压缩感知课程课件

Lecture 6: Stable
1
Recovery
Comments on x∗ − x
2
¯ ˜ ≤ C · + C · k −1/2 x − xk
1.
1. Suppose
= 0; focus on the last term;
2
≤ .
Theorem
Assume that δ3k + 3δ4k < 2. The solution of the BPDN model returns a solution x ∗ satisfying x∗ − x
2
¯ ˜ ≤ C · + C · k −1/2 x −stants C and C . Proof sketch: Similar to the previous one, except x is no longer k-sparse and eS 1 > eS 1 is no longer valid. ¯
2
≤ .
Theorem
Assume that δ3k + 3δ4k < 2. The solution of the BPDN model returns a solution x ∗ satisfying x∗ − x
2
¯ ˜ ≤ C · + C · k −1/2 x − xk
1
¯ ˜ for some constants C and C . Proof sketch: Similar to the previous one, except x is no longer k-sparse and eS 1 > eS 1 is no longer valid. Instead, we get ¯ e S 1 + 2 x − xk 1 > e S 1 . ¯
Lecture 6: Stable
Consider A sparse x
1
Recovery
Noisy CS measurements b ← Ax + z, where z Apply the BPDN model: min x
1
2
≤
s.t. Ax − b
2
≤ .
Theorem
Assume that δ3k + 3δ4k < 2. The solution of the BPDN model returns a solution x ∗ satisfying x∗ − x 2 ≤ C · for some constant C . Proof sketch: Let e = x ∗ − x. S = {i : largest 4k |x|(i) }. One can show e
Consider A sparse x
1
Recovery
Noisy CS measurements b ← Ax + z, where z Apply the BPDN model: min x
1
2
≤
s.t. Ax − b
2
≤ .
Theorem
Assume that δ3k + 3δ4k < 2. The solution of the BPDN model returns a solution x ∗ satisfying x∗ − x 2 ≤ C · for some constant C .
Lecture 6: Stable
Consider A sparse x
1
Recovery
Noisy CS measurements b ← Ax + z, where z Apply the BPDN model: min x
1
2
≤
s.t. Ax − b
2
≤ .
Lecture 6: Stable
2
≤ C1 eS
2
≤ C2 Ae
2
≤C· .
Lecture 6: Stable
Consider A sparse x
1
Recovery
Noisy CS measurements b ← Ax + z, where z Apply the BPDN model: min x
1
2
≤
s.t. Ax − b
2
≤ C1 eS
2
≤ C2 Ae
2
≤C· .
1st inequality essentially from eS 1 > eS 1 , ¯ 2nd inequality essentially from the RIP; 3rd inequality essentially from the constraint.
Lecture 6: Stable
Consider
1
Recovery
a nearly sparse x = xk + w , xk is the vector x with all but the largest (in magnitude) k entries set to 0, CS measurements b ← Ax + z, where z
Lecture 6: Stable
1
Recovery, Comments
Theorem
Assume that δ3k + 3δ4k < 2. The solution of the BPDN model returns a solution x ∗ satisfying x∗ − x 2 ≤ C · for some constant C . Comments: The result is universal and more general than exact recovery; The error bound is order-optimal: knowing supp(x) will give C · at best;
2
≤ .
Theorem
Assume that δ3k + 3δ4k < 2. The solution of the BPDN model returns a solution x ∗ satisfying x∗ − x
2
¯ ˜ ≤ C · + C · k −1/2 x − xk
1
¯ ˜ for some constants C and C .
2
≤ .
Theorem
Assume that δ3k + 3δ4k < 2. The solution of the BPDN model returns a solution x ∗ satisfying x∗ − x 2 ≤ C · for some constant C . Proof sketch: Let e = x ∗ − x. S = {i : largest 4k |x|(i) }. One can show e
Lecture 6: Robust Compressive Sensing
In order to be practically powerful, CS must deal with Nearly sparse signals Measurement noise or both Goal: To obtain accurate reconstructions from highly undersampled measurements, or in short, stable recovery.
Lecture 6: Stable
1
Recovery, Comments
Theorem
Assume that δ3k + 3δ4k < 2. The solution of the BPDN model returns a solution x ∗ satisfying x∗ − x 2 ≤ C · for some constant C . Comments: The result is universal and more general than exact recovery; The error bound is order-optimal: knowing supp(x) will give C · at best; x ∗ is almost as good as if one knows where the largest k entries are and directly measure them.
2
≤ .
Theorem
Assume that δ3k + 3δ4k < 2. The solution of the BPDN model returns a solution x ∗ satisfying x∗ − x
2
¯ ˜ ≤ C · + C · k −1/2 x − xk
1
¯ ˜ for some constants C and C . Proof sketch: Similar to the previous one, except x is no longer k-sparse and eS 1 > eS 1 is no longer valid. Instead, we get ¯ eS 1 + 2 x − xk 1 > eS 1 . Then, ¯ e 2 ≤ C1 e4k 2 + C k −1/2 x − xk 1 , ......
Lecture 6: Stable
Consider
1
Recovery
a nearly sparse x = xk + w , xk is the vector x with all but the largest (in magnitude) k entries set to 0, CS measurements b ← Ax + z, where z