Electronics > Beginners
Op amp beginner's questions
rstofer:
Or, one pole versus two poles and 20dB/decade versus 40dB/decade rolloff.
Look at the two transfer functions given above. Note how the first has a linear denominator (s + ...) and the second has a quadratic denominator (s2 + ...).
Poles are determined by the denominator of the transfer function. It the denominator is linear, it is a first order filter. If the denominator is quadratic, it is a second order filter. And so on...
We find the location of the poles by solving for the denominator going to 0. Those roots can be real, imaginary or even 0. Root Locus comes to mind and I really don't want to go there.
If you want to chase the math down a rathole, you need Calc I, Calc II, Differential Equations and Laplace Transforms. And that's WHY I don't want to go there. Yes, I took the classes, way back when.
You might have to throw in the Fourier Transform and the Inverse Fourier Transform to convert between the time and frequency domains and back.
Mattjd:
if the region of convergence of the laplace transform contains the imaginary axis then you can just substitute in the definition of s with sigma = 0.
I.e to go from S to frequency set S = sigma+j*omega
Be careful with going backwards, as the converse is not always true.
i.e.
L{H(t)}|(s=j*w) = H(w) <=/=> H(w)|(j*w = s)
but
L{H(t)}|(s=j*w) = H(w) <=> Imaginary axis is an element of ROC{L{H(t)}}
nForce:
--- Quote ---Or, one pole versus two poles and 20dB/decade versus 40dB/decade rolloff.
--- End quote ---
How do you know, that the first one has 20dB/decade rolloff? How can we calculate this?
From this, I can say that we can achieve a low-pass filter programmatically, why should we use then op-amp active filter?
rstofer:
--- Quote from: nForce on November 27, 2018, 04:32:25 pm ---
--- Quote ---Or, one pole versus two poles and 20dB/decade versus 40dB/decade rolloff.
--- End quote ---
How do you know, that the first one has 20dB/decade rolloff? How can we calculate this?
--- End quote ---
There's a fair bit of math involved in proving that so here is a web page. There are MANY others. The roll-off is kind of a universal constant, like 'e' or 'pi' and it is derived from the definition of the Bel and deciBel (dB).
https://en.wikipedia.org/wiki/Roll-off
Here's a video re: Bode' Plots and is just the first of a series related to drawing the plots by hand. Obviously, we aren't going to do it that way given MATLAB or some other solver like Maxima or Octave.
--- Quote ---From this, I can say that we can achieve a low-pass filter programmatically, why should we use then op-amp active filter?
--- End quote ---
But you really can't achieve the true analog nature of any continuous function with a program. The best you can do is create a discrete time function which is, by definition, not continuous. Sure, it's a good rendition if the CPU is fast enough and the ADC and DACs have enough bandwidth but they are still discrete time, not continuous, functions. There a whole field of EE math involved with discrete time and it comes up frequently when building PID loops. Here's a video (one of a series) on discrete time control systems:
And every time that DAC value changes, even by one bit, it generates harmonics. Just like a square wave does. They may not be significant but you can't avoid the math. It happens... There is a step change in voltage level in a short period of time. Fourier Series comes up a lot about now!
One problem with the passive LPF is the loss of amplitude. That is the big thought behind active filters - restore the output level with an op amp. Nothing much changes in terms of breakpoint and roll-off but the output isn't attenuated.
https://www.electronics-tutorials.ws/filter/filter_5.html
Will every circuit you construct with spare parts exhibit the exact characteristics assumed by the math? Probably not! It isn't that the math is wrong, it's that the parts aren't, and probably can't be, modeled perfectly.
We know resistors have inductance, we know capacitors have resistance, we know inductors have resistance and so on.
rstofer:
Remember that roll-off is in terms of dB per decade (usually). Assume a corner frequency of 1 Hz and let's step away from the corner. Between 10 Hz and 100 Hz is 1 decade but only 90 Hz spread and the attenuation is 20 dB. Between 100 Hz and 1 kHz is one decade but 900 Hz spread and the attenuation is 20 dB. Same between 1 kHz and 10 kHz, 9000 Hz spread, and so on.
The reason this looks like a straight line on the Bode' Plot is because the horizontal axis is logarithmic, not linear. We use that scaling because we can use a straightedge to plot the roll-off. And, of course, dB is logarithmic. In the voltage case dB = 20 log10(Vo/Vi). In the power case it is dB = 10 log10(Pout/Pin).
https://en.wikipedia.org/wiki/Decibel
BTW: 20 dB/decade is 6 dB/octave where an octave is just a doubling in frequency. Sometimes you will see dB/octave in datasheets.
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