Why is the Steady State Response described as steady state despite being multiplied to a negative exponential?

Question

I'm evaluating Newton's Temperature Model $$dT/dt = k(T_e-T)$$ to find the response of the system.$T_e$ and $T$ are both functions of t.

The response evaluates to $$T=e^{-kt}\int e^{ks} T_e(s)ds + T_0 e^{-kt}$$$$T(0)=T_0$$

where $T_0 e^{-kt}$, as my professor stated, is Transient Response as it gradually declines because of the $e^{-kt} \rightarrow$ 0 as $t \rightarrow \infty$ while $e^{-kt}\int e^{ks} T_e(s)ds$ is steady-state response as it tends to be steady as $t \rightarrow \infty$.

My question is the steady-state also has $e^{-kt}$ multiplied to it so shouldn't this $\rightarrow 0$ as $t \rightarrow \infty$. Then how can this be called steady-state solution.

Also how does steady state response $\rightarrow 0$ as $t \rightarrow 0$ and the intial response depends only on the transient resposne.

Answer 1

That's the same behaviour as an series RC circuit.

The point is that the input signal is not just $T_e(t)$, but actually $0$ till $t<0$ and then $T_e(t)$ for $0 \le t$, i.e. $H(t) T_e(t)$ with $H(t)$ being the Heaviside function.

The initial response is heavily dominated by the response to the step $T_e(0)$, which declines with time (transient) evidencing the response (steady state) to $T_e(t)$ as if it had been applied starting from $t= -\infty$.

Answer 2

There is an extra $t$ hiding in the low-quality notation used here. That should be $$e^{-kt}\int_0^t e^{ks} T_e(s)ds + T_0e^{-kt}$$

The $t$ upper limit can be confirmed by differentiating to check it against the differential equation. The $0$ lower limit is necessary for the claim $T(0) = T_0$ to be true.

Note that if $T_e(s)$ were constant, then the first term would also be constant, which is why it gets called "steady-state".