I know that generally the impulse response of a system is the output when input is an impulse. It is also the differentiation of a step response. So if we have a system described by thi differential equation: $$\frac{d^2 y(t)}{dt^2}+5\frac{\text{d}y(t)}{\text{d}t} +6y(t)=-\frac{\text{d}x(t)}{\text{d}t}$$ where $x(t)$ is the input, $y(t)$ is the output, how do I find the impulse response?

My first idea was to let $x(t)$ be a ramp function so that it's derivative would be a step function then I could easily solve the differential equation and differentiate what I got to get the step response then differentiate it again to get the impulse response. But now I'm doubting this path as the ramp function is not a bounded function so the response might be unstable. I'm not sure how to approach this.

  • $\begingroup$ Can you use the Laplace transform, or are you supposed to solve it in the time domain? $\endgroup$
    – Matt L.
    Commented Jan 13, 2017 at 17:39
  • $\begingroup$ @MattL. I can solve in time or frequency domain $\endgroup$
    – user281270
    Commented Jan 13, 2017 at 17:40
  • $\begingroup$ Impulse response is the inverse Laplace transform of H(s) = Y(s)/X(s), where X(s) and Y(s) are the Laplace transforms of x(t) and y(t), respectively. You can also use some properties related to time derivatives and Laplace transforms. en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems $\endgroup$
    – Atul Ingle
    Commented Jan 13, 2017 at 17:41
  • $\begingroup$ Assuming that you specifically want a time-domain solution for finding the impulse response of a continuous time system, I will replicate one of my previous answer which gives you the necessary steps in finding so. $\endgroup$
    – Fat32
    Commented Jan 13, 2017 at 17:41
  • $\begingroup$ The course notes linked to on this MIT OCW page give a very good treatment of time domain methods for deriving impulse and step responses of LTI systems. $\endgroup$
    – Matt L.
    Commented Jan 14, 2017 at 9:22

2 Answers 2


Let me provide a method, applied only for finding the impulse response $h(t)$ of an LTI system characterised by an LCCDE of the form $ \sum_{k=0}^{N}{ a_k {{d^k y(t)}\over {dt^k}}} = \sum_{k=0}^{M}{ b_k {{d^k x(t)}\over {dt^k}}}$ by using the classical time domain approach. A method which is ignored as it's replaced by transform domain techniques instead...

Note that the input $x(t) = \delta(t)$ is formally a problematic signal (function) and therefore mathematically oriented classical books on ODEs, do generally avoid any discussion of such functions and their solutions, unless the scope of the book specifically includes generalised functions (distributions), usually to be used in some engineering or physical fields.

The procedure: Consider an LTI system which is causal with initial rest conditions. We say that when the input to this system is $x(t)=\delta(t)$ an impulse, then its output $y(t)=h(t)$ is the impulse-response of the system. We find this solution $h(t)$ in two steps, by breaking the LCCDE into two parts as inspired by a serial (cascade) implementation of two LTI systems corresponding to the right and left sides of the LCCDE.

I/O relationship of Part-I is given by the LCCDE: $$\sum_{k=0}^{N}{ a_k {{d^k y(t)}\over {dt^k}}} = x(t)$$ which represents the first stage with its input $x(t)=\delta(t)$ and we denote its solution (stage1-output) as $h_0(t)$ which is actually the impulse response of Part-I.

And the I/O relationship of the Part-II is given by the equation: $$ y(t) = \sum_{k=0}^{M}{b_k{{d^k x(t)}\over {dt^k}}} $$ which requires nothing but simple summation of its input, $x(t) = h_0(t)$ ,and its derivatives to compute the output as $$h(t) = \sum_{k=0}^{M}{b_k {{d^k h_0(t)}\over {dt^k}}}$$ Therefore we need to find $h_0(t)$ of the Part-I to simply compute the impulse response $h(t)$ of the complete system.

To find the solution of part-I: $\sum_{k=0}^{N}{ a_k {{d^k y(t)}\over {dt^k}}} = x(t)$ whose solution $y(t)$ is what we call $h_0(t)$ , we we make the following observation that $y(t)$ is composed of two parts as $$y(t) = y_{h}(t) + y_{p}(t)$$ where $y_{h}(t)$ is the homogeneous solution corresponding to $x(t)=0$ for all t, and $y_{p}(t)$ is the particular solution corresponding to input $x(t)$ for $ t > 0$, in particular as what remains as input when time goes to infinity.

Now, we can see that for the input $x(t)=\delta(t)$ the particular solution $y_p(t)$ for $t>0$ is zero as the input is zero for $t > 0$. Then the output simplifies to $h_0(t) = y(t) = y_h(t)$

Now, we have to find the homogeneous part of $y(t)$ as the solution of the equation $$\sum_{k=0}^{N}{ a_k {{d^k y(t)}\over {dt^k}}} = 0$$ following the usual procedure by finding the roots of its characteristic polynomial in complex-s: $$p(s) = \sum_{k=0}^{N}{a_k s^k}$$

which is obtained when we simply insert $y_h(t) = Ke^{st}$ as a solution canditade to the homogeneous differential equation $\sum_{k=0}^{N}{ a_k {{d^k y(t)}\over {dt^k}}} = 0$. The solution is then hypothesised to be $$y_h(t) = \sum_{k=0}^{N}{ A_k e^{st}}$$ for the case when all the roots are distinct. If they are not, you know what to do right ;) The solution therefore now requires to determine the N unknown coefficients $A_k$ from N-inital conditions, at time t=0 for the framework of this method, possesed by the output $y(t)$ and its derivates up to order N-1.

Remember that this system was defined to be under inital rest with complete zero inital conditions prior to the application of any excitation. Hence if we were to solve the homogeneous solution $y_h(t)$ for an ordinary (formally valid) input signal $x(t)$ to the system, the homogeneous solution will be identically zero, as the initial conditions would all be. (all the coefficients $A_k$ of the exponential terms turn out to be zero). But the magic of the generalised input $\delta(t)$ is that it will set one of those inital conditions to non-zero, thereby, enabling a non-zero homogeneous response to exist, which will become the solution of the part-I as well.

It can be shown that the inital conditions of the part-I due to the excitation $\delta(t)$ will be set as $y(t)=0 ~,~ y(t)' = 0 ~,~ y(t)''=0 ~,~... ~,~ y(t)^{N-1}=1/a_N$ all at $t=0$.

From this information you can proceed to solve the algebraic equaitons as usual and find $h_0(t)$ from which you can compute the impulse response $h(t)$ for all $t > 0$. What happens for all $t <0 $? As we stated, this system was LTI and causal with initial rest condition. For such a system if the input $x(t)$ is zero for all $t<0$ then that output is also zero for all $t<0$ which is achieved by multiplying $h_0(t)$ via $u(t)$ to represent the solution of Part-I for all t.

It is instructive to apply this procedure to your example: $$y'' + 5 y' + 6y = -x'$$

Lets first solve Part-I: $$y'' + 5 y' + 6y = x$$ To find $h_0(t)$ when the input is $x(t)$ = $\delta(t)$ first solve the characteristic equation: $s^2 + 5s + 6 = 0$ which yields $s=-2$ and $s=-3$.

The proposed homogeneous solution is therefore $$y_h(t) = A_1 e^{-2t} + A_2 e^{-3t}$$

For which we need 2 inital conditions to find $A_1$ and $A_2$. These initial conditions are set by the impulse $\delta(t)$ to be: $y(0)=0$ and $y(0)'=1/1=1$. Using these initial conditions we find $A_1$ and $A_2$ to be $A_1=1$ and $A_2=-1$ and this yields the solution to part-I as $$ h_0(t) = [ e^{-2t} - e^{-3t}] $$ for all $t > 0$.

Also note that this system was LTI and causal with initial rest and input $x(t)=\delta(t)$ is zero for all $t<0$ implies that output is also zero for all $t<0$ which is simplified by using a unit step $u(t)$ function to format the output as: $$ h_0(t) = [ e^{-2t} - e^{-3t}]u(t)$$ for all t.

Finally, to solve part-II and to get the impulse response of the overall system, apply this $h_0(t)$ as an input to Part-II whose output is simply

$$y(t)= -x(t)' \longrightarrow h(t)= -h_0(t)'$$

$$h(t) = -[ -2 e^{-2t} + 3 e^{-3t}]u(t) - [e^{-2t} - e^{-3t}]\delta(t) $$
$$h(t) = -[ -2 e^{-2t} + 3 e^{-3t}]u(t) - [1 - 1]\delta(t) $$ $$h(t) = -[ -2 e^{-2t} + 3 e^{-3t}]u(t) - 0\delta(t) $$ $$h(t) = [2 e^{-2t} - 3 e^{-3t}]u(t)$$

Note that the impulse at the output is cancelled by the property $f(t) \delta(t-a) = f(a)\delta(t-a)$

  • $\begingroup$ Honestly, I don't really see the point of copying a previous answer as an answer to this question, instead of just linking to it in a comment. Of course, the same method can be used here too, but that's what links are for. The specific solution for $h(t)$ shown in this answer is of course not equal to the one solving the current question, which might cause confusion. $\endgroup$
    – Matt L.
    Commented Jan 14, 2017 at 8:52
  • $\begingroup$ I've changed the example solution to the OP's specific LCCDE. $\endgroup$
    – Fat32
    Commented Jan 14, 2017 at 12:33

You can solve such problems directly in the time domain (as shown in this answer to a related question), or by using the Laplace transform. I'll show you that latter approach. The Laplace transform of the given differential equation is


where $X(s)$ and $Y(s)$ are the Laplace transforms of the input $x(t)$ and the output $y(t)$, respectively. Rearranging $(1)$ gives


where $H(s)$ is the system's transfer function. The impulse response $h(t)$ is the inverse Laplace transform of the transfer function $H(s)$. $H(s)$ can be decomposed as follows:


from which $h(t)$ can be easily derived as



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