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I was wondering if the following system is time-invariant or time-varying $$ y(t)=e^{-t}\int_{0}^{t}e^{\sigma}x(\sigma+1)d\sigma$$ I know that it's not causal, and I think it's time-varying because of that \$e^{-t}\$ before the integral that makes the input delay and the delay of the output not equal, I am not sure if I am right.

Thank you.

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  • \$\begingroup\$ \$e^{-t}\$ is not a delay, and does not render the system time variant (confusing with \$e^{-sT}\$)? What function of \$\sigma\$ is \$x(\sigma +1)\$? \$\endgroup\$ – Chu Apr 19 at 9:32
  • \$\begingroup\$ @Chu x(t) is the input, the variable is formally changed to a dummy integration variable to avoid double meanings. Variable t is used as integration limit. \$\endgroup\$ – user287001 Apr 19 at 10:16
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Homework, I guess => only guidance is available.

No doubt, this is a linear system. Calculate the response y(t) when x(t) is the impulse function del(t). With that input the integration is trivial.

Then calculate the response with delayed impulse input del(t-A). The integration is still trivial. Compare the responses. The system is time-varying if there's more difference between the responses than the used time delay A. If you find even one A which causes more difference than just the time delay =A, the system is proven to be time variant. To prove the system to be time independent, you must show that every A, positive or negative, causes amount A delayed response.

BTW. You have some dim areas to be cleared at first. As commented, factor exp(-t) is NOT a time delay. You are now in time domain. Integrations stay non-trivial until you understand the sampling property of the impulse function.

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