# Z-transform of the Unit Step and DTFT

In class we showed the the z transform of the unit step only exists for |z|>1 but we also calculated the DTFT of the unit step. Does convergence on the unit circle imply the DTFT exists but not the other way around? What are interesting cases of this relationship?

Assuming that the $$\mathcal{Z}$$-transform $$X(z)$$ of a sequence $$x[n]$$ exists, there are three cases we need to distinguish when considering the relation between $$X(z)$$ and the corresponding DTFT $$X_F(e^{j\omega})$$. Let the region of convergence (ROC) of $$X(z)$$ be $$r<|z| with $$0\le r:

1. $$r<1$$ and $$R>1$$, i.e., the unit circle is inside the ROC. In that case we simply obtain the DTFT from the $$\mathcal{Z}$$-transform by replacing $$z$$ by $$e^{j\omega}$$: $$X_F(e^{j\omega})=X(e^{j\omega})$$

2. $$r>1$$ or $$R<1$$, i.e., the unit circle is not inside the ROC. In this case, the DTFT doesn't exist.

3. $$r=1$$ or $$R=1$$, i.e., the ROC is limited by the unit circle, and there are $$K$$ simple poles on the unit circle: $$p_k=e^{j\omega_k}$$, $$k=0,1,\ldots K-1$$. In that case, the DTFT is given by

$$X_F(e^{j\omega})=\pi\sum_{k=0}^{K-1}a_k\delta(\omega-\omega_k)+X(e^{j\omega}),\qquad -\pi\le\omega<\pi\tag{1}$$

where $$a_k$$ are the residues of the poles.

The unit step sequence is an example of case $$3$$ above with just one simple pole at $$z=1$$ with residue $$a_0=1$$ (i.e., $$K=1$$ and $$\omega_0=0$$). Consequently, with

$$X(z)=\frac{1}{1-z^{-1}}$$

we obtain from $$(1)$$ for the DTFT of the unit step sequence

$$X_F(e^{j\omega})=\pi\delta(\omega)+\frac{1}{1-e^{-j\omega}},\qquad -\pi\le\omega<\pi\tag{2}$$

Equation $$(1)$$ can be generalized to poles on the unit circle with multiplicity greater than $$1$$. In that case, the DTFT contains derivatives of the Dirac impulse.

See this question and its answers for different ways to derive the DTFT of the unit step sequence.

EDIT (for Robert):

We know that the real part of the DTFT of a sequence equals the DTFT of its even part. The even part of the unit step sequence $$u[n]$$ is

$$\frac12\big(u[n]+u[-n]\big)=\frac12+\frac{\delta[n]}{2}\tag{3}$$

The DTFT of $$(3)$$ is $$\pi\delta(\omega) + \frac12$$. This must equal the real part of the DTFT given by Eq. $$(2)$$. To see this, let's rewrite $$(2)$$:

\begin{align*} \pi\delta(\omega)+\frac{1}{1-e^{-j\omega}} & = \pi\delta(\omega)+\frac12\cdot\frac{1+1}{1-e^{-j\omega}} \\ &= \pi\delta(\omega)+\frac12\cdot\frac{1-e^{-j\omega}+1+e^{-j\omega}}{1-e^{-j\omega}} \\ &= \pi\delta(\omega)+\frac12\left[1+\frac{1+e^{-j\omega}}{1-e^{-j\omega}}\right] \\ &= \pi\delta(\omega)+\frac12\left[1-j\cot\left(\frac{\omega}{2}\right)\right]\tag{4} \end{align*}

We see that the real part of $$(4)$$ equals $$\pi\delta(\omega)+\frac12$$, as expected.

As a side result we obtain the DTFT of the discrete-time signum sequence:

$$\operatorname{sgn}[n]=\begin{cases}1,&n>0\\0,&n=0\\-1,&n<0\end{cases}\tag{5}$$

The odd part of $$u[n]$$ is

$$\frac12\big(u[n]-u[-n]\big)=\frac12\operatorname{sgn}[n]\tag{6}$$

Its DTFT is $$j$$ times the imaginary part of $$(4)$$. Hence,

$$\operatorname{DTFT}\big\{\operatorname{sgn}[n]\big\}=-j\cot\left(\frac{\omega}{2}\right)\tag{7}$$

• Matt, what is the value of $u[0]$? Commented May 2 at 22:16
• @robertbristow-johnson: We're in discrete time, so the only definition I've ever come across is $u[0]=1$. Commented May 4 at 11:01
• That's normally the case for me. Now what happens when you add $u[n] + u[-n]$? Does that give you the result you want regarding the Fourier or Z transform $\mathscr{F}\{u[n]\} + \mathscr{F}\{u[-n]\}$ ? So $$u[n] + u[-n] = 1 + \delta[n]$$ where $\delta[n]$ is the Kronecker delta. Commented May 4 at 16:19
• @robertbristow-johnson: Tell me, what's bothering you? That's faster and easier than me answering your test questions :) Commented May 4 at 21:07
• @robertbristow-johnson: Check out my updated answer. Commented May 5 at 11:00