# 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 ...

8

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. ...

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 ...

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 ...

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

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 $t=-2$.

5

No it does not satisfy the condition. Simply take an example: $$n = 1 \implies y = x$$ 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

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 ...

5

HINT: Use a negative time $t$ to see that $y(t)$ depends on future input values.

4

My answer implicitly refers to ideal brickwall lowpass filters which do have infinetely sharp transition bands. For other possible interpretations, refer to other answers. In practice, neither analog, nor digital filters can have infintely sharp (ideal) transition band frequency responses, whether causal or not. The frequency response of an ideal filter can ...

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

This is three years later, but since I don't see the real answer posted here, I will post it. The correct answer is that if we are literally interpreting the original statement as a purely mathematical claim, taken at face value, then it is incorrect. There do exist causal filters, even minimum-phase ones with nice closed-form Fourier domain expressions, ...

4

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 ...

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

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: $$H(s)=\frac{s^... 4 A transfer function is called realizable if it can be implemented by a causal and stable system. The given frequency response is continuous and doesn't have any impulses, so the corresponding system is stable. The transfer function (as a function of s) is given by$$H(s)=\frac{a}{b-s^2}e^{-st_0}\tag{1}$$The given frequency response is obtained form (1)... 4 They are independent of each other. Continuous systems: For stability, the ROC (region of convergence) must include the jw-axis of the s-plane. Causal systems have a ROC which is a right-sided plane, with Re(s)>\alpha. Here \alpha is the real part of the "most to the right" pole. Due to this, for a continuous system to be causal and stable, all its ... 4 Purely by inspection of the block diagram the system is causal, because the output is the sum of the current input sample and stuff that's delayed -- there's no z blocks in there to predict the future, just z^{-1} block to react to the past. Also by your method of finding the transfer function, the system is causal -- with a 3^{rd} order numerator and ... 3 How can we calculate output at 0 input if the system depends on future or past or both? Well, surely zero input just means;$$ 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 $$So the system is homogeneous. Well, as @Dilip points out, this ... 3 one thing about a non-minimum phase system (with a rational transfer function), is that it can be thought of as the series concatenation (or cascade) of a minimum-phase system, having identical magnitude response as the given non-min-phase filter, with an all-pass filter. the APF will have a poles that cancels specific zeros of the min-phase system that are ... 3 The Paley-Wiener criterion defines a condition on the magnitude spectrum of a causal time-domain function. So if the Paley-Wiener criterion is satisfied for a given A(\omega)=|H(\omega)|, we know that there is a causal function with magnitude spectrum A(\omega). It should be noted that the Paley-Wiener criterion is only applicable to square-integrable ... 3 An LTI system (or even a system that isn't L or TI) that is "causal" has a prayer of being realized in real time whereas an "acausal" system cannot ever be realized in real time because an acausal system responds to input from the future. A causal system responds only to input of the present and/or input from the past. For LTI systems, it can be shown that ... 3 A confusion may arise, for causal systems, from mistaking "having negative signal values (amplitudes)" and "depending on negative time indices". A strict memory-less system does depend neither on past (for the causal case) nor on future values, only the current ones, to determine the current value of the output. The output at n only depends on inputs at ... 3 Also consider the somewhat simpler "identity system", given a continuous signal x(t):$$\begin{align}y(t) &= x(t)\tag{1}\\ &= \lim_{\Delta t\to0^-}x(t + \Delta t)\tag{2}\\ &= \lim_{\Delta t\to0^+}x(t + \Delta t)\tag{3}\\ &= \lim_{\Delta t\to0}\frac{x(t - \Delta t) + x(t+\Delta t)}{2}.\tag{4}\end{align}$$This might be interpreted as the ... 3 The proof is quite straightforward. With$$x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})e^{jn\omega}d\omega\tag{1}$$and with X(e^{j\omega})=X^*(e^{j\omega}) (i.e., a real-valued DTFT) we get$$\begin{align}x[-n]&=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})e^{-jn\omega}d\omega\\&=\frac{1}{2\pi}\left[\int_{-\pi}^{\pi}X^*(e^{j\omega})e^{jn\...

3

The statement is not correct. In a causal system, the cause (i.e. input) does indeed precede (come before in time) the corresponding effect (output). The correct statement is the cause doesn't follow the corresponding effect. "Cause creates the effect", so in a causal system, Cause ($x[n]$) will always come first and then the corresponding effect ($y[n]$). ...

3

Is there a way to make this filter non-causal? Remember that non-causal filters aren't possible to interpret in any case, because "non-causal" literally means there's output caused by input that comes later. Just because you can buffer something if latency doesn't matter doesn't mean you've actually built a non-causal system - your buffering adds ...

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

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-\...

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