ECE 6540, Lecture 19ece6540/slides/2016_ece6540_class019.pdf · 2016-04-01 · ECE 6540, Lecture 19...

ECE 6540, Lecture 19Matched Filters

Last Time Bayes Detection

MAP Detection

Maximum Likelihood Detection

Multiple Hypothesis Testing

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Multiple Hypothesis Testing(M-ary Hypothesis Testing)

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Multiple Hypothesis Testing The MAP Estimator is: Choose ℋ𝑘𝑘 if

𝑃𝑃 ℋ𝑘𝑘|𝒙𝒙 > 𝑃𝑃 ℋ𝑖𝑖|𝒙𝒙 for all 𝑖𝑖 ≠ 𝑘𝑘

Or𝑃𝑃 𝒙𝒙|ℋ𝑘𝑘

𝑃𝑃 𝒙𝒙|ℋ𝑖𝑖>𝑃𝑃 ℋ𝑖𝑖

𝑃𝑃 ℋ𝑘𝑘for all 𝑖𝑖 ≠ 𝑘𝑘

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Likelihood ratio

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

p(Li

kelih

ood

ratio

)

0

0.005

0.01

0.015

0.02

0.025

0.03

Multiple Hypothesis Testing More general Bayes Risk

𝑅𝑅 = 𝐸𝐸 𝐶𝐶 𝑖𝑖, 𝑗𝑗 = ∑𝑖𝑖=0𝑀𝑀−1∑𝑗𝑗=0𝑀𝑀−1𝐶𝐶𝑖𝑖𝑗𝑗𝑃𝑃 𝐻𝐻𝑖𝑖|𝐻𝐻𝑗𝑗 𝑃𝑃 𝐻𝐻𝑗𝑗 Use a minimize probability of error / MAP rule:

𝐶𝐶𝑖𝑖𝑗𝑗 = �0 for 𝑖𝑖 = 𝑗𝑗1 for 𝑖𝑖 ≠ 𝑗𝑗

After some derivations: The M-ary test is:

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𝑖𝑖 = arg max𝑖𝑖𝑃𝑃 ℋ𝑖𝑖|𝒙𝒙

Choose ℋ𝑖𝑖 if

Choose the hypothesis that maximizes the posterior probability

Detection Theory Consider the hypotheses 𝑥𝑥 = 𝜇𝜇 +𝑤𝑤 𝑤𝑤~𝒩𝒩 0,𝜎𝜎2

ℋ0 ∶ 𝜇𝜇 = 0 Pr ℋ0 = 𝑝𝑝0 ℋ1 ∶ 𝜇𝜇 = 𝜃𝜃 Pr ℋ1 = 1− 𝑝𝑝0

Determine the MAP detector

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Determine the MAP detector

𝑥𝑥 > lnp0

1− 𝑝𝑝0𝜎𝜎2/𝜃𝜃 +

𝜃𝜃2

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What is the probability of error?

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𝜃𝜃2



𝜃𝜃2

|ℋ0 𝑝𝑝0

𝑃𝑃𝑒𝑒 = Pr𝑥𝑥 − 𝐴𝐴𝜎𝜎2

< lnp0

1−𝑝𝑝0𝜎𝜎2

𝜃𝜃−

𝜃𝜃2 𝜎𝜎2

|ℋ1 1−𝑝𝑝0 + Pr𝑥𝑥𝜎𝜎2

> lnp0


𝜃𝜃+

𝜃𝜃2 𝜎𝜎2

|ℋ0 𝑝𝑝0

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𝑃𝑃𝑒𝑒 = 1 − 𝑄𝑄 lnp0

1 − 𝑝𝑝0𝜎𝜎2

𝜃𝜃−

𝜃𝜃2 𝜎𝜎2

1 − 𝑝𝑝0 + 𝑄𝑄 lnp0


𝜃𝜃+

𝜃𝜃2 𝜎𝜎2

𝑝𝑝0






𝜃𝜃2



𝜃𝜃2

|ℋ0 𝑝𝑝0

𝑃𝑃𝑒𝑒 = Pr𝑥𝑥 − 𝐴𝐴𝜎𝜎2

< lnp0


𝜃𝜃−

𝜃𝜃2 𝜎𝜎2

|ℋ1 1−𝑝𝑝0 + Pr𝑥𝑥𝜎𝜎2

> lnp0


𝜃𝜃+

𝜃𝜃2 𝜎𝜎2

|ℋ0 𝑝𝑝0

Question: What do we get when 𝑝𝑝0 = 1/2?

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𝑃𝑃𝑒𝑒 = 1 − 𝑄𝑄 lnp0


𝜃𝜃−

𝜃𝜃2 𝜎𝜎2

1 − 𝑝𝑝0 + 𝑄𝑄 lnp0


𝜃𝜃+

𝜃𝜃2 𝜎𝜎2

𝑝𝑝0




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𝑃𝑃𝑒𝑒 = 1 − 𝑄𝑄 ln 1𝜎𝜎2

𝜃𝜃−

𝜃𝜃2 𝜎𝜎2

1 − 𝑝𝑝0 + 𝑄𝑄 ln 1𝜎𝜎2

𝜃𝜃+

𝜃𝜃2 𝜎𝜎2

𝑝𝑝0

𝑃𝑃𝑒𝑒 = 1 − 𝑄𝑄−𝜃𝜃

2 𝜎𝜎21 − 𝑝𝑝0 + 𝑄𝑄

𝜃𝜃2 𝜎𝜎2

𝑝𝑝0

𝑃𝑃𝑒𝑒 = 𝑄𝑄𝜃𝜃

2 𝜎𝜎21 − 𝑝𝑝0 + 𝑄𝑄

𝜃𝜃2 𝜎𝜎2

𝑝𝑝0

𝑃𝑃𝑒𝑒 = 𝑄𝑄𝜃𝜃

2 𝜎𝜎2= 𝑄𝑄

𝜃𝜃2

4𝜎𝜎2= 𝑄𝑄

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𝑆𝑆𝑆𝑆𝑅𝑅




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𝑃𝑃𝑒𝑒 = 𝑄𝑄12


Matched FiltersDetection of deterministic signals

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Detection Theory Consider the hypotheses ℋ0 ∶ 𝒙𝒙 = 𝒘𝒘

ℋ1 ∶ 𝒙𝒙 = 𝒔𝒔+𝒘𝒘 𝒘𝒘~𝒩𝒩 𝟎𝟎,𝜎𝜎2𝑰𝑰

What is the Neyman-Pearson Detector?

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𝑝𝑝 𝒙𝒙|ℋ0 =1

2𝜋𝜋𝜎𝜎2 𝑁𝑁/2 exp −1

2𝜎𝜎2𝒙𝒙𝑇𝑇𝒙𝒙



2𝜎𝜎2𝒙𝒙 − 𝒔𝒔 𝑇𝑇 𝒙𝒙 − 𝒔𝒔

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𝑝𝑝 𝒙𝒙|ℋ1

𝑝𝑝 𝒙𝒙|ℋ0= exp −

12𝜎𝜎2

𝒙𝒙 − 𝒔𝒔 𝑇𝑇 𝒙𝒙 − 𝒔𝒔 − 𝒙𝒙𝑇𝑇𝒙𝒙

= exp −1

2𝜎𝜎2𝒙𝒙𝑇𝑇𝒙𝒙 − 2𝒙𝒙𝑇𝑇𝒔𝒔+ 𝒔𝒔𝑇𝑇𝒔𝒔 − 𝒙𝒙𝑇𝑇𝒙𝒙

= exp −1

2𝜎𝜎2−2𝒙𝒙𝑇𝑇𝒔𝒔+ 𝒔𝒔𝑇𝑇𝒔𝒔 > 𝛾𝛾

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12𝜎𝜎2

−2𝒙𝒙𝑇𝑇𝒔𝒔+ 𝒔𝒔𝑇𝑇𝒔𝒔 > 𝛾𝛾

2𝒙𝒙𝑇𝑇𝒔𝒔 − 𝒔𝒔𝑇𝑇𝒔𝒔 > ln 𝛾𝛾 2𝜎𝜎2

𝒙𝒙𝑇𝑇𝒔𝒔 > ln 𝛾𝛾 𝜎𝜎2 +12𝒔𝒔𝑇𝑇𝒔𝒔

𝒙𝒙𝑇𝑇𝒔𝒔 > 𝛾𝛾′

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Question: What is this? How does it work?

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Matched filter / correlator

Detection Theory Implementations of this detector


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×𝒙𝒙

𝒔𝒔

�𝒏𝒏=𝟏𝟏

𝑵𝑵

(�)> 𝜸𝜸

< 𝜸𝜸

ℋ1

ℋ0

𝒙𝒙 > 𝜸𝜸

< 𝜸𝜸

ℋ1

ℋ0ℎ 𝑛𝑛

FIR Filter

Sample when n = N-1(assuming h[n] is non-zero for n=0 to n=N-1)

Correlator implementation

Matched filter implementation



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×𝒙𝒙

𝒔𝒔

> 𝜸𝜸

< 𝜸𝜸

ℋ1

ℋ0

𝒙𝒙 > 𝜸𝜸

< 𝜸𝜸

ℋ1

ℋ0ℎ 𝑛𝑛

FIR Filter




How should we define the impulse response h[n]???

�𝒏𝒏=𝟏𝟏

𝑵𝑵

(�)



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×𝒙𝒙

𝒔𝒔

> 𝜸𝜸

< 𝜸𝜸

ℋ1

ℋ0

𝒙𝒙 > 𝜸𝜸

< 𝜸𝜸

ℋ1

ℋ0ℎ 𝑛𝑛

FIR Filter




ℎ 𝑛𝑛 = 𝑠𝑠 −𝑛𝑛

�𝒏𝒏=𝟏𝟏

𝑵𝑵

(�)



What is the threshold for a given probability of false alarm 𝛼𝛼?


26





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𝑃𝑃𝐹𝐹𝐹𝐹 = Pr 𝒙𝒙𝑇𝑇𝒔𝒔 > 𝛾𝛾′;ℋ0 = �𝛾𝛾′

∞𝑝𝑝 𝒙𝒙𝑇𝑇𝒔𝒔;ℋ0 𝑑𝑑𝑥𝑥 = 𝛼𝛼

What is the distribution of 𝒙𝒙𝑇𝑇𝒔𝒔?




28




𝒙𝒙𝑇𝑇𝒔𝒔 ~ 𝒩𝒩 0,𝜎𝜎2𝒔𝒔𝑇𝑇𝒔𝒔𝒙𝒙𝑇𝑇𝒔𝒔 ~ 𝒩𝒩 0,𝜎𝜎2 𝒔𝒔 2




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𝑃𝑃𝐹𝐹𝐹𝐹 = 𝑄𝑄𝛾𝛾′

𝜎𝜎2 𝒔𝒔 2= 𝛼𝛼

𝛾𝛾′ = 𝑄𝑄−1 𝛼𝛼 𝜎𝜎2 𝒔𝒔 2




𝒙𝒙𝑇𝑇𝒔𝒔 > 𝑄𝑄−1 𝛼𝛼 𝜎𝜎2 𝒔𝒔 2

Or

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𝒙𝒙𝑇𝑇𝒔𝒔𝜎𝜎 𝒔𝒔

> 𝑄𝑄−1 𝛼𝛼



What is the probability of detection?

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𝒙𝒙𝑇𝑇𝒔𝒔𝜎𝜎 𝒔𝒔





𝑃𝑃𝐷𝐷 = Pr 𝒙𝒙𝑇𝑇𝒔𝒔 > 𝛾𝛾′;ℋ1 = �𝛾𝛾′

∞𝑝𝑝 𝑥𝑥;ℋ1 𝑑𝑑𝑥𝑥 = 𝛼𝛼

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33


𝒙𝒙𝑇𝑇𝒔𝒔 ~ 𝒩𝒩 𝒔𝒔𝑇𝑇𝒔𝒔,𝜎𝜎2𝒔𝒔𝑇𝑇𝒔𝒔𝒙𝒙𝑇𝑇𝒔𝒔 ~ 𝒩𝒩 𝒔𝒔 2,𝜎𝜎2 𝒔𝒔 2






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𝑃𝑃𝐷𝐷 = 𝑄𝑄𝛾𝛾′ − 𝒔𝒔 2

𝜎𝜎2 𝒔𝒔 2= 𝑄𝑄

𝑄𝑄−1 𝛼𝛼 𝜎𝜎2 𝒔𝒔 2 − 𝒔𝒔 2

𝜎𝜎2 𝒔𝒔 2

= 𝑄𝑄 𝑄𝑄−1 𝛼𝛼 −𝒔𝒔 2

𝜎𝜎2 𝒔𝒔 2= 𝑄𝑄 𝑄𝑄−1 𝛼𝛼 −

𝒔𝒔𝜎𝜎





𝑃𝑃𝐷𝐷 = 𝑄𝑄 𝑄𝑄−1 𝛼𝛼 − 𝑆𝑆𝑆𝑆𝑅𝑅

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What is significant about SNR being here? What did we show in prior homework?

Generalized Matched FiltersDetection of deterministic signals

36


ℋ1 ∶ 𝒙𝒙 = 𝒔𝒔+𝒘𝒘 𝒘𝒘~𝒩𝒩 𝟎𝟎,𝑪𝑪

What is the Neyman Pearson Detector?

37





2𝜋𝜋 𝑁𝑁/2 𝑪𝑪 1/2 exp −1

2𝜎𝜎2𝒙𝒙𝑇𝑇𝑪𝑪−1𝒙𝒙


2𝜋𝜋 𝑁𝑁/2 𝑪𝑪 1/2 exp −1

2𝜎𝜎2𝒙𝒙 − 𝒔𝒔 𝑇𝑇𝑪𝑪−1 𝒙𝒙 − 𝒔𝒔

38






12

𝒙𝒙− 𝒔𝒔 𝑇𝑇𝑪𝑪−1 𝒙𝒙 − 𝒔𝒔 − 𝒙𝒙𝑇𝑇𝑪𝑪−1𝒙𝒙

= exp −12𝒙𝒙𝑇𝑇𝑪𝑪−1𝒙𝒙 − 2𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔+ 𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔 − 𝒙𝒙𝑇𝑇𝑪𝑪−1𝒙𝒙

= exp −12−2𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔+ 𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔 > 𝛾𝛾

39






12−2𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔+ 𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔 > 𝛾𝛾

2𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔 − 𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔 > ln 𝛾𝛾 2

𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔 > ln 𝛾𝛾 +12𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔

𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔 > 𝛾𝛾′

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𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔 > 𝛾𝛾′

Question: What is this? How does it work?

41





42




43

𝑃𝑃𝐹𝐹𝐹𝐹 = Pr 𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔 > 𝛾𝛾′;ℋ0 = �𝛾𝛾′

∞𝑝𝑝 𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔;ℋ0 𝑑𝑑𝑥𝑥 = 𝛼𝛼

What is the distribution of 𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔?




44




𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔 ~ 𝒩𝒩 0, 𝒔𝒔𝑇𝑇𝑪𝑪−1𝑪𝑪𝑪𝑪−1𝒔𝒔𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔 ~ 𝒩𝒩 0, 𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔




45




𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔 ~ 𝒩𝒩 0, 𝒔𝒔𝑇𝑇𝑫𝑫𝑇𝑇𝑫𝑫 𝑫𝑫𝑇𝑇𝑫𝑫 −1𝑫𝑫𝑇𝑇𝑫𝑫𝒔𝒔𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔 ~ 𝒩𝒩 0, 𝒔𝒔′𝑇𝑇𝑫𝑫 𝑫𝑫𝑇𝑇𝑫𝑫 −1𝑫𝑫𝑇𝑇𝒔𝒔′

𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔 ~ 𝒩𝒩 0, 𝒔𝒔′𝑇𝑇𝑰𝑰𝒔𝒔′

𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔 ~ 𝒩𝒩 0, 𝒔𝒔′𝑇𝑇𝒔𝒔′




46



𝑃𝑃𝐹𝐹𝐹𝐹 = 𝑄𝑄𝛾𝛾′

𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔= 𝛼𝛼

𝛾𝛾′ = 𝑄𝑄−1 𝛼𝛼 𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔




𝒙𝒙𝑇𝑇𝒔𝒔 > 𝑄𝑄−1 𝛼𝛼 𝒔𝒔𝑇𝑇𝑪𝑪𝒔𝒔

Or

47

𝒙𝒙𝑇𝑇𝒔𝒔𝒔𝒔𝑇𝑇𝑪𝑪𝒔𝒔





48

𝒙𝒙𝑇𝑇𝒔𝒔𝒔𝒔𝑇𝑇𝑪𝑪𝒔𝒔





𝑃𝑃𝐷𝐷 = Pr 𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔 > 𝛾𝛾′;ℋ1 = �𝛾𝛾′


49





𝑃𝑃𝐷𝐷 = Pr 𝒙𝒙𝑇𝑇𝑪𝑪−1𝒔𝒔 > 𝛾𝛾′;ℋ1 = �𝛾𝛾′


50


𝒙𝒙𝑇𝑇𝒔𝒔 ~ 𝒩𝒩 𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔, 𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔






51

𝑃𝑃𝐷𝐷 = 𝑄𝑄𝛾𝛾′ − 𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔

𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔= 𝑄𝑄

𝑄𝑄−1 𝛼𝛼 𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔 − 𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔

= 𝑄𝑄 𝑄𝑄−1 𝛼𝛼 −𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔𝒔𝒔𝑇𝑇𝑪𝑪−1𝒔𝒔





𝑃𝑃𝐷𝐷 = 𝑄𝑄 𝑄𝑄−1 𝛼𝛼 − 𝑆𝑆𝑆𝑆𝑅𝑅

52

Multiple SignalsBinary Case

53

Detection Theory Consider the hypotheses ℋ0 ∶ 𝒙𝒙 = 𝒔𝒔0 +𝒘𝒘 𝑝𝑝 ℋ0 = 1/2

ℋ1 ∶ 𝒙𝒙 = 𝒔𝒔1 +𝒘𝒘 𝑝𝑝 ℋ1 = 1/2

𝒘𝒘~𝒩𝒩 𝟎𝟎, 𝑰𝑰𝜎𝜎2

What is the Likelihood Ratio Detector?

54


ℋ1 ∶ 𝒙𝒙 = 𝒔𝒔1 +𝒘𝒘 𝑝𝑝 ℋ1 = 1/2





2𝜎𝜎2𝒙𝒙 − 𝒔𝒔0 𝑇𝑇 𝒙𝒙 − 𝒔𝒔0



2𝜎𝜎2𝒙𝒙 − 𝒔𝒔1 𝑇𝑇 𝒙𝒙 − 𝒔𝒔1

55


ℋ1 ∶ 𝒙𝒙 = 𝒔𝒔1 +𝒘𝒘 𝑝𝑝 ℋ1 = 1/2


What is the Likelihood Ratio Detector? 𝑝𝑝 𝒙𝒙|ℋ1


12𝜎𝜎2

𝒙𝒙 − 𝒔𝒔1 𝑇𝑇 𝒙𝒙 − 𝒔𝒔1 +1

2𝜎𝜎2𝒙𝒙 − 𝒔𝒔0 𝑇𝑇 𝒙𝒙 − 𝒔𝒔0 > 1

− 𝒙𝒙− 𝒔𝒔1 𝑇𝑇 𝒙𝒙 − 𝒔𝒔1 + 𝒙𝒙− 𝒔𝒔0 𝑇𝑇 𝒙𝒙 − 𝒔𝒔0 > 0𝒙𝒙− 𝒔𝒔1 𝑇𝑇 𝒙𝒙 − 𝒔𝒔1 < 𝒙𝒙− 𝒔𝒔0 𝑇𝑇 𝒙𝒙 − 𝒔𝒔0

𝒙𝒙 − 𝒔𝒔1 2 < 𝒙𝒙− 𝒔𝒔0 2

56


ℋ1 ∶ 𝒙𝒙 = 𝒔𝒔1 +𝒘𝒘 𝑝𝑝 ℋ1 = 1/2



Choose the hypothesis ℋ𝑖𝑖 for which the data 𝒙𝒙 is closest (in terms of squared error) to the expected signal 𝒔𝒔𝑖𝑖

57

Minimum distance receiver𝒙𝒙 − 𝒔𝒔1 2 < 𝒙𝒙 − 𝒔𝒔0 2

𝐷𝐷1 < 𝐷𝐷0

𝑖𝑖 = arg min𝑖𝑖

𝒙𝒙 − 𝒔𝒔𝑖𝑖 2


ℋ1 ∶ 𝒙𝒙 = 𝒔𝒔1 +𝒘𝒘 𝑝𝑝 ℋ1 = 1/2



58




ℋ1 ∶ 𝒙𝒙 = 𝒔𝒔1 +𝒘𝒘 𝑝𝑝 ℋ1 = 1/2



Choose the hypothesis ℋ𝑖𝑖 for which the data 𝒙𝒙 is closest (in terms of squared error) to the expected signal 𝒔𝒔𝑖𝑖

59




𝒙𝒙 − 𝒔𝒔𝑖𝑖 2 How is this relate to our NP matched filter?


ℋ1 ∶ 𝒙𝒙 = 𝒔𝒔1 +𝒘𝒘 𝑝𝑝 ℋ1 = 1/2



60


𝒙𝒙 − 𝒔𝒔𝑖𝑖 2 = arg min𝑖𝑖𝒙𝒙𝑇𝑇𝒙𝒙 − 2𝒙𝒙𝑇𝑇𝒔𝒔𝑖𝑖 + 𝒔𝒔𝑖𝑖𝑇𝑇𝒔𝒔𝑖𝑖

= arg min𝑖𝑖−2𝒙𝒙𝑇𝑇𝒔𝒔𝑖𝑖 + 𝒔𝒔𝑖𝑖𝑇𝑇𝒔𝒔𝑖𝑖

= arg max𝑖𝑖𝒙𝒙𝑇𝑇𝒔𝒔𝑖𝑖 −

12

𝒔𝒔𝑖𝑖 2

Matched filterSignal energy


ℋ1 ∶ 𝒙𝒙 = 𝒔𝒔1 +𝒘𝒘 𝑝𝑝 ℋ1 = 1/2



Question: What is the solution when both signal have equal energy?

61


𝒙𝒙 − 𝒔𝒔𝑖𝑖 2 = arg min𝑖𝑖𝒙𝒙𝑇𝑇𝒙𝒙 − 2𝒙𝒙𝑇𝑇𝒔𝒔𝑖𝑖 + 𝒔𝒔𝑖𝑖𝑇𝑇𝒔𝒔𝑖𝑖

= arg min𝑖𝑖−2𝒙𝒙𝑇𝑇𝒔𝒔𝑖𝑖 + 𝒔𝒔𝑖𝑖𝑇𝑇𝒔𝒔𝑖𝑖

= arg max𝑖𝑖𝒙𝒙𝑇𝑇𝒔𝒔𝑖𝑖 −

12

𝒔𝒔𝑖𝑖 2

Matched filterSignal energy


62

×

𝒙𝒙 𝒔𝒔0 ℋ𝑖𝑖

𝑖𝑖 = arg max𝑖𝑖𝒙𝒙𝑇𝑇𝒔𝒔𝑖𝑖 −

12

𝒔𝒔𝑖𝑖 2

×

𝒔𝒔1

Find max

12

𝒔𝒔0 2

12

𝒔𝒔1 2

-

-

�𝒏𝒏=𝟏𝟏

𝑵𝑵

(�)

�𝒏𝒏=𝟏𝟏

𝑵𝑵

(�)

Multiple SignalsM-ary Case

63

Detection Theory Consider the hypotheses ℋ0 ∶ 𝒙𝒙 = 𝒔𝒔0 +𝒘𝒘 𝑝𝑝 ℋ0 = 1/𝑀𝑀

ℋ1 ∶ 𝒙𝒙 = 𝒔𝒔1 +𝒘𝒘 𝑝𝑝 ℋ1 = 1/𝑀𝑀

ℋ2 ∶ 𝒙𝒙 = 𝒔𝒔2 +𝒘𝒘 𝑝𝑝 ℋ2 = 1/𝑀𝑀

⋮

ℋ𝑀𝑀 ∶ 𝒙𝒙 = 𝒔𝒔𝑀𝑀 +𝒘𝒘 𝑝𝑝 ℋ𝑀𝑀 = 1/𝑀𝑀



64


⋮




𝑝𝑝 𝒙𝒙|ℋ𝑖𝑖 =1


2𝜎𝜎2𝒙𝒙 − 𝒔𝒔𝑖𝑖 𝑇𝑇 𝒙𝒙 − 𝒔𝒔𝑖𝑖

65


⋮




66

𝒙𝒙 − 𝒔𝒔0 2 < 𝒙𝒙 − 𝒔𝒔1 2

𝒙𝒙 − 𝒔𝒔0 2 < 𝒙𝒙 − 𝒔𝒔2 2

To choose 𝓗𝓗𝟎𝟎

𝒙𝒙 − 𝒔𝒔0 2 < 𝒙𝒙 − 𝒔𝒔𝑀𝑀 2⋮

𝒙𝒙 − 𝒔𝒔1 2 < 𝒙𝒙 − 𝒔𝒔0 2

𝒙𝒙 − 𝒔𝒔1 2 < 𝒙𝒙 − 𝒔𝒔2 2

To choose 𝓗𝓗𝟏𝟏


𝒙𝒙 − 𝒔𝒔2 2 < 𝒙𝒙 − 𝒔𝒔0 2

𝒙𝒙 − 𝒔𝒔2 2 < 𝒙𝒙 − 𝒔𝒔1 2

To choose 𝓗𝓗𝟐𝟐



⋮




67

𝒙𝒙 − 𝒔𝒔𝑖𝑖 2 < 𝒙𝒙 − 𝒔𝒔𝑗𝑗2

𝐷𝐷𝑖𝑖 < 𝐷𝐷𝑗𝑗

To choose 𝓗𝓗𝒊𝒊

For all 𝑖𝑖 ≠ 𝑗𝑗 Minimum distance receiver


⋮




68






For all 𝑖𝑖 ≠ 𝑗𝑗 Minimum distance receiver


⋮




69






For all 𝑖𝑖 ≠ 𝑗𝑗


12

𝒔𝒔𝑖𝑖 2

Minimum distance receiver


70


12

𝒔𝒔𝑖𝑖 2

×

𝒙𝒙 𝒔𝒔0

ℋ𝑖𝑖×

𝒔𝒔1

Find max

12

𝒔𝒔0 2

12

𝒔𝒔1 2

-

-

×

𝒔𝒔𝑀𝑀12

𝒔𝒔𝑀𝑀 2-

⋮

�𝒏𝒏=𝟏𝟏

𝑵𝑵

(�)

�𝒏𝒏=𝟏𝟏

𝑵𝑵

(�)

�𝒏𝒏=𝟏𝟏

𝑵𝑵

(�)

ECE 6540, Lecture 19ece6540/slides/2016_ece6540_class019.pdf · 2016-04-01 · ECE 6540, Lecture 19...

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Transcript of ECE 6540, Lecture 19ece6540/slides/2016_ece6540_class019.pdf · 2016-04-01 · ECE 6540, Lecture 19...