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# DSP MinimumPhase AllPass Decomposition Digital Signal Processing MinimumPhase AllPass Decomposition D

Richard Brown III D Richard Brown III 1 6 brPage 2br DSP MinimumPhase AllPass Decomposition MinimumPhase AllPass Decomposition Suppose we have a causal stable rational transfer function with one or more zeros outside the unit circle We denote the z

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## DSP MinimumPhase AllPass Decomposition Digital Signal Processing MinimumPhase AllPass Decomposition D

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## Presentation on theme: "DSP MinimumPhase AllPass Decomposition Digital Signal Processing MinimumPhase AllPass Decomposition D"— Presentation transcript:

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DSP: Minimum-Phase All-Pass Decomposition Digital Signal Processing Minimum-Phase All-Pass Decomposition D. Richard Brown III D. Richard Brown III 1 / 6
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DSP: Minimum-Phase All-Pass Decomposition Minimum-Phase All-Pass Decomposition Suppose we have a causal stable rational transfer function with one or more zeros outside the unit circle. We denote the zeros outside the unit circle as ,...,c We can form a minimum phase system with the same magnitude respo nse by reﬂecting these poles to their conjugate symmetric locatio ns inside the unit circle, i.e.,

min ) = {z unit-magnitude all-pass ﬁlter It should be clear that min and have the same magnitude response. Moreover, we have the decomposition ) = min min ap where ap is an all-pass ﬁlter. D. Richard Brown III 2 / 6
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DSP: Minimum-Phase All-Pass Decomposition Example 1 (part 1 of 2) Suppose ) = 1+ This system is clearly not minimum phase since it has a zero at = 2 . We can reﬂect this zero inside the unit circle with an all-pass ﬁlte r to write min ) = 1+ hence ) = 1+ {z min {z ap D. Richard Brown III 3 / 6
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DSP: Minimum-Phase All-Pass

Decomposition Example 1 (part 2 of 2) Note that, just requiring the zeros to be inside the unit circ le does not uniquely specify min . For example, we could also write ) = 1+ {z min {z ap where for both systems. Does it matter which one we pick? To satisfy the minimum phase delay property, recall that we r equire min ) = 0 . Note that min ) = 1+ and min ) = 1+ hence min ) = and min ) = 0 . The correct answer is to choose the min ap decomposition. D. Richard Brown III 4 / 6
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DSP: Minimum-Phase All-Pass Decomposition Example 2 Suppose ) = (1+3 )(1 (1+ This is also clearly not

minimum phase due to the zero at . We can reﬂect this zero inside the unit circle to write min ) = +3 1+3 +3)(1 (1+ 3(1 Recall that minimum phase ﬁlters must be causal. Since min ) = 3 is clearly not causal, we can factor out the denominator term (putting it in the all-pass ﬁlter since it does not aﬀect the magnitude response) to arrive at ) = 3 {z min +3 (1+3 {z ap D. Richard Brown III 5 / 6
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DSP: Minimum-Phase All-Pass Decomposition Equalization of Nonminimum Phase Channel Suppose ) = 4)( +5) +0 5)( 3) with ROC We form the inverse system ) = +0 5)(

3) 4)( +5) . Note that there is no causal stable inverse here. One approach in this case is to factor into a causal stable minimum phase ﬁlter and a causal stable allpass ﬁlter, i.e. ) = min ap ) = (4 1)(5 +1) +0 5)( 3) {z minimum phase 4)( +5) (4 1)(5 +1) {z allpass ROC : and invert just the minimum phase component, i.e. ) = +0 5)( 3) (4 1)(5 +1) Then = 1 but rather ) = ap . Hence, the equalizer corrects the magnitude distortion, but leaves some residual phase distortion. D. Richard Brown III 6 / 6