If A=\(\frac{Ω_1 (Ω_u-Ω_l)}{-Ω_1^2+Ω_u Ω_l}\) and B=\(\frac{Ω_2 (Ω_u-Ω_l)}{-Ω_2^2+Ω_u Ω_l}\), then which of the following is the backward design equation for a low pass-to-band stop transformation?(a) ΩS=Max{|A|,|B|}(b) ΩS=Min{|A|,|B|}(c) ΩS=|B|(d) ΩS=|A|This question was addressed to me by my college professor while I was bunking the class.This key question is from Frequency Transformations in the Analog Domain topic in division Digital Filters Design of Digital Signal Processing

If A=\(\frac{Ω_1 (Ω_u-Ω_l)}{-Ω_1^2+Ω_u Ω_l}\) and B=\(\frac{Ω_2

1.	If A=\(\frac{Ω_1 (Ω_u-Ω_l)}{-Ω_1^2+Ω_u Ω_l}\) and B=\(\frac{Ω_2 (Ω_u-Ω_l)}{-Ω_2^2+Ω_u Ω_l}\), then which of the following is the backward design equation for a low pass-to-band stop transformation?(a) ΩS=Max{\|A\|,\|B\|}(b) ΩS=Min{\|A\|,\|B\|}(c) ΩS=\|B\|(d) ΩS=\|A\|This question was addressed to me by my college professor while I was bunking the class.This key question is from Frequency Transformations in the Analog Domain topic in division Digital Filters Design of Digital Signal Processing
Answer» The correct option is (B) ΩS=Min{\|A\|,\|B\|} The explanation: If Ωu and Ωl are the upper and lower cutoffpass BAND frequencies of the desired band stop filter and Ω1 and Ω2 are the lower and upper cutoffstop band frequencies of the desired band stop filter, then the BACKWARD design equation is ΩS= Min{\|A\|,\|B\|} where, =\(\FRAC{Ω_1 (Ω_u-Ω_l)}{-Ω_1^2+Ω_u Ω_l}\) and B=\(\frac{Ω_2 (Ω_u-Ω_l)}{-Ω_2^2+Ω_u Ω_l}\).

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