# Impulse response of a continuous time system

Question

The impulse response of a continuous time system is $h(t) = e^{-t}u(t)$. When $x(t) = u(t - 1)$ is input to this system what is the value of the output signal $y(t)$ at $t = 2.34$.

My Attempt

For convolution $y(t) = h(t) \star x(t)$ \begin{align} x(2.27) &= u(2.34-1) = 1\\ h(2.27) &= e^{-2.34}\cdot u(2.34) = 0.09632763823\\ \Longrightarrow y(t) &= 1 \cdot 0.09632 = 0.096 \end{align}

But the answer is wrong. Not too sure where my mistake is. Could someone correct my error?

Second Attempt \begin{align} y(t) &= ∫x(\tau)h(t-\tau)d\tau \quad\text{with limits negative and positive infinity}\\ &= ∫u(\tau-1)e^{-\tau}u(t-\tau)d\tau \\ &= ∫e^{-\tau}u(\tau-1)u(t-\tau)d\tau \\ &= ∫e^{-\tau}d\tau \quad \text{with limits 1 to 2.34}\\ &= 0.2715518029 \end{align}

Though the answer still appears to be wrong.

• You need to do the convolution integral and then substitute $t=2.34$ into it, not before. – Peter K. May 8 '16 at 16:27
• @PeterK. That's the "hard way" of doing it and likely to end in disaster. All the OP needs to do is calculate $$y(2.34) = \int_{-\infty}^\infty h(t)x(2.34-t)\,\mathrm dt = \int_0^\infty e^{-t}x(2.34-t)\,\mathrm dt = \int_0^? e^{-t}\,\mathrm dt,$$ which is a lot easier if the OP can figure out why the lower limit got changed to $0$ and what that $?$ in the upper limit should be in the integral on the right. – Dilip Sarwate May 8 '16 at 17:58
• @Dilip Sarwate I have edited my post after attempting the method you described but not to clear as to why your using 0 over 1. Also if the ? is supposed to be 2.34 i have calculated using both 0 and 1 and have gotten the wrong answer – sky knight May 8 '16 at 18:09
• Sorry, I had typos in what I wrote. Please replace $u(2.34-t)$ by $x(2.34-t)$ in two places in the displayed equation. – Dilip Sarwate May 8 '16 at 18:17
• @DilipSarwate I've edited your previous comment with that correction. – Peter K. May 8 '16 at 18:40

Given a continuous time LTI system with impulse response $h(t) = e^{-t}\text{u}(t)$ the output $y(t)$ for any input $x(t)$ is found via the convolution-integral: $$y(t) = x(t) \star h(t) = \int_{-\infty}^{\infty} {x(\tau)h(t-\tau)} d\tau$$
Specifically when $x(t) = \text{u}(t-1)$, we can proceed with the most general approach by first taking the integral and then inserting t=2.34, or else as @DilipSarwate suggests, you can first substitude t=2.34 into the integrand and then take the integral. I prefer the first here: \begin{align} y(t) &= \int_{-\infty}^{\infty} {\text{u}(\tau-1)e^{-(t-\tau)} \text{u}(t-\tau)d\tau}\\ y(t) &= \int_{\tau=1}^{\tau=t} {e^{-(t-\tau)}d\tau}\\ y(t) &= 0 &\scriptstyle{\text{for t less than 1}}\\ y(t) &= e^{-(t-\tau)} \big\vert_{\tau=1}^{\tau=t} = 1 - e^{-(t-1)} &\scriptstyle{\text{ for t greater than 1, hence:}}\\ y(t) &= [1 - e^{-(t-1)}] \text{u}(t-1) &\scriptstyle{\text{, for all t}}\\ \end{align}
I hope it's clear how the integral limits are modified by the arguments of the $\text{u}(t)$ step function.
And inserting $t=2.34$ yields:
$$y(2.34) = [1 - e^{-(2.34-1)}] \text{u}(2.34-1) = [1 -e^{-1.34}]\text{u}(1.34) = 0.7382$$