Z transform

Z transform can be considered as the discrete-time equivalent of the Laplace transform. If we sample a continuous signal \(x(t)\) with a discrete sampling rate \(\frac{1}{T}\), we can substitute the discrete signal \(x_d[n]\) by the underlying signal \(x_d(t)\) on continuous domain, \[ x_d(t) = \sum_{n=-\infty}^{\infty} x(nT) \delta(t - nT) \] Now we can take the Laplace transform on both sides with respect to time \(t\), \[ \mathcal{L}[x_d(t)] = \sum_{n=-\infty}^{\infty} x(nT) \, \mathcal{L}[\delta (t - nT)] = \sum_{n=-\infty}^{\infty} x(nT) e^{-nTs} = X(e^{Ts}) \] Let \(z=e^{Ts}\), thus the Z transform is given by, \[ \begin{equation} X_d(z) = X(e^{Ts}) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} \label{z-transform} = \mathcal{Z}\{x[n]\} \end{equation} \] where \(z=e^{Ts}\) is also a complex variable. If we let \(z=r \exp(i \omega)\) and let \(r=1\), the Z transform would degenerate to the discrete Fourier transform, \[ X(\omega) = \mathcal{F} \{ x[n] \} = \sum_{n=-\infty}^{\infty} x[n] \exp(-i \omega n) \label{Fourier} \] Thus, the discrete Fourier transform is a specialized Z transform which is evaluated on the unit circle in the complex \(z\) domain.

s-domainRe(s)Im(s)z=eTsz-domainRe(z)Im(z)

As we can see, the inner product of unit circles in vector space \(<e^{i\theta}, e^{i\theta} >_{\theta}\) is equal to \(2\pi\), thus normalization is required to ensure conservation.

Normalization of Fourier transform

The \(\sin(\theta)\) and \(\cos(\theta)\) functions and the corresponding exponential form \(e^{i \theta}\) have a periodicity of \(2\pi\), which makes \(e^{i \theta}\) as vector bases in function space are not orthonormal (the inner product \(< e^{i \theta}, e^{i \theta} > = \int_{-\pi}^{\pi} e^{i \theta} e^{-i \theta} \, d\theta= 2 \pi\)). So normalization is needed between forward and backward Fourier transform.

continuous

The function \(f(t)\) defined on \([-\frac{T}{2}, \frac{T}{2}]\) can be expressed by a set of bases in vector space of functions, \[ \begin{equation} f(t) = \sum_{\omega=-\infty}^{\infty} A_{\omega} e^{i \omega t} \label{vector_sum} \end{equation} \] where \(e^{i \omega t}\) is the vector basis and \(A_{\omega}\) is the projected length of \(f(t)\) on the basis \(e^{i \omega t}\), \[ \begin{equation} A_{\omega} = \frac{< f(t), e^{i \omega t} >}{< e^{i \omega t}, e^{i \omega t} >} \label{projection} \end{equation} \] The Fourier transform \(\widetilde{F}(\omega)\) is defined exactly as \(< f(t), e^{i \omega t} > = \int_{-\frac{T}{2}}^{\frac{T}{2}} f(t) e^{-i \omega t} \, dt\). However, bases \(e^{i \omega t}\) are not orthonormal, \[ < e^{i \omega t}, e^{i \omega t} > = \int_{-\frac{T}{2}}^{\frac{T}{2}} e^{i \omega t} e^{-i \omega t} \, dt = T = \frac{2\pi}{\delta \omega} \] where \(T\) is the periodicity of basis on \(t\)-domain and \(\delta \omega\) is the angular sampling frequency. Substituting \(\eqref{projection}\) into \(\eqref{vector_sum}\), we will find the normalization factor \(\frac{1}{2\pi}\) in the inverse Fourier transform, \[ \begin{align*} f(t) &= \sum_{\omega=-\infty}^{\infty} \lim_{T \to \infty} \frac{\color{blue}{\int_{-\frac{T}{2}}^{\frac{T}{2}} f(t') e^{-i \omega t'} \, dt'}}{T} e^{i \omega t} \\ &= \sum_{\omega=-\infty}^{\infty} \lim_{\delta \omega \to 0} \frac{\color{blue}{\widetilde{F}(\omega)}}{2\pi} \delta \omega e^{i \omega t} \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \widetilde{F}(\omega) e^{i \omega t} \, d\omega \\ \end{align*} \] So the normalization factor comes from the periodicity of vector bases that we choose. If we use \(e^{i 2 \pi \omega t}\) as bases, then the Fourier transform and inverse transform are normalized by \(1\), \[ \begin{align} \widetilde{F}(\omega) &= \int_{-\infty}^{\infty} f(t) e^{-i 2\pi \omega t} \, dt \\ f(t) &= \int_{-\infty}^{\infty} \widetilde{F}(\omega) e^{i 2\pi \omega t} \, d\omega \label{cft} \\ \end{align} \] where \(d\omega=\frac{1}{T}\) is sampling frequency here.

discrete

Supposing the \(N\)-element discrete signal on continuous domain is given by, \[ f_d(t) = \sum_{n=0}^{N-1} f(nT) \delta(t-nT) \] where \(T\) is the sampling interval and \(T = 1/N\) if we treat the signal as one cycle and make time unit. The Fourier transform is given by, \[ \mathcal{F}[f_d(t)] = \sum_{n=0}^{N-1} f(nT) \mathcal{F}[\delta(t-nT)] = \sum_{n=0}^{N-1} f(nT) e^{-i \omega nT} \] The \(N\)-element signal can be evaluated as one cycle at discrete angular frequency, \[ \omega = \frac{2 \pi}{N} k \] where \(k=0,1,...,N-1\). Thus the discrete Fourier transform is given by, \[ \widetilde{F}[k] = \sum_{n=0}^{N-1} f[n] e^{-i 2 \pi k n T / N} \] Similarly, the inner product of vector bases \(e^{i 2 \pi k n T / N}\) is \(N\), so the inverse discrete Fourier transform is, \[ \begin{align} f[n] &= \sum_{k=0}^{N-1} \color{blue}{\frac{< f[n'], e^{i 2 \pi k n' T / N} >}{< e^{i 2 \pi k n' T / N}, e^{i 2 \pi k n' T / N} >}} e^{i 2 \pi k n T / N} \\ &= \frac{1}{N} \sum_{k=0}^{N-1} \color{blue}{\widetilde{F}[k]} e^{i 2 \pi k n T / N} \label{dft} \\ \end{align} \] When \(N \to \infty\), Eq. \(\eqref{dft}\) will converge to Eq. \(\eqref{cft}\), in which \(d\omega = 1/N\).