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

[anix]

Can someone prove to me how the backward-difference system is a high pass filter:

y[n] = x[n]-x[n-1]

 

[albbg]

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

 

[anix]

Thanks a LOT albbg for your help..i couldnt have figured it out that easily..

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

 

[zaxahmed]

It would be really helpful if you could show the maths that was involved while deriving the answer. I'm have a similar situation as the question asked and it will help me a lot if you could show the maths. Thank you for your time.

 

[albbg]

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)

 

[anix]

I am guessing we need to square the frequency response (H(jω)²) to get rid of the imaginary part of the e^jω ? Am I right?

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)

 

[albbg]

The frequency response is the square of the modulus |H(jω)|²; it is a vector of real numbers.

The square of a complex number is a complex number, in fact consider y = a + jb

y² = (a + jb)² = (a + jb)•(a + jb) = a² - b² +2jab

 

[anix]

Thank you again albbg