0
$\begingroup$

I have a signal of the form $x^{*}(-n+2)$. To derive the signal's Fourier transform I use The following properties of DFT:

$x^{*}(n)=X^{*}(-\omega)$ and $x(n-k) = e^{-j\omega k}X(\omega)$

so the signal's DFT will have the form: $e^{j\omega2}X^{*}(\omega)$.

I am not sure if this derivation is correct or not.

I also get stuck with another question stating that whether we can prove this sequence does not satisfy the Dirichlet sufficient conditions:

$(1-\frac{|t|}{2014})(u(t+2014)-u(t-2014))$

can someone help me with this, please, much appreciated!

$\endgroup$
1
$\begingroup$

I'll help you out with your first question. The second question is totally unrelated, so it should be asked as a separate question.

First of all, it's important to distinguish between the discrete Fourier transform (DFT), and the discrete-time Fourier transform (DTFT). The first one transforms a finite number of (time-domain) samples to a finite number of (frequency domain) samples. So the independent variable is discrete in both domains. On the other hand, the DTFT transforms sequences of generally infinite length, and the frequency variable is continuous. From your example, I conclude that you're talking about the DTFT.

Let $y[n]=x^*[-n+2]$ with DTFT $Y(\omega)$. From the properties of the DTFT we know that $Y^*(\omega)$ corresponds to the sequence $y^*[-n]$. Consequently,

$$Y^*(\omega)\Longleftrightarrow x[n+2]\tag{1}$$

and

$$Y^*(\omega)e^{-2j\omega}\Longleftrightarrow x[n]\tag{2}$$

I.e.,

$$Y^*(\omega)e^{-2j\omega}=X(\omega)\tag{3}$$

and

$$Y(\omega)=\Big[X(\omega)e^{2j\omega}\Big]^*=X^*(\omega)e^{-2j\omega}\tag{4}$$

So you got the sign of the exponent wrong. This is a common problem with this type of exercises, so just work through them systematically, and cross-check your result.

$\endgroup$
1
  • $\begingroup$ Oh, my bad, I forgot to take conjugation for the exponential term. I will make a separate post for the second question. Thank you for your help! $\endgroup$ Aug 2 at 12:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.