# Tag Info

23

If you google for 555 square wave generator you'll get thousands of hits for circuits based on a 555 chip that produce a square wave. There's a square wave calculator here, which should allow you to experiment with the calculations. Plus as an added bonus 555 chips are dirt cheap. Or look at 556 chips which are basically two 555s on the same chip. Amos

19

Sinusoidal waves don't have harmonics because it's exactly sine waves which combined can construct other waveforms. The fundamental wave is a sine, so you don't need to add anything to make it the sinusoidal signal. About the oscilloscope. Many signals have a large number of harmonics, some, like a square wave, in theory infinite. This is a partial ...

15

Pentium100's answer is quite complete, but I'd like to give a much simpler (though less accurate) explanation. The reason because sinewaves have (ideally) only one harmonic is because the sine is the "smoothest" periodic signal that you can have, and it's therefore the "best" in term of continuity, derivability and so. For this reason is convenient to ...

12

You can decompose any waveform into an infinite series of sine waves added together. This is called Fourier analysis (if the original waveform is repeating) or Fourier transform (for any waveform). In case of a repeating waveform (like a square wave), when you do Fourier analysis you find that all the sines that compose the waveform have frequencies that ...

11

For a simple oscillator people often think immediately of a 555 timer IC. This ciruit is even simpler: The 74HC1G14 is the single gate version of the more common 74HC14 in SOT-23 package.

10

The Fourier series: $V_t = \dfrac{a_0}{2} + \displaystyle \sum_{i=1}^{\infty}[a_i sin(i \omega_0 t) + b_i cos(i \omega_0 t) ]$ The term $\dfrac{a_0}{2}$ is a constant, that's the DC level. It could also have been written without dividing by two, but this is the convention. The terms of the infinite sum are the sum of a weighted sine and a ...

10

The frequency of your square wave is the same as your sine, it's not half. The following solution focuses on getting the 45° phase difference between sine and pulse. Differentiate the sine to get a cosine. (Note that a simple RC differentiator won't give you the required 90° phase shift. The opamp differentiator will.) Pass both sine and cosine though a ...

9

Please note: the circuit you linked to uses a comparator, not an op-amp. You can use op-amps in comparator circuits but they aren't up to the job for various reasons: op-amps are optimized for amplification applications where the inputs are driven to the same voltage through feedback, and may take a long time to recover from saturation when their inputs zoom ...

9

If the frequency for both waves is going to be 100 kHz with the same amplitude, you could construct a narrow bandpass filter at 200 kHz to put the signal through. In theory a pure squarewave should only have odd harmonics, so there should not be much output at the second harmonic frequency. On the other hand, a sawtooth wave has booth even and odd ...

8

An outline of a solution: Maybe run it through a differentiator. The derivative of a square wave will be alternating positive-going and negative-going spikes, whereas the derivative of a sawtooth should be more or less constant at a low value in one polarity during the rampy bits, with periodic larger valued spikes in the opposite polarity when the sawtooth ...

8

If you are wanting to build an A-stable vibrator, then the circuit you chose is fine. You will want to keep the R value from loading the op amp. This means selecting R so that it doesn't load an op amp. I would suggest that mean staying in the 10k-100k region of resistance keeps you safe if you use a baseband op amp like a TL072(FET) or an LM358(BJT). With ...

6

Loudness is roughly correlated to RMS amplitude, not to peak amplitude, so you need to sample the input regularly, recording the samples to memory take a chunk of samples and square each take the average of all the resulting values square root You can probably simplify this depending on how much accuracy you need. Oh wait, you want analog output. ...

6

Can't you just use a standard crystal oscillator circuit, like the Pierce oscillator, and drive the piezo at its natural resonant frequency? Here's a circuit for an ultrasonic cleaner, which would seem to be the same principle as your vaporizer. You can also look at patents for things like ultrasonic humidifiers, atomizer, nebulizer, etc. No matter what ...

6

If you check out the wikipedia page: Transverse Mode You'll see at the start of the second paragraph they say "Transverse modes occur because of boundary conditions imposed on the wave by the waveguide" This accounts for the discrepancy you have observed. Most diagrams of electromagnetic waves depict waves in free space, far away from any objects, and ...

6

The solid lines in the graph follow the squares (i.e. have low resolution, for instance 8-bit), while the dashed lines have about 3 bits higher resolution, giving 8 levels per square. We see that the dashed line follows the curve closer than the solid line, even when the sampling rate doesn't change; we have 1 sample per 2 squares for both curves. The ...

5

Say you have a 100Hz fundamental frequency. That means it repeats every 10ms, the blue curve in the graph. This signal may have a 3rd harmonic (purple curve), so that's at 300Hz and hence repeats every 3.3ms. It has to be an exact multiple of the ground frequency so that it also starts a new cycle when the fundamental starts a new cycle, namely after ...

5

You have to find the protocol being used somehow. Preferably you find a document that tells you outright. You might be able to reverse engineer it, but less likely for someone that has to ask about basic decoding here. You say the protocol is standard, so you obviously know something about it. If it's truly standard, you should be able to look up the ...

5

Similar questions were asked here and here. In this answer I talk about DDS, direct digital synthesis, which has replaced classical analog oscillators like Wien bridge. The DDS technique is crystal-based so has the same stability and accuracy. Here you'll find a design for a simple DDS. DDSs which use special function ICs typically achieve a wide ...

5

Ancient but superb oscilloscope! Oscilloscope probe may need adjusting but othewise the scope is well capable of handling this signal. The waveform shows what would probably be expected from a digital generator running near the top end of its frequency range. Try adjusting probe used to give a correct square wave response. You will find a small ...

5

The derivative - rate of change - of a sinusoid is another sinusoid at the same frequency, but phase-shifted. Real components - wires, antennas, capacitors - can follow the changes (of voltage, current, field-strength, etc.) of the derivatives as well as they can follow the original signal. The rates of change of the signal, of the rate-of-change of the ...

5

One way is to use a peak detector, but this has a diode voltage drop, which gives an error in the result. You don't say anything about frequency, but I'll assume it is much lower than the ADC's sampling frequency. I would use a resistive voltage adder to bring the full signal within the ADC's range. The resistive adder has the advantage over the ...

5

One way is to essentially build a couple of stages of digital inverter from discrete parts: simulate this circuit – Schematic created using CircuitLab U4 is a linear regulator programmed for 13 V output or a bit higher. Keeping the input power voltage close (but not too close) to 13 V will minimize power lost in the regulator. The value of R1 ...

5

Any signal (of any shape) can be the lowest frequency single sinewave imaginable or, a massive concoction of billions of individual sinewaves. What you have drawn (a, b or c) is reproducable (and analysable) as just this - a bunch of sinewaves. If you can draw limits around it where it might repeat then it's realizable mathematically as a series of ...

4

First something Olin noticed as well: the levels are the reverse of what a microcontoller usually outputs: Nothing to worry, we'll see that we can read it this way too. We just have to remember that on the scope a start bit will be a 1 and the stop bit 0. Next, you have the wrong time base to read this properly. 9600 bits per second (more ...

4

Every realizable analog signal, anything you can think of or draw legitimately on a voltage vs. time graph can be expressed in mathematical terms as the sum of an infinite number of sine waves of different frequencies - something of this form: any_signal(t) = A*sin(f1*t) + B*sin(f2*t) + C*sin(f3*t) .... Different signals are constructed by changing the ...

4

I agree with Scott above: A micro is the way to go here, unless you're just playing about with the specific intention of learning oscillators. Making the amplitude adjustable could be a bit tricky, though. Can you tell us more about that? Does it need to be adjusted once (or very infrequently) for calibration purposes, or do you need to be able to change ...

4

Convert your sound file to a WAV file: mplayer -ao pcm music.mp3 Make an unsigned mono 8bit version at the desired sample rate: sox audiodump.wav -c 1 -r 8000 -u -1 converted.wav Convert the samples to a C header file (get wav2c): wav2c converted.wav sounddata.h sounddata (For your BASIC Stamp, you'll need to convert this array into some other ...

4

What you need to do is simply remove the DC offset all together, not supply a negative one. This is known as AC coupling. If you run the output of your square wave generator through series capacitor, it should do what you need. This will however be at the expense of making the square wave less square. An example circuit is shown below for you: And the ...

4

Capacitive coupling has been suggested, but this has two big disadvantages: Your signal is no longer a square wave It will only center your signal around 0V if the duty cycle is 50%; you'll see the signal go up and down if you play with the duty cycle A good function generator will have a potmeter to set an offset to the signal. One way to do this ...

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