20

There are two issues here: Bandwidth and Measurement technique. Bandwidth: Your measured signal is bandwidth limited. A 30MHz square wave has lots of harmonic content and your scope has a relatively low bandwidth (200MHz) so the higher frequencies are attenuated and you only see the sum of the first few odd-order harmonics (30MHz, 90MHz, 150MHz, ...). ...


11

Yes, the TSOP1738 will do at this short distance. The 0.65 relative responsitivity means that at 36 kHz your IR LED needs to be \$\sqrt{0.65}\$ = 0.8 times closer to see the same signal strength, due to the inverse-square law. So if your TSOP1738 sees a certain level for 38 kHz at 1 m, you'll have to hold the transmitter at 80 cm to get the same signal ...


10

Most common audio equipment is set to match human hearing, 20Hz to 20kHz is their operating range. So you want to measure ultrasound, If you want only to measure high frequencies, you can use a bat detector or something to go a bit above 20kHz, it wont be able to detect "normal" sound though.If you want everything from 20Hz to 100kHz or something like that ...


9

As I see, no answer was accepted. Let me offer another answer. Most modern ICs use so-called Pierce Oscillator to generate stable clocks using crystals. Here is the main circuit configuration: As one can see, the circuit is not symmetrical: the right side is output of some driver (usually designated as XO), and the left side is input to an inverting ...


8

By definition of the audio range for humans, beyond 20kHz is beyond audio. There are audio equipment designers such as Rupert Neve who believe the range could be wider, but this is not necessarily based on strong evidence. The vast majority of audio equipment therefore is limited to this maximum frequency, and you will never be able to do it. Some ...


7

Monostable turns OFF when voltage on capacitor reaches 2/3 of VCC. 1.1×R1×C1 is time required for capacitor to charge to 2/3 VCC.


7

I had a similar debugging issue once with my Atmel ATMEGA328P controllers, the 8Mhz ceramic resonators were not seemingly working. I had a Rigol dual channel cheapo oscilloscope, and had scoped a working board I made previously and the nice 8Mhz signal was easily seen, with no problem due to loading by the probes. You should not worry about the probe's ...


7

The external components make the oscillator look symmetrical, but there is an amplifier on the chip that is anything but. The oscillator pin that is on the output side will have a lower impedance and putting the scope probe there will not affect it nearly as much as probing the input pin. If it's oscillating, the output will have a larger amplitude than ...


7

The first thing I noted when I saw your circuit is that the VAS (Voltage Amplification Stage, as you call it) is directly connected to the feedback network and this implies that its gain and its bandwidth are heavily influenced by the value of the equivalent impedance presented by this network: this is the main cause of its "strange" high frequency behavior. ...


6

In the astable configuration, the low time is given by 0.693×R2×C and the high time is given by 0.693×(R1+R2)×C. The factor of 0.693 is the scale factor (relative to the RC time constant) for the time that the capacitor takes to charge or discharge exponentially between 1/3 Vcc and 2/3 Vcc. The precise value is -ln(1/2) = 0.693147... ...


6

The actual measurement of Spectrum Analyzers is "power / filter bandwidth". Your spectrum analyzer has a filter bandwidth that you can typically configure. For example, if your filter bandwidth is 10 kHz and you measure -12dBm, that's -12 dBm/10 kHz = -12 dBm - 40 dBHz = -52 dBm/Hz. why is it frequency dependent? Because you're using a spectrum analyzer!...


6

For some perspective, consider that optical signals are still too high frequency for the instantaneous electric field to be sampled and measured, but there are still lots of different kinds of measurements we can do on an optical signal. With a power sensor (a photodiode or even an LDR) we can measure the power of the signal. With a prism or diffraction ...


6

Not really. 500MHz is too far above the nominal bandwidth for any aliasing frequencies to get through to the digitiser within it. You can use your scope as a 'back end' for two different types of front end. The first is a heterodyne downconverting mixer. You would need a local oscillator, close to 500MHz, and a mixer. The scope would measure the IF, the ...


6

Yes, such devices exists and are called spectrum analyzers. You will also need to buy an near field probe kit, if you want to work with lower frequencies. They do not propagate well through air, so you need a somewhat sensitive "antenna" that must be put near the source. The range of frequencies you need is quite wide, be ware that not all makes/models ...


5

You need to look at the datasheet and the specific products in hand to make any valid judgment. A badly damaged HP anything probably isn't going to perform better than even a cheap-o new frequency counter which at least works. However, let's assume for some reason you happen to find a used HP2053181A which is in good condition (i.e. everything works and it ...


5

What is x-y graph? What does each axis represent? XY mode is used when you have two inputs (channel 1 and channel 2) and you want to represent the phase relationship between them differently: - Typical is when Y lags X by 90 degrees and is the same frequency - it produces a circle as per the top line above. Channel 1 normally drives the Y axis and channel ...


5

EDIT: Because of the low precision of the HSI - 0.1%, the clock needs to be switched to HSE. I'm doing all of this in cubeMX so I won't be writing out the bits, you can find them in the reference manual -http://www.st.com/web/en/resource/technical/document/reference_manual/DM00043574.pdf After some experimenting I figured it out. For simplicity, i'll ...


5

In a two-port network, S21 is the complex number that describes the linear power gain of the network. (it's kinda backward - S2 is output, S1 is input). It's a complex number because it is used to characterize both magnitude and phase changes to the input. There is a different gain number for every frequency of interest, because, in general, gain and phase ...


5

It is very difficult to see, but as far as I can tell it looks like you have the wrong pinout for the 555 timer. The pins on the right-hand side should be numbered 5-8 from the bottom up, not from the top down: If you have it connected the way the picture shows, I expect that's why it's giving you problems.


4

the oscillator is designed to drive a high-impedance IC clock input In this case, you will want to add a buffer chip to be able to drive the 50 Ohm coax and the spectrum analyzer input. The appropriate chip depends on what type of load the oscillator is designed to drive (TTL, CMOS, ...?), the frequency, the available power supply voltages, etc. Possibly ...


4

Your output filter response should basically be the 20 dB/decade response of the capacitor. The ESR zero will depend on the number and type of capacitors you're using. It's the natural corner frequency formed by the effective ESR and total capacitance: \$ F_z = \dfrac{1}{2 \pi \times R_{esr} \times C_{out}} \$ The gain slope will swing more positive at ...


4

Edit: Upon reflection I don't think this is the culprit in this case, I think Dave Tweed's comment is where the issue is, but I'll leave this in case someone runs into a spikier noise problem. I suspect that a switching power supply is causing the problem. Here is the schematic (from Jan Panteltje) of a cheap eBay supply. It's not perfect, Jan got the 1M ...


4

The steep response you want will require a multistage filter. The most commonly used active filter configuration is Sallen–Key topology, which provides two stages per amp using just 2 resistors and 2 capacitors. Various filter responses can be created by varying the component values. Here is an LTSpice simulation of a 4 stage Butterworth high-pass filter ...


4

The Pi is not really the best tool for the job. You have an entire operating system vying for processing time, with many processes running at the same time as yours. The only way you would be able to do it reliably would be to directly interface with the hardware and use timers and such to do the job. Certainly trying to do anything where reading rapidly ...


4

The question here is what is meant by DC and what is meant by AC. There are two contexts, where DC and AC mean completely different things. One perspective is that DC is any static voltage that doesn't change - i.e., a 0Hz signal. AC is any voltage which changes over time. The other perspective is that DC is any voltage that can change but remains the ...


4

If you know what frequencies you're looking for, run a Goertzel for each of those frequencies. https://en.wikipedia.org/wiki/Goertzel_algorithm This is much more efficient than an FFT. This algorithm is used, for example, to detect tones in touch-tone phone systems.


4

Am only going to address your question about reading the count output from a '393 counter. You have two problems to overcome. This counter is asynchronous, which means that a clock input ripples through the individual flip-flops to its next count state. If you happen to read the count value on-the-fly very soon after a clock transition, you might get a ...


4

First, you can easily make a nice square wave from any duty cycle pulse train by running it thru a toggle flip-flop. That's basically using it as a 1-bit counter. That divides the frequency by 2, so you compensate by doubling the pulse train frequency. To your question, every sharp edge contains lots of high frequencies. Especially if the pulse frequency ...


4

I was studying tube designs as a child, long before I had any clue how a grounded cathode amplifier might be designed. Which means I don't remember that much. But one thing I do recall, which had bothered me a lot back then, is the ability of a grid to self-bias. You might look up "grid leak resistor," or something similar. A vacuum tube's cathode may ...


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