Horizontal and Vertical). The entire thing is combined into a grand complex wave.

For completeness, circular and elliptical waves shift the polarization between horizontal and vertical at a constant rate. This adds a time(phase) factor to the H,V transition.

The wave speed is completely frequency independent and mostly constant and is close to the speed of light. This speed is irrelevant of frequency, only medium. Although it does change between different mediums air/wire/vacuum... and with movement velocity(Doppler effect, only noticeable from spacecraft at RF frequencies).

All frequencies are combined at the transmitting antenna with the the existing wave, either constructively or destructively to a form single complex wave.

So to answer your question, the tuner is a (LC)filter network that produces a low impedance to the desired signal frequency range and a high impedance to all other frequencies causing a reflection to be either re-radiated or consumed as heat, through recirculation between the inductor and capacitor.

The bandwidth allowed is determined by the Q factor of the inductor and capacitor chosen and the driven frequency.The Q factor is the ratio of bandwidth to center frequency. A higher Q indicates a narrower filter and less insertion loss. The Q is determined by the losses and efficiency of the materials used to construct the inductors and capacitors and interconnection between them.

The selection of values for the inductance and capacitance form an oscillator with a resonant frequency. The signals close to the resonant frequency pass through with little loss as a purely resistive impedance with no reactance, while frequencies further away are attenuated by the lack of balance between the inductor and capacitor; thereby increasing the impedance to signals outside of the pass-band by increasing the reactance component.

This question forms the basis of software defined radios: The sample rate defines the maximum bandwidth, in the case of direct-sampled radios the maximum frequency detectable, but with decimation lower frequencies can be ascertained.


"The antenna at the end is coiled up and feeds the inductor which is part of the LC circuit. It looks like a transformer to me." Antennas that are electrically short present a capacitive reactance that is canceled out by the inductor; the capacitor cancels out the overshoot due to winding limitations of the inductor. In this form, this is a matching transformer and not a filter per se (although some filtering will occur). Its purpose is to transform the reactive component to -j0 (zero reactance) to ensure maximum power transfer.

I have left many things out to simplify this answer. I would be glad to elaborate.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Your second-to-last sentence is incorrect: the sample rate of a device limits its bandwidth, not its maximum frequency. SDRs exploit that using mixers and aliasing (going for Zero-IF, Superhet or subsampling architectures, or combinations of these). You don't need a 12 GHz sampling rate to digitize Wifi at 5.8 GHz – a sampling rate sufficient to sample the bandwidth of the signal is sufficient, given, indeed, adequate analog filtering. $\endgroup$ – Marcus Müller Oct 27 at 12:09
  • $\begingroup$ What is the Q factor? I can maybe see it picking up a specific frequency but how does it pick up a range? $\endgroup$ – Derpy Oct 27 at 15:09
  • $\begingroup$ @MarcusMüller, I have clarified the conditions to specify direct sampling. I was trying not to get into Nyquist frequencies and aliasing. The point was that multiple frequencies were present in the digitized samples. $\endgroup$ – KX4UQ Oct 27 at 15:16
  • $\begingroup$ @Derpy, I have updated my answer to include Q and the range reasons. $\endgroup$ – KX4UQ Oct 27 at 19:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.