# Tag Info

Initial conditions are always given at $t=0^-$, because they define the state of the system before any input is applied, and - by definition - the input is applied at $t=0$. The state at $t=0^+$ is determined by the initial conditions as well as by the input signal. The unilateral Laplace transform can be used to solve LCCDEs with initial conditions $y(0^-), ... 1 If you add "dissipative" to your passivity requirement, then yes to both. A perfectly lossless LC system (or any other perfectly lossless resonant oscillator) is neither BIBO stable, nor is it Lyapunov stable. (I had to look it up, but Lyapunov stability basically says that if the system is stable around an equilibrium point, then over time it'll ... 2 A counter-example would be an ideal LC-circuit, which is passive but not BIBO stable: if excited at its resonance frequency, the output grows without bounds. Such systems are sometimes called marginally stable, because they have poles on the stability boundary (the imaginary axis). Note that if a system is not lossless, i.e., if there is a positive ... 1 A second-order system without damping (such as an ideal LC circuit) will not produce any output without input and with zero initial conditions. It will produce output only if there is either a non-zero input signal or non-zero initial conditions. One very common definition of linearity in system theory is that if$y_1(t)$and$y_2(t)$are the responses to ... 1 An ideal LC circuit has no resistance, and hence no resisitive loss mechanism. Therefore if it starts to oscillate (due to, for example, a non-zero initial voltage or current stored in the L or C), then it will oscillate indefinetely on its own, yielding a situation as described in your question. Zero input response of a linear system is the response under ... 2 Clearly, for negative values of$t$, the system needs to know the future in order to determine its output. Hence, the system can't be causal. Since the system is also time-varying (show it!), its response to an impulse doesn't say much about its general behavior, unlike it would be the case for a linear time-invariant (LTI) system. So the given system's ... 3 I think your understanding is correct. The steady state response of an LTI system is the part of the response that is caused by a steady excitation at the input, like a sinusoidal signal, or any other periodic signal. The transient response is caused by changes at the input, like switching on a signal, or changing the parameters of a periodic input signal, ... 2 Sequences that are "time-invariant" must be constants (not varying with time at all). What the OP is actually asking about is a (discrete-time) system whose response$y$to the input$x$is defined to be $$y[n] = \begin{cases}\sum_{k=n_0}^n x[k], & n \geq n_0,\\??? & n < n_0\end{cases}$$ where the$???\$ is there because according to some ...