- 2080 Bhadra · 4 marks
Hadamard matrix H8
Short answer
Using Sylvester construction, H8 = [H4 H4; H4 -H4] in ±1 form, with H4 = [1 1 1 1; 1 -1 1 -1; 1 1 -1 -1; 1 -1 -1 1]. Rows are mutually orthogonal, so H8 H8^T = 8I.
Construct Hadamard matrix for H8.
A Hadamard matrix of order N is a square matrix whose rows are mutually orthogonal. It is constructed recursively by the Sylvester method:
H₂ₙ = [ Hₙ Hₙ ; Hₙ −Hₙ ] in ±1 form, or H₂ₙ = [ Hₙ Hₙ ; Hₙ Hₙ′ ] in 0/1 form, where Hₙ′ is the complement of Hₙ.
Step 1: H₁ = [+1]
Step 2:
H2 = | +1 +1 |
| +1 -1 |
Step 3:
H4 = | H2 H2 | = | +1 +1 +1 +1 |
| H2 -H2 | | +1 -1 +1 -1 |
| +1 +1 -1 -1 |
| +1 -1 -1 +1 |
Step 4:
H8 = | H4 H4 |
| H4 -H4 |
| +1 +1 +1 +1 +1 +1 +1 +1 |
| +1 -1 +1 -1 +1 -1 +1 -1 |
| +1 +1 -1 -1 +1 +1 -1 -1 |
| +1 -1 -1 +1 +1 -1 -1 +1 |
| +1 +1 +1 +1 -1 -1 -1 -1 |
| +1 -1 +1 -1 -1 +1 -1 +1 |
| +1 +1 -1 -1 -1 -1 +1 +1 |
| +1 -1 -1 +1 -1 +1 +1 -1 |
In binary form (+1 → 0, −1 → 1):
c1 c2 c3 c4 c5 c6 c7 c8
W0: 0 0 0 0 0 0 0 0
W1: 0 1 0 1 0 1 0 1
W2: 0 0 1 1 0 0 1 1
W3: 0 1 1 0 0 1 1 0
W4: 0 0 0 0 1 1 1 1
W5: 0 1 0 1 1 0 1 0
W6: 0 0 1 1 1 1 0 0
W7: 0 1 1 0 1 0 0 1
Check: the inner product of any two different rows is 0 (e.g. row 2 · row 3 = 1 − 1 − 1 + 1 + 1 − 1 − 1 + 1 = 0), and H8·H8ᵀ = 8·I. These 8 rows are the Walsh codes used for channelization in CDMA.
✓ Answer checked by Claude Opus 5.5 (high effort)
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