This question stems from a point of confusion that I still have about the causality, linearity, and time-invariance in LCCDEs. I wanted to use the capacitor as an example.

Consider a capacitor with capacitance $C$. Taking the current $i(t)$ to be the input to the system and the voltage $v(t)$ to be the output we have $$i(t) = C \frac{\mathrm{d} v(t)}{\mathrm{d}t}$$

This differential equation can be solved to obtain $$v(t) = v(t_0) + \frac{1}{C} \int_{t_0}^{t} i(\tau) \mathrm{d} \tau$$

My first question is: isn't this mathematically valid for all $t$? In other words, does this give us the response for all $t$ or is it only valid for $t > t_0$? If it is valid for all $t$, including the $t < t_0$ case, doesn't this make the system non-causal since it anticipates future input and output values? Are we allowed to integrate backwards in time?

My second question relates to the assertion that a for the LCCDE to describe a linear system, the initial conditions must be zero. Suppose $t_0 = 0$ such that $$v(t) = v(0) + \frac{1}{C} \int_{0}^{t} i(\tau) \mathrm{d} \tau$$

With $v(0) = 0$ the system is linear. But the choice of $t_0 = 0$ is arbitrary, since for example $$v(t) = v(0) + \frac{1}{C} \int_{0}^{t} i(\tau) \mathrm{d} \tau = v(2) + \frac{1}{C} \int_{2}^{t} i(\tau) \mathrm{d} \tau$$

Why shouldn't we require that $v(2) = 0$ as well for that matter? What am I missing here? Thank you in advance.


3 Answers 3


Your first equation is indeed valid for all $t$, but this has nothing to do with causality. You have


If $t_0>t$ you simply subtract from the first integral the contribution from the interval $[t,t_0]$. But this is still causal, you don't need the future input, you just need to subtract it in case you added it in the first integral. The output could of course be computed without it.

Concerning your second question, you choose the time $t_0$ as the beginning of your processing. The value of $t_0$ doesn't matter but once you choose it, it's fixed. The system is linear if the initial conditions at $t=t_0$ are zero.

  • $\begingroup$ Why is it (mathematically) that $\frac{1}{C}\int_{-\infty}^{t_0}i(\tau)d\tau = v(t_0)$? Also, could you please elaborate on your second point regarding linearity? $\endgroup$
    – user33568
    Commented Feb 17, 2018 at 15:55
  • $\begingroup$ In other words: $v(t) = v(t_0) + \frac{1}{C} \int_{t_0}^{t} i(\tau) \mathrm{d} \tau$, where $v(t_0)$ is an unknown initial condition. How/why is it possible to express it as an integeral? $\endgroup$
    – user33568
    Commented Feb 17, 2018 at 16:07
  • $\begingroup$ I would agree with everything except the last sentence. $\endgroup$
    – user28715
    Commented Feb 17, 2018 at 18:01
  • 1
    $\begingroup$ @StanleyPawlukiewicz: A system for which part of its response is independent of the input cannot be linear. It cannot be described by an impulse response. I consider $i(t)$ as the input. The internal state is not an input, unless one wishes to define it as such. $\endgroup$
    – Matt L.
    Commented Feb 17, 2018 at 19:31
  • 1
    $\begingroup$ @StanleyPawlukiewicz: The idea of initial conditions is that they are given and that they have nothing to do with the input signal that you choose to be the input to the system. A capacitor could be initially charged because you took it out of a different circuit before using it in your circuit. In discrete time it often happens that people allocate memory and forget to initialize it, which will result in non-zero initial conditions which are unrelated to your input signal. Such a system is not LTI, even though I agree that in practice this hardly matters as long as the system is stable. $\endgroup$
    – Matt L.
    Commented Feb 18, 2018 at 10:44

capacitor voltages and inductor currents are state variables.

A system is linear if it is both zero state linear ( initial conditions are zero and nonzero input) and zero input ( nonzero initial condition with zero input) linear.

All possible configurations are a superposition of the zero input and zero state outputs.

Most Signal Processing texts use zero state linearity as a definition but the lower limit of the integral is minus infinity, or in other words, the total input history including the history prior to $t=0$.

Control Theory texts tend to break down zero state and zero input linearity because Control Systems are Causal in nature, while in Signal Processing one can process a recorded signal in reverse time as well as forward time.

For zero input linearity $i(t)=0 \;\text{for}\; t >0 $:

$$ v_{i}^{-}(t_0)=\frac{1}{C}\int_{-\infty}^{t_0}i(t)dt $$

$$ v_{i_{1}}(t_0)=\frac{1}{C}\int_{-\infty}^{t_0}i_{1}(t)dt \quad\text{also} \quad v_{i_{2}}(t_0)=\frac{1}{C}\int_{-\infty}^{t_0}i_{2}(t)dt $$ Let $ i(t)=\alpha i_1(t) + \beta i_2(t)$ $$ v_{i}^{-}(t_0)=\frac{1}{C}\int_{-\infty}^{t_0}\alpha i_{1}(t)+\beta i_2(t) dt = \alpha\frac{1}{C}\int_{-\infty}^{t_0}i_{1}(t)dt +\beta \frac{1}{C}\int_{-\infty}^{t_0}i_{2}(t)dt $$ which is zero input linear For zero state linearity $i(t)=0 \;\text{for}\; t \le 0 $:

$$ v_{i}^{+}(t_0)=\frac{1}{C}\int_{-\infty}^{t_0}i(t)dt $$

$$ v_{i_{1}}(t)=\frac{1}{C}\int_{t_0}^{t}i_{1}(t)dt \quad\text{also} \quad v_{i_{2}}(t_0)=\frac{1}{C}\int_{t_0}^{t}i_{2}(t)dt $$ Let $ i(t)=\alpha i_1(t) + \beta i_2(t)$ $$ v_{i}^{+}(t)=\frac{1}{C}\int_{t_0}^{t}\alpha i_{1}(t)+\beta i_2(t) dt = \alpha\frac{1}{C}\int_{t_0}^{t}i_{1}(t)dt +\beta \frac{1}{C}\int_{t_0}^{t}i_{2}(t)dt $$ which is zero state linear and since (linear and time invariant)$$ i(t)=i^{-}(t)+i^{+}(t) \Rightarrow v(t)=v^{-}(t)+v^{+}(t)$$

A capacitor is linear.

Essentially, a capacitor has memory and all inputs prior to $t_0$ are contained in $v(t_0)$. If $i(t)$ was random, we would call this a Markov property but it is also true in the deterministic case.

I might add that you asked a very similar question Linear Constant Coefficient Differential Equations: Zero-Input and Zero-State responses

It would seem that you are fishing for an answer


Your first equation: $$ i(t) = C \frac{dv_c}{dt} $$ is the fundamental governing equation of the system. To solve for $v_c(t)$, you must integrate both sides of the equation, but the choice of integration limits is entirely up to the discretion of the person doing the integration. In other words, both sides of the equation are still equal, regardless of the limits you choose.

This, however, does not always produce a useful solution. You must choose limits that are appropriate to the problem at hand. You nearly always choose $t_0$ to be the time instant where your initial conditions are known at, and start your time axis at this point, i.e. $t_0 = 0$. The upper limit is traditionally chosen to be the free variable $t$, such that $t > 0$ to study the response at times after the initial conditions are given.

For example, when the Laplace transform is used to solve initial value problems, the one-sided Laplace transform is used because it starts its integration at $t_0 = 0$.


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