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Option 4 : | H(ω) | = constant; -∞ < ω < ∞
Pole zero plot is redrawn as
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As observed from the given pole-zero plot, poles & zeros are located symmetrically.
Therefore, it is all pass filter.
As we know that, from the property of all pass filter is,
- |H(jω)| = constant for all values of ω
- ∠H(jω) = Linearly vary with ω
Hence, the correct option is (4)
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