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Okay. I think i got it after some clarification from Dan. Thanks Dan. ! I think what i was confused with was the step size in my frequency vector.. i had thought to have the final result in db/mhz.. the steps should be in 1 Mhz.. but i now realize that the freq vector could be in whatever steps i like but if its in Hz then i need to have my final answer ...

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You can't implement the transfer function H(z) directly, you need to convert it to a difference equation. However, the process is trivial, so once you understand it you'll see the connection between the diagram and transfer function better. First, we need to unroll the summation. For example, we get this with m=2, for a second-order equation: $H(z) = \... 1 To the extent you can factor the transfer function into individual integrator sections of the general form$\frac{1}{s}$you can make this substitution, which is an approximation of the Matched-$z$Transform where you substitute every$s$for$s=\frac{\ln(z)}{T}$. (map from$s$to$z$using$z =e^{sT}$). This results in first order forms given by$H_\mathrm{...

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It would help to apply some specificity to the general equation: $H(z) = \frac{b_0 + b_1z^{-1}+b_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}$ Rearrange into a difference equation, skipping the steps for brevity: $y[n] = b_0x[n] +b_1x[n-1]+b_2x[n-1]-a_1y[n-1]-a_2y[n-2]$ Does the difference equation form show now that the Direct Form structure is derived purely by ...

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Another way to see how the forward Euler method approximates a continuous-time system is by considering the "ideal" mapping of the $s$-plane to the $z$-plane (why?): $$z=e^{sT}\tag{1}$$ For frequencies that are much smaller than the sampling frequency (i.e., $|s|T\ll 1$) we can approximate $e^{sT}$ by its first order Taylor series: $$z\approx 1+sT\tag{2}$$...

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Multiplication with $s$ in the Laplace transform domain equals differentiation in the time domain. In the discrete-time domain we can approximate differentiation by the equation $$y[n]=\frac{x[n+1]-x[n]}{T}\tag{1}$$ where $T$ is the sampling interval. In the Z-transform domain, Eq. $(1)$ becomes $$Y(z)=X(z)\frac{z-1}{T}\tag{2}$$ I.e., the transfer ...

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Assuming that $$H(z) = A\frac{\prod_k (1-c_kz^{-1})}{\prod_l (1-d_l z^{-1})}, \: \: R_H$$ you can perform Partial Fraction Expansion (PFE) to quickly get your impulse response $h[n]$ (what you probably call anti-transformation) using Z-transform properties and tables of Z-transform pairs. If your transfer function is not rational, such as $$H(z) = \mathrm{... 1 First of all - I myself am not a pro in Control theory, but a mathematician - so I write what I think might be what your prof means. Part where i am quite certain: Not to be polynomial means, that there exists NO polynomial which describes the transfer function. E.g. let H be your transfer function, then there exists no n\in\mathbb{N} and p\in P^n[X] ... 0 Different approach: Let's start with rewriting using s = j \omega and RC = 1/\omega_0. We get$$H(\omega) = \frac{j\omega}{j\omega+\omega_0}$$Then we look at the most "relevant" frequency first. In this case that's \omega = \omega_0. At this frequency your magnitude is -3dB. For frequencies much larger then \omega_0 you simply have |H(\omega)|=1 ... 3$$H(s) = \frac{R C s}{R C s + 1}$$If R C s is much lower than 1 (i.e., R C s \ll 1, in Math), then you can make an approximation:$$H(s) \simeq \frac{R C s}{1} Basically, the effect of the $R C s$ in the denominator becomes insignificant. So, just pick a radian frequency well below $1 / R C$ and compute the magnitude of $H(s)$ at that point -- mark ...

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See the image (found in this article). Inject a signal into the loop somewhere (I show it below as going between the controller and the plant). Then measure the input to the plant and the output from the plant. Note that the image is a bit confusing -- it assumes that the normal command to the loop is 0 or some constant. It would be better to actually ...

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or the transfer function from the input 𝑢 to the output, how can I do this ? Is that not what your $G$ is? If you want to find the transfer function from just one element of $u$ to the output, then either delete the columns of $B$ that don't pertain to that element of $u$ and get your transfer function, or just look at the column of the transfer function ...

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