The Fourier transform of a constant exists. Can anyone please tell me what the $\mathcal{Z}$-transform of a constant is? Thanks in advance.


2 Answers 2


if you define the constant like this: $$ x[n] = C ~~~,~~~\text{ for all } n $$

Then its $\mathcal{Z}$ transform $$ X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} = \sum_{n=-\infty}^{\infty} C z^{-n} $$

does not converge for any value of $z$. Hence its $\mathcal{Z}$ transform does not exist.

Note that $\mathcal{Z}$ transform must be an analytic function with continuous derivatives of all orders unlike the DTFT which is not necesarily an analytic function and which would admit the use of impulses (or non-continuous functions) in its expressions. Then this would allow the DTFT of the constant $x[n] =C$ to be $X(e^{j\omega}) =C 2\pi \delta(\omega)$. But as stated, impulses (which are not analytic functions) are not allowed to exist in $\mathcal{Z}$ transforms.

Had you defined the constant as the practical signal like $$ y[n] = C u[n]$$ where $u[n] = \begin{cases} 1 &,& \text{ for } n \geq 0 \\ 0 &,& \text{ for } n < 0\\ \end{cases} $ is the unit step function, then its $\mathcal{Z}$-Transform would exist: $$ Y(z) = \sum_{n=-\infty}^{\infty} y[n]z^{-n} = \sum_{n=0}^{\infty} C z^{-n} = \frac{C}{ 1- z^{-1} } ~~~, ~~~ ROC~~ |z|>1$$

The distinction is clear I suppose.

  • $\begingroup$ A constant does have a DFT, wouldn't that imply that Z transform is defined on the circle? $\endgroup$
    – user28715
    Oct 28, 2017 at 13:49
  • $\begingroup$ @StanleyPawlukiewicz No it would not. Existance of Fourier transform does not imply existance of Z-transform, but the converse is true; i.e., existance of Z-transform (may) imply existing of Fourier transform (which is found by evaluating Z-transform on the unit circle) which requires that ROC includes unit circle. For example ideal lowpass filter $h[n] = \frac{\sin(w_c n)}{\pi n}$ does have a FT but not a ZT, just as $x[n]=\cos(w_0 n)$. So their DTFT expressions $X(e^{j\omega})$ do not result from an evaluation of their (nonexistant) ZTs on the unit circle. $\endgroup$
    – Fat32
    Oct 28, 2017 at 14:02
  • $\begingroup$ @Fat32 might add x(n)=C u(-n) as a mention $\endgroup$
    – user28715
    Oct 28, 2017 at 17:45

Instead of considering a constant signal, let us consider a rectangular window of length $2 \ell + 1$

$$w (n) = \begin{cases} 1 & \text{if } |n| \leq \ell\\ 0 & \text{otherwise} \end{cases}$$

whose $\mathcal Z$-transform is

$$W (z) = z^{-\ell} \left( 1 + z + z^2 + \cdots + z^{2 \ell} \right) = \dfrac{1}{z^\ell} \left(\dfrac{z^{2\ell+1} - 1}{z-1}\right)$$

which has a simple pole at $z = 1$, an $\ell$-th order pole at the origin, and $2 \ell + 1$ simple zeros on the unit circle at angles that are integer multiples of $ \frac{2 \pi}{2 \ell + 1}$. The pole at $z = 1$ cancels the zero at $z = 1$ and we are left with $2 \ell$ zeros on the unit circle. Note that

$$W (1) = 2 \ell + 1$$

As $\ell \to \infty$, the window $w (n)$ approaches a constant signal of unit amplitude, the unit circle of the complex plane gets "perforated" with more and more zeros, and $W (1) \to \infty$.


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