(Brian Thomas rhea) |
|||
| Line 1: | Line 1: | ||
==Part 1== | ==Part 1== | ||
| − | + | I choose Christen Juzeszyn's function <math>f(t)=5 \cos(2t)</math>. | |
| − | + | ||
| − | < | + | |
| − | + | ||
| − | </ | + | |
| − | === | + | One DT signal that could be generated from this would be <math>f[n]=5 \cos(2n)</math>. This DT signal is '''non-periodic'''. <math> \text{For }k,n \in \mathbb{N}, f[n+k] = 5 \cos(2(n+k)) = 5 \cos(2n+2k)).</math> For f to be periodic, there needs to exist some natural number k (k > 0) such that f[n+k]=f[n]. i.e., <math>5 \cos(2n+2k) = 5 \cos(2n).</math> For this to be true, we would need to find k such that for all n, <math>2n+2k = 2\pi (2n) \implies k=2\pi (n-1).</math> However, <math>2\pi (n-1) \notin \mathbb{N}</math> for any n, therefore no valid k exists. |
| − | = | + | Another DT signal that could be generated from this would be <math>f[n]=5 \cos(2\pi n)</math>. This DT signal is '''periodic'''. For instance, let <math>k=1 (k\in\mathbb{N})</math> be the period. Then, <math>f[n+k]=f[n+1]=5 \cos(2\pi (n+1))=5 \cos(2\pi n+2\pi)=5 \cos(2\pi n)=f[n]\ \forall n\in\mathbb{Z}.</math> |
| − | + | ||
| − | + | ||
| − | < | + | |
| − | + | ||
| − | </ | + | |
| − | == | + | |
| + | ==Part 2== | ||
I chose Ben Horst's function <math>x(t)=\frac{\sin(t)}{t}</math>. The function <math>\sum_{k \in \mathbb{Z}} x(t+2 \pi k) = \sum_{k \in \mathbb{Z}} \frac{\sin(t+2 \pi k)}{t+2 \pi k} = \sum_{k \in \mathbb{Z}} \frac{\sin(t)}{t+2 \pi k}</math> is periodic. | I chose Ben Horst's function <math>x(t)=\frac{\sin(t)}{t}</math>. The function <math>\sum_{k \in \mathbb{Z}} x(t+2 \pi k) = \sum_{k \in \mathbb{Z}} \frac{\sin(t+2 \pi k)}{t+2 \pi k} = \sum_{k \in \mathbb{Z}} \frac{\sin(t)}{t+2 \pi k}</math> is periodic. | ||
Latest revision as of 17:53, 10 September 2008
Part 1
I choose Christen Juzeszyn's function $ f(t)=5 \cos(2t) $.
One DT signal that could be generated from this would be $ f[n]=5 \cos(2n) $. This DT signal is non-periodic. $ \text{For }k,n \in \mathbb{N}, f[n+k] = 5 \cos(2(n+k)) = 5 \cos(2n+2k)). $ For f to be periodic, there needs to exist some natural number k (k > 0) such that f[n+k]=f[n]. i.e., $ 5 \cos(2n+2k) = 5 \cos(2n). $ For this to be true, we would need to find k such that for all n, $ 2n+2k = 2\pi (2n) \implies k=2\pi (n-1). $ However, $ 2\pi (n-1) \notin \mathbb{N} $ for any n, therefore no valid k exists.
Another DT signal that could be generated from this would be $ f[n]=5 \cos(2\pi n) $. This DT signal is periodic. For instance, let $ k=1 (k\in\mathbb{N}) $ be the period. Then, $ f[n+k]=f[n+1]=5 \cos(2\pi (n+1))=5 \cos(2\pi n+2\pi)=5 \cos(2\pi n)=f[n]\ \forall n\in\mathbb{Z}. $
Part 2
I chose Ben Horst's function $ x(t)=\frac{\sin(t)}{t} $. The function $ \sum_{k \in \mathbb{Z}} x(t+2 \pi k) = \sum_{k \in \mathbb{Z}} \frac{\sin(t+2 \pi k)}{t+2 \pi k} = \sum_{k \in \mathbb{Z}} \frac{\sin(t)}{t+2 \pi k} $ is periodic.
Consider the period $ P = 2 \pi k $. $ x(t+P) |_{P=2 \pi} = \sum_{k \in \mathbb{Z}} \frac{\sin(t+P)}{(t+P)+2 \pi k}= \sum_{k \in \mathbb{Z}} \frac{\sin(t+2 \pi)}{(t+2 \pi)+2 \pi k}|_{P=2 \pi} $
$ = \sum_{k \in \mathbb{Z}} \frac{\sin(t)}{t+2 \pi (k+1)} = \sum_{k \in \mathbb{Z}} \frac{\sin(t)}{t+2 \pi k} = x(t) \forall t \in \mathbb{R} $
