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Hot answers tagged step-response

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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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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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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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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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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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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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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 ...

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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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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^{... 1 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. 1 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 ... 1 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 ... 1 with ordinary linear differential equations, you end up solving it essentially the same way as you do with the Laplace transformed counterparts. you could come up with an operation that breaks a higher-order differential equation into two lower-order diff eqs. in much the same manner as you factor a high-order polynomial of s into two lower-order ... 1 Why we use Transfer Functions, when we can get a system's output by just solving it's differential equation? Because differential equations are unwieldy and hard to deal with, and you can't see the behaviour on different frequencies from these, whereas transfer functions just give you the behaviour of an LTI system given an excitation of given property. ... 1 HINT: For t\neq 0 both solutions are identical and correct. So you only need to consider t=0 to find out which of the two solutions is correct. Does the step response jump at t=0 or is it continuous? What would a jump at t=0 imply for the impulse response? Alternatively, you should try to compute the step responses corresponding to the two impulse ... 1 The given impulse response is unbounded for n\rightarrow -\infty, and, consequently, the corresponding step response is also unbounded for all n, as you've shown. The step response given in the answer corresponds to a totally different impulse response:$$h[n]=2^nu[n]\tag{1}$$which is also unbounded for n\rightarrow\infty, but since the ... 1 Computing the partial fraction expansion,$$H (s) = \frac{2.302 s + 0.3548}{s^{3} + 0.739 s^{2} + 3.223 s + 0.3548} \approx \frac{2.2861 - 0.0309306 s}{s^2 + 0.626454 s + 3.1525} + \frac{0.0309306}{s + 0.112546}$$For the time being, let us neglect the step response of the 1st order subsystem. Note that the 2nd order subsystem has the following form$$\pm ...

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Your approach is correct. Rewrite $H(s)$ as $$H(s)=\frac{2s}{s+3}=2\frac{s+3-3}{s+3}=2-\frac{6}{s+3}\tag{1}$$ and use basic Laplace transform identities to obtain $h(t)$ from $(1)$. Note that you don't need to use the Laplace transform. A time domain approach as suggested in oxuf's answer is even more straightforward.

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When feeding a filter with a unit step, you are actually integrating the impulse response of the filter. So $$h(t) = \frac{dy(t)}{dt} = -6e^{-3t}\cdot u(t) + 2e^{-3t}\cdot \delta(t) = -6e^{-3t}\cdot u(t) + 2 \delta(t)$$ is its impulse response.

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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 $g(t)$: $$\displaystyle\lim_{t\to \infty}g(t)=\displaystyle\lim_{s\to 0}s\cdot \frac{3}{s}\cdot\frac{K(s^2 + s - 20)}{s^3 + ... 1 You can solve such problems directly in the time domain (as shown in this answer to a related question), or by using the Laplace transform. I'll show you that latter approach. The Laplace transform of the given differential equation is$$Y(s)(s^2+5s+6)=-sX(s)\tag{1} where $X(s)$ and $Y(s)$ are the Laplace transforms of the input $x(t)$ and the output \$y(...

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