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In this description of transfer functions on the z-plane (image linked), I'm confused by equation 1.49, which says that $H(f)=v_{out}(f)$ when $v_{in}(f)= 1 * e^{j 2 \pi (f/f_s)}$. (For another matter I'm confused that they're kind of conflating the frequency response and the Z-transform, but since the point of this section is evaluating the Z-transform on the unit circle to get the frequency response I guess I see what they're trying to say).

Equation 1.49 is confusing to me since it's saying that for this given input, the output is the same as the transfer function. But I don't see how this is the case here. I would see how this would hold for an input whose phasor is equal to 1 but I don't think that's what's going on here. My understanding of how the output of the system could be equal to the transfer function is that it's true for an input whose phasor is 1, like in this description:

Suppose I have an LTI system with freq. response $H(\omega)$. If I input a complex exponential signal $x(t) = e^{j \omega t}$, then the output is the same complex exponential but scaled by the complex frequency response, i.e. $y(t) = H(\omega)e^{j \omega t}$. Hence, the output $y(t)$ is not exactly equal to the input, but a scaled version of the input.

However, expressing the same thing in terms of phasors, then $x(t) = (1e^{j0}) e^{j \omega t}$, where the phasor of $x(t)$ is $X = 1e^{j0}$. This means that the phasor of $y(t)$ is $Y = H(\omega) = |H(\omega)| e^{ j\angle H(\omega)}$. Now you can say that the output phasor is equal to the transfer function, since the input phasor is equal to 1.

But in the example above, the input is not a phasor equal to one since it has nonzero phase.

Am I missing something here?

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  • $\begingroup$ It's just stating that if you stimulate the system at frequencies f of unit amplitude, then then its response is the frequency response. H = Y/X where X=1. $\endgroup$ – jaket Jan 28 at 8:39
  • $\begingroup$ @jaket I think I'm confused about how the input X is actually 1 in this case, since like the picture points out it has magnitude 1 but nonzero phase. $\endgroup$ – knzy Jan 28 at 22:51

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