what a comment. well said.

+Paul G indeed, I couldn't agree more and Brenner's comment got it right too!

I completely agree with you, this video is like revealing a top secret, all grey areas in rf mixer completely demystified.

How is it a damning indictment of the human race that completely different videos reaching completely different audiences would have completely different numbers of views? Radio is interesting to some and of no interest to others. What are you saying exactly?

@@jimcervantes5659 I think it's damning that a large portion of the human race gets over-excited about an unboxing of pretty much _anything_ regardless of what's actually in the box. At least, that's the gist as I understand it. A bit like reading your comment, a waste of my time. As was writing this reply.

Ehsan Tofigh Wow, that's a prestigious list to be grouped into! :) Really glad you're enjoying the videos!

Esi haha. Definitely

Seriously you guys are doing great work. You are educating engineers and students all around the world in a real sense. In a country where we do not have easy access to these instruments, your videos are blessings for us. Together with good classroom lectures, these videos make a great combination to educate an entire generation of privileged or not so privileged audience.

Agreed.

Уравнение вида a*sin(x)+b*cos(x)=c из учебника алгебры 11 класса сейчас такое "Дада, пошел я нахрен" 😆😆😆😆

Benny De Wachter How so? This video didn't discuss demodulation. That is what happens later on with the IQ samples.

I saw this video after watching two other vids by w2aew (#170 and #171). As far as I remember it was the combination of these 3 videos that made things clear for me.

Thank you an excellent lesson 4z4RE.

The two LO lines come from a binary counter that's clocked by the LO. The example of switching thro the phases is confusing, I think he mentions the RF signal and LO are in phase, but of course they aren't in actual use. Suppose he meant for the example given.

No you don't need a phase lock to the RF only your s0 and s1 signals to get the I and Q signals have to be in exact 90° phase!

Remember that the capacitor (C = 0.1u) and cable impedance (R = 50oHm x4 = 200oHm) form a LPF with cut-off frequency 8kHz, cutting off any high frequency harmonics (10MHz,. 20MHz, 30MHz,...)

Ehsan Tofigh That is correct, superheterodyne receivers solve the image problem by first converting the RF signal to an intermediate frequency (IF). Let's say I use a 14MHz LO to convert the 10MHz signal down to 4MHz instead of 1kHz; the image frequency is then 14+4=18MHz, which is 8MHz away from the desired signal, which is far enough away that it can be easily filtered out by the RF front-end filter. The 4MHz signal can then be further filtered/amplified if desired, and then converted down to 1kHz with a second mixer and LO running at 4.001MHz. The second mixer/LO has its own image frequency at 4.002MHz of course, but again, that is 6MHz away from the desired 10MHz signal, and is easy for the RF front end filter to reject, ensuring that any RF signals at 4.002MHz never make it to the IF stage.

devttys0 I should also mention that while a 10.002MHz signal will still be converted to 4.002MHz in that example, thus causing an image frequency at the second mixer, very tight IF filters can be made at 4MHz (e.g., using crystal lattice filters) that can reject that unwanted signal before it gets to the second mixer.

+coffeecuppepsi To reject an unwanted image signal, you need to introduce an additional 90 degree phase shift between your I and Q signals, then take the sum or difference of I and Q, depending on which image you want to reject. Let’s focus first on the image you want to reject at 1.1MHz, which is 100kHz above your LO. Since it is above your LO, we know that the I signal will be leading the Q signal by 90 degrees in the output from your quadrature mixer. Now, let’s say you delay the I signal by an additional 90 degrees; this means that I and Q are now 180 degrees out of phase. Now, if we feed these 180 degree out of phase signals into a summing amplifier, their sum will be 0 because they cancel each other out, and you get no output from your summing amplifier. You’ve just rejected the unwanted transmission at 1.1MHz As for the desired image at 0.9MHz, because it is below your LO, we know that its I signal will be lagging the Q signal by 90 degrees in the output of the quadrature mixer. Thus, delaying the I signal by another 90 degrees puts its I and Q signals in phase (no phase shift), and the sum of two otherwise identical in phase signals will be double the original amplitude. So you’ve just doubled your desired transmission, while simultaneously rejecting the undesired transmission. Now, if you wanted the opposite (i.e., reject the 0.9MHz signal, but keep the 1.1MHz signal), you would use a differential amplifier instead of a summing amplifier. The difference between two in phase signals is 0 (this rejects the 0.9MHz transmission), while the difference between two 180-out-of-phase signals is 2x (this amplifies the 1.1MHz transmission). Or, you could still use a summing amplifier, but delay the Q signal instead of the I. In the real world of course, there are always errors in phase, amplitude, etc., so the image rejection is not perfect. It can be made very good though, good enough for most applications.

+devttys0 oh wow it's so elegant, thank you for explaining that :)

+Shyam You're correct of course, I'll have to add an annotation to the video; the resistance should be (50 ohms + switch Ron resistance) x 4. Thanks for the correction!

+devttys0 Thanks.

Thanks!

+Oskar Ceso If you have a nice clean signal like in the video, and all you want to do is tell if the signal is above or below your local oscillator frequency, it's very simple. You just need to determine if the I signal is leading or lagging the Q signal. A simple way to do this would be to feed the I and Q signals into two I/O pins of a microcontroller. The micro could monitor those pins to see which one goes high first, and then you would know if I is leading Q, or vice versa. Of course, if you have many input signals at once, or a lot of noise (both common with RF receivers), you'd need some filtering/signal conditioning to make this reliable. Typically the I and Q signals are digitized with an ADC, and the signals are then digitally filtered/processed in order to determine the phase relationship between I and Q.

+devttys0 Thank you very much! :)

+devttys0 about the "see which one goes high first", there isn't really a reference to determine that. I think you'd need to measure the delay between three instances, and pick the smallest of the two delays (ie find a 90deg case and a 270 deg case, and focus on the 90 deg case). Once you have the two signals which are 90deg apart, see which one came first. BTW, what's the impact of the time it takes for the mux to close one channel and open the next? Is it a make before brake or vice versa? Which is best?

+mdesm2005 I'm not sure what you mean by "there isn't really a reference to determine that". The reference is time, just like on an oscilloscope. By definition, when you have two signals (in this case I and Q) that are out of phase, one will be leading the other, meaning that one will cross a microcontroller's logic high threshold first. Hence, there will be a state when one of the signals will be high and the other is still low. If I goes high while Q is still low, then I must be leading Q and the RF signal coming into your mixer is above your local oscillator frequency; if I goes high and Q is already high, then Q is leading I, and the RF signal coming into your mixer is below your local oscillator frequency. It's pretty easy to configure an interrupt on the rising edge of the pin that the I signal is connected to, and have the interrupt routine check the state of the Q pin to determine this relationship. It's not a robust solution for most real-world scenarios, but it's simple and demonstrates the basic principle; I've done this myself, and I can assure you that it does work. I've not seen any literature, or done any experimentations myself, regarding make-before-break vs break-before-make in this application. However, since each output is intended to effectively be a separate sample and hold circuit, I'd suspect that break-before-make is preferable. Intuitively, the more time it takes to switch between outputs, the more of the signal you will lose (assuming break-before-make), so I'd say that generally faster is better, but the impact of a high-speed multiplexer's switching time should be negligible. Dan Tayloe's paper states that the maximum operating frequency is limited by the FET switches, and should be in the neighborhood of 10GHz: www.wparc.us/presentations/SDR-2-19-2013/Tayloe_mixer_x3a.pdf.

devttys0 thanks

Wayne Marsh If you just had a single on/off switch as a mixer, it would. As I mentioned at 2:58, not all mixers work that way, but I used it as an example because it is conceptually easy to understand, and is also fundamental to the operation of the Tayloe mixer which I discuss later on in the video. The TUF-1 mixer that I used as a single mixer example doesn't work that way; it's a diode ring mixer which multiplies the RF signal by 1 and -1, not 1 and 0 like a simple switch would. Note at 3:40 how the signal seems be be continually inverted causing sharp "points" in the waveform, rather than hard edges with periods of no signal. (See W2AEW's video on diode ring mixers: kzfaq.info/get/bejne/oNueqKio0re6gWw.html) Although the Tayloe mixer does use on/off switching, the capacitors on the output lines hold the voltage when the switch is disconnected, so the output doesn't suddenly drop to zero when the switch is turned off. The capacitors, in conjunction with the source input impedance, also form a low pass filter which help to filter out higher frequency distortion.

devttys0 That makes a lot of sense. I figured something along those lines. I think I mentioned in another video, I really enjoy your videos as I do everything completely in software, and it's cool to see how it's done in actual physical circuits.

Maybe because "i" could be mixed up with "current" in electronics?

@@hopkinskong There's no maybe. That's exactly why it is used. Lowercase 'i' is used to denote AC current in electronics. So as a result mathematical 'i' got renamed to 'j' to avoid the confusion.

120 kHz? Mentioned at 31:47.

It is both a defunct electronics store AND a place ham radio operators set up their equipment to keep from annoying their wives