Finding whether a filter is low/high/band pass

  • Thread starter Thread starter t.kirschner99
  • Start date Start date
  • Tags Tags
    Filter
t.kirschner99
Messages
18
Reaction score
0

Homework Statement


Consider the IIR filter yn = xn - yn-2

State whether the filter is low, high, or band pass.

Homework Equations


The Z-transform: $$H(z) = \frac {1} {1+z^2},$$

Subbing ##z = e^{2πiw}## : $$H(w) = \frac {1} {1+e^{4πiw}},$$

Amplitude response: $$|H(w)| = \sqrt{{(\frac {1} {1 + cos{(4πw)}})}^2 + {(\frac {1} {sin{(4πw)}})}^2},$$

The Attempt at a Solution


I set up the domain ##w = [0,1]##

##w = 0##

Amplitude response would equal ##\frac {1} {2}##

##w = \frac {1} {2}##

Amplitude response would equal ##\frac {1} {2}##

##w = 1##

Amplitude response would equal ##\frac {1} {2}##

Doesn't this not prove anything? I know the answer is band pass, but I would expect zeroes on both ends and a number greater than zero in the middle.
 
Physics news on Phys.org
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top