Express a signal by another signal?

I am having trouble understanding and getting the solution to this question:

Let $x : \mathbb Z \to \mathbb C$ denote a discrete-time sequence and for $N \in \mathbb N$ let $x |_N$ be the restrictin of $x$ to $\{0, \ldots, N-1\}$ i.e., the mapping $$x|_N : \{0, \ldots, N-1\} \to \mathbb C, \quad n \mapsto x[n].$$ Further, $X(e^{j\theta}) = \operatorname{DTFT} \{ x [n]\}$ denotes the DTFT of $x[n]$ and $X[k] = \operatorname{DFT}_N \{ x|_N[n] \}$ is the $N$-point DFT of $x|_N[n]$.

(a) Assume that the support* of the signal $x[n]$ is a subset of $\{ 0, \ldots, N-1 \}$ for some fixed $N$. Express the signal $x[n]$ by the signal $x|_N [n]$ (and some kind of zero-padding). Further, express the DFT $X[k]$ by the DTFT $X(e^{j \theta})$.

*The support of a function is the set of indices $n$ for which $x[n]$ is nonzero.

From what I understand, that is $x[n]=x|_N[n]$ for $n= 0,1, \ldots, N-1$. But there is $x[n] \in N$. So I have to define the value for $x[-1]$ or $x[N+1]$ and fill that up with zeros.

But I don't know if it is correct or how to write that correctly. Hope someone can help me with that.

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I don't like much the notation, but..

You have two functions, $x[n]$ is defined for every integer (it's the typical discrete time signal). And $x|_N[n]$ is only defined for $n=0 \cdots N-1$ (discrete and finite domain - sometimes it's seen more as a 'vector' than as a signal).

As we know that the second is the restriction of the first , and that $x[n]$ is zero outside that range (I don't get what you mean by 'there is $x[n] \in N$'), we can simply write

$$x[n] = \begin{cases} x|_N[n], & \text{if } n = 0\cdots N-1, \\ 0, & \text{otherwise}. \end{cases}$$

What it's left is the usual relation of the (finite) Discrete Fourier transform (takes a signal as vector and returns a finite set of transformed numbers $X(k)$ another vector) and the Discrete Time Fourier Transform (takes a discrete infinite signal and returns a continuous function in $[-\pi,\pi]$ : $X(e^{j\theta}$). In your case, you must just show that $X(k)$ equals $X(e^{j\theta})$ evaluated as some equispaced points - just a matter of writing boths Fourier formulas down and equating terms.

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Well, now i have the DFT and the DTFT formulas: $x[k] = \sum_{n=0}^{N-1} x[n]e^{-\frac{2\pi j}{N} k n}$ and $x[\omega] = \sum_{n=-\infty}^{\infty} x[n]e^{-j \omega n}$. Should i now calculate the values from 0 to N-1 and compare both vectors ? – madmax May 24 '11 at 21:51
There's no need to calculate, really. By inspection, comparing terms, you get the answer. – leonbloy May 24 '11 at 21:55