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I am self studying Alan Opennheim's course Signals and Systems. I am a math major and have no background in EE.

I understand that for a linear constant-coefficient difference equation (LCCDE) system to be linear its auxiliary conditions must be 0.

I also understand that for the system to be causal every output $y(t)$ corresponding to input $x(t)$ s.t. $x(t) = 0$ for $t<t_0$ must satisfy $y(t)=0$ for all $t<t_0$.

What i am having trouble understanding is why this imposes the initial condition $y(t_0)=0$ for the response to the unit step function. Couldn't we have set the initial condition to be $y(t_1)=0$ for $t_1>t_0$ and kept the system causal and LTI?

In the 6th lecture in the video series Alan Oppenheim finds the unit step function response to the causal LTI system described by $y'(t) +ay(t) = x(t)$. He imposes the initial condition $y(0)=0$, as i mentioned above i can't understand why this is imposed by the properties of the system. He then goes on to find the unit impulse response, and finds that it is $e^{-at}u(t)$ where $u(t)$ is the unit step. Clearly this function does not satisfy the initial condition which were imposed on the unit step response, I am having trouble understanding why this is valid.

Any help is appreciated, Thanks!

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If you are given an input $x(t)$ with $x(t)=0$ for $t<t_0$, and you specify an initial condition $y(t_1)=0$ for $t_1>t_0$, then the resulting system is generally non-causal, because we already know the system's response at $t_1>t_0$, regardless of the input signal in the interval $[t_0,t_1]$.

For a system to be linear (in the sense used by Oppenheim) and causal, we need to specify the initial condition at the point $t_0^-$, i.e., we require that the left-sided limit $y(t_0^-)$ satisfies $y(t_0^-)=0$.

For the given example, if an impulse is applied at $t=0$, we require $y(0^-)=0$ for the system to be linear and causal. Note that there is no requirement on the right-sided limit $y(0^+)$.

Clearly, the response $y(t)=e^{-at}u(t)$ satisfies the initial condition $y(0^-)=0$.

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    $\begingroup$ Thanks for the reply. With an initial condition at $t_1 > t_0$, wouldn't the system still be causal because the response at $t_1$ depends only past values of $t$. Also can you recommend a reference, Oppenheim makes no mention of one sided limits. $\endgroup$ – Adi Oct 10 '19 at 23:25
  • $\begingroup$ Also, is the system time-invariant because we define it by its impulse / unit response? $\endgroup$ – Adi Oct 12 '19 at 11:11
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From my understanding for any system to be time invariant , it must not grow with time before the system is excited with an input. If the initial conditions are not zero i could shift my input , and in the meantime the system would have evolved to a different state and hence the new output will not just be the shifted version of the old output so the system will not be time invariant.

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  • $\begingroup$ Given a system with a time-dependent output function and a time-dependent input function the system will be considered time-invariant if a time-delay on the input directly equates to a time-delay of the output function. $\endgroup$ – Filipe Pinto Jan 29 at 0:03
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If the LCCDE is casual and the input is $f(t)u(t-t_0)$, then you can say that before $t_0$ (excluding $t_0$), the output is zero, by the equivalence of initial rest condition and casuality for LTI systems.

But when we try to solve the LCCDE with such an input signal, we are actually discussing two occasions separately, 1) when $t<t_0$, the input is 0; 2) when $t>t_0$, the input is $f(t)$. We solve the LCCDE and get its general solutions in 2nd case, $t<t_0$. Hence we can't substitute like "$x(t_0-1)$" into the unsettled general solution and let it be zero, to solve the coefficients.

The only time that can be utilized is the $t_0$.

P.S. Yes, I have said "excluding $t_0$". However, it only disturbs us when $f(t)$ is abnormal, such as delta function.

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