# DSP - Are real exponential signals still eigen functions of LTI systems?

#### Hawx

Joined Nov 17, 2022
1
First sorry for context missmatching, when I'm adding this thread I cannot find DSP in contexts. My question is about Digital Signal Processing.

I know that complex exponentials are eigen functions of LTI systems for example e^{j2t}, e^{-j5t} , e^{j8t} .

If we can define complex exponential as e^{st} where s is a complex number. Can we say that e^t,e^{2t},e^{(2-j4)t} are still complex exponentials and they are eigenfunctions to LTI systems ?

I know that complex number includes real numbers and 1 , 2 and 2-j4 are indeed complex numbers. What I'm curious about is , when 2 or 2-j4 is exponent i.e. e^{2t},e^{(2-j4)t} are eigenfunctions to LTI systems?

Thanks for be interested.

#### Papabravo

Joined Feb 24, 2006
19,854
First sorry for context missmatching, when I'm adding this thread I cannot find DSP in contexts. My question is about Digital Signal Processing.

I know that complex exponentials are eigen functions of LTI systems for example e^{j2t}, e^{-j5t} , e^{j8t} .

If we can define complex exponential as e^{st} where s is a complex number. Can we say that e^t,e^{2t},e^{(2-j4)t} are still complex exponentials and they are eigenfunctions to LTI systems ?

I know that complex number includes real numbers and 1 , 2 and 2-j4 are indeed complex numbers. What I'm curious about is , when 2 or 2-j4 is exponent i.e. e^{2t},e^{(2-j4)t} are eigenfunctions to LTI systems?

Thanks for be interested.
Yes, they are. Real systems with bounded outputs will have complex exponentials with negative real parts. In the control engineering business we have a saying that: "The positive real part must disappear or the system will". Also note that using the law of exponents we can write the following:

$$e^{st}\;=\;e^{\sigma +j\omega t}\;=\;e^{\sigma}e^{j\omega t}$$

You should always think of the real part as representing a decaying exponential in a real system.

• Hawx