Sampling, Z-Transform, Eigenfunctions, and the DTFT
10.2.1 Sampling, Z-Transform, Eigenfunctions, and the DTFT
The connection of the DTFT with sampling, eigenfunctions, and the Z-transform can be shown as follows:
Sampling and the DTFT. When sampling an analog signal x(t), the sampled signal x s ( t) can be written as
s ( t) =
x(nT s )δ( t − nT s )
Its Fourier transform is then
F [x s ( t)] =
x(nT s )F [δ(t − nT s ) ]
= X x(nT
s ) e −jnT s
Letting ω = T s , the discrete frequency in radians, the above equation can be written as
X s ( e jω
) = F[x s ( t)] =
x(nT s ) e −jnω
coinciding with the DTFT of the discrete-time signal x(nT s ) = x(t)| t =nT s or x[n]. At the same time, the spectrum of the sampled signal can be equally represented as
which is a periodic repetition, with period 2π/T s , of the spectrum of the analog signal being sampled. Thus, sampling converts a continuous-time signal into a discrete-time signal with a periodic spectrum varying continuously in frequency.
Z-transform and the DTFT. If in the above we ignore T s and consider x(nT s ) a function of n, we can see that
(10.5) That is, it is the Z-transform computed on the unit circle. For the above to happen, X(z) must
X s ( e jω ) = X(z)| z =e jω
have a region of convergence (ROC) that includes the unit circle. There are discrete-time sig- nals for which we cannot find their DTFTs from the Z-transform because they are not absolutely summable—that is, their ROCs do not include the unit circle. However, any discrete-time signal x[n], of finite support in time, has a Z-transform X(z) with a region of convergence the whole z-plane, excluding either the origin or infinity, and as such its DTFT X(e jω ) is computed from X(z) by letting z
=e jω .
C H A P T E R 10: Fourier Analysis of Discrete-Time Signals and Systems
Eigenfunctions and the DTFT. The frequency representation of a discrete-time linear time-invariant (LTI) system is shown to be the DTFT of the impulse response of the system. Indeed, according to the eigenfunction property of LTI systems, if the input of such a system is a complex exponential, x[n]
=e jω 0 n , the steady-state output, calculated with the convolution sum, is given by
X y[n] X = h[k]x[n
− k] = jω h[k]e jω 0 ( n −k) jω n =e 0 H(e 0 ) (10.6)
where
H(e jω 0 ) X
= k h[k]e −jω 0 (10.7)
or the DTFT of the impulse response h[n] of the system computed at ω =ω 0 . As with continuous- time systems, the system needs to be bounded-input bounded-output (BIBO) stable. Without the stability of the system, there is no guarantee that there will be a steady-state response.
■ Example 10.1
Consider the noncausal signal x[n] =α |n| with |α| < 1. Determine its DTFT. Use the obtained DTFT to find
The Z-transform of x[n] is
X(z) =
1 − α(z + z −1 ) +α 2 where the first term has an ROC of |z| > |α|, and the ROC of the second term is |z| < 1/|α|. Thus,
1 − αz
1 − αz
the region of convergence of X(z) is
1 ROC: |α| < |z| < |α|
and that includes the unit circle. Thus, the DTFT is
X(e jω
) − 2α cos(ω)
10.2 Discrete-Time Fourier Transform 575
According to the formula for the DTFT at ω = 0, we have that
x[n]e j0n =
and according to Equation (10.8), equivalently we have