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I have a problem with these signals: $x[n]$ is given: $ x[n] = 3 \delta [n] - 2 \delta [n-1] + \delta [n-1] \text{, } n = 0,...,N \text{, N=8 }$
I computed the DFT and afterwards the 5 IDFT's of the resulting signal.
But now i have to explain what to the original signal x[n] happened in each of these cases. And i don't really know what to tell about these signals.

$y_4[n]$ looks like a shifted version of $x[n]$.
$y_3[n]$ looks the same like $y_5[n]$.
I don't know what else to tell...

$$ y_1[n] = IDFT_N \{|X[k]|^2\} $$ $$ y_2[n] = IDFT_N\{\Re X[k]\} $$ $$ y_3[n] = IDFT_N\{ X[(-k)\mod N]\} $$ $$ y_4[n] = IDFT_N\{(-1)^k X[k]\} $$ $$ y_5[n] = IDFT_N\{X^*[k]\} $$

Maybe someone can help me a bit... enter image description here

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This (I guess) is a question oriented towards teaching/exemplifing the basic properties of the Discrete Fourier transform (DFT) - which are mostly analogous from those of the continuous transform. If you already are supposed to know them (perhaps not), you should be able to identify (or predict) each inverse 'modified' transform, thinkig of it as the transform of a 'modified' sequence.

To give a basic example (simpler than any of your five cases). Given the pair $x[n] \leftrightarrow X[k]$, you probably know that if we multiply the signal by a constant $a$, the DTF results multiplied by the same constant. So that if you think about that modified transform pair $y[n] \leftrightarrow Y[k]$ with $y[n]= a \; x[n]$, you know that $Y[k] = a X[k]$, or more compactly : $a \; x[n] \leftrightarrow a \;X[k]$.

This means that, if somebody asks you to compute the IDFT of $a X[k]$ , you know what you are going to get: the same original signal multiplied by $a$

Now, other modifications of signal-transform pairs are less trivial than this (and the five cases are such). For example, take the last one. What transformation of a signal results in the transform being conjugated? It's easy to see that for the DTFT it's time reversal (see eg here), which for the DFT should correspond to some 'ciclical' time reversal. And if you look at the result, you see that the obtained signal is esentially the original reversed in time.

For the first case you recall $x[n] * y[n] \leftrightarrow X[k] Y[k]$ (convolution) , so $x[n] * x[-n] \leftrightarrow |X[k]|^2$ So that the modified signal is a (cyclic) convolution of the original with itself reversed.

For the second, recall that that the DFT is real if the signal is even (mod N). Decompose it in a a sum of even/odd signals, and you're done.

And so on.

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Thank you, this explanation helps a lot! But i am not sure with $y3$ and $y4$. For $y3$ i found this DTFT example $x^*[n]$. For $y4$ i didn't found a DTFT pair, i assume that it is $(-1)^n x[n]$. But i don't think it is correct because of the exponent $n$. –  madmax Jun 4 '11 at 19:41
    
And what happens to the signal when i double the $N$? I think X[k] gets a higher resolution and so x[n] gets filled up with zeros. –  madmax Jun 4 '11 at 19:46
    
I think that you should try to demonstrate each one of the properties -"guessed" from the results- working with the DTF formula, not just look them up in a table. For example (for the case 4), you see that the signal appears to be the original shifted by $N/2$. Then, ask yourself what is the DFT of $x[n-N/2]$ (assuming the DTF of $x[n]$ is $X[k]$), write down the formula and you'll get what you want. Same for the others. –  leonbloy Jun 4 '11 at 23:18
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