Discrete Wavelet Transform - Comparison With Fourier Transform

Comparison With Fourier Transform

See also: Discrete Fourier transform

To illustrate the differences and similarities between the discrete wavelet transform with the discrete Fourier transform, consider the DWT and DFT of the following sequence: (1,0,0,0), a unit impulse.

The DFT has orthogonal basis (DFT matrix):

1 1 1 1 1 0 –1 0 0 1 0 –1 1 –1 1 –1

while the DWT with Haar wavelets for length 4 data has orthogonal basis in the rows of:

1 1 1 1 1 1 –1 –1 1 –1 0 0 0 0 1 –1

(To simplify notation, whole numbers are used, so the bases are orthogonal but not orthonormal.)

Preliminary observations include:

  • Wavelets have location – the (1,1,–1,–1) wavelet corresponds to “left side” versus “right side”, while the last two wavelets have support on the left side or the right side, and one is a translation of the other.
  • Sinusoidal waves do not have location – they spread across the whole space – but do have phase – the second and third waves are translations of each other, corresponding to being 90° out of phase, like cosine and sine, of which these are discrete versions.

Decomposing the sequence with respect to these bases yields:

\begin{align}
(1,0,0,0) &= \frac{1}{4}(1,1,1,1)+\frac{1}{4}(1,1,-1,-1)
+ \frac{1}{2}(1,-1,0,0)
\qquad\text{Haar DWT}\\
(1,0,0,0) &= \frac{1}{4}(1,1,1,1)
+ \frac{1}{2}(1,0,-1,0) + \frac{1}{4}(1,-1,1,-1)
\qquad\text{DFT}
\end{align}

The DWT demonstrates the localization: the (1,1,1,1) term gives the average signal value, the (1,1,–1,–1) places the signal in the left side of the domain, and the (1,–1,0,0) places it at the left side of the left side, and truncating at any stage yields a downsampled version of the signal:

\begin{align}
&\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)\\
&\left(\frac{1}{2},\frac{1}{2},0,0\right)\qquad\text{2-term truncation}\\
&\left(1,0,0,0\right)
\end{align}

The DFT, by contrast, expresses the sequence by the interference of waves of various frequencies – thus truncating the series yields a low-pass filtered version of the series:

\begin{align}
&\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)\\
&\left(\frac{3}{4},\frac{1}{4},-\frac{1}{4},\frac{1}{4}\right)\qquad\text{2-term truncation}\\
&\left(1,0,0,0\right)
\end{align}

Notably, the middle approximation (2-term) differs. From the frequency domain perspective, this is a better approximation, but from the time domain perspective it has drawbacks – it exhibits undershoot – one of the values is negative, though the original series is non-negative everywhere – and ringing, where the right side is non-zero, unlike in the wavelet transform. On the other hand, the Fourier approximation correctly shows a peak, and all points are within of their correct value, though all points have error. The wavelet approximation, by contrast, places a peak on the left half, but has no peak at the first point, and while it is exactly correct for half the values (reflecting location), it has an error of for the other values.

This illustrates the kinds of trade-offs between these transforms, and how in some respects the DWT provides preferable behavior, particularly for the modeling of transients.

Read more about this topic:  Discrete Wavelet Transform

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