[SOLVED] Backward-difference system is high pass filter?

albbg said:

Mathematically you can z-transform the sequence having:

y(z) = x(z) - z^-1*x(z) = x(z)*(1-z^-1)

since z^-1 = e^jω

H(jω) = 1 - e^jω

Then, after some math, we have the frequency response:

|H(jω)|² = 4*sin²(ω/2) defined in the rage [-Π,Π]

you can see easily that it is increasing with the frequency.
In a more intuitive way, at zero frequency x = x(n-1) ==> y = 0.
For low frequency the variation of x is slow so that x ≈ x(n-1) ==> y = small value
For high frequency the variation of x is fast so that x ≠ x(n-1) ==> y = high value

By the way x - x(n-1) is the discrete derivative of x

Click to expand...

albbg said:

OK, math is:

|H(jω)|² = |1 - e^jω|²

using Euler formula that is e^jω = cos(ω)+j•sin(ω) we have:

|1 - e^jω|² = |1 - cos(ω) - j•sin(ω)|² = [1-cos(ω)]² + sin²(ω) hence:

|1 - e^jω|² = 1 + cos²(ω) - 2•cos(ω) + sin²(ω) = 2•[1 - cos(ω)]

but, remembering that sin²(x) = 1/2 - cos(2•x)/2 we will have:

|1 - e^jω|² = 2•[1 - cos(ω)] = 2•[2•sin²(ω/2)] = 4•sin²(ω/2)

Click to expand...