WAVELETS AND MULTIRATE DIGITAL SIGNAL PROCESSING

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WAVELETS AND MULTIRATE DIGITAL SIGNAL PROCESSINGLecture 7: Frequency domain behaviour of Haar filter banksProf.V. M. Gadre, EE, IIT Bombay1IntroductionSo far we have looked at the structure of the Haar Analysis and synthesis filter bank. In thislecture, the frequency domain behaviour of the Haar MRA filter banks is explored.2Haar filter banksFigure 1: Haar analysis filter bankFigure 2: Haar synthesis filter bankThe reason we use in the synthesis filter bank is due to a slight ambiguity to determinewhere to place sum sample and difference sample. If ‘ ’ sign in used, sum sample would getplaced at even location and difference sample at odd location. If ‘-’ sign is used, the reversewould happen.Important: Note the the analysis and synthesis filter banks are almost the same(except forthe scaling factor).Haar filter banks are not the ideal filter banks. It will soon be understood why. However,understanding Haar filter banks leads to clarification of many concepts of Multiresolutionanalysis.7-1

3Frequency domain behaviourThe frequency domain behaviour can be determined if we substitute z ejω . To do this , wemust first ensure that the unit circle lies within the Region of Convergence (ROC) the Ztransform.3.1Region of ConvergenceThe region of convergence of any Z-transform lies within two concentric circles of radius R1and R2 as shown in the figure 3.Figure 3: The ROC for any Z-transformIn general, R1 could be and R2 could be 0. The boundary circles may or may not be includedin the ROC. If the circle with radius 1, i.e unit circle, is included in the ROC, the system issaid to have a frequency response i.e., we say the sequence has a DTFT(Discrete time fouriertransform). To determine the frequency response, we substitute z ejω . Notice that z 1,i.e. we are evaluating the Z-transform on the unit circle.3.2Analysis filter bankSubstitute z ejω in1 z 1,2we getjω jω e 2 e1 e jω e 222 jωω2 e cos( )2 jω2(1) jωThe magnitude of this response is given by cos( ω2 ) (because e 2 1). Similarly the phaseresponse is given by ωas the cos( ω2 ) doesn’t contribute to phase as it is real and positive.2The graph of magnitude response is shown in the figure.Important: We plot only for positive ω noting that magnitude is a response is an evenfunction of ω and hence the complete spectrum will also involve a mirror image of the spectrumin figure about the Y-axis. Similarly, the phase response is an odd function of ω and thus H( Ω0 ) H(Ω0 ).We see that this response approximates a crude low pass filter. For comparison, the frequencyresponse of an ideal lowpass filter is shown in figure 5.If we plot the phase response of the above filter we will get a filter as shown in figure 6.We see that the phase response is a straight line passing through the origin.7-2

Figure 4: Magnitude response of filterFigure 5: Magnitude response of ideal lowpass filter3.2.1Importance of having a linear phaseIf we apply a signal A0 cos(Ω0 t φ0 ) to a system with frequency response given by H(Ω), thenthe output is given byOutput H(Ω0 ) A0 cos(Ω0 t φ0 H(Ω0 ))hi0) H(Ω0 ) A0 cos Ω0 (t H(Ω) φ0Ω0(2)where H(Ω0 ) is the angle introduced by system function H(Ω0 ). We see that this has resultedin a time shift in signal, which is dependent on signal frequency Ω0 . Thus here we see thatphase is a necessary evil. For without phase, the system would not be causal, and phaseintroduces a time shift, which is, in general, different for different frequencies. If we want topreserve shape of the waveform, we can at least try that all frequencies are shifted by the sametime i.e. H(Ω0 ) τ0 (independent of Ω0 )Ω0This implies that H(Ω0 ) Ω0 τ0This is an equation of a straight line passing through origin, hence called as linear phase.7-3

Figure 6: Phase response of ideal lowpass filter3.3Second filter in analysis filter bankSubstitute z ejω in the expression 12 (1 z 1 ), we get jω1ω(1 e jω ) je 2 sin( )22The magnitude and phase responses are shown in the figures 7 and 8 respectively.Figure 7: Magnitude response of second analysis filterNote that in calculating phase response we get an extra term of π2 due to presence of . Theexpression for phase response is thus given byπ ωphase(ω) 22It is thus seen that although the graph is still a straight line, it no longer passes throughthe origin. This phase response is thus called pseudo-linear response. To summarize, wewill again see the complete phase and magnitude responses of both filters. Figure 9 shows theresponse of the first filter and figure 10 refers to the response of the second filter. The phaseresponse is antisymmetric about origin because the expression is an odd function of ω(due topresence of sin( ω2 )).Note that in the the phase response of the second filter, at ω 0 has two values namely π2and π2 . However, since value of magnitude at ω 0 is zero, the phase doesn’t matter.If we add the functions of both filters, we get11(1 z 1 ) (1 z 1 ) 1227-4

Figure 8: Phase response of second analysis filterFigure 9: Magnitude and Phase response of first analysis filterThe meaning of this result is that if we pass a sine wave through both filters and add thefilter outputs, we get back the original sine wave.This property is called the Magnitudecomplimentarity property of the filtersIf we pass a wave of frequency ω0 through a filter of transfer function H(ω), the power outputis given by H(ω0 ) 2 . If we add the power outputs of both filters, we get cos(ω0 2ω0) sin( ) 2 122Thus, we see that even the sum of powers from both filters is conserved and and on additionof the filter outputs, we get the same power back. This property is known as the Powercomplimentarity property.7-5

Figure 10: Magnitude and Phase response of second analysis filter7-6

WAVELETS AND MULTIRATE DIGITAL SIGNAL PROCESSING Lecture 7: Frequency domain behaviour of Haar ﬁlter banks Prof.V. M. Gadre, EE, IIT Bombay 1 Introduction So far we have looked at the structure of the Haar Analysis and synthesis ﬁlter bank. In this lecture, the frequency domain beha

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