Question

It is desired to extract a constant signal s(n)= s from the noisy measured signal x(n)= s(n)+v(n)= s + v(n), where v(n) is zero-mean white Gaussian noise of variance ϭv2. For that purpose, the following IIR lowpass filter is used: H(z)= b/(1- az-1), H(ejω)= b/(1- a e-jω) , |H(ejω)|2 = b2/(1- 2a cos(ω) + a2) where a and b are arbitrary constants (0 < a < 1). Estimate the value of the cut-off frequency ωc if a = 0.08.

Answer 1

Given that the IIR low pass filter

In terms of w,

The magnitude is expressed as,

The cut-off frequency is also called as half-power frequency, when we take square of the magnitude.

Thus, the cut-off frequency occurs when the power is half of its peak value. That is,

As varies from 0 to , the only term which affects the equation is and this term varies from -1 to +1.

When , we get maximum value because the affecting term has negative impact in the denominator. So, we get maximum value.

When , we get minimum value because the affecting term has positive impact in the denominator. So, we get minimum value.

Let us find the maximum and the minimum value for the magnitude,

The maximum value of occurs at and the minimum value of   occurs at .

Thus, the maximum value is

Put a = 0.08 in the above equation,

The minimum value is

Put a = 0.08 in the above equation,

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Let us find the cut-off frequency of half-power frequency

If we compare, the minimum value () reached by the filter and the cut-off frequency value (). The minimum value of the filter is greater than the cut-off frequency value. Hence, we don't get any cut-off frequency for the IIR lowpass filter used to remove the noise.

We can conclude that the used IIR lowpass filter is in fact an all pass filter. In case of digital filter, the cut-off frequency will be half of the sampling frequency.

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We may ask, when do we get a cut off frequency, for which, we need to take the ratio of minimum and maximum that falls below 0.5.

Taking square root on both the sides

Cross-multiplying

otherwise

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