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The transfer function is $H(s) = \frac{16.94s + 579.5}{s^2 + 507.2s + 1224}$ This transfer function has 2 poles, one slow pole at -2.4248 and a fast pole at -504.7752. The function has a slowish zero at -34.2. Good news, your poles and zero are all in the left-half plane. It is much easier to control a system with zeroes and poles in the left-half plane ...


3

Two suggestions to move forward: Reduce $K_i$ to the point of an acceptable overshoot (this will provide the bottom line answer for comparison to the computations. Do system identification (Bode plots) on the open loop system and individual components to isolate the difference between implementation and loop model; the computations were compared below ...


2

You can split up the real and imaginary part of the state into their own seperate states. Namely by defining $x_r=\mathrm{Re}(x)$, $x_i=\mathrm{Im}(x)$, $A_r=\mathrm{Re}(A)$, $A_i=\mathrm{Im}(A)$, $u_r=\mathrm{Re}(u)$, $u_i=\mathrm{Im}(u)$, $B_r=\mathrm{Re}(B)$ and $B_i=\mathrm{Im}(B)$ then the differential equation can also be written as $$ \dot{x}_r+i\,\...


2

Where are the poles of your filter? If they are close to the unit circle, then quantizing them can cause some of them to cross beyond the unit circle, causing the quantized filter to be unstable. You should check the poles of your filter before quantizing and after quantizing. See how close they match. Secondly, it's risky, albeit not impossible, to ...


2

Your system H(s) is continuous $$ H(s) = \frac{s}{s+1} $$ Using a discrete impulse does not make a lot of sense. I suspect Simulink added a "zero-order hold" or ZOH block between your discrete impulse and your system $H(s)$. As such, your impulse response will be affected by the sampling period of your discrete impulse source. If you decrease the sampling ...


2

If the filter has a fixed order then increasing the sampling rate will make its response scale with frequency, thus the transition width increases relative to the sampling frequency. For example: if $f_s=1\;\mathrm{kHz}$ and $\omega_{tw}=10\;\mathrm{Hz}$ for a given order then keeping the order and increasing $f_s$ tenfold will make $\omega_{tw}$ increase ...


2

1 There's a mistake in the PID connection. You must feed the quadrature component, i.e $U_q$ to the PID, not $U_d$. The setpoint of your PLL is $U_q = 0$ because you want your PLL to be in phase with your 3-phase input i.e. $U_d = 1, U_q = 0$. 2 - Perhaps there are hidden delays in the block you instantiated ? 3 - Notice the error equation of your PLL is ...


1

The problem is that you can't use a Dirac delta impulse as an input, which would be the theoretically correct thing to do. Since you only have a discrete impulse, which has some finite value, and since Simulink probably uses something like a zero-order hold to make a continuous-time signal out of that impulse you actually use two different input signals: a ...


1

However, when I add an AWGN channel, I get pretty low BER of 10^-5 in passband as compared to around 10^-4 BER in baseband under same AWGN channel of SNR 10 dB. Why is it happening? You've got a bug in either passband channel or baseband channel model, or in how you transform your noise or your signal – end of story! I thought that there would be more ...


1

Hi I am confused about the implementation of AGC in Simulink example. I am unable to relate the given implementation to the algorithm given by MATLAB. Kindly help me in understanding the reason behind the difference in algorithm and implementation. Best Regards Sunny


1

From the diagram in the Algorithms section of the documentation you can see how the different quantities are computed: Note that $z$ in the diagram is an estimate of the output power.$^1$ The error signal $e$ is computed by comparing the reference value $A$ to $\ln(z)$. So if you choose $$A=\ln(P)\tag{1}$$ then the average output power will be adjusted to ...


1

You are seeing the "Digital IF” carrier assuming you are modelling what you want to implement digitally in VHDL. The sine and cosine blocks together with the multipliers and combiner perform a frequency translation from DC to the carrier mentioned. Thus the constellation at this point is really not observable as it will be "spinning" at this carrier rate. (...


1

I am trying to perform an FFT of a non-uniformly sample signal. My input comes from Simulink and PLECS which uses variable-time solver. Assuming that the first column in the data provided is your time variable ($t$) in units of seconds, the variable-time solver introduces a very small jitter. (All code in Octave) x = csvread("somefile.csv"); time_variable ...


1

The input at each array will be complex, because the wireless channel "spreads" the energy of the in-phase and quadrature parts of the signal into each other. One simple way to think about this is transmitting a pure in-phase carrier: $$s(t) = \cos(2\pi f_c t).$$ The wireless channel will add a random phase, so the received signal is $$r(t) = \cos(2\pi f_c ...


1

You can see this example of Sigma-Delta A/D Conversion.


1

I figured it out coming back to basic maths, I was making a mistake when integrating my sin function that Matlab block wasn't doing, I was missing behind the $\omega$ variable $$ve[n] = -B \cdot \cos[\Omega\cdot n] = \int A\sin(\omega\cdot t) = (A/\omega)(-\cos(\omega\cdot t)) + Const$$ $$\mathbf {B = (A/\omega) }$$


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If you have the Control System toolbox, you can use a LTI System block to implement the transfer function.


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You will have to implement the transfer function in pieces, rather than as a single LTI block. See http://www.mathworks.com/support/solutions/en/data/1-VPJDU


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