Calculation of the frequency response by filter coefficients

Question

I can not find clear information on this topic. On Dutch wikipedia, I discovered that you can apply a z-transform that gives a formula in this form:
www.music.mcgill.ca/~gary/618/week1/img15.gif

This FIR filter is used as an example:
upload.wikimedia.org/math/b/9/e/b9e2ed5184f98621922f716e5216f33d.png

With z conversion:
upload.wikimedia.org/math/4/d/6/4d6621be8fabf4db8816c12f34ed9877.png

And in this example, e ^ it (the natural logarithm raised to an imaginary unit, and t = theta) is replaced by z: upload.wikimedia.org/math/0/6/e/06eada8fedfb492bd63bb50491b042aa.png

Then the graph of this function is used and is considered as a frequency response. I believe this method was a simple way to calculate the frequency response of a filter. However, is this method valid? When I thought about a small delay (which “blocks” the original signal), it occurred to me that the frequency response should be 1 for each frequency, since the signal does not change, it is only delayed, but using this method I calculated that the frequency response will be:

y(n) = 0*x(n) + 1*x(n-1)  

Z-Transformation

H(z) = 0 + 1z^-1

Substituting e ^ it (with t = theta):

H(e^it) = 0 + 1 * e^-it

Since this creates a sine wave as a frequency response, I have to do something wrong or misunderstand something. I would be very happy if someone could help me!

Answer 1

, H . , cos [ωn] = cos [2πfn], (f) cos [2πfn + Φ (f)], a (f) = | H (f) | Φ (f) = (H (f)). 1, , . -ω, ω - angular .

, Stack Overflow, , , , - .

h [n] = δ [n-1] ( δ [n] - -), , , 1 . , . 0,5 (.. 2 ) - , []. [1, -1,...]. 1, [-1,1,...], .. Cos [πn - π] = cos [π (n - 1)], .. , -π (-180 ). 0,25 (.. 4 ) - , [0.5πn]. [1,0, -1,0,...]. [0,1,0, -1,...], .. Cos [0,5πn - 0,5π] = cos [0,5π (n - 1)], .. - π/2 (-90 ). , , cos [0,25πn] cos [0,25πn - 0,25π] = cos [0,25π (n - 1)], .. , -π/4 (-45 ) .. ..

, angular ω (, 0,5π), Φ = -ω. , , , . angular 0,5π , 4 : 0, 0,5π, π, 1,5π. 0 . , -0,5π .

H (f), , , exp (-i2πf) = exp (-iω). , 2, h [n] = δ [n-2] H (f) = exp (-i4πf) = exp (-2iω) - ​​2 . , / , , .

FIR (.. , [MA]), , (, ) . IIR (.. , [AR]) , .

Answer 2

: , , , " ", FIR, IIR-.

. , :

• .
• .
• , N.
• D.
• N/D.

, .

Answer 3

Matlab:)

y(n) = 0*x(n) + 1*x(n-1)  

b=[ 0 1 ];
a = 1;
freqz(b,a)

Answer 4

. . . : (e ^ jw - e ^ -jw)/j2

, , , . , x [n] = cos (pi/3 * n). y [n] = H (e ^ jw) * x [n]

, , pi/3 - . cos (pi/3 * n) = (e ^ pi/3 * n + e ^ -pi/3 * n)/2. , pi/3, - -pi/3. : e ^ -j (pi/3) * e ^ (pi/3 * n) + e ^ j (pi/3) * e ^ (- pi/3 * n) 2 * cos (-pi/3 * n - pi/3). , .

In addition, there is nothing wrong with the fact that there is a sine wave as the frequency response.