How to find the impulse response with input and output given?

Question

The Question:

A CT signal x(t), which is non-zero only over the time interval, t = [-2,3] is applied to an LTIC system with impulse response h(t). The output y(t) is observed to be non-zero only over the time interval t = [-5,6]. Determine the time interval in which the impulse response h(t) of the system is possibly non-zero.

I understand how to find the output given the input and impulse response h(t) with the convolution integral either through integration or the graphical method, but I can't figure out how you can find out the impulse response given just this information. Am I just suppose to guess?

Answer 1

Hint: I guess you realize that you don't need to find the impulse response, but you just need to find the interval where it is possibly non-zero. From your experience with convolution you probably know that the length of the result is the sum of the lengths of the two convolved functions. In your example, the length of the result is $6-(-5)=11$, and the length of the input signal is $3-(-2)=5$. This means that the length of the impulse response is $11-5=6$. Now you just need to find out where it starts. In order to do this draw a (simple) signal between $-2$ and $3$ and graphically convolve it with a (simple) signal of length $6$. Look at the interval where the result is (possibly) non-zero. From this little experiment you will be able to find out where $h(t)$ needs to start to get the desired non-zero interval of the output signal.

Answer 2

Hint: note that $$ y(t) = \int_{-\infty}^\infty x(\tau)h(t-\tau)\,d\tau $$ For what values of $t$ can we guarantee that the product $x(\tau)h(t-\tau)$ is equal to zero for all values of $\tau$? That is, for what values of $t$ can we guarantee that, for every $\tau$, either $x(\tau) = 0$ or $h(t-\tau) = 0$?.

Answer 3

I will use the convolution theorem as $x(t)$ is non zero within t = [-2,3]. $x(t)*h(t)=y(t)$, from the convolution theorem: convolution in time domain (your case) and multiplication in frequency domain; imply: $\mathcal{F}\big\{x(t)*h(t)\big\}=\mathcal{F}\big\{x(t)\big\}\times\mathcal{F}\big\{h(t)\big\}$ then $h(t)=\mathcal{F}^{-1}\Bigg\{\frac{\mathcal{F}\big\{y(x)\big\}}{\mathcal{\mathcal{F}\big\{x(x)\big\}}}\Bigg\}$. $\mathcal{F}$ is the Fourier transform and $\mathcal{F}^{-1}$ its inverse.