New answers tagged

0

In order to filter a discrete-time signal with a linear and time-invariant (LTI) filter, you need to implement a difference equation: $$y[n]=-a_1y[n-1]-a_2y[n-2]-\ldots -a_Ny[n-N]+\\+b_0x[n]+b_1x[n-1]+\ldots+b_Nx[n-N]\tag{1}$$ where $x[n]$ is the input sequence, $y[n]$ is the output sequence, and $a_i$ and $b_i$ are the filter coefficients, which determine ...


0

What you need to do is find a digital filter $H(z)$ that behaves similar to the $H(s)$ of your Butterworth filter. That is, you need to read about IIR implementation. There are several methods described in signal processing books about this subject. I recommend you to check for example "Essentials of Digital Signal Processing" from Lathi and Green. Two of ...


0

As an addition to Dilip's answer I'll show you how to derive that result: $$\begin{align}R_{y_1,y_2}(\tau)&=E[y_1(t+\tau)y_2(t)]\\&=E\left[\int_{-\infty}^{\infty}x(\alpha)h_1(t+\tau-\alpha)d\alpha\int_{-\infty}^{\infty}x(\beta)h_2(t-\beta)d\beta\right]\\&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}E[x(\alpha)x(\beta)]h_1(t+\tau-\alpha)h_2(t-\...


1

There are a lot of misconceptions in the way that the problem has been posed (in particular, $X(\omega)$ as defined by the OP in the first version of his question -- he has since then deleted the definition -- ) has nothing to do with the matter), but when $\{x(t)\}$ is a wide-sense-stationary (WSS) process, then the processes $\{y_1(t)\}$ and $\{y_2(t)\}$ ...


Top 50 recent answers are included