# Fourier Transform Notation in Signal Processing

I have a question about the notation adopted for the Fourier Transform of both continuous-time signals and discrete-time sequences in Signal Processing.

In the textbook I am using (Signals and Systems by Alan Oppenheim), the Fourier Transform of the signal $x(t)$ is denoted by $X(j\omega)$ and is defined as $$X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \mathrm{d}t.$$ Its discrete-time counterpart (for a sequence $x[n]$ is denoted by $X(e^{j\omega})$ and is given by $$X(e^{j\omega}) = \sum_{n = -\infty}^{\infty} x[n] e^{-j\omega n}.$$

By virtue of the inherent periodicity in frequency of discrete-time periodic signals, the discrete-time Fourier Transform is periodic, but the continuous-time Fourier Transform is not periodic in general. Now, although we denoted the continuous-time Fourier Transform by $X(j\omega)$, couldn't have we just as well used $X(e^{j\omega})$? Except, the problem with this notation is that it implies that $X(e^{j(\omega + 2\pi)}) = X(e^{j\omega})$, which is not true in the continuous-time case. I know that the notation doesn't work, but could someone explain to me mathematically why it is invalid to do so?

Apologies for using $j$ to denote the imaginary unit.

## 1 Answer

The continuous time Fourier Transform's result can be thought of as having its independent variable being on the $j\omega$ axis of the s-plane. (The same s-plane that the Laplace transform maps to.)

The discrete time Fourier Transform's result can be thought of as having its independent variable being on the unit circle of the z-plane. (The same z-plane that the Z Transform maps to.)

Hence the difference in notation of the independent variable.

The $e^{j\omega}$ notation indicates the discrete time Fourier transform result wraps around the unit circle and appears periodic when unwrapped.

The continuous time Fourier transform maps to an infinitely long axis and doesn't wrap around and is not periodic. Hence the $j\omega$ notation for the independent variable.

• I understand, but why is it wrong to define $H(j\omega) = G(e^{j \omega})$ for the continuous-time Fourier Transorm? Why does it yield a wrong result? – 0MW Feb 7 '18 at 1:41
• The discrete time input $x[n]$ is inherently bandlimited by virtue of being sampled at a certain rate. It's transform can never have frequencies outside of $(-F_s/2, F_s/2]$. The continuous time signal is not necessarily bandlimited and can contain frequencies in $(-\infty, \infty)$. – Andy Walls Feb 7 '18 at 1:49
• I guess the short answer is "because sampling", but I suppose Shannon's or Nyquist's papers on the subject of sampling give much better answers than that. – Andy Walls Feb 7 '18 at 1:51