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[![enter image description here][1]][1]

For this problem: Actually,i think that the question asks us to get the IDFT from the DFT,so I did it..but I am not sure,till now, whether this problems asks us to do transformations or? enter image description here

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  • $\begingroup$ please take a higher-contrast picture that doesn't cut off formulas to the left, or just use the built-in LaTeX-formula typesetting to get the formulas into your question. Why did you strike out the first sentence from that book copy? If you have to do an exercise, say so, we don't mind, as long as you explain what you've tried and what your problem is (which you seem to do – just hard to read). Trying to hide something from the people you ask for help feels wrong, I hope. Please write a good title "question, not sure" is not a title for a question;what about "Constructing the IDFT from XYZ"? $\endgroup$ Dec 19 '16 at 0:58
  • $\begingroup$ I have editted the problem,sorry. it is the first time for me to type the answer in this forum. wouls you spend some time taking a lookin this question? $\endgroup$
    – jenn
    Dec 19 '16 at 1:45
  • $\begingroup$ Sure, but it was better with the calculations you made, because they would have explained what you've tried, it's just that they were unnecessarily hard to read due to bad lighting! $\endgroup$ Dec 19 '16 at 1:46
  • $\begingroup$ I posted my answer,wait $\endgroup$
    – jenn
    Dec 19 '16 at 1:47
  • $\begingroup$ It is my answer for this problems,but I think it isn't right.... $\endgroup$
    – jenn
    Dec 19 '16 at 1:52
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What you derive in your answer is that the iDFT of the DFT of a signal is the signal itself. The missing step is that $$ \sum_{k=0}^{N-1}\exp(j2\pi uk/N)=\sum_{m\in\mathbb{Z}}N\delta[u-mN].$$ Then, you get the identity $f[n] = \sum_m f[m]\delta[n-m]$.

However, this does not help for your problem. You need to express the IDFT in terms of DFT operations. So, you need to get

$$ X[m] = \frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}x[k]\exp(j2\pi km/N) $$

from a function, which can only calculate

$$ A[m] = \frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}a[k]\exp(-j2\pi km/N). $$

If you conjugate both sides of the first equation, you get:

$$ X^*[m]=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}x[k]^*\exp(-j2\pi km/N). $$

This equation is of the structure of the second equation, when you substitute $A[m]=X^*[m]$ and $a[k]=x^*[k]$. Hence, if you perform the DFT the conjugated signal, you get the conjugated of the IDFT of signal:

$$DFT(X^*)=IDFT(X)^*.$$

So, conjugating both sides yields $$DFT(X^*)^*=IDFT(X)$$ which is exactly what you need to do to solve your problem:

  1. Perform elementwise conjugation of the signal
  2. Perform DFT of the signal
  3. Perform elementwise conjugation of the result

Then, you get back the IDFT of the signal. It's analoguous in the 2D-case (i.e. images). Also, note that you might need to consider a different normalization factor (which would correspond to elementwise division by a scalar), depending on your definition of the DFT. If it is with $\frac{1}{\sqrt{N}}$, then normalization is OK as it is.

Edit: As Royi has pointed out, if your original signal is real, you can omit the 2nd conjugation (since the signal is already real after the DFT, and conjugation won't do anything). As images are most often real-valued, this simplication is very likely to hold here.

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  • $\begingroup$ actually (more precisely) $$\sum_{k=0}^{N-1}e^{j2\pi nk/N} = N \sum_{m=-\infty}^{\infty}\delta[n-mN]$$ you're including only the term on the right where $m=0$ and you're missing the scaler $N$ on the right. $\endgroup$ Dec 19 '16 at 7:21
  • $\begingroup$ Great answer. Just one note, if the signal is real then its DFT is Hermitian which means its Conjugate is also Hermitian hence applying DFT is Real. Namely for Real signal - $ DFT \left( {X}^{\ast} \right) = IDFT \left( X \right) $. $\endgroup$
    – Royi
    Dec 19 '16 at 7:39

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