# Finding impulse response by a non-Laplace-transform method

The input function is $$x(t)=e^{-t}u(t)$$ and the corresponding output function of a linear system is $$y(t)=10e^{-t}\cos(4t)u(t)$$ where $$\u(t)\$$ is unit step function.

I understand how to find the impulse response of the system using transfer function of Laplace transform. The other way would be to experimentally try to determine the impulse response.

My question is that if there is any other way other than Laplace transform or other transform to find the impulse response of system. So, by using some time domain tool.
Thank you.

Edit: This is not a homework related problem. Actually I was doing self-study and was motivated by this example problem: http://imageshack.com/a/img923/6360/i1N0U3.jpg

• You can use any orthogonal transform, but Laplace transforms are the easiest for IMPULSE response, because Laplace is the impulse transform. Apr 23, 2020 at 4:24
• Isn't it possible to use some time domain tool rather than any transform? Apr 23, 2020 at 6:15
• How would you solve $10e^{-t}\cos(4t)u(t) = \int_{-\infty}^{\infty} e^{-\tau}u(\tau) h(t - \tau) d\tau$ for $h(t)$? Apr 23, 2020 at 7:30
• @Huisman By taking derivative of both sides? We need to get rid of integral on the right side and as far as I know integration and differentiation cancel each other. Apr 23, 2020 at 18:22
• @PG1995 it won't be as simple as that, the derive w.r.t. $t$ will not give you much to go on, and neither will the derivative w.r.t. $\tau$, the Laplace transform is the easiest way.
– jDAQ
Apr 23, 2020 at 18:28

Great question. Of course, transforms are the best way to solve this, which you already know. But maybe when you start with a bunch of numbers from an oscilloscope instead of a nice, neat formula, your perspective changes a bit. Here is another way to approach the problem numerically. It is really best suited if you have some a priori info to bring to the problem, but that is often the case for such problems.

In the time domain, the output is $$\y(t)=\int_{0}^tx(t')h(t-t')dt'\$$. In this case, you know y(t) and x(t) and want to know h(t-t'). You can treat it as a very important class of problems called Inverse Problems. Now, if you knew nothing at all about $$\h\$$, this might be quite a bit more difficult, but if you can narrow down the space of solutions a bit, you have a good chance. The idea is to parameterize a potential impulse-response function, and then optimize the solution over those parameters. If you can come up with an initial guess to launch the optimization calculation, you have a good chance, but no guarantee, of success.

Since LTI systems so often have solutions such as exponentials and decaying oscillations, it gives you a basis for an educated guess. For instance, in this case, you see that the output y(t) involves an oscillatory component with frequency $$\\omega = 4\$$, so I don't think it requires any great leap to guess a second-order system solution of a decaying exponential with frequency $$\\omega=4\$$, such as $$\h_o(t)=Ce^{-at}cos(\omega t + \phi)\$$, with $$\C\$$, $$\a\$$,$$\\phi\$$, and maybe $$\\omega\$$ as optimization parameters.

Next you numerically calculate an output $$\z(t)\$$ to the known input $$\x(t)\$$, using a starting guess for the optimization parameters, and compute a mean-square error $$\\int (y(t)-z(t))^2dt\$$. You run a non-linear optimization over the parameters $$\C\$$, $$\a\$$, and $$\\phi\$$ (and maybe $$\\omega\$$) to get the solution that minimizes this error.

There is no guarantee you come up with a good solution this way, and it is dependent upon coming up with a decent 'guess' parameterization, but for many problems, this is not so unreasonable.

For more complicated system response functions, such as higher-order systems, your chances of solving the optimization problem diminish. But this is actually not all that difficult a method to implement, and could succeed for a good number of problems.

Of course, all this described as such is really nothing more than fitting data to a potential solution. However, looking at it as the more general inverse problem that it is motivates the use of other techniques from inverse problems to help deal with noise and overfitting, such as regularization.

• I really appreciate your help. So, the only other way to find the impulse response is numerical optimization. I'm not sure if there is any time-domain analytical tool available. On the other hand, even to apply numerical optimization method, one needs to have an understanding of impulse response and equation, y(t)=∫x(t')h(t-t')dt'; {limit 0 to t} . You said: "Since LTI systems so often have solutions such as exponentials and decaying oscillations,.." IMHO, I'd rather say, "Since LTI systems so often have solutions such as decaying exponentials and decaying oscillations". Cont'd May 4, 2020 at 1:27
• It's more of a note to self and I know I'm splitting hairs. Once again, thanks. May 4, 2020 at 1:27
• @PG1995 I won't say there is no analytical tool available, but I don't know of one. As for decaying exponentials/oscillations, those are the only $\textit{stable}$ solutions. Growing exponentials are solutions, but not stable. Non-decaying oscillations are marginally stable solutions. Thanks for raising an interesting question. Cheers. May 4, 2020 at 1:32

I am sorry I got confused in my previous answer. Yes the system is LTI and there is a way to find the impulse response of a system from just time domain method. But you obviously need the differential equations for the system.
Consider the system which you have: $$\frac{d^2y(t)}{dt^2}+2\frac{dy(t)}{dt}+17y(t) = 10x(t)$$ To find the impulse response of this system, you solve this ODE for step response and differentiate the result to get the impulse response.
Find the natural response of the ODE, given by:
$$y_n(t) = Ae^{-t+4jt} + Be^{-t-4jt}$$ Forced response for unit step input will be: $$y_f(t) = \frac{10}{17}$$ Giving the total response: $$y(t) = Ae^{-t+4jt} + Be^{-t-4jt} + \frac{10}{17}$$ Assuming the system starts from rest, i.e., y(0) = 0 and y'(0) = 0, evaluate A and B, giving, $$y(t) = (\frac{-5}{17}+j\frac{1.25}{17})e^{-t+4jt}+(\frac{-5}{17}+j\frac{-1.25}{17})e^{-t-4jt} + \frac{10}{17}$$ $$y(t) = \frac{-10e^{-t}}{17}cos(4t) - \frac{2.5e^{-t}}{17}sin(4t) + \frac{10}{17}$$ The impulse response will be the differential of the step response, so: $$h(t) = y'(t) = (\frac{-5}{17}+j\frac{1.25}{17})(-1+4j)e^{-t+4jt}+(\frac{-5}{17}+j\frac{-1.25}{17})(-1-4j)e^{-t-4jt}$$ $$h(t) = -\frac{21.25j}{17}e^{-t}(e^{j4t} - e^{-j4t}) = 2.5e^{-t}sin(4t)$$

• OP clearly says $u(t)$ is the step function (as expected by the notation, though not necessarilly true), not some other input $x(t)$. Huisman gave the right hint in the comments (and it's best left to be a hint, because this looks like a homework). Apr 23, 2020 at 15:13
• I think what Huisman is saying is not valid here. Apr 23, 2020 at 15:19
• @sarthak if the system is "not LTI" the impulse would depend on $t$ and would not be linear, impulse of size 2 is not twice the impulse of size 1. I wouldn't be useful to know the impulse response in that case. What makes you think it is not LTI? Also, why are you characterizing the system as LPTV based on the description of the OP? (also, the tag itself is LTI-system)
– jDAQ
Apr 23, 2020 at 18:08
• @PG1995 The start of this answer is incorrect. The gain is not y(t)/(x(t). The gain is Y(s)/X(s). That's why people use the Laplace domain because then you can 'simply' divide output by input. In time domain, this approach is not valid. Apr 24, 2020 at 5:29
• Three main problems, OP is trying to find the ODE from $x(t)$ and $y(t)$ so I don't think you can assume to have it. Second, where did you get that impulse response? It seems oddly constant/simple for an impulse response. Third, if you assume the system starts "at rest" shouldn't your natural response be zero?
– jDAQ
Apr 24, 2020 at 7:17