# What is the steady state response of a system of two exponentials to sinusoidal excitation?

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:

Example:

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: ????

Thanks!!!!

• Please format your equations – LJSilver Dec 7 '16 at 20:22
• Anyway you just have to take the convolution among the impulse response and the input signal – LJSilver Dec 7 '16 at 20:23
• 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. – Maximilian Matthé Dec 7 '16 at 20:31
• For the sake of correctness, "double exponential" is not what you have written. It is something like $a^{b^{c}}$. – msm Dec 7 '16 at 23:43

## 1 Answer

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:

inverselaplacetransform((c/(s+1/m)+d/(s+1/n))*(a/s+bw/(s^2+w^2)))