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I got that system \$x_n \to x_n-x_{n-1}\$, so \$h_n=[.....,0,1,-1,0,...]\$, with \$h_0=1\$ and \$h_1=-1\$, so the transfer function given by:

$$H(\omega)=\sum_{i=-\infty}^{\infty} h_ne^{-jwn} = h_0e^{-jw(0)}+h_1e^{-jw(1)}=1-e^{-jw}$$. How to knwo the type of filter, if it is high pass filter, low pass filter, etc.

I appreciate your help.

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closed as off topic by clabacchio Apr 4 '13 at 11:28

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    \$\begingroup\$ As a first level of understanding, it should be obvious that if all the coefficients sum to 0, its gain at DC is zero, then it won't be a LPF. \$\endgroup\$ – Brian Drummond Apr 3 '13 at 8:52
  • \$\begingroup\$ What you mean by all coefficients sum to 0 \$\endgroup\$ – Sebastian Valencia Apr 3 '13 at 8:56
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    \$\begingroup\$ Simply add them. In your list above, I can see 0, -1, 1, 0 : the sum of these four is 0. If the same is true for the full set of coeffs, it isn't a LPF. It should be obvious why this is so. \$\endgroup\$ – Brian Drummond Apr 3 '13 at 9:01
  • \$\begingroup\$ Is that a theorem, i mean where can i read more about that, i'm just taking an Oppenheim level Signals course, so i do not know much about digital filters \$\endgroup\$ – Sebastian Valencia Apr 3 '13 at 10:53
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    \$\begingroup\$ @BrianDrummond Saying something is obvious is generally in poor form. I was once taught when you tell someone something is obvious either it is, and you are saying something useless, or it is not, and you are insulting them. \$\endgroup\$ – Kortuk Apr 3 '13 at 23:52
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A completely formal answer would be: given the z-transform of an impulse response $$h = (0, \ldots, 0, 1, -1, 0, \ldots, 0)$$ which is $$H(z) = \sum^{\infty}_{n=-\infty} h_n z^{-n} = h_0 + h_1 z^{-1}$$ we study the frequency response $$H(z=e^{j 2 \pi f T}) = h_0 + h_1 e^{-j 2 \pi f T} = h_0 + h_1 \cos(2 \pi f T) - j h_1 \sin(2 \pi f T)$$ Clearly, $$|H(z=e^{j 2 \pi f T})| = 0$$ is the same as $$|H(z=e^{j 2 \pi f T})|^2 = 0 \Leftrightarrow (h_0 +h_1 \cos(2 \pi f T))^2 + (h_1 \sin(2 \pi f T))^2 = 0$$ $$h^2_0 + h^2_1 (\cos(2 \pi f T)^2+ \sin(2 \pi f T)^2) + 2 h_0 h_1 \cos(2 \pi f T) = h^2_0 + h^2_1 + 2 h_0 h_1\cos(2 \pi f T) = 0$$ So the zeros of this transfer function are at $$f_0 = \frac{k}{2 \pi T} \text{arccos}\left(- \frac{h_0^2 h_1^2}{2 h_0 h_1}\right), k \in \mathbb{Z}$$ If $$h_0 = 1, h_1 = -1$$ then this is a digital differentiator. For $$f \in [0, 1/T] = [0, f_s]$$ the frequencies at which the amplitude response is 0 are $$f_0 = \frac{1}{2 \pi T} \text{arccos}(1/2)= \frac{k}{T}, k \in \mathbf{Z}$$ so practically $$f_0 = \{0, f_s\}$$ which makes this filter the simplest highpass response. As for the phase response it is simply $$ \angle H(z = e^{j 2 \pi f T}) = \frac{\pi}{2} - \pi f T$$ The dual case is the digital integrator, i.e. $$h_0 = h_1 = 1$$

Another way to derive the differentiatior response is $$H(z = e^{j 2 \pi f T}) = 1 - e^{- j 2 \pi f T} = (e^{j \pi f T} - e^{-j \pi f T} ) e^{-j \pi f T} = j 2 \sin(\pi f T) e^{-j \pi f T}$$ We can also define $$G(f) = |H(z = e^{j 2 \pi f T})| = 2 \sin(\pi f T)$$ with $$f \in [0, f_s]$$. It is even simpler to see that $$G(f) = 2 \sin(\pi f T) = 0 \Leftrightarrow f = \{0, f_s = \frac{1}{T}\}$$ In this way it is also easier to see that the phase response is $$\angle H(z = e^{j 2 \pi f T}) = \angle j + \angle G(f) - \pi f T = \frac{\pi}{2} + 0 - \pi f T $$

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One of the so many available solutions is that just expand the transfer function and write its real and imaginary parts separately. Then consider the magnitude of the complex term. Now find Magnitude by replacing 'w' with 0(zero) and '2pi'. Depending on the values obtained you can decide which filter it is.

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  • \$\begingroup\$ So if H(0)=0, H($\infty$)=1, thn is HPF? \$\endgroup\$ – Sebastian Valencia Apr 3 '13 at 10:44
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    \$\begingroup\$ Yes only if it is a first order filter which will be having one pole and one zero. But if it is a higher order filter then will have to examine the Magnitude value for more 'w's. \$\endgroup\$ – Durgaprasad Apr 3 '13 at 17:08

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