How Do You Calculate Poles and Zeros of a Digital Filter?

[BriWel]

1. Given equation:
y(n) = x(n) + x(n - 2) - 0.81y(n-2)

a) Find the poles and zeros
b) Determine its z transfer function

2. Z transfer:
H(z) = (SUMOF(a(m)*z(^-m))) / (1 - SUMOF (b(m)*z (^-m)))



3. I get the feeling that this is pretty easy, I just don't know the required method to work it out. I know how to calculate poles and zeros of the z-transfer, but am not sure how to do it for the pre-transfer filter equation ( I assume ypu don't have to first carry out a z-transfer as this is the next part of the question)

I know the formula for z-transfer but don't have any experience in using it. I can't seem to find any suitable examples on line to help me either

Thanks in advance for any help

 

[Valkarie]

you can give.Answer: To find the poles and zeros of the pre-transfer filter equation, we use the characteristic equation method. The characteristic equation is defined as the equation obtained by setting the denominator of the transfer function (in this case, the equation y(n) = x(n) + x(n - 2) - 0.81y(n-2)) to 0. Thus, the characteristic equation for this equation is: 1 - 0.81z^(-2) = 0Solving this equation yields the roots z = 0.9 and z = -0.9, which are the poles. Since the equation is linear and time-invariant, it has no zeros.The Z-transfer function is then given by: H(z) = (x(n) + x(n - 2)) / (1 - 0.81z^(-2))which can be further simplified toH(z) = x(n) + 0.81z^(-2)x(n)

 

[Mene]

[/b]

I can understand your confusion about the poles and zeros of the given filter equation. To find the poles and zeros, we need to first convert the given equation into its z-transfer function. This can be done by taking the z-transform of both sides of the equation.

a) To find the poles and zeros, we need to determine the values of z for which the denominator of the z-transfer function becomes zero. These values are known as the poles of the filter. In this case, the denominator of the z-transfer function is 1 - z^-2, which becomes zero when z = ±1. Therefore, the poles of the filter are z = ±1.

Similarly, the values of z for which the numerator of the z-transfer function becomes zero are known as the zeros of the filter. In this case, the numerator is 1 + z^-2, which becomes zero when z = ±i. Therefore, the zeros of the filter are z = ±i.

b) The z-transfer function of the given filter is:

H(z) = (1 + z^-2) / (1 - z^-2)

This can be simplified to:

H(z) = (z^2 + 1) / (z^2 - 1)

Therefore, the z-transfer function of the given filter is (z^2 + 1) / (z^2 - 1).

To summarize, the poles of the filter are z = ±1 and the zeros are z = ±i. These values can help us understand the behavior of the filter and how it affects the input signal. I hope this explanation helps you understand the method for finding poles and zeros of a filter.