# Difference between the two forms of input-output relationships of an LTI system

$$y(t)=f(t)*h(t)\tag{1}$$

$$y(t)=H(s)e^{st}\tag{2}$$

$$H(s)=\int_{-\infty}^{\infty}h(t)e^{-st}dt\tag{3}$$

Let $$f(t)$$ in Eqn $$(1)$$ be $$e^{st}$$. In many worked out examples, I have found that the two equations/formulae give different responses. To be specific, the actual convolution (Eqn $$(1)$$) is giving the zero-state response, while the Eqn $$(2)$$ involving $$H(s)$$ is giving the forced response. Recognising that Forced response and Zero-State response are not equal (Z-S responses may involve characteristic modes of the system, while the Forced Response is the non-characteristic mode part of the Total response), I am confused what is the intuitive difference between the two equations? Which equation should be primarily used for finding the response to an input to a system? Note that in the examples, the inputs $$e^{st}$$ were multiplied by $$u(t)$$.

• Two things: first, please start using Latex for the formulas in your questions and second, please provide an example where the two give different results. – Matt L. Sep 1 at 18:19
• Sorry but I am on a mobile device.. Is there any way to access this Latex on mobile? I am unaware actually – Nullbyte Sep 1 at 18:21
• Just enclose the commands between dollar signs. Check my edit. – Matt L. Sep 1 at 18:29
• Thank you sir!! I'll use it from now.. Sorry fr the inconvenience..😅 – Nullbyte Sep 1 at 18:32

For an LTI system with impulse response $$h(t)$$ and transfer function $$H(s)=\mathcal{L}\{h(t)\}$$, an input signal $$x(t)=e^{st}$$ (note: no multiplication with the unit step function $$u(t)$$), results in an output
\begin{align}y(t)&=(x\star h)(t)\\&=\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau\\&=\int_{-\infty}^{\infty}h(\tau)e^{s(t-\tau)}d\tau\\&=e^{st}\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}d\tau\\&=e^{st}H(s)\tag{1}\end{align}
If the input signal is given by $$x(t)=e^{st}u(t)$$ then the output is not given by $$(1)$$, but it needs to be determined from the convolution integral:
\begin{align}y(t)&=\int_{-\infty}^{\infty}h(\tau)e^{s(t-\tau)}u(t-\tau)d\tau\\&=\int_{-\infty}^th(\tau)e^{s(t-\tau)}d\tau\\&=e^{st}\int_{-\infty}^th(\tau)e^{-s\tau}d\tau\end{align}