Discrete-Fourier transform of $$u[-n+2]$$

Question

I'm Trying to Find the fourier transform in discrete time for $$u[-n+2]$$ .

My steps :

Time-Reversal Property : $$ u[(-n+2)] \{\omega\} = u[-(-n+2)] \{-\omega\} = u[n-2] \{-\omega\} $$
Time-Shifting Property : $$ u[n-2] \{-\omega\} =e^{-(-2)j\omega} \cdot u[n] \{-\omega\} = e^{2j\omega} \cdot u[n] \{-\omega\}$$

I know That :

$$ u[n] \Rightarrow \frac{1}{1-e^{-j\omega}}$$

So by replacing the $\omega$ with $-\omega$ the answer would be :

$$e^{-2j\omega} \cdot \frac{1}{1-e^{j\omega}}$$

My Professor has given the answer as such :

$$ F\{u[-n+2]\}=F\{u[(-n-1)+3]\}=F\left\{z^3 \cdot Z\{u[-n-1]\}\right\}=-e^{+j 3 \omega} \frac{1}{1-e^{-j \omega}} $$

Which Does not seem to match my answer for some reason .

would appreciate any kind of help !

Matt L. · Accepted Answer · 2024-09-07 20:16:04Z

First of all, you're mixing $\mathcal{Z}$-transform and discrete-time Fourier transform (DTFT) here. Just because

$$\mathcal{Z}\big\{u[n]\big\}=\frac{1}{1-z^{-1}}\tag{1}$$

doesn't mean that

$$\textrm{DTFT}\big\{u[n]\big\}\stackrel{?}{=}\frac{1}{1-e^{-j\omega}}\tag{2}$$

The correct expression is

$$\textrm{DTFT}\big\{u[n]\big\}=\pi\delta(\omega)+\frac{1}{1-e^{-j\omega}}\tag{3}$$

[Cf. this question and its answers.]

Apart from that, you are right and your professor is wrong. Your professor's expression should look like this:

$$-e^{-3j\omega}\frac{1}{1-e^{-j\omega}}\tag{4}$$

(note the negative sign in the exponent), because he didn't replace $n$ by $n+3$ but by $n-3$, hence the multiplication with $e^{-3j\omega}$.

Now it's straightforward to show that his (corrected) solution and yours are identical:

$$-\frac{e^{-3j\omega}}{1-e^{-j\omega}}=-\frac{e^{-2j\omega}}{e^{j\omega}-1}=\frac{e^{-2j\omega}}{1-e^{j\omega}}\tag{5}$$

With $(3)$ we have

$$\textrm{DTFT}\big\{u[-n+2]\big\}=\pi\delta(\omega)+\frac{e^{-2j\omega}}{1-e^{j\omega}}\tag{6}$$

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