# Tag Info

## Hot answers tagged causality

13

Of course they don't exist. But we can stop time and use systems that would be non-causal if we hadn't stopped time. Stop time? Yes, just store your data and work offline / non-realtime. Or work on data that are not temporal but something else, for instance spatial. Of course, non-causal is a misnomer in that case, but the term is still used, in analogy with ...

6

This has absolutely nothing to do with causality. The frequency response of a real-valued filter (i.e., one with a real-valued impulse response) is (conjugate) symmetric, i.e., the negative frequencies are redundant. That's why it is sufficient to show the frequency response at non-negative frequencies only. You can easily see that symmetry as follows. The ...

5

Causality is a necessary condition for realizability. Stability (or, at least, marginal stability) is also important for a system to be useful in practice. For linear time-invariant (LTI) systems, which are fully characterized by their transfer function, we get realizability constraints on the transfer function. For continuous-time LTI systems, if we work ...

5

No it does not satisfy the condition. Simply take an example: $$n = 1 \implies y[1] = x[2]$$ Hence the output value at the present time $n=1$ depends on a future value of the input at time $n=2$. This violates the causality principle.

5

If the derivative exists at the given point, then it doesn't matter if you look (infinitesimally) into the future or into the past, you can do both, because both will give the same result: $$x'(t)=\lim_{h\to 0}\frac{x(t+h)-x(t)}{h}=\lim_{h\to 0}\frac{x(t)-x(t-h)}{h}\tag{1}$$ So a differentiator can be (theoretically) implemented by a causal system. ...

4

The transfer function $$H(z)=-z^{-2} -2z^{-1} +2z +z^2$$ can be written as \begin{align} H(z)&=z^2z^{-2}\left(-z^{-2} -2z^{-1} +2z +z^2\right)\\ &=z^2\left(-z^{-4} -2z^{-3} +2z^{-1} +1\right)\\ &=H_1(z)H_2(z) \end{align} It can now be seen more easily that $H_1(z)$ has two zeros at $z=0$ (and two poles at $\pm\infty$) . $H_2(z)$ has four ...

4

Because for negative values of $t$ you have, for example, $y(-2) = x(-1)$ which depends on a future value of $x(t)$ at $t=-1$ for the current value of $y(t)$ at $t=-2$. Note that $t=-1$ represents a future time for $t=-2$.

4

If the number of finite zeros is not greater than the number of finite poles then the transfer function is proper, i.e., the degree of the numerator polynomial is not greater than the degree of the denominator polynomial. If the degree of the numerator polynomial were greater than the degree of the denominator polynomial, we would get at least one pole at ...

4

I don't have a concrete proof for this one. However, I can tell you this... Consider a perfect low pass filter. The time domain representation is a sinc. And for any system to have a sharp transition band, a base signal has to be multiplied with a rectangular waveform in the frequency domain. Which implies that, the time domain signal of the same has to be ...

4

An improper system cannot be causal and stable. If the order of the numerator is greater than the order of the denominator, you'll always have at least one pole at infinity. Consequently, not all poles are in the left half-plane (or inside the unit circle in the case of discrete-time systems). The system in your example is clearly unstable: $$H(s)=\frac{s^... 4 They are independent of each other. Continuous systems: For stability, the ROC (region of convergence) must include the jw-axis of the s-plane. Causal systems have a ROC which is a right-sided plane, with Re(s)>\alpha. Here \alpha is the real part of the "most to the right" pole. Due to this, for a continuous system to be causal and stable, all its ... 3 Neither analog, nor digital filters can have infintely sharp (ideal) frequency responses. Those ideal waveforms can only be defined in mathematical terms and no exact physical counterpart is possible. However sufficient approximations will be realized. Most typically the reason is that the ideal waveforms would require an infinetely long signal interval ... 3 HINT: Use a negative time t to see that y(t) depends on future input values. 3 How can we calculate output at 0 input if the system depends on future or past or both? Well, surely zero input just means;$$ x(t) = 0~~~~\forall t $$and the \forall t means for all time: positive and negative. Substituting that into the equation:$$ y(t) = x(t+1) + x(t-1) = 0 + 0 = 0 $$So the system is homogeneous. Well, as @Dilip points out, this ... 3 one thing about a non-minimum phase system (with a rational transfer function), is that it can be thought of as the series concatenation (or cascade) of a minimum-phase system, having identical magnitude response as the given non-min-phase filter, with an all-pass filter. the APF will have a poles that cancels specific zeros of the min-phase system that are ... 3 The Hilbert transform \mathcal{H}\left\{f(\omega)\right\} with$$f(\omega)=-\frac12\log(1+\omega^2)\tag{1}$$can be calculated in the following way. First, note that$$\frac{df(\omega)}{d\omega}=-\frac{\omega}{1+\omega^2}\tag{2}$$From this table we know that$$\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}=\frac{\omega}{1+\omega^2}\tag{3}$$We also ... 3 An LTI system (or even a system that isn't L or TI) that is "causal" has a prayer of being realized in real time whereas an "acausal" system cannot ever be realized in real time because an acausal system responds to input from the future. A causal system responds only to input of the present and/or input from the past. For LTI systems, it can be shown that ... 3 A confusion may arise, for causal systems, from mistaking "having negative signal values (amplitudes)" and "depending on negative time indices". A strict memory-less system does depend neither on past (for the causal case) nor on future values, only the current ones, to determine the current value of the output. The output at n only depends on inputs at ... 2 I think the "causal signal" is simply borrowed from the "causal system". For a system, the "causal" constrain is meaningful and fundamental, i.e., if the input does not occur, the system should not produce any output with respect to the input values. Then, for a LTI system, the "causal" nature means h(t) =0, \text{ for } t<0. Then, the concept is ... 2 The Paley-Wiener criterion defines a condition on the magnitude spectrum of a causal time-domain function. So if the Paley-Wiener criterion is satisfied for a given A(\omega)=|H(\omega)|, we know that there is a causal function with magnitude spectrum A(\omega). It should be noted that the Paley-Wiener criterion is only applicable to square-integrable ... 2 In general, from the positiveness of the phase nothing can be concluded about the causality of the corresponding system. Note that the phase of the given system is$$\phi(\omega)=\arctan(\alpha\omega\tau)-\arctan(\omega\tau)\tag{1}$$which is positive for all \omega>0 if \alpha>1. Yet the system is causal, which can be easily seen by calculating ... 2 Convolution is what happens when a signal passes through an LTI system.$$y(t)=x(t)\star h(t)=\int_{-\infty}^{\color{red}\infty}x(\tau)h(t-\tau)d\tau$$The keyword in this question is real-time. The associated keyword in the answer would be weather the LTI system is causal or not. If the system is causal, then h(t)=0, \ \forall t<0, and therefore h(t-\... 2 HINT: Prove that$$y(t)=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{x(\tau)}{t-\tau}d\tau=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{x(t-\tau)}{\tau}d\tau\tag{1}$$Use the right-most expression in (1) to show that the system is time-invariant. Equivalently, you can directly show that the original input-output relation is a convolution integral, from which ... 2 Any practical couple of a discrete signal and filter has, at some point, one being of finite length. Because a causal filter does not know about the future by definition, and any signal is unknown in some past (before the big-bang for instance). So the "infinite" convolution formula:$$\sum_{-\infty}^{\infty} h_{n-k}x_k$$is necessarily trimmed on ... 2 The transition bandwidth of a filter is inversely proportional to the filter kernel length. The approximate equation is given as follows filter_kernel_length ≈ 4 / transition_bandwidth; //(roll-off) The above equation is taken from this online book ( http://www.dspguide.com/ch16.htm ) but it's only approximate because it doesn't take into account the ... 2 As you've pointed out, inversion leads to poles at locations of the zeros of the original transfer function and vice versa. Assuming that G(z) is causal and stable (i.e., it has all its poles inside the unit circle), we have to distinguish 3 cases: G(z) has at least some zeros outside the unit circle. This means its inverse has some poles outside the ... 2 Since nobody has answered my question yet. I will try to partially answer it myself. However, even though I accept it as the final answer I can revoke this if another user elaborates a much better answer, which I think is possible. After studying the proposal given by the above mentioned article about how to obtain a piece-wise positive group delay in a ... 2 Every symbol in the channel will be affected by ISI from previous symbols and next symbols. To understand this you must notice that there's a propagation delay from transmitter(TX) to receiver(RX). That means that it is possible for several symbols to be present in the channel at the same time before the first of them actually arrives at the receiver. When a ... 2 A transfer function is called realizable if it can be implemented by a causal and stable system. The given frequency response is continuous and doesn't have any impulses, so the corresponding system is stable. The transfer function (as a function of s) is given by$$H(s)=\frac{a}{b-s^2}e^{-st_0}\tag{1}$$The given frequency response is obtained form (1)... 2 You are correct that y[n] only depends on past input values for n \gt 0, but when n \lt 0, that is not the case --- then y[n] is dependent on future input values, regardless of what x is. To see this:$$ y[-2] = x[-(-2)] = x[2]  so the output at time $n=-2$ is dependent on the input at time $n=2$.

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