Monday, October 27, 2014
Upcoming Filter Webinar
I will be teaching a filter webinar through PDHengineer.com on December 30. The six hour course can be taken all together or either half can be take separately. Join me if you can!
Thursday, July 17, 2014
DFFE is #1!
Thanks to all of you, Digital Filters for Everyone is currently number 1 in signal processing books on Amazon. I am touched and honored. Thank you!
Wednesday, February 12, 2014
Slide and Divide
Am I the only one who didn't know the slide and divide method?
If you've read my book, or the digital filter courses in this blog, you know that I know a thing or two about math in general, and polynomials in particular. But I hadn't heard of "Slide and Divide" until last week. The excellent math teacher at one of the places where I volunteer as a tutor, e-mailed me to Google on it and learn how to do it. I did, and have now been able to help several of her students with it.
You could also Google it, but I'll show it here for those who haven't seen it. To begin with, let's make an example that we know will be factorable:
(3x + 2)(5x - 4) = 15x^2 -2x -8. (Did I do that right?)
The first thing we do is replace c (the -8 in this case) with the product of a and c = -8 x 15 = -120.
Now, we write a new polynomial: x^2 -2x -120. Note that it is NOT equivalent to the one we're trying to factor. It's just a new polynomial that will help us with our job. So let's factor it. It's easier since a = 1.
(x + 10)(x - 12)
Now, we have to divide the roots by the 15 by which we multiplied c above. (Repeat caveat about equivalence.)
(x + 10/15)(x - 12/15) = (x + 2/3)(x - 4/5)
Now, in the case of fractional roots such as these, we multiply each factor by its root's denominator: 3(x + 2/3) and 5(x-4/5). (Repeat caveat about equivalence.)
Finally, we simplify those: (3x + 2)(5x - 4)
Ordinarily we could FOIL it back out to check, but there's no need since we started with the same factors above.
Pretty slick right? The fact that it violates equivalence along the way was bothersome at first, but that's really just an accounting problem. One keeps track of what's equivalent and not, and uses the rest as intermediate calculations. In the end it's easy to demonstrate that the result is correct.
If you've read my book, or the digital filter courses in this blog, you know that I know a thing or two about math in general, and polynomials in particular. But I hadn't heard of "Slide and Divide" until last week. The excellent math teacher at one of the places where I volunteer as a tutor, e-mailed me to Google on it and learn how to do it. I did, and have now been able to help several of her students with it.
You could also Google it, but I'll show it here for those who haven't seen it. To begin with, let's make an example that we know will be factorable:
(3x + 2)(5x - 4) = 15x^2 -2x -8. (Did I do that right?)
The first thing we do is replace c (the -8 in this case) with the product of a and c = -8 x 15 = -120.
Now, we write a new polynomial: x^2 -2x -120. Note that it is NOT equivalent to the one we're trying to factor. It's just a new polynomial that will help us with our job. So let's factor it. It's easier since a = 1.
(x + 10)(x - 12)
Now, we have to divide the roots by the 15 by which we multiplied c above. (Repeat caveat about equivalence.)
(x + 10/15)(x - 12/15) = (x + 2/3)(x - 4/5)
Now, in the case of fractional roots such as these, we multiply each factor by its root's denominator: 3(x + 2/3) and 5(x-4/5). (Repeat caveat about equivalence.)
Finally, we simplify those: (3x + 2)(5x - 4)
Ordinarily we could FOIL it back out to check, but there's no need since we started with the same factors above.
Pretty slick right? The fact that it violates equivalence along the way was bothersome at first, but that's really just an accounting problem. One keeps track of what's equivalent and not, and uses the rest as intermediate calculations. In the end it's easy to demonstrate that the result is correct.
Sunday, February 9, 2014
Amazon's Top 100
Amazon recently published a list of 100 books to read before you die. I checked, and Digital Filters for Everyone wasn't on it. Could that be right? Hey, get Amazon on the phone...
Surely we're 101, and didn't quite make the cut? Well, not quite. Surprising though it may seem to those of us who do, there are people in the world who will never bother to design a digital filter. It seems like such a waste.
But I am grateful to my loyal followers who have made Digital Filters for Everyone a top seller within its category. I am humbled by your support.
Surely we're 101, and didn't quite make the cut? Well, not quite. Surprising though it may seem to those of us who do, there are people in the world who will never bother to design a digital filter. It seems like such a waste.
But I am grateful to my loyal followers who have made Digital Filters for Everyone a top seller within its category. I am humbled by your support.
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