Please, check the following discrete periodic sequence when the period $N=2$.
$x[k]=\exp(j\frac{2\pi}{N}k), N=\text{period}$
$..., x[0]= 1, x[1]= -1, x[2]= 1, x[3]= -1, ... , N=2$
According to my DTFS calculation, DTFS coefficient of above sequence is as follows.
$..., D_{0}=0, D_{1}=1, D_{2}=0, D_{3}=1, ... , N=2 $
If I shift above DTFS by 1 in the frequency domain like this,
$..., D_{0}=1, D_{1}=0, D_{2}=1, D_{3}=0, ...$
Inverse DTFS of above coefficient is as follows according to my calculation.
$..., x[0]=1, x[1]=1, x[2]=1, x[3]=1, ... N=2$
Actually, if DTFS of $x[k]$ is $D_{n}$, then DFTS of $\exp(ja\frac{2\pi}{N}k)x[k]$
will be $D_{n-a}$ by DTFS frequency shifting definition.
So, DFTS of $\exp(j(-1)\frac{2\pi}{N}k)x[k]$ will be $D_{n+1}$.
In this case, $\exp(j(-1)\frac{2\pi}{2}k)x[k]$ is $\exp(-j\frac{2\pi}{2}k)$ $\exp(j\frac{2\pi}{2}k)=1$.
Shifting by $-1$ shows the same result in this case because $N=2$.
So, I think above result is correct.
I know DTFS of $x[k]=1$ with $N=1$ is as follows.
$..., x[0]=1, x[1]=1, x[2]=1, x[3]=1, ... N=1$
My conclusion is as follows.
$x[k]=..., 1, -1, 1, -1,... \longleftrightarrow D_{n}=..., 0, 1, 0, 1,... N=2$ case 1
$x[k]=..., 1, 1, 1, 1,... \longleftrightarrow D_{n}=..., 1, 0, 1, 0,... N=2$ case 2
$x[k]=..., 1, 1, 1, 1,... \longleftrightarrow D_{n}=..., 1, 1, 1, 1,... N=1$ case 3
I can't understand the last 2 lines. I think case 2 and case 3 should have
different sequences in the time domian. What did I miss?
