Tag Info

New answers tagged power-spectral-density

0

The original paper uses Wiener filtering, which usually requires knowledge of the expected spectrum. I haven't read the details of the algorithm (included below), but I suspect the PSD input allows the basic estimate to be avoided.

1

Assuming the machines are LTI (Linear Time Invariant) systems, they are completely determined by their transfer function. In theory $FFT(y)/FFT(x)$ will give you this but that's not that easy in practice. You need enough good enough signal to noise ratio at ALL frequencies. If the input has low or no energy at some frequencies you'll end up with a "...

2

$Y(t)$ is not a wide-sense-stationary (WSS) process even though $X(t)$ is, and so it does not have an autocorrelation function with a single argument $\tau$ as you have written it; there are two arguments $t$ and $\tau$, or $t+\tau$ if you prefer. So the concept of PSD as you have learned it does not apply in this instance. Note that there is nothing random ...

1

This looks like homework, so I will just give a few hints . The power spectrum is simply the magnitude squared of the spectrum so $PSD_x(f) = |X(f)|^2$ A modulator is NOT an LTI system. There is no transfer function and there is no impulse response. You can't apply LTI methods to this problem. Multiplication in the time domain is equivalent to convolution ...

1

Are there situations in signal processing where one would like to identify e.g. a time-domain signal from only it's power spectrum? Well, your described optical algorithm doesn't "only" depend on the power spectrum, but on the power spectrum under different phase shifts – which is a way of estimating the Fourier transform of the time (or spatial) ...

2

Upgraded to full answer. The diff function implements the difference equation $$y[n] = x[n]-x[n-1]$$ The transfer function is simply $$H(z) = 1 - z^{-1}$$ or $$H(\omega) = 1 - e^{-j\omega}$$ where $\omega$ is the normalized frequency. We can write this as H(\omega) = e^{-j\omega /2} \left( e^{+j\omega/2} - e^{-j\omega/2}\right) = e^{-j\omega /2} \cdot 2 ...

Top 50 recent answers are included