Determine the difference equation of the filter described by the following transfer function H(z):
H(z) = (0.065z^2 + 0.13z + 0.065) / (z^2 – 1.12z + 0.37)
b0 = 0.065
b1 = 0.130
b2 = 0.065
a1 = – 1.12
a2 = 0.37
The general form of a difference equation for a 2nd order filter is:
y[n] = b0*x[n] + b1*x[n-1] + b2*x[n-2] - a1*y[n-1] - a2*y[n-2]
Substitute the given coefficients into the general equation, we get:
y[n] = 0.065*x[n] + 0.13*x[n-1] + 0.065*x[n-2] + 1.12*y[n-1] - 0.37*y[n-2]
can you show me the steps of how you got: y[n] = 0.065*x[n] + 0.13*x[n-1] + 0.065*x[n-2] + 1.12*y[n-1] - 0.37*y[n-2]
To derive the specific difference equation using the provided filter transfer function H(z), follow these steps:
1) Express the transfer function H(z) as the ratio of two polynomials, one with the numerator coefficients (b0, b1, b2) and the other with the denominator coefficients (a1, a2):
H(z) = Y(z) / X(z) = (0.065z^2 + 0.13z + 0.065) / (z^2 - 1.12z + 0.37)
2) Rewrite the numerator and denominator polynomials in terms of the backward shift operator z^-1:
Y(z) = 0.065z^2 + 0.13z + 0.065
X(z) = z^2 - 1.12z + 0.37
3) Convert the polynomials Y(z) and X(z) into their corresponding time-domain sequences y[n] and x[n]:
y[n] = 0.065 * x[n] + 0.13 * x[n-1] + 0.065 * x[n-2]
4) Perform a similar conversion for the denominator polynomial to obtain the recursive terms involving y[n]:
x[n] = 1.12 * y[n-1] - 0.37 * y[n-2]
Combining these linear combinations gives the difference equation:
y[n] = 0.065 * x[n] + 0.13 * x[n-1] + 0.065 * x[n-2] + 1.12 * y[n-1] - 0.37 * y[n-2]
Therefore, the difference equation representing the filter described by H(z) is:
y[n] = 0.065 * x[n] + 0.13 * x[n-1] + 0.065 * x[n-2] - 1.12 * y[n-1] + 0.37 * y[n-2]