00:01
All right, so in this question, we're dealing with a square wave.
00:04
And we are kind of being walked through the process of determining the coefficients to the four -year transform over the square wave.
00:12
So to start with, we are given kind of the ideal formula for the square wave, which is y of t, is equal to a piecewise function.
00:24
The first part is an amplitude a.
00:27
And that's true over a range from zero is less than t, which is less than capital t over two.
00:36
And then the second part is a negative a.
00:39
And that is the case from capital t over two is less than lowercase t is less than t, where capital t is the period and lower case t is time.
00:52
And, you know, this makes sense because it'll have a, constant single amplitude for half the period and then for the other half the period it'll switch to the same magnitude just on the negative side and then we are also given the kind of general form of a foyer equation which is the sum overall n of a n of a n times sine of n omega t plus bn times cosine of n omega t and so this is our starting point so these two equations will be equal to each other um and in part a we're asked to multiply both sides of we're multiplied the both sides of this second equation here the y equals t or y of t is equal to the four year sum of m omega t and then we're going to integrate both sides of that equation over the full period of the wave and then we are supposed to take a look at the left hand side and consider what happened and we're supposed to show that when m is even all the the left hand sign is equal to zero and and then when m is odd, we're supposed to show that the left -hand side is equal to a constant, which is 4a over m times omega.
02:49
So that's all that we're doing in part a, which is a lot of different things.
02:53
So let's go ahead and get started.
02:54
This problem is pretty long.
02:56
So i'm just going to go ahead and do the different parts in different colors.
02:59
So i'll do blue for part a.
03:02
So first thing we're doing is going to multiply both sides by sine of m omega -t.
03:09
And put it into an integral over the period of the wave.
03:15
So for our left -hand side, we'll end up with the equation or with the expression, integral from 0 to t of y of t times sine of m omega -t d -t.
03:36
So that's our left -hand side of the equation.
03:38
And then the right -hand side becomes a much longer thing.
03:43
So i'm going to put it on the bottom, or just the next line.
03:48
And so since there's two terms, the sign of m -o -m -t will get distributed to both, and i will go ahead and just do that all in one step.
03:56
So we'll have the integral from 0 to t once again.
04:00
And this will be the sum of a -n times sine n -on -t, t times sine m omega t plus that doesn't look like plus plus bn cosine and omega t times sine m omega t d t just barely fit everything in there so that is our right hand side of the equation so that's the first part of part a now we are going to look exclusively at the left -hand side, so the integral of y of t times sine of m omega -t d t.
05:00
And we're supposed to consider, you know, what happens when m is even and when an m is odd.
05:05
So the first way that we're going to do that is we're actually going to recall that y of t is a piecewise function, where it's equal to a from zero to t over two, and it's where it's equal to negative a from t over two.
05:19
To t.
05:21
So we're actually going to break the left -hand side into two integrals, one from 0 to t over 2 and one from t over 2 to t, just based on the piecewise definition.
05:34
So then the left -hand side becomes an integral from 0 to t over 2, and then for that interval, y of t is just equal to capital a, the amplitude of the wave.
05:52
And so that will be multiplied by sine of m omega -t and d -t.
06:01
And then the second integral will be negative a sign of m -omega -t -d -t.
06:08
But i'm going to go ahead and just bring out the negative sign.
06:12
Normally we'd add the two integrals together, but i'll bring out the negative sign, so we'll just subtract them.
06:19
And then this integral, will go from t over 2 to t and then it will be a sign of m omega t d t that parentheses should be over here there we go so this makes things actually pretty nice because now we have two integrals that are exactly the same just with different limits of integration so now it's a little bit now we can start considering what happens when a is even or when m is even and when it's odd so if m is even if we're integrating a sign function over half its period if m is even then the first and this the integral for the first and the second half of the period will have the same magnitude and the same sign so then that value which for lack of a better symbol i'm just going to go ahead and call capital e that value e will be subtracted from itself because over the first half of the period it is equal to e and then the second half has the same magnitude and sign so we'd have so by evaluating these two integrals the left hand side would be equal to e minus e which would be equal to zero so if m is even the left hand side equals zero but if m is odd then this first uh this first integral from zero to t over two is going to have some value and then the second integral is going to have the same value, but instead of the same magnitude, but instead of having the same sign, it will have opposite sign.
08:15
So it will be negative e.
08:18
So by evaluating the left -hand side, you would have e minus negative e, which would instead be equal to 2e.
08:32
So if m is odd, left -hand side is equal to a constant.
08:42
And now we want to evaluate what that constant is.
08:47
So now we're going to consider that m is odd.
08:55
But instead of evaluating two integrals that are practically the same, we can go ahead, we've established that they're going to have the same value.
09:06
So the first integral has the same magnitude as the second integral, and that this whole expression is going to have a value of 2e, where e is the value of the first one.
09:21
So we can go ahead and rewrite this expression saying that the left -hand side is going to be equal to two times the integral from zero to t over two of a sine of m -omega -t, d -t.
09:40
And this is actually a very easy integral to deal with.
09:44
So we can go ahead, since a is a constant, we can bring it out of the integral, and then we're just integrating sine of m -omega -t, which due to chain rule will bring out a 1 over m omega to the front and then the sign will become a negative cosine.
10:01
So then out of the front of this equation we're going to have a 2a over m omega and then that is going to be multiplied by evaluating from 0 to t over 2.
10:27
M omega t evaluated from zero to t over two and so we go ahead and plug in zero and t over two for those so we have 2 a over m omega and then we have that is multiplied by negative cosine of m omega of m omega times t over two plus because we would have minus another negative cosine so we have plus cosine of zero.
11:08
Cosson of zero, of course, is just one.
11:12
And then in order to evaluate the negative cosine of m omega -t over 2, we have to remember that omega -t is defined as being 2 -pi over t.
11:25
So omega times t over 2 is just going to be equal to pi.
11:32
So then this expression becomes 2a over m omega times the negative cosine of m pi plus one.
11:48
And now we need to consider, we need to remember that m is odd.
11:54
We've made that our condition for pursuing this evaluation.
11:58
So m is odd, which means that the negative cosine is of pi or 3 pi or 5 pi.
12:05
Or negative pi.
12:06
The point is it's an odd value of pi, which means that if you take the negative cosine of that, that will always be equal to one because you're just doing two pi rotations, you know, regardless of whatever value m is.
12:24
So the negative cosine of m pi, because um is odd, has to be equal to one.
12:30
So inside the parentheses here, we just end up with, uh, one plus one.
12:36
So 2a over m omega times 2.
12:42
And so the left -hand side is equal to 4a over m -omega.
12:51
And so i'm going to box this because this is very important and we'll come back to it much later.
12:57
So that's the end of part a.
13:00
In part b, which i'm going to do in red, we're supposed to use a trig identity to evaluate a the bn terms of the right -hand side...