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Sums of harmonic cosines The Hann window is a sum of a rectangular function and a truncated cosine. As an impulse response it has the qualities you asked for: Figure 1. Hann window as impulse response, and the corresponding step response. The step response is valued $x - \frac{\sin(2\pi x)}{2\pi}$ over $0 \le x \le 1$, and is constant outside that range, ...

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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 ...

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The final value of the step response is the DC gain of the closed-loop transfer function, which is generally different from the open-loop DC gain. Assuming unity gain feedback, the feed-forward transfer function $G(s)$ equals the open-loop transfer function, and the closed-loop transfer function is given by $$C(s)=\frac{G(s)}{1+G(s)}\tag{1}$$ The final value ...

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The short answer is that the step response is the inverse Laplace Transform of $\frac{1}{s}G(s)$. Here are some hints that should help in being able to solve the problem. Since it is likely a homework problem I wanted to provide more details as relevant background and steps toward a solution rather than detail the mathematics for the specific solution. A ...

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As explained in Tendero's answer everything you've calculated is correct. Matlab is supposed to complain because the system with transfer function $$H(s)=1-sT\tag{1}$$ has (assuming causality) a step response $$a(t)=\mathcal{L}^{-1}\left\{\frac{1-sT}{s}\right\}=u(t)-T\delta(t)\tag{2}$$ where $\delta(t)$ is the Dirac delta impulse. The system is unstable ...

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There is no misunderstanding at all. The Padé approximant you found is correct. The "problem" is that you chose $M$ and $N$ such that the transfer function you get to approximate the delay is improper. Namely, the order of the numerator exceeds the order of the denominator. In Control Theory, improper systems are not too useful because they cannot be ...

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If your system is an ideal differentiator with input-output relation $$y(t)=\frac{dx(t)}{dt}\tag{1}$$ then its transfer function is $$H(s)=\frac{Y(s)}{X(s)}=s\tag{2}$$ From the equation in your question you obtain for its step response $$A(s)=1\tag{3}$$ which in the time domain corresponds to a Dirac delta impulse: $$a(t)=\delta(t)\tag{4}$$ This is ...

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Dan's answer -- to compute $H(s)$ as normal, and then compute $1/H(s)$ -- is equivalent to your suggestion of swapping the $x$ and $y$ (or doing it in one step by solving for $H_{yx}(s) = \frac{X(s)}{Y(s)}$). In the Laplace domain it's justified by noting that $$\frac{H(s)}{H(s)} = H(s)H_{yx}(s) = 1$$ In theory this means that a system followed by its ...

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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 ...

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Open loop gain at DC is -3dB or .707 and 0 degrees. We don’t know the forward gain but assuming it is the open loop gain, the closed loop gain would be $.707/(1+.707)= .4148$, matching the first plot. (With 60 degrees of phase margin however I would have expected a response closer to the third plot, but as explained in the comments this is due to my ...

3

You can use the step-invariant transformation, which makes sure that the step response of the discrete-time system matches the step response of the continuous-time system at the sampling instants. I suppose this is what you mean by "exact solution". Sampling the given continuous-time step response at $t=nT$ gives the desired step response of the discrete-...

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In the given example you have 3 types of systems: the ideal integrator (D) with a step response that grows infinitely (fig. 1); that's the obvious one, as you've found out by yourself. underdamped systems with complex conjugate poles (A, B, and C): their step response oscillates critically damped systems with a double real pole (E and F): no oscillation in ...

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Let me provide a method, applied only for finding the impulse response $h(t)$ of an LTI system characterised by an LCCDE of the form $\sum_{k=0}^{N}{ a_k {{d^k y(t)}\over {dt^k}}} = \sum_{k=0}^{M}{ b_k {{d^k x(t)}\over {dt^k}}}$ by using the classical time domain approach. A method which is ignored as it's replaced by transform domain techniques instead......

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It's a bad translation of step response, i.e., the filter's response to a unit step $u[n]$ at the input. Note that the step response is the convolution of the unit step with the filter's impulse response: $$s[n]=\big(u\star h\big)[n]=\sum_{k=0}^nh[k],\qquad n\ge 0\tag{1}$$ where I've assumed causality, i.e., $h[n]=0$ for $n<0$. From the given step ...

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A general approach would be to take the Laplace Transform of the equation and put it in form of a transfer function: $$H(s) = \frac{Y(s)}{X(s)}$$ And then invert that and solve for the unit step response from $1/H(s)$ Note that the inverse Laplace transform of $1/H(s)$ would give the impulse response, since that implies an impulse is at the input to the ...

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The transfer function is $$H(s)=\frac{K(s - 4)(s + 5)}{(s + 3)(s + 6)(s + 10)}=\frac{K(s^2 + s - 20)}{s^3 + 19s^2 + 108s + 180}$$ So we need to find $K$. The LT of input step is $\frac{3}{s}$. Using the FVT for the step response $g(t)$: $$\displaystyle\lim_{t\to \infty}g(t)=\displaystyle\lim_{s\to 0}s\cdot \frac{3}{s}\cdot\frac{K(s^2 + s - 20)}{s^3 + ... 2 You have to judge yourself if for your purposes you need to get rid of the first few output values. The phenomenon you observe is determined by two factors. The first is the delay of the filter (which is usually frequency dependent). This delay is a consequence of the causality of the filter. For a linear phase FIR filter that delay is independent of ... 2 I can help with some of your multiple questions. First, a cascade of n buffered RC low pass filters (LPFs), a so-called n-th order synchronous LPF, has the impulse response and step response shown in the screenshot below: This is a screenshot from my paper 1 referenced at the bottom. All R values are the same and all C values are the same. Buffering simply ... 2 The simplest (imho) explanation is this. Consider a time-domain signal which has narrow pulses or sudden jumps in value or on/off switches at certain (unknown!) time instants. Such a signal is said to carry its information in those specific time instants at which such sudden changes occur. A PWM (pulse width modulation) signal is an excellent example for ... 2 Your system H(s) is continuous$$ H(s) = \frac{s}{s+1} $$Using a discrete impulse does not make a lot of sense. I suspect Simulink added a "zero-order hold" or ZOH block between your discrete impulse and your system H(s). As such, your impulse response will be affected by the sampling period of your discrete impulse source. If you decrease the sampling ... 2 The closed loop poles are the roots of the polynomial$$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}$$which leaves step response C as the only option. 2 The step response of your BPF has a fast decay, so you may eliminate it by setting the filter states. Have a look at the difference equation of your biquad BPF:$$ y[n] = 0.0377 x[n] + 0 x[n-1] - 0.0377 x[n-2] + 1.9240 y[n-1] - 0.9246 y[n-2] $$Since b_0=-b_2 and we have x[n] = 0.65 for all n\geq 0, if we set x[n-1]=x[n-2] = 0.65 and y[n-1]=y[n-2]=... 1 In an LTI system, any linear operation on inputs, is directly imposed on the outputs, that is, if an LTI system responses to \delta[n] as h[n], then it responses to u[n]=\sum_{k=-\infty}^{n}\delta[k] as s[n]=\sum_{k=-\infty}^{n}h[k]. This is also clear from the convolution since$$s[n]=u[n]*h[n]=\sum_{k=-\infty}^\infty h[k]u[n-k]=\sum_{k=-\infty}^n ...

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If you have complex conjugate poles (i.e., $\text{Im}\{p_i\}\neq 0$) you have oscillations in the step response. Only real-valued poles will give you a monotonic step response as shown in the figure. Take a look at this related answer.

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It does matter that $a=b=2$, because this gives a relatively nice looking solution. As mentioned in a comment, you should split up the given transfer function $H(s)$. But first, let's introduce a normalized variable $p$: $$p=\frac{s}{\omega_n}\tag{1}$$ Now we can write the given transfer function as $$\hat{H}(p)=\frac{2p^2+2p+1}{p^3+2p^2+2p+1}\tag{2}$$ ...

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The answer (How to calculate the impulse response of an RC circuit using time-domain method) provides a direct time-domain solution of an RC circuit for the impulse reponse $h(t)$. Now this new answer modifies it to solve for the step-response $s(t)$ instead and then computes the impulse response according to : $$h(t) = s(t)'$$ The differential equation ...

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So, when you remember how the step response, impulse response and frequency response are related, you'll notice that if you, instead of an actual impulse integral (i.e. a step) use something that is wider, then the frequency response simply gets windowed to lower frequencies. In other words, and to put it as an information-gathering problem: if the ...

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Here is a messy way it can be done. I started it and then it got too involved for me to finish, but I can see it is solvable. First I simplified the equation to make the algebra less messy: $$y(t) = C + A e^{\alpha t} + B e^{\beta t}$$ You should be able to see the equivalence. Then I repeated it for two different domain values:  y(t-1) = C + A e^{...

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There are two problems here that need to be solved. The first is the delay of the measured step response, i.e. how many samples from the beginning you need to discard. This is critical in your example because you chose a second-order system to model the data, and such a system is not very flexible in adding or subtracting delay. Delay could be added by ...

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Because handling polynomials are much easier than differential operators. Also multiplication of polynomials are again much easier than convolution of signals. The only reason you learn partial fraction expansion is actually to separate the terms of a compound fraction to simpler items and obtain the eigenmodes, e.g., say you have the following for the ...

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