I want to find the ETFE (Empirical Transfer Function Estimate) of the system $G(e^{j\omega})$:
Where $H(e^{j\omega})$ is some filter that zero-mean white Gaussian noise $e(k)$ passes through. Let's assume that the input $u(k)$ is periodic with period $M$. Taking the Discrete Fourier transform, I have:
$$ Y_N(e^{j\omega_n})=\sum_{k=0}^{N-1}{y(k)e^{-j\omega_nk}}\quad\text{where }\omega_n=\frac{2\pi n}{N}\quad n=0,1,\ldots,N-1 \tag{1}$$
$$ U_N(e^{j\omega_n})=\sum_{k=0}^{N-1}{u(k)e^{-j\omega_nk}}\quad\text{where }\omega_n=\frac{2\pi n}{N}\quad n=0,1,\ldots,N-1 \tag{2}$$
$$ V_N(e^{j\omega_n})=\sum_{k=0}^{N-1}{v(k)e^{-j\omega_nk}}\quad\text{where }\omega_n=\frac{2\pi n}{N}\quad n=0,1,\ldots,N-1 \tag{3}$$
Therefore, I have:
$$ Y_N(e^{j\omega_n})=G(e^{kj\omega_n})U_N(e^{j\omega_n})+V_N(e^{j\omega_n})+R_N(e^{j\omega_n}) \tag{4}$$
Where $N=r\cdot M$ where $r\in \mathbb Z$ is an integer number of periods. I also added $R_N(e^{j\omega_n})$ as the transient part of the response (it would die out for example after the first period of the response). The ETFE is then:
$$ \hat G_N(e^{kj\omega_n})=\frac{Y_N(e^{j\omega_n})}{U_N(e^{j\omega_n})}=G(e^{kj\omega_n})+\frac{R_N(e^{j\omega_n})}{U_N(e^{j\omega_n})}+\frac{V_N(e^{j\omega_n})}{U_N(e^{j\omega_n})} \tag{5}$$
Let's define the error between the ETFE and the true transfer function as:
$$ E_N(e^{j\omega_n})=G(e^{kj\omega_n})-\hat G_N(e^{kj\omega_n}) \tag{6}$$
In my System Identification course, the professor presented two case scenarios. One where input $u(k)$ is zero-mean Gaussian non-periodic noise of length $N$, and one where $u(k)$ is some zero-mean Gaussian noise sequence of length $M$ repeated $r$ times - which makes a periodic signal of sorts. Here is the plot of the norm of $E_N(e^{j\omega_n})$ in the non-periodic case:
And here is the plot of the norm of $E_N(e^{j\omega_n})$ in the periodic case:
In the second picture, the period $M=128$ so for example for $N=128$ we have $r=1$ (only one period was considered) and for $N=512$ we have $r=4$ (four periods were considered).
Also in both of the above figures, the professor took the expectation of the error $E_N$, i.e. he actually ran the experiment 1000 times and took the average $E_N$ for each frequency, which is what helps to give the "smooth" error curve.
My question: I do not understand what makes the error $E_N$ stay the same as we increase $N$ for the non-periodic input case, but makes $E_N$ decrease in the periodic case as we increase the number of periods that we collect.
Thank you for your help!
