Question

What is the factor to be multiplied to the dc gain of the filter to obtain filter magnitude at cutoff frequency?

(a) 1

(b) √2

(c) 1/√2

(d) 1/2

I got this question in an online interview.

I would like to ask this question from Butterworth Filters in chapter Digital Filters Design of Digital Signal Processing

Answer 1

Correct answer is (c) 1/√2

The best I can explain: The dc gain of the filter is the filter magnitude at Ω=0.

We know that the filter magnitude is given by the equation

|H(jΩ)|=\(\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}\)

At Ω=ΩC, |H(jΩC)|=1/√2=1/√2(|H(jΩ)|)

Thus the filter magnitude at the cutoff frequency is 1/√2 times the dc gain.