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Since the unit step $u[t]$ is not absolutely summable, it has no Fourier Transform.

In the DSP book (Proakis), the Fourier Transform of the unit step is formed by evaluating its $z$-Transform on the unit-circle except at $z=1$ where the pole is located.

The $z$-Transform of the unit step is given as $$ F(z)=\frac{z}{z-1} $$ evaluating it on the unit circle except at $z=1$ $$ F(\omega)=\frac{e^{\frac{j\omega}{2}}}{2j\sin(\frac{\omega}{2})} $$ (How?)

How did the author get that result? I know that we simply substitute $z=re^{j\omega}$ where $r=|z|=1$ and $j\omega=\angle z$. Then why did the author have $z=e^{\frac{j\omega}{2}}$?

How do you get the value of $\omega$ and $r$ if the $z$-transform is given and $z=1$ is excluded?

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As you have said, the unit-step $u[n]$ is neither absolutely, nor square summable, and thus, it does not have a convergent Fourier transform which can be obtained by evaluating its $z$-Transform $$U(z) = \frac{1}{1- z^{-1}} = \frac{z}{z- 1}$$ on the unit circle.

However, we also know that the Fourier transform of the unit-step is:

$$ U(\omega) = \frac{1}{1- e^{-j\omega}} + \pi \delta(\omega). \tag{1}$$

You can check the same book for its derivation.

I don't know what the author really wanted to show, but if you want to skip the impulse at the origin, and evaluate $U(z)$ on the unit circle except at $\omega = 0$ (which corresponds to $z=1$ on the z-plane) then you will get

$$U_0(\omega) = \frac{1}{1- e^{-j\omega}} \tag{2}$$

which is identical to $U(\omega)$, ignoring the impulse at $\omega = 0$.

It's quite easy to show that $U_0(\omega)$ is what you have posted as $F(\omega)$:

$$ \begin{align} U_0(\omega) &= \frac{1}{1- e^{-j\omega}} \\ \\ &= \frac{1}{ e^{-j \frac{\omega}{2}} (e^{j \frac{\omega}{2}}- e^{-j \frac{\omega}{2}})} \\\\ &= \frac{e^{j \frac{\omega}{2}}}{ (e^{j \frac{\omega}{2}}- e^{-j \frac{\omega}{2}})} \\\\ &= \frac{e^{j \frac{\omega}{2}}}{ 2 j \sin( \frac{\omega}{2}) } \tag{3}\\ \end{align} $$

Hope that the complex algebra is clear.

NOTE: This Fourier transform $U_0(\omega)$ can be considered as the effective frequency response of an accumulator (unstable with an impulse response $u[n]$) on an input signal $x[n]$ which does not have a DC component; i.e., $X(0)=0$, so that the impulse at $\omega=0$ of $U(\omega)$ is ineffective at the output, as shown:

$$ \begin{align} Y(\omega) &= X(\omega) U(\omega) \\\\ &= X(\omega) ( U_0(\omega) + \pi \delta(\omega) ) \\\\ &= X(\omega)U_0(\omega) + X(\omega) \pi \delta(\omega) \\\\ &= X(\omega)U_0(\omega) + \pi X(0) \delta(\omega) \\\\ &= X(\omega)U_0(\omega) + \pi 0 \delta(\omega) \\\\ &= X(\omega)U_0(\omega) \tag{4} \\\\ \end{align} $$

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    $\begingroup$ THIS SOLVES EVERYTHING, I CAN NOW SLEEP IN PEACE $\endgroup$
    – qcpz
    Oct 29 '21 at 11:56
  • $\begingroup$ @mkcpz ok then :-) $\endgroup$
    – Fat32
    Oct 29 '21 at 11:57

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