7 deleted 251 characters in body edited May 20 at 13:15 oliver 11855 bronze badges I hope that I have not misunderstood something terribly wrong, but the continuous derivative $$D=d/dt$$ can be considered a transfer function in Laplace space $$D(s) = s$$, right? So when I try to discretize it using the bilinear transform (Tustin's method) I trivially get $$D(z) = \frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}}$$ When I apply this to a series containing one discrete impulse, the response oscillates at about the Nyquist frequency and the numeric spectrum of the response seems to diverge at Nyquist (see image as an example of a 10kHz sampled impulse-signal of length 20 seconds being treated like that). Even worse, the spectrum around $$\omega=0$$ is quadratic and not $$\sim i\omega$$ like it would be expected from the derivative. So it can be safely claimed that D(zEdit: the latter was just due to roundoff error, because the low-freq amplitude got swamped by the Nyquist-peak) doesn't show any simple or obvious properties of a derivative. Although I know of course how well the bilinear transform works for discretizing all kinds of filters, it is hard for me to understand, why it is considered superior if it seems to fail so miserably for something as simple as the derivative, which can otherwise be easily represented by first order finite differences: $$D(z)=\frac{1-z^{-1}}{T}$$ or even second order (symmetric) finite differences $$D(z)=\frac{1-z^{-2}}{2Tz^{-1}}$$ I am sure it all has a very simple explanation, but I can't see it. PS: What is all the more confusing: when I apply the bilinear D(z) to a step function, the result is (correctly) a single peak. Consequently the inverse of the bilinear D(z) applied to an impulse yields the step function, like it has to be. What is going on there? I hope that I have not misunderstood something terribly wrong, but the continuous derivative $$D=d/dt$$ can be considered a transfer function in Laplace space $$D(s) = s$$, right? So when I try to discretize it using the bilinear transform (Tustin's method) I trivially get $$D(z) = \frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}}$$ When I apply this to a series containing one discrete impulse, the response oscillates at about the Nyquist frequency and the numeric spectrum of the response seems to diverge at Nyquist (see image as an example of a 10kHz sampled impulse-signal of length 20 seconds being treated like that). Even worse, the spectrum around $$\omega=0$$ is quadratic and not $$\sim i\omega$$ like it would be expected from the derivative. So it can be safely claimed that D(z) doesn't show any simple or obvious properties of a derivative. Although I know of course how well the bilinear transform works for discretizing all kinds of filters, it is hard for me to understand, why it is considered superior if it seems to fail so miserably for something as simple as the derivative, which can otherwise be easily represented by first order finite differences: $$D(z)=\frac{1-z^{-1}}{T}$$ or even second order (symmetric) finite differences $$D(z)=\frac{1-z^{-2}}{2Tz^{-1}}$$ I am sure it all has a very simple explanation, but I can't see it. PS: What is all the more confusing: when I apply the bilinear D(z) to a step function, the result is (correctly) a single peak. Consequently the inverse of the bilinear D(z) applied to an impulse yields the step function, like it has to be. What is going on there? I hope that I have not misunderstood something terribly wrong, but the continuous derivative $$D=d/dt$$ can be considered a transfer function in Laplace space $$D(s) = s$$, right? So when I try to discretize it using the bilinear transform (Tustin's method) I trivially get $$D(z) = \frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}}$$ When I apply this to a series containing one discrete impulse, the response oscillates at the Nyquist frequency. Even worse, the spectrum around $$\omega=0$$ is quadratic and not $$\sim i\omega$$ like it would be expected from the derivative. (Edit: the latter was just due to roundoff error, because the low-freq amplitude got swamped by the Nyquist-peak) Although I know of course how well the bilinear transform works for discretizing all kinds of filters, it is hard for me to understand, why it is considered superior if it seems to fail so miserably for something as simple as the derivative, which can otherwise be easily represented by first order finite differences: $$D(z)=\frac{1-z^{-1}}{T}$$ or even second order (symmetric) finite differences $$D(z)=\frac{1-z^{-2}}{2Tz^{-1}}$$ I am sure it all has a very simple explanation, but I can't see it. PS: What is all the more confusing: when I apply the bilinear D(z) to a step function, the result is (correctly) a single peak. Consequently the inverse of the bilinear D(z) applied to an impulse yields the step function, like it has to be. What is going on there? Became Hot Network Question occurred May 20 at 12:18 6 added 10 characters in body edited May 20 at 10:22 oliver 11855 bronze badges I hope that I have not misunderstood something terribly wrong, but the continuous derivative $$D=d/dt$$ can be considered a transfer function in Laplace space $$D(s) = s$$, right? So when I try to discretize it using the bilinear transform (Tustin's method) I trivially get $$D(z) = \frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}}$$ When I apply this to a series containing one discrete impulse, the response oscillates at about the Nyquist frequency and the numeric spectrum of the response seems to diverge at Nyquist (see image as an example of a 10kHz sampled impulse-signal of length 20 seconds being treated like that). Even worse, the spectrum around $$\omega=0$$ is quadratic and not $$\sim i\omega$$ like it would be expected from the derivative. So it can be safely claimed that D(z) doesn't show any simple or obvious properties of a derivative. Although I know of course how well the bilinear transform works for discretizing all kinds of filters, it is hard for me to understand, why it is considered so superior if it seems to fail so miserably for something as simple as the derivative, which can otherwise be easily represented by first order finite differences: $$D(z)=\frac{1-z^{-1}}{T}$$ or even second order (symmetric) finite differences $$D(z)=\frac{1-z^{-2}}{2Tz^{-1}}$$ I am sure it all has a very simple explanation, but I can't see it. PS: What is all the more confusing: when I apply the bilinear D(z) to a step function, the result is (correctly) a single peak. Consequently the inverse of the bilinear D(z) applied to an impulse yields the step function, like it has to be. What is going on there? I hope that I have not misunderstood something terribly wrong, but the continuous derivative $$D=d/dt$$ can be considered a transfer function in Laplace space $$D(s) = s$$, right? So when I try to discretize it using the bilinear transform (Tustin's method) I trivially get $$D(z) = \frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}}$$ When I apply this to a series containing one discrete impulse, the response oscillates at about the Nyquist frequency and the numeric spectrum of the response seems to diverge at Nyquist (see image as an example of a 10kHz sampled impulse-signal of length 20 seconds being treated like that). Although I know of course how well the bilinear transform works for discretizing all kinds of filters, it is hard for me to understand, why it is considered so superior if it seems to fail so miserably for something as simple as the derivative, which can otherwise be easily represented by first order finite differences: $$D(z)=\frac{1-z^{-1}}{T}$$ or even second order (symmetric) finite differences $$D(z)=\frac{1-z^{-2}}{2Tz^{-1}}$$ I am sure it all has a very simple explanation, but I can't see it. PS: What is all the more confusing: when I apply the bilinear D(z) to a step function, the result is (correctly) a single peak. Consequently the inverse of the bilinear D(z) applied to an impulse yields the step function, like it has to be. What is going on there? I hope that I have not misunderstood something terribly wrong, but the continuous derivative $$D=d/dt$$ can be considered a transfer function in Laplace space $$D(s) = s$$, right? So when I try to discretize it using the bilinear transform (Tustin's method) I trivially get $$D(z) = \frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}}$$ When I apply this to a series containing one discrete impulse, the response oscillates at about the Nyquist frequency and the numeric spectrum of the response seems to diverge at Nyquist (see image as an example of a 10kHz sampled impulse-signal of length 20 seconds being treated like that). Even worse, the spectrum around $$\omega=0$$ is quadratic and not $$\sim i\omega$$ like it would be expected from the derivative. So it can be safely claimed that D(z) doesn't show any simple or obvious properties of a derivative. Although I know of course how well the bilinear transform works for discretizing all kinds of filters, it is hard for me to understand, why it is considered superior if it seems to fail so miserably for something as simple as the derivative, which can otherwise be easily represented by first order finite differences: $$D(z)=\frac{1-z^{-1}}{T}$$ or even second order (symmetric) finite differences $$D(z)=\frac{1-z^{-2}}{2Tz^{-1}}$$ I am sure it all has a very simple explanation, but I can't see it. PS: What is all the more confusing: when I apply the bilinear D(z) to a step function, the result is (correctly) a single peak. Consequently the inverse of the bilinear D(z) applied to an impulse yields the step function, like it has to be. What is going on there? 5 added 10 characters in body edited May 20 at 10:15 oliver 11855 bronze badges I hope that I have not misunderstood something terribly wrong, but the continuous derivative $$D=d/dt$$ can be considered a transfer function in Laplace space $$D(s) = s$$, right? So when I try to discretize it using the bilinear transform (Tustin's method) I trivially get $$D(z) = \frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}}$$ When I apply this to a series containing one discrete impulse, the response oscillates at about the Nyquist frequency and the numeric spectrum of the response seems to diverge at Nyquist (see image as an example of a 10kHz sampled impulse-signal of length 20 seconds being treated like that). Although I know of course how well the bilinear transform works for discretizing all kinds of filters, it is hard for me to understand, why it is considered so superior if it seems to fail so miserably for something as simple as the derivative, which can otherwise be easily represented by first order finite differences: $$D(z)=\frac{1-z^{-1}}{T}$$ or even second order (symmetric) finite differences $$D(z)=\frac{1-z^{-2}}{2Tz^{-1}}$$ I am sure it all has a very simple explanation, but I can't see it. PS: What is all the more confusing: when I apply the bilinear D(z) to a step function, the result is (correctly) a single peak. Consequently the inverse of the bilinear D(z) applied to an impulse yields the step function, like it has to be. What is going on there? I hope that I have not misunderstood something terribly wrong, but the continuous derivative $$D=d/dt$$ can be considered a transfer function in Laplace space $$D(s) = s$$, right? So when I try to discretize it using the bilinear transform (Tustin's method) I get $$D(z) = \frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}}$$ When I apply this to a series containing one discrete impulse, the response oscillates at about the Nyquist frequency and the numeric spectrum of the response seems to diverge at Nyquist (see image as an example of a 10kHz sampled impulse-signal of length 20 seconds being treated like that). Although I know of course how well the bilinear transform works for discretizing all kinds of filters, it is hard for me to understand, why it is considered so superior if it seems to fail so miserably for something as simple as the derivative, which can otherwise be easily represented by first order finite differences: $$D(z)=\frac{1-z^{-1}}{T}$$ or even second order (symmetric) finite differences $$D(z)=\frac{1-z^{-2}}{2Tz^{-1}}$$ I am sure it all has a very simple explanation, but I can't see it. PS: What is all the more confusing: when I apply the bilinear D(z) to a step function, the result is (correctly) a single peak. Consequently the inverse of the bilinear D(z) applied to an impulse yields the step function, like it has to be. What is going on there? I hope that I have not misunderstood something terribly wrong, but the continuous derivative $$D=d/dt$$ can be considered a transfer function in Laplace space $$D(s) = s$$, right? So when I try to discretize it using the bilinear transform (Tustin's method) I trivially get $$D(z) = \frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}}$$ When I apply this to a series containing one discrete impulse, the response oscillates at about the Nyquist frequency and the numeric spectrum of the response seems to diverge at Nyquist (see image as an example of a 10kHz sampled impulse-signal of length 20 seconds being treated like that). Although I know of course how well the bilinear transform works for discretizing all kinds of filters, it is hard for me to understand, why it is considered so superior if it seems to fail so miserably for something as simple as the derivative, which can otherwise be easily represented by first order finite differences: $$D(z)=\frac{1-z^{-1}}{T}$$ or even second order (symmetric) finite differences $$D(z)=\frac{1-z^{-2}}{2Tz^{-1}}$$ I am sure it all has a very simple explanation, but I can't see it. PS: What is all the more confusing: when I apply the bilinear D(z) to a step function, the result is (correctly) a single peak. Consequently the inverse of the bilinear D(z) applied to an impulse yields the step function, like it has to be. What is going on there? 4 added 163 characters in body edited May 20 at 10:10 oliver 11855 bronze badges 3 added 163 characters in body edited May 20 at 10:04 oliver 11855 bronze badges 2 added 163 characters in body edited May 20 at 9:58 oliver 11855 bronze badges 1 asked May 20 at 9:50 oliver 11855 bronze badges