[proofplan] The proof is the standard zero-initial-condition transfer-function computation. Take the [Laplace transform](/page/Laplace%20Transform) of the linear error equation, solve algebraically for the transformed error in terms of the transformed measurement noise, and read off the multiplier. [/proofplan] [step:Transform the error equation] For transfer functions we use zero initial condition $\tilde{x}(0)=0$. Let $\widetilde{X}(s)$ denote the Laplace transform of $\tilde{x}$ and let $N(s)$ denote the Laplace transform of $n_{\mathrm{meas}}$. Applying the Laplace transform to \begin{align*} \dot{\tilde{x}}(t)=(A-HC)\tilde{x}(t)-Hn_{\mathrm{meas}}(t) \end{align*} gives \begin{align*} s\widetilde{X}(s)=(A-HC)\widetilde{X}(s)-HN(s). \end{align*} Equivalently, \begin{align*} (sI_n-A+HC)\widetilde{X}(s)=-HN(s). \end{align*} [/step] [step:Solve for the input-output multiplier] Since $A-HC$ is Hurwitz, every eigenvalue of $A-HC$ has negative real part. Hence $sI_n-(A-HC)=sI_n-A+HC$ is invertible for every $s$ in the usual closed-loop transfer-function domain away from the poles. Multiplying by its inverse gives \begin{align*} \widetilde{X}(s)=-(sI_n-A+HC)^{-1}HN(s). \end{align*} Therefore the transfer function multiplying the measurement-noise transform $N(s)$ is \begin{align*} G_{n\to \tilde{x}}(s)=-(sI_n-A+HC)^{-1}H. \end{align*} This is the claimed observer error noise map. [guided] A transfer function is computed with zero initial condition. Under that convention, the Laplace transform of $\dot{\tilde{x}}$ is $s\widetilde{X}(s)$. Transforming the differential equation gives $s\widetilde{X}(s)=(A-HC)\widetilde{X}(s)-HN(s)$. Moving the state term to the left gives $(sI_n-A+HC)\widetilde{X}(s)=-HN(s)$. Wherever the resolvent exists, solve for $\widetilde{X}(s)$ by multiplying by $(sI_n-A+HC)^{-1}$. The multiplier of $N(s)$ is exactly $-(sI_n-A+HC)^{-1}H$. [/guided] [/step]