The diagram gives H(z)=β₀z⁻¹+β₁. For the positive-β₀,negative-β₁ high-pass choice, |H(π)|=β₀−β₁=1 and |H(0)|=β₀+β₁=.5. Adding and subtracting yields β₀=.75, β₁=−.25, A. The resulting high-frequency gain is twice its low-frequency gain.
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