What is the two exponential steady state response?

Here is an example solution for sinusoidal excitation of a system having a single exponential response to a impulse excitation:


Excitation: impulse: $L(t) = \delta(t)$

Impulse response of system: $I(t) = I_o e^{-t/\lambda}$

Excitation: sinusoid: $L(t) = a + b \sin(\omega t)$

Sinusoid response of system: $I(t) = I_o \lambda \left(a + \frac{b}{\sqrt{1 + (\omega \lambda)^2}} \cos(\omega t - θ)\right)$

$\tan(\theta) = \omega \lambda$

Question: What is a similar solution for a system that has a double exponential impulse response?

Excitation: impulse: $L(t) = \delta(t)$

Impulse response of system: $I(t) = I_{o1} e^{-t/\lambda_1} + I_{o2} e^{-t/\lambda_2}$

Excitation: sinusoid: $L(t) = a + b \sin(\omega t)$

Sinusoid response of system: ????


  • 1
    $\begingroup$ Please format your equations $\endgroup$
    – LJSilver
    Dec 7, 2016 at 20:22
  • $\begingroup$ Anyway you just have to take the convolution among the impulse response and the input signal $\endgroup$
    – LJSilver
    Dec 7, 2016 at 20:23
  • 1
    $\begingroup$ Since you have the solution for a single exponential already given, you can easily exploit the linearity of the system: The output is the sum of the responses of the two exponentials. $\endgroup$ Dec 7, 2016 at 20:31
  • $\begingroup$ For the sake of correctness, "double exponential" is not what you have written. It is something like $a^{b^{c}}$. $\endgroup$
    – msm
    Dec 7, 2016 at 23:43

1 Answer 1


The system with impulse response: $$I(t)=I_1e^{-t/\lambda_1}+I_2e^{-t/\lambda_2}$$

with Laplace Transform: $$I(s)=I_1\frac{1}{s+1/\lambda_1}+I_2\frac{1}{s+1/\lambda_2}$$

and the input: $$L(t)=a+bsin(\omega)$$

with Laplace Transform: $$L(s)=\frac{a}{s}+\frac{bw}{s^2+w^2}$$

will obtain the following output, in Laplace Domain: $$IL(s)=I(s)L(s)=(I_1\frac{1}{s+1/\lambda_1}+I_2\frac{1}{s+1/\lambda_2}) (\frac{a}{s}+\frac{bw}{s^2+w^2})$$

with this very simple Inverse Laplace Transform, as the final solution: $$IL(t)= -\frac{I_1 \lambda_1 e^{-t/\lambda_1} (a \lambda_1^2 w^2 + a - b \lambda_1 w)}{\lambda_1^2 w^2 + 1} - \frac{I_2 \lambda_2 e^{-t/\lambda_2} (a \lambda_2^2 w^2 + a - b \lambda_2 w)}{\lambda_2^2 w^2 + 1} + a I_1 \lambda_1 + a I_2 \lambda_2 + \frac{\{b (sin(t w) (I_1 (\lambda_1 \lambda_2^2 w^2 + \lambda_1) + I_2 \lambda_2 (\lambda_1^2 w^2 + 1)) - w cos(t w) (I_1 \lambda_1^2 (\lambda_2^2 w^2 + 1) + I_2 \lambda_2^2 (\lambda_1^2 w^2 + 1)))\}}{(\lambda_1^2 w^2 + 1) (\lambda_2^2 w^2 + 1)}$$

As you can see, this kind of symbolisms can be readily solved in Wolfram Alpha through this:


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.