6

Note that in this case you can see that the system is causal only from the given implementation. It's important to understand that you can't see it from the difference equation (if no initial conditions are given), and in general you can't see it from the transfer function either (if no region of convergence is given). The only case for which the expression ...


6

In your original code you defined the sampling interval to be $T_s=0.05$ (idd1 = iddata(ddout, ddin, 0.05);). Yet, according to the data file, the time step between sample points appears to be $T_s=0.003$. Changing $T_s$ from $0.5$ to $0.003$ should compress the step response by a factor of $0.05/0.003=16.67$, which compares favorably to the relation between ...


5

What you are looking for are what we, in the audio space, call comb filters. Comb filters may or may not have a feedback path, just like FIR and IIR filters. In fact, there is a generalized theory about designing comb filters based simply on designing any digital filter. In fact, you can think of every LTI digital filter as being a comb filter where the ...


4

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 ...


4

Purely by inspection of the block diagram the system is causal, because the output is the sum of the current input sample and stuff that's delayed -- there's no $z$ blocks in there to predict the future, just $z^{-1}$ block to react to the past. Also by your method of finding the transfer function, the system is causal -- with a $3^{rd}$ order numerator and ...


4

This is unfortunately a difficult problem. A microphone doesn't have a single frequency response. It has a different one for each direction of incidence. Especially for second order microphones (dipoles, cardioids, etc.), the frequency response also depends heavily on the distance and radiation impedance of the source. So first you need to decide: what ...


4

Or another way, visually, with the plot of the magnitude response of your filter shown below you see that it goes to zero at $\pi/2$. So what you could do is use the relationship between the angular frequency $\omega$ in [radians/sample] and the relative frequency $f$ in [cycles/sample] $$ \omega = 2\pi f = 2\pi \left(\frac FF_s\right) $$ Then compute $F_s$ ...


3

The general approach would be to find the zeros of the filter's transfer function. If you do things right it will turn out that there are two complex conjugate zeros on the unit circle at $\pm j$, i.e., at half the Nyquist frequency. Now you need to choose the sampling frequency such that $80$ Hertz corresponds to half the Nyquist frequency.


3

Basically the transfer function is given by (Continuous time case): $$ H \left( j \omega \right) = K \frac{ \left( j \omega - {q}_{1} \right) \left( j \omega - {q}_{2} \right) }{ \left( j \omega - {p}_{1} \right) \left( j \omega - {p}_{2} \right) } $$ Now, given your frequency data sampled an $ N $ points $ {\left\{ {\omega}_{i} \right\}}_{i = 1}^{N} $ we ...


3

If the OP is actually interested in selecting only one individual frequency from the even or odd harmonics, then a moving average filter (MAF) would be ideal since this can provide a null at every other frequency when the frequencies are harmonically related. The low-pass MAF with all the coefficients as 1 will pass DC at $f=0$ and provide nulls spaced by $1/...


3

You can simplify the z-transform further, $ H(z) = \frac{Y(z)}{X(z)}= \frac{0.5-0.5z^{-2}}{1-z^{-1}}$ $ H(z) = \frac{(1-z^{-1})(1+z^{-1})}{2(1-z^{-1})} $ canceling the common pole and zero $ H(z) = \frac {1+z^{-1}}{2}$ take inverse z-transform $ Y(n) = \frac{x(n)+x(n-1)}{2} $


3

There is already a good answer, but here is some extra explanation. This is called "pole/zero cancellation". That means you have a pole, say, $z_p$ but you also have a zero at $z_p$ so the zero cancels the pole. One particular example where this can be useful is a moving average filter of length $N$, The transfer function is $$H(z) = \frac{1}{N} \...


3

It has to do with the Bilinear Transform, as Hilmar already stated. The theoretical function is defined for the analog domain, so inherently there will be differences when you convert the response to the digital domain. However, you can still generate the "analog" frequency response and yield the expected result. I took your code and added/modified ...


3

That depends on what you want to plot: The response of the continuous time system or the response of a sampled discrete time system, which will always depend on the choice of sample rate. Your system is not band-limited, so it can't be sampled without some amount of aliasing. Do I just evaluate s at iω If you just want to look at the approximate frequency ...


2

A transfer function describes an LTI system. As such, the given system can be described by a transfer function. However, if there are non-zero initial conditions, the system is no longer linear because there's a contribution in the output that does not depend on the input signal but only on the initial conditions. Consequently, the transfer function cannot ...


2

Your filter is a 5-tap delay line FIR filter. The FIR filter's coefficients are [2 6 1 5 1]. If x is your filter's input sequence and you execute the following MATLAB command: h = [2 6 1 5 1]; you can then implement the filter (to produce a y output sequence) using the following MATLAB command: y = filter(h, 1, x); You can see your filter's frequency ...


2

The closed loop poles are the roots of the polynomial $$D(s)=s^2+2s+2+K\tag{1}$$ and, according to the root locus plot, they are $s_{1,2}=-1\pm 2j$. Consequently, we get $$2+K=|1+2j|^2=5\quad\Longrightarrow\quad K=3\tag{2}$$ With $K=3$ we obtain $$H(0)=\frac{K}{2+K}=\frac35\tag{3}$$ which leaves step response $C$ as the only option.


2

The system is causal, provided that the recursion is forward; i.e., it's recursed for increasing $k$. Seeing that you are confused about causality tests, let me elaborate on it. Let's put the definition of causality from Oppenheim's Signals & Systems book : A system is causal if the output at any time depends only on values of the input at the present ...


2

$$\begin{align} Y(z) &= E(z) + \frac{1}{1-z^{-1}}(X(z) - z^{-1} Y(z)) \\ &= E(z) + \frac{1}{1-z^{-1}}X(z) - \frac{z^{-1}}{1-z^{-1}} Y(z) \\ \\ Y(z) + \frac{z^{-1}}{1-z^{-1}} Y(z) &= E(z) + \frac{1}{1-z^{-1}}X(z) \\ \left(1 + \frac{z^{-1}}{1-z^{-1}}\right) Y(z) &= E(z) + \frac{1}{1-z^{-1}}X(z) \\ \frac{1}{1-z^{-1}} Y(z) &= E(z) + \frac{1}...


2

You can use a window as an FIR filter (convolve with window in time), such that the zero crossings of the window transform are at $\omega_0 = 2\pi * 2k*200$ to filter the even harmonics, and phase shift the window by transform $\frac{\pi}{2}$ to filter the odd harmonics. For example, a rectangular window of length $M$ will have zero crossings at $\omega = \...


2

The slope of the lowpass is 20dB/decade or 6dB/octave, that means it's a simple first order lowpass filter. At the corner frequency the gain is -3dB. Same of the phase. It goes from 0 to -90, so it's a first order filter. At the corner frequency, it's 45 degrees. Looking at both phase and level we conclude that it's a first order lowpass with $f_c = 1000 rad/...


2

Considering an LTI filter, one can define its frequency response by evaluating its transfer function H(z) on the unit circle H(ejω). Is this definition correct? This is only correct if the LTI filter discrete in time. But if it is: yes. I also came across the second definition, which says that the filter's frequency response is the Fourier Transform of its ...


1

UPDATE Python's signal.csd routine, as it is used in my code, don't work miracles, at least in the sense you would expect ("but it is possible that with the short sequence used here (255) there is no averaging/windowing being done" -- as if it does averaging/windowing for >255 sequences). From the code you can see that, when presented only the ...


1

Assuming both black boxes are LTI systems, that's reasonably easy to do. Let's call the transfer function of the first box $H_A(\omega)$ and the second one $H_B(\omega)$. The open look transfer function is simply the product of the, i.e. $H_O(\omega) = H_A(\omega) \cdot H_B(\omega)$ A sufficient stability condition is $|H_O(\omega)| < 1$. For an actual ...


1

The OP has a mistake in the numerator vector as there is an additional space between the minus sign and the 1.027e14 coefficient which will cause that to operate as an expression (-2.58e13 - 1.027e14) instead of negate the sign. Had the system been correct, the resulting closed loop system with the compensator and gain chosen is underdamped. This can be ...


1

The filter is a discrete FIR filter, so the transfer function would be given as: $$H(z) = .0287 + 0.1430z^{-1} + 0.3283z^{-2} + 0.3283z^{-3}+0.1430z^{-4}+0.0287z^{-5}$$ With the frequency response given by using the unit circle for the complex variable $z$, as in $z = e^{j\omega}$ for $\omega \in [0, 2\pi)$, with the sampling rate normalized to $\omega = 2\...


1

You have 2 right-half plane zeroes : 0.012 and 18. The zero at +18 is "fast" and will not affect the performance much, but your slow zero at 0.012 will severely limit your performance. You can't cancel this right-half plane zero with a right-half plane pole in your controller, your controller output will be unbounded. Is this homework or a real ...


1

I don't think it's a good idea to add random noise to frequency response. In general, noise is added to the input or output signal. You should be clear that at which stage the noise is introduced. According to the different stages of noise introduction, there are two main transfer function estimators, $H_1$ estimator and $H_2$ estimator. $H_1$ assumes that ...


1

Assuming you have an input signal $ u = A cos(2\pi ft) $ and you measure an output signal $y = B cos(2\pi ft + \theta) $ The $\frac{B}{A}$ ratio is the gain and $\theta$ is the phase shift for frequency $f$. You could simply demodulate $y$ by multiplying by $x = cos(2\pi ft) - jsin(2\pi ft)$ $$ z(t) = y*x = \frac{B}{2}(cos(4\pi ft) + cos(\theta) + j(sin(4\pi ...


1

For $|\zeta| \le 1$, let $\zeta= \cos\theta$, so $\theta=\mathrm{arccos}\,\zeta$ $$\begin{align*}g(s) &= K\frac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}\\ \\ &= K\frac{\omega_n^2}{s^2 + 2\omega_n s \cos\theta+ \omega_n^2(\cos^2\theta +\sin^2\theta)}\\ \\ &= K\frac{\omega_n^2}{(s + \omega_n \cos\theta)^2+ \omega_n^2\sin^2\theta}\\ \\ &...


Only top voted, non community-wiki answers of a minimum length are eligible