Homework help with concepts of Low/High Pass circuit.

[steven2410]

Here is the problem: .
For the 1st one, It's the low pass filter config. However, I think it's not possible to tell so just based on the Input and Output Voltage graph.
2nd one, i actually have no clue.
3rd question, would the Probe be placed on the left hand side of the resistor?
Thank you.

 

[BradtheRad]

Here is the problem: View attachment 105388.
For the 1st one, It's the low pass filter config. However, I think it's not possible to tell so just based on the Input and Output Voltage graph.

In a sense you are correct. By applying a squared pulse, it does not particularly show that it passes lows and diverts highs. Instead by applying a square wave, it shows it acting as an integrator.

To show it acting as a low-pass filter, it would clarify the lesson better by applying a frequency sweep.

2nd one, i actually have no clue.
3rd question, would the Probe be placed on the left hand side of the resistor?
Thank you.

Although it is a low pass filter according to the position of the probes, if you place the probes across the resistor, then it would be a high pass filter.

I believe that is what Q 2 & 3 are getting at. You'll observe the complementary, or 'filtered-out', waveform across the resistor.

(Puzzling over the term 'filtered component'. I wonder if that means the waveform which remains after filtering, or the waveform which is filtered out?)

 

[FvM]

I don't agree with the comment. The response to a step function contains all information about a linear circuit. For the present square wave case, an additional assumptions must be made to make it equivalent to a step function: All circuit time constants are sufficiently short compared to the square wave period. Apparently this is the case.

The circuit doesn't act as an integrator (would involve a pole at frequency zero, respectively long time constant). It's exactly a first order low pass, it's time constant is lower than 1/10 of the square wave period. The nature of the circuit can be clearly seen from the exponential response in time domain.

 

[BradtheRad]

I don't agree with the comment. The response to a step function contains all information about a linear circuit. For the present square wave case, an additional assumptions must be made to make it equivalent to a step function: All circuit time constants are sufficiently short compared to the square wave period. Apparently this is the case.

Yes, applying a square wave displays the familiar charge/discharge curve (governed by the RC time constant).

However this is not exactly the same role as filtering out high frequencies. I believe the OP sensed this. Although it is a low-pass filter, a different concept is displayed here... The RC network softens each transition from one DC level to the other.

Square waves are created by adding odd harmonics to the fundamental. These high frequencies create the sharp corners (right angles) where the peak was previously mound-shaped.

However notice the RC network cannot filter out highs just before the negative-going edge (right angle).

What we can do is to speed up the frequency, so the waveform starts to resemble a triangle wave. Eventually there are no sharp corners.

However we have not filtered out all the highs. The remainder of my explanation is below.

The circuit doesn't act as an integrator (would involve a pole at frequency zero, respectively long time constant). It's exactly a first order low pass, it's time constant is lower than 1/10 of the square wave period. The nature of the circuit can be clearly seen from the exponential response in time domain.

So I'm thinking how the square wave consists of odd harmonics (in addition to the fundamental frequency). The RC low-pass filter may reduce these, but we will never see the waveform turned into a sinewave fundamental.

Instead it will become more of a triangle.

Fourier analysis shows that the triangle wave is created by adding the even harmonics. Thus where we had odd harmonics, we now have even harmonics. The slope of response has changed. This is more along the lines of a calculus function... namely integration.

As you know, integrator is another name for this filter.

 

[KlausST]

Hi,

Fourier analysis shows that the triangle wave is created by adding the even harmonics

afaik triangle is created with odd harmonics. with alternating sign, and with squared attenuation.


btw.
Filtering (in the meaning of lowpass, highpas...) will never add any frequencies. but it can attenuate or amplify the given amplitudes of the existing frequencies.

From my experience:
I find it hard to analyse the signals frequency components just by a view of a graph. It looks so different when a phase shift is added to the frequency components.

Klaus

 

[BradtheRad]

afaik triangle is created with odd harmonics. with alternating sign, and with squared attenuation.

Yes, you are correct. I was mistaken. More than once I have heard audiophiles praise old-fashioned tubes, on the premise that their distortion is more listenable because it contains even harmonics, creating triangle waves. (I guess one of these premises is right, the other wrong.)
Solid state components, OTOH, distort by creating square waves, containing harsh-sounding odd harmonics.

Wikipedia refreshed my memory about a triangle wave being created by the combining of odd harmonics.

https://en.wikipedia.org/wiki/Triangle_wave

Nevertheless I found it has this statement, regarding a triangle wave being the integral of a square wave.

btw.
Filtering (in the meaning of lowpass, highpas...) will never add any frequencies. but it can attenuate or amplify the given amplitudes of the existing frequencies.

Yes.
However I continue to juggle two ways of looking at these frequencies: (1) Are they really and truly present in the square wave, or (2) are they a mathematical construct for building the square wave in a theoretical sense?

Looking at post #1, I interpret the square wave as a sudden transition in DC levels... more as a transient, an instigator.

Now, to an experienced eye, the scope trace is telling us there are components attenuating the highs. The net effect is that of a low-pass filter...

However the waveform in post #1 illustrates chiefly a shaping effect, more so than it illustrates a low-pass filter. I imagine the OP discerned this. Hence the question.

From my experience:
I find it hard to analyse the signals frequency components just by a view of a graph. It looks so different when a phase shift is added to the frequency components.

Klaus

Yes, the triangle wave has some of the harmonics phase-inverted. These same harmonics are of the same phase in the square wave.

Therefore what harmonics should we say are present when the waveform is partially a square wave and partially a triangle? Perhaps two same-frequency components cancel, due to one being in phase, the other phase-inverted?

 

[KlausST]

Hi,

Nevertheless I found it has this statement, regarding a triangle wave being the integral of a square wave.

I fully agree with that.
There is an attached picture to this. The sentence is corresponding, but the formula is only for a sine input and not a square wave.


The discussion is getting off topic.
Therefore i started a new discussion named "discussing waveforms, frequencies, FFT, and so on..." in "Miscellaneous Engineering" .

Klaus

 

[Audioguru]

Take a perfect squarewave that has no even harmonics and filter it with an 8th-order switched capacitor Butterworth lowpass filter like I have done. The output is a good sinewave with very low distortion. The output also has no even harmonics.

Then make an over-sampled stepped "sine" waveform with 10 steps and filter it with the 8th-order filter like I have done. The output sinewave is so good and its distortion is so low that it is difficult to measure.

 

[FvM]

The Falstad fourier series applet is quite instructive in visualizing the harmonic composition of common waveforms. https://www.falstad.com/fourier/

 

[BradtheRad]

The Falstad fourier series applet is quite instructive in visualizing the harmonic composition of common waveforms.
https://www.falstad.com/fourier/

Yes, that's right. Thanks. I'd forgotten it was there.
I tried it years ago. Now I'm trying it anew.

- - - Updated - - -

Take a perfect squarewave that has no even harmonics and filter it with an 8th-order switched capacitor Butterworth lowpass filter like I have done. The output is a good sinewave with very low distortion. The output also has no even harmonics.

Then make an over-sampled stepped "sine" waveform with 10 steps and filter it with the 8th-order filter like I have done. The output sinewave is so good and its distortion is so low that it is difficult to measure.

Yes, remarkable.
I find that even with just a few RC stages, I get a reasonable sine wave output from a square wave input (simulated).



The right-hand scope trace compares the output to a sinewave.

Of course the filters cause a drop in signal level.

- - - Updated - - -

There is an attached picture to this. The sentence is corresponding, but the formula is only for a sine input and not a square wave.

There's the term 'sgn'. It's a roundabout way to obtain a square wave.

The 'sgn' function seems to be an informal math symbol, originally used as a keyword in programming languages such as BASIC.
It returns a 1 or 0 or -1, to show the sign of a number.

Thus by taking the sign of the sine of a number, the result is a square wave.

 

[KlausST]

Hi

There's the term 'sgn'. It's a roundabout way to obtain a square wave.

Thanks for the information, i didn´t know that.

Klaus