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Hyperspace2

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One I know that of op amp adder is that they can be used as audio mixer .

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- Thread starter Hyperspace2
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In summary, an integrator can be used to measure phase difference, and an inverter can be used as a Tea Party filter.

- #1

Hyperspace2

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One I know that of op amp adder is that they can be used as audio mixer .

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berkeman

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Hyperspace2 said:

One I know that of op amp adder is that they can be used as audio mixer .

Good thought on the adder/mixer. What would be some uses that you can think of for the other opamp building blocks that you mention?

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Averagesupernova

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schip666!

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Differentiator == high pass filter

Inverter == Tea Party Filter... just kidding, sorry...

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berkeman

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schip666! said:Inverter == Tea Party Filter... just kidding, sorry...

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Hyperspace2

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Ok I got it.Averagesupernova said:

Yeah, I also came to learn about the low pass filter and high pass filter too. But I think of them now as a independent circuit elements . Aren't they just combination of Resisitor and capacitors? I don't think they need op amp. Active High pass filter are just the combination of filter and op amp for High pass filter and amplification respectively. Similar in the case of the active low pass filter. Amn't I correct?schip666! said:

Differentiator == high pass filter

Inverter == Tea Party Filter... just kidding, sorry...

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Averagesupernova

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An active filter has advantages over a passive. Have you done the math and made some comparisons?

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waht

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Something I've recently worked with using an integrator. Perhaps might be of help.

The goal is to measure phase between two square waves. If you feed the two signals into a XOR gate, then one can show that the output will have different width or delta t which is directly proportional to a phase between the square waves.

The output looks something like this:

[PLAIN]http://www.pcsilencioso.com/cpemma/graphics/swave.gif

The wider the signal, the more phase difference there is. Now, what you want to do is to get DC voltage output that is proportional to the phase difference, and all you have is this rectangular pulse wave. How would you recover the width?

The answer is with an integrator. We known that the amplitude is constant. If you integrate this rectangular pulse wave, then you get the area under the rectangle which is (Amplitude * Delta t) and the delta t is the width you are after which is proportional to phase difference.

The goal is to measure phase between two square waves. If you feed the two signals into a XOR gate, then one can show that the output will have different width or delta t which is directly proportional to a phase between the square waves.

The output looks something like this:

[PLAIN]http://www.pcsilencioso.com/cpemma/graphics/swave.gif

The wider the signal, the more phase difference there is. Now, what you want to do is to get DC voltage output that is proportional to the phase difference, and all you have is this rectangular pulse wave. How would you recover the width?

The answer is with an integrator. We known that the amplitude is constant. If you integrate this rectangular pulse wave, then you get the area under the rectangle which is (Amplitude * Delta t) and the delta t is the width you are after which is proportional to phase difference.

Last edited by a moderator:

- #9

Hyperspace2

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No I haven't tried it yet. I would like to look at it now.Averagesupernova said:An active filter has advantages over a passive. Have you done the math and made some comparisons?

waht said:

The goal is to measure phase between two square waves. If you feed the two signals into a XOR gate, then one can show that the output will have different width or delta t which is directly proportional to a phase between the square waves.

The output looks something like this:

[PLAIN]http://www.pcsilencioso.com/cpemma/graphics/swave.gif

The wider the signal, the more phase difference there is. Now, what you want to do is to get DC voltage output that is proportional to the phase difference, and all you have is this rectangular pulse wave. How would you recover the width?

The answer is with an integrator. We known that the amplitude is constant. If you integrate this rectangular pulse wave, then you get the area under the rectangle which is (Amplitude * Delta t) and the delta t is the width you are after which is proportional to phase difference.

Thanks for sharing . I am the begineer . I have to go long way ahead to view yours. I think you have many things for sharing.

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An op amp, short for operational amplifier, is an electronic device that amplifies the difference between two input signals. It consists of a non-inverting input, an inverting input, and an output. The output is determined by the difference between the two inputs, amplified by a very high gain.

An op amp can be used for a variety of applications, including amplification, filtering, mixing, and signal conditioning. Its high gain and versatile inputs and outputs make it a useful tool in many electronic circuits.

An op amp subtractor is a circuit that subtracts one input signal from another. It is commonly used in audio mixers, electronic filters, and voltage regulators. The output of the subtractor is proportional to the difference between the two input signals, with the gain determined by the op amp's feedback resistors.

An op amp integrator is a circuit that performs mathematical integration on an input signal. It is commonly used in analog computers, audio filters, and signal generators. The output of an integrator is proportional to the integral of the input signal, and the gain is determined by the op amp's feedback capacitor.

An op amp differentiator is a circuit that performs mathematical differentiation on an input signal. It is commonly used in frequency modulation, high-pass filters, and signal detection. The output of a differentiator is proportional to the derivative of the input signal, with the gain determined by the op amp's feedback resistor.

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