next up previous print clean

Inverse Z-transform

Fourier analysis is widely used in mathematics, physics, and engineering as a Fourier integral transformation pair:
B(\omega)&=&\int^{+\infty}_{-\infty} b(t)\, e^{i\omega t}\, dt
\int^{+\infty}_{-\infty} B(\omega)\, e^{-i\omega t}\, d\omega\end{eqnarray} (26)
These integrals correspond to the sums we are working with here except for some minor details. Books in electrical engineering redefine $e^{i\omega t }$as $e^{-i\omega t }$.That is like switching $\omega$ to $-\omega$.Instead, we have chosen the sign convention of physics, which is better for wave-propagation studies (as explained in IEI). The infinite limits on the integrals result from expressing the Nyquist frequency in radians/second as $\pi/\Delta t$.Thus, as $\Delta t$ tends to zero, the Fourier sum tends to the integral. When we reach equation (31) we will see that if a scaling divisor of $2\pi$ is introduced into either (26) or (27), then b(t) will equal $\bar b (t)$.

The Z-transform is always easy to make, but the Fourier integral could be difficult to perform, which is paradoxical, because the transforms are really the same. To make a Z-transform, we merely attach powers of Z to successive data points. When we have B(Z), we can refer to it either as a time function or a frequency function. If we graph the polynomial coefficients, then it is a time function. It is a frequency function if we evaluate and graph the polynomial $B(Z = e^{i\omega})$ for various frequencies $\omega$.

If the Z-transform amounts to attaching powers of Z to successive points of a time function, then the inverse Z-transform must be merely identifying coefficients of various powers of Z with different points in time. How can this mere ``identification of coefficients'' be the same as the apparently more complicated operation of inverse Fourier integration? Let us see. The inverse Fourier integral (27) for integer values of time is  
b_t \eq {1 \over 2\pi} \int^{+\pi}_{-\pi} B(\omega)\, e^{-i\omega t}\,
 d\omega\end{displaymath} (28)
Substituting (21) into (28), we get  
b_t \eq {1 \over 2\pi} \int^{\pi}_{-\pi} (\cdots + b_{-1}e^{...
 ... + b_0
 + b_1 e^{+i\omega} + \cdots)\, e^{-i\omega t}\, d\omega\end{displaymath} (29)
Since sinusoids have as much area above the axis as below, the integration of $e^{in\omega}$over ${-}\pi \leq \omega < {+}\pi$gives zero unless n = 0, that is,
{1 \over 2\pi} \int^{\pi}_{-\pi} e^{in\omega} \, d\omega &= & {...
 ...\  0 & \mbox{if $n =$\space non-zero integer}
 \end{array} \right.\end{eqnarray}
Of all the terms in the integrand (29), we see from (30) that only the term with bt will contribute to the integral; all the rest oscillate and cancel. In other words, it is only the coefficient of Z to the zero power that contributes to the integral, so (29) reduces to  
b_t \eq {1 \over 2\pi} \int^{+\pi}_{-\pi} \ b_t \ e^{-i0} \ d\omega\end{displaymath} (31)
This shows how inverse Fourier transformation is just like identifying coefficients of powers of Z. It also shows why the scale factor in equation (28) is $2\pi$.


  1. Let B(Z) = 1 + Z + Z2 + Z3 + Z4. Graph the coefficients of B(Z) as a function of the powers of Z. Graph the coefficients of $\left[ B(Z) \right]^2$.
  2. As $\omega$ moves from zero to positive frequencies, where is Z and which way does it rotate around the unit circle, clockwise or counterclockwise?
  3. Identify locations on the unit circle of the following frequencies: (1) the zero frequency, (2) the Nyquist frequency, (3) negative frequencies, and (4) a frequency sampled at 10 points per wavelength.
  4. Given numerical constants $\Re Z_0$ and $\Im Z_0$,derive $\omega_0$ and $\rho$.
  5. Sketch the amplitude spectrum of Figure 9 from 0 to $4\pi$.

next up previous print clean
Stanford Exploration Project