25

Am I missing some important reason why you would do the signal conditioning in this order? Yes you are... The front-end differential amplifier will be chosen such that it has a common-mode rejection level of many tens of dB, quite possibly in the region of 80 dB. This diff amp converts a differential signal into a single-ended signal and any common-mode ...


23

Forget sampling rate for a few seconds... Think about sampling period for a second, which is the time interval between two consecutive samples. This time can be an integer or any real number (as long as it’s positive, of course). Sampling rate is simply the inverse of sampling period. Does it make more sense this way?


14

Yes, the sampling rate can be any number you want. But you obviously would not get partial samples in the end, you just have to round down. In your example the first sample is taken at \$ \frac{1}{15.5}s \$ = 64.5 ms and then at every multiple from that. This means you get your last sample at 6,966 s. That is the 108's sample. So at 7 s you still have taken ...


13

Some things are always an integer. Samples are always integer. You can take 108 or 109 samples. Sample rate can be a floating point number, or more generally a rational, or even a real. You calculate the sample rate by dividing the number of samples (less one to get the number of periods between samples) by the time it takes to obtain those samples. ...


12

Capacitors block DC and pass AC. You can use a series capacitor into an opamp with whatever gain you need. Even better might be a simple RC high-pass filter...One capacitor (series) and one resistor (to ground) in front of your amplifier. Like this: simulate this circuit – Schematic created using CircuitLab R2 and R3 set your gain. C1 and R1 set ...


8

I'll second @Andy's answer, and I'd like to add one thing. A passive low-pass filter between the electrodes and the InAmp is called for. I put the cutoff frequency somewhere in the kHz region. InAmps have a great CMRR at low frequencies, but the CMRR degrades at higher frequencies (above 3kHz-10kHz depending on the chip). Rectification at high ...


8

Andy and Nick offered great answers. Let me try to beef them up just a little. First, the math says that amplifying then filtering is equivalent to filtering then amplifying. This, of course, applies to the ideal situation, so lets discuss the non-idealities. The BIG one here, IMO, is saturation. If the noise is so big that it saturates your amplifier,...


8

The spikes seem to have a few common heights that are roughly related by factors of 2, which strongly suggests that they are noise-induced single-bit errors in the binary data. One good way to address this is to take the standard deviation of all the data, and then simply throw away any samples that are more than, say, 2σ from the mean. Another ...


8

1960's standards convertors did involve CRTs and cameras, with quite a lot of tweaking to minimise artefacts. This description of the state of the art at the time, from BBC Engineering, goes into a lot of detail about the process. Quartz crystals used as delay lines were just around the corner. By 1970 the BBC had built an entire field store based convertor ...


7

There are quite a few advantages. Taking a look at a typical superhet (up to the IF): The input signal at the RF input is small (as low as -122dBm in some narrow band voice systems I have worked on - that is about 6.3fW) To amplify a signal at a high RF (say a few GHz) is expensive compared to doing that amplification at a lower frequency. A few dB of RF ...


7

The only filtering you ever do before the first amplifier is that related to shape of antenna/waveguide. And that only applies to microwave and higher frequencies. Conventional passive filters add noise -- you want the signal to be as large compared to that added noise as possible. Even if it means you are also amplifying interfering signals, you aren't ...


7

This may be a boring answer for electrical engineers, but it is all about the optics. The optical system for a CD looks something like this. source As can be seen from the figure, the light traveling to the disc needs to go one way and the reflected light needs to be directed towards the detector. This is achieved with the polarizing prism, which forms an ...


6

Note that this answer is skewed toward analog radio reception. The rules are different for software-defined radios, and for digital services. The biggest drawback to direct conversion is sideband suppression. If you use a single mixer, a signal at \$f_c + f_s\$ is indistinguishable from a signal at \$f_c - f_s\$, where \$f_c\$ is the carrier and \$f_s\$ ...


6

Here's something inspired by the first 2 answers. Make a 10-second low pass filter of the input signal and feed that into an op-amp's non-inverting input (+). Then take a 1-second high pass filter of the same input signal, and feed that into the inverting (-) input of the same op-amp. Fluctuations get subtracted from the average and amplified a lot. If ...


6

There isn’t enough information in 10 mS of sampled audio to reliably tell the difference in frequency between a low E and a low F or D# notes. 10 mS is barely more than 1 full pitch cycle at those low frequencies, and there’s tons of non-pitched noise from the attack (pluck or strum) interfering with any measurement of the rate of phase change of the higher ...


5

The simple answer is: yes, in many cases a simple voltage source & resistor (Thevenin) model is enough to model the output of a circuit (there are theorems to that effect). Formally, you would need a voltage source & a complex impedance, to also be able to represent the frequency behavior. But it can be even more complicated than that. Some of the ...


5

The idea is fine, but 220k seems rather high. You should check the specs of the ADC module, there should be an indication of the maximum input impedance (typically around 50k for ADC integrated to most MCUs). Use resistors with values lower than this. Hint: you could also use a potentiometer instead of the lower resistors (closer to ground), so you can ...


5

You can think of any perfectly bandlimited signal as the superposition of a set of \$\frac{\sin(t)}{t} = \text{sinc}(t)\$ curves, with their peaks positioned uniformly along the time axis. Their spacing is \$\frac{2}{BW}\$. sinc(x) also happens to be the time-domain response of a perfect low-pass filter, and it explains how the continuous-time ...


5

Since your guitar string can only produce a handful of pitches, a full FFT is wasteful. Take the lower E string. The tones are approximately 5Hz apart. Assume a sampling rate of 11025 Hz. To get your FFT bins spaced closely enough together, you will need an FFT length of at least 2048 resulting in 1024 bins. You calculate 1024 bins, then discard 1000 ...


5

Could an op-amp also be used for this, and how? No need to use an op-amp if your signal is audio. Shifting a biased audio signal of 0 - 4 volts to an unbiased voltage requires only a RC high pass filter. It will remove the DC offset: - You decide on the values of R and C so that the lowest frequency you wish to pass is: - $$F_{LOW} = \dfrac{1}{2\pi RC}$$


4

Getting a precise analogue of a digital moving average filter is somewhere between very difficult and impossible using analogue circuitry, as it involves delaying a signal for a long time. There are various approximations you could use, which might (or might not) be acceptable to your particular application. The first is a simple lowpass RC filter. Instead ...


4

Yes, direct-conversion receivers exist, but they require special care, especially with certain kinds of modulation. For example, with SSB modulation, in order to reject the unwanted sideband, your baseband demodulator must be able to distinguish between "positive frequency" and "negative frequency". This is not trivial, and is only really practical using ...


4

Direct conversion is conceptually simple, but requires considerable engineering to do it right. Besides Dave's & Tim's answers, there is a subtle pernicious potential problem with direct conversion... Most mixers (even doubly-balanced ones) leak local oscillator power to both RF port and IF port. Power leaking backwards through to the RF port to the ...


4

The output of the fft function will go from 0 Hz to fs/2, then -fs/2 to almost 0. If you ignore the 2nd half (for purely real signals), the following works well: For n samples at a sampling rate of fs, you can create a frequency axis, f with the following formula: f = (0:n-1)/n*fs This gives you freqs from fs/2 to almost fs, but if you overlook that... ...


4

For standard IO the STM32F103 datasheet section 5.3.13 says: \$ V_{IL(max)} = 0.28 (V_{DD} -2 \ \text V) + 0.8 \ \text V \$. For a 3.3 V supply this is 1.16 V. \$ V_{IH(min)} = 0.41 (V_{DD} -2 \ \text V) + 1.3 \ \text V \$. For a 3.3 V supply this is 1.83 V. Offset DC about 800mV, Vmax = 2.4V. 0 1 2 3 3.3 V | ...


4

You can find the harmonics of a signal by computing the Fourier series of the signal, given that the signal is periodic. You have a non-periodic signal here . However, you can still calculate Fourier series between the full finite interval \$T\$ of the signal, assuming that the signal is 'periodic' with \$T\$, ie., assuming that the signal repeats over and ...


3

Yes, you are be able to simply reverse the inputs to invert the signal. CML, like LVDS, uses symmetrical drivers for the P and N lines, which means there is electrically no difference between the lines. In fact this is commonly done in PCIe which uses a CML variant. Routing can frequently be made easier by crossing the P and N lines of the data serial ...


3

$$\mathcal{X_1}=\mathcal{N}(0,\sigma^2)$$ $$\mathcal{X_2}=\mathcal{N}(0,\sigma^2)$$ $$\theta=\arctan(\mathcal{X_1},\mathcal{X_2})$$ Since arctan ranges from \$-\frac{\pi}{2}\$ to \$-\frac{\pi}{2}\$ there can only be one answer and that is c, you could do the math but the answer is c These questions are designed to be tricky, simple relationships can save ...


3

UPDATE Imagine this is a $1k competition to a $trillion global power infrastructure problem with ageing grid transformers and switchgear that can explode from dissolved combustible gasses from H2 to Acetylene that requires your data analysis skills. Here you can see a unique impulse (PD) response of the system due to PD events in high voltage After all ...


3

anywhere from 10-10kHz. Would I need to use a band pass? Sounds like you at the very least would want to use a low-pass filter; that'd remove the noise power from above 10 kHz. In noisy, low voltage situations like this, how do I go about measuring this signal? You'd apply as much gain as necessary for your measurement device or ADC to measure ...


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