[Andy:] On the forums, we saw a lot of confusion about homeworks 2.5 and 2.6.
Specifically, questions about what was going on with the linear algebra.
I want to talk about some of that today, and I want to do it by comparing the 2D case --
and that's the case we talked about in lecture--
with the 4D case, which is what you're asked about on the homework.
So in the 2-dimensional case, I want to first talk about this f matrix
that Sebastian was calling the "state transition matrix."
The idea behind this matrix was that we wanted to take some old beliefs,
some old state, which in the 2-dimensional case was represented by x and ẋ
where ẋ is our velocity and x is our position,
and from that we want to extract our predict some new state,
which was called x-prime and ẋ-prime.
The question was what do we fill in here to get the proper values for x-prime and ẋ-prime.
Let's think.
What should our position be--our predicted position after some time has elapsed?
Well, we want to include our old position, right?
Lets first write out these formulas.
We expect that x-prime will be composed of our old position
plus whatever motion was occurring due to the velocity.
That is going to be dt--the time elapsed--times our velocity.
This is just velocity times time, which tells us how much our position has changed.
Now, in matrix terms how do we express that?
We're talking about x-prime, so that means we're going to think about this top row here.
We want to keep x, which means a 1 goes here.
We want to multiply ẋ by dt, so that means dt goes here.
Just like that we figured out the first row of our F matrix.
I'll label it here--F. Now what about the second row?
Let's do a similar thing for ẋ prime to figure out where we should go in the second row.
After our prediction, we said that we're just
going to assume that the velocity hasn't changed.
Velocity after equals velocity before.
That means we don't want to have anything to do with this x, meaning a 0 goes here.
We want everything to do with this ẋ--we want to keep this--so we put a 1 here.
Okay. This kind of gives us some intuition for how this works in 2 dimensions.
Let's see if we can generalize to 4.
Now, again, we're going to some new state,
and we're doing that by multiplying a state-transition matrix by some old belief.
But now instead of x and ẋ, we also have y coordinates.
So we have x, y, ẋ, and ẏ.
Here we're going to, of course, get x, y, ẋ, ẏ, and all of those are prime,
because they indicate after our prediction.
Now, I'm not going to fill in this 4 x 4 matrix for you,
but I think using similar reasoning to what we did in the 2-dimensional case,
you can come up with what these formulas should be,
and from that fill in this matrix appropriately,
remembering that this entry corresponds to the first row,
this entry the second, and so on. Good luck.
フォーラムを見ると課題の5問目と6問目で 混乱した人が多かったようです
特に線形代数に関する質問が多くありました
今回はその部分を説明します
講義で説明された二次元の場合と 課題に出た四次元の場合との
比較をしながら説明したいと思います
まず二次元の行列Fから 説明していきたいと思います
スラン教授は状態遷移行列と呼んでいました
この行列の基本となる考え方は
二次元の時にxとxドットで表していた 古い信念や状態を使うということです
xドットは速度でxは位置です
そこから新たな状態の予測を引き出したいので
この部分をx′、xドット プライムと呼びます
ここに何を入れればx’とxドット プライムの 適切な値を求められるでしょうか
考えてみてください
位置はどうなるでしょうか？ 時間経過後の予測位置です
古い位置情報も含めます まず数式を書いてみましょう
x’は古い位置を基に計算されると考えられます
そこへ速度に応じて生じる動作を加えます
つまり時間経過のdtに速度を掛けるのです
速度×時間です 位置がどのくらい変化したかを示します
これを行列ではどう表せばよいでしょう？
x’に関することなので まずは上段を考えていきます
xを残したいのでここに1を入れます
xドットにdtを掛けたいのでdtをここに入れます
行列Fの1行目はこのようになりますね
上にFと書きます では2行目はどうなるでしょう？
今と同じ要領でxドット プライムと 行列の2行目を考えてみます
問題に出たように予測のあとも 速度が変わらないことを前提とします
予測の前後で速度は等しいのです
xには何もしないので0をここに入れます
xドットは残しておきたいので1をここに置きます
これで二次元でどのように機能するか 何となく理解できたと思います
四次元へ一般化してみましょう
四次元でも新たな状態がいくつかあります
状態遷移行列に古い信念を掛けて 新たな状態を計算します
今回はxとxドットだけでなくy座標もあります
なのでこのようになります
こちらにはプライムの記号をつけます
予測後であると示すためです
ここではこの4×4の行列を 埋めるつもりはありません
ですが二次元の時と同じ考え方をすれば 数式を導き出せるでしょう
そしてこの行列を埋めた時のように
この要素は1行目これは2行目に対応すると 考えながら解いてください
頑張ってください