Discrete Fourier Transform interpreted in terms of sampling period I understand that $$X(k) = \sum_{n=0}^{N-1}x_ne^{-i2\pi \frac{k}{N} n}$$ and $$x(n) = \frac{1}{N}\sum_{k=0}^{N-1}X_k e^{i2\pi \frac{k}{N} n}$$ are the discrete Fourier transform and inverse discrete Fourier transform. How do I translate/interpret/apply these formulas in terms of sampling period and frequency. For example, if I sample $11$ times so that $N=11$, and 1.5 seconds is the sampling period. Then from what little I understand, I should be able to write something that resembles sort of like, (ignoring the sine terms), $$y(t)=A_1\cos(2\pi ft) + A_2\cos(2\pi(2f)t)+\ldots$$ where $y(0),y(1.5),y(2*1.5),\ldots, y(10*1.5)$ take on the values of those $11$ sampled values and $t$ is time in seconds, $f$ is some lowest frequency needed in the representation.
Now somehow using the inverse discrete Fourier transform $x(n)$ I'm suppose to be able to get at $y(n)$. However, everything in $X(k)$ and $x(n)$ is based around integers (i.e. $n$ and $k$ are integers). Whereas $y(t)$ deals with values like $t=1.5$ and $f$ that are not integers. Can someone show the baby steps to get from $x(n)$ to $y(t)$.
 A: The discrete Fourier transform and its inverse do not include any specifications of the continuous-time signal (and its Fourier transform) from which the sample values are obtained.  The vector $(x(0), x(1), \ldots, x(N-1))$ can represent 
sample values spaced $T$ seconds apart for your choice of $T$ and thus 
the $y(t)$ that you seek cannot be determined until you specify $T$.
In your example, you chose $N=11$, $T = 1.5$ seconds and so you know the values of
$y(t)$ for $t = 0, 1.5, 3.0, \ldots, 15.0$.
Your signal $y(t)$ is unknown except for its values at $11$ points on the
time axis: $y(1.5n) = x(n), 0 \leq n < 11$. 
Anything more that could be said about $y(t)$ depends on what
assumptions you are willing to make.  One common assumption is
that (for the case when $N$ is odd)

$y(t)$ can be represented on the interval $[0, NT] = [0,16.5]$
  by the finite Fourier series
  $$y(t) = \sum_{k=-(N-1)/2}^{(N-1)/2} 
Y_k \exp(i 2\pi f_0 kt) $$
  where $f_0 = 1/NT$ Hz.  

Note that this means that $y(t)$ has no frequencies higher than
$((N-1)/2)f_0 < Nf_0/2 = (2T)^{-1}$, that is, half the sampling
rate $T^{-1}$.
Now, the Fourier coefficients cannot be computed in the
usual manner as 
$Y_k = \frac{1}{16.5}\int_{0}^{16.5}y(t)\exp(-i2\pi kf_0 t)
\,\mathrm dt$ 
because $y(t)$ is not known except at the sample points.
But, when $N = 11$,
$$\begin{align*}
y(nT) &= \sum_{k=-5}^{5} 
Y_k \exp(i2\pi kf_0 nT) = \sum_{k=-5}^{5} 
Y_k \exp(i 2\pi kn/11)\\
&= \sum_{k=0}^{5} Y_k \exp(i 2\pi kn/11) 
+ \sum_{k=6}^{10} Y_{k-11} \exp(i 2\pi kn/11)\\
\text{But,}\qquad \qquad
y(nT) &= x(n)\\
&= \frac{1}{N} \sum_{k=0}^{11} X_k \exp(i 2\pi kn/11)
\end{align*}$$
and so we have that 
$Y_k = \begin{cases}\frac{1}{N}X_k, & 0 \leq k \leq 5,\\
\frac{1}{N} X_{k+11}, & -5 \leq k < 0.\end{cases}$
So the Fourier coefficients can be determined from the
known values of the discrete Fourier transform.  Note that
I have discussed the case when $N$ is odd. When $N$ is
even, a slightly different calculation is used to
account for the relationship between $Y_{\pm N/2}$ and
$X_{N/2}$.
