# Find Fourier Transform of Unit Step using the $z$-Transform [duplicate]

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?

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}

• THIS SOLVES EVERYTHING, I CAN NOW SLEEP IN PEACE
– Han
Oct 29, 2021 at 11:56
• @mkcpz ok then :-) Oct 29, 2021 at 11:57