# Tag Info

1

MATLAB has built-in functions taking care of the steps you mention in 2. You can check out the pwelch function (here) which uses Welch's method for PSD estimation.(here) The choice of segment length, number, overlap and windowing function presents a trade-off between bias and variance.

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For stationary stochastic signal, taking limit after expectation works as explained below. You have a windowed random process $x_T(t)$ from which you want to derive the PSD $S_{xx}(f)$ by letting $T \rightarrow \infty$. There are two approaches to this. (1) You take the $lim_{T \rightarrow \infty}$ on $x_T(t)$ , then compute $|X_T(f)|^2$ and then take ...

2

It is not the expectation operator that makes sure that the limit exists. The expectation just results in an ensemble average, which we need to obtain a deterministic function $S(f)$ for the power spectrum. Assume we're given a deterministic power signal $x(t)$, i.e., a signal with finite non-zero power, and, consequently, infinite energy. Its Fourier ...

2

See section 2.4.3 of this reference https://web.stanford.edu/~dntse/Chapters_PDF/Fundamentals_Wireless_Communication_chapter2.pdf If the doppler spectrum has to be gaussian, the auto-correlation of tap gains should be gaussian (which is correctly mentioned in other answer but I somehow felt more details were not captured). For a coherence time $T_c$, ...

1

To spread in frequency with a Gaussian shape is to convolve the frequency domain on the waveform with the Gaussian shape. To convolve in frequency is to multiply in time the respective Fourier Transforms. The Fourier Transform of a Gaussian is a Gaussian; so therefore you would multiply in time by a Gaussian window. In MATLAB you can use the "gaussian" ...

1

Important Information : Sampling at $f_{s}$ will map $[-\frac {f_{s}}{2}, \frac {f_{s}}{2}]$ to digital frequency $\omega=[-\pi, \pi]$, and similarly sampling at $2f_{s}$ will map $[-f_{s}, f_{s}]$ to digital frequency $\omega=[-\pi, \pi]$. Also, we need to look only at digital frequency $\omega = [-\pi, \pi]$ as the digital spectrum is a $2\pi$-periodic ...

1

Ultimately we are trading how much signal we lose versus how much noise (phase noise dominantly) we are rejecting along with being able to track dynamics in the overall model between trasnmitter and receiver. If I was optimizing this I would do a simulation with the actual parameters for all noise sources and expected dynamics in my system, but I start with ...

2

You are attempting to simulate an RF signal directly at it’s carrier frequency which will require much more processing so is not recommended. You can achieve the same results by simulating everything at its equivalent complex baseband signal due to the linear properties of the underlying frequency translation. Equivalently everything in your simulation model ...

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The fourier transform of the two dimensional autocorrelation function should do it.

1

I see that the difference falls down as 20dB/dec. Because you high pass also only has a 20 dB/octave slope. For first order filters the upwards slope of the high pass is symmetrical to the downwards slope of (1 - highpass) other than bilinear distortion. Your difference cannot fall faster than the high pass rises. How can I make it fall faster? You need ...

1

Zero padding a signal in time domain will not increase resolution in the frequency domain. Consider N< M, It may seem that increase in length by zero padding (adding M-N zeros) increases the resolution, but it simply is observing the original N DFT coefficients interpolated by lagrangian interpolation and now observing this interpolation at M equally ...

3

The reason is the decorrelation due to the varying phase shift between the input and output. See the plot below showing the phase of the OP's high pass filter, and how closely it follows the cancellation when plotted on a log log scale (the higher frequency components of the two signals are better aligned in time and therefore have higher correlation leading ...

1

I would recommend a different approach without zero padding. Choose your DFT frame sizes based a common time duration and do different sized DFTs on your different signals with a 1/N (or 2/N if you prefer) normalization factor. The rationale for this is that the bins in a DFT correspond to frequencies in units of cycles per frame making the same bin index ...

1

Fft normalization is often an overlooked topic and many articles dealing with FFT just say to divide the fft ouput by N (divide by either N or N/2 depending on the actual FFT algorithm used). Real FFT usually discards the upper part of the spectrum (due to FFT redundancy) so the normalization coefficient here is N/2, rather than N. So, my question is: ...

0

Whether you want to normalize or not depends on whether you want to know the level or the energy of the DFT input. IIRC, the SciPy FFT returns energy (complies with Parseval’s relation). A signal N times as long at the same level has N times more energy. So you could divide by N to get an estimation of a level instead of energy.

1

Your decision to normalize or not does not change the accuracy of your answer, as it is simply a scaling factor. If you use the common scaling of $1/N$, then the output for each DFT bin will represent the average of the portion of the input signal that is at the frequency defined by that bin, scaled to the same units as the input. So that is convenient and ...

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The secret to proving the limit is to not convert the $sin$ to $sinc$. This leaves the $T$ in the argument and not outside. No need to worry about the complex conjugate as the values are real. The $sin$ values are bounded by -1 and 1. At that point it becomes: \begin{align} 0 \le S_{xx}(f) &= \lim_{T\rightarrow\infty}\frac{1}{T}\left|X_{T}(f) \... 1 You don't need to make it so complicated as to use two variables t and t^\prime. \begin{align} X_T(f) &= \int_{-T/2}^{T/2} x(t)\exp(-j2\pi ft)\,\mathrm dt\\ &= \int_{-T/2}^{T/2}\frac{\exp(j2\pi f_0t)+\exp(-j2\pi f_0t)}{2}\exp(-j2\pi ft)\,\mathrm dt\\ &= \frac 12\int_{-T/2}^{T/2} \exp(j2\pi (f_0-f)t)+\exp(-j2\pi (f_0+f)t)\,\mathrm dt\\ &= ... 1 The OP is correct in their dimensional analysis |X(f)|^2 is NOT the power spectral density, despite what other authors might claim. Other authors probably call this the power spectral density because it is close to right and it captures most of the important features without having to delve into technicalities. Power has dimensions of [\text{signal}^2]. ... 1 Below is a function which I wrote long back, when I needed to generate AWGN time-domain samples given Noise PSD in dBm/Hz. AWGN_NOISE() : Generates Additive White Gaussian Noise of PSD power in dBm/Hz AWGN has Gaussian PDF with 0 mean and \sigma^{2} = N_{o}/2 Noise Spectrum in dBm/Hz = 10*log_{10}(BW*N_o/2), hence \sigma = \sqrt{No/2} = \sqrt{10^{... 1 Starting with from the linked question:S_{xx}(\omega)=\lim\limits_{T\to \infty}\mathbf{E} \left[ | \hat{x}_T(\omega) |^2 \right]  = \lim\limits_{T\to \infty}\mathbf{E} \left[ \frac{1}{T} \int\limits_0^T x^*(t) e^{i\omega t}\, dt \int\limits_0^T x(t') e^{-i\omega t'}\, dt' \right] = \lim\limits_{T\to \infty}\frac{1}{T} \int\limits_0^T \int\limits_0^T \...

2

This is not really an answer, but a different angle: Physically speaking, the power of a single signal is not well defined. Physical power (or intensity) is always the product of two root-power quantities (which used to be called filed quantities). Voltage times current, force times velocity, etc. Let's say you have an impedance with a circuit with a ...

2

I am curious about the same issue the OP raised: sometimes the units of power spectral density seem not quite right. I assume it is just me, so I always go back to my touchstone reference 1. Blinchikoff and Zverev use the definitions of Fourier transform and inverse transforms I have always used and preferred [1, p. 294]: and they give the units of the ...

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THIS IS NOT YET A COMPLETE ANSWER BUT THE CONTINUATION OF THE OP's QUESTION IN MY ATTEMPT TO ANSWER and like the OP, I would welcome a short concise bottom line answer that addresses this. Update: After the OP pushed me through the mud on this I ended up siding with his same level of questioning and concluding that in a strict sense $|X(f)|^2$ is en energy ...

2

It has more to do with the simple nature of analysis of the data present. For example: given a sequence, it is easier to calculate the DTFT to get the Fourier transform then to calculate the autocorrelation and calculate the Fourier transform of the autocorrelation to get the PSD. One could easily interpolate the DTFT, assuming a sampling time to get the ...

0

That seems fine. If you integrate your PSD over all frequencies you get a $1$ at $-f_0$ and $+f_0$ and zero everywhere else. $1+1 = 2$ so the total integral will come out to be $A^2/2$ which matches your time domain number. Yes, the PSD is also the magnitude squared of the Fourier Transform, i.e. $$PSD(f) = X(f) \cdot X^*(f)$$ where $X(f)$ is the Fourier ...

0

@jithin 's answer gave me an idea that I think answers my question. I'm going to define a new window function: $$W_{\Delta t}(t) = \theta\left(t-\frac{\Delta t}{2}\right)\theta\left(\frac{\Delta t}{2} - t\right)$$ This is the same as $W_{\Delta t}(t)$ in the post except for the missing factor of $\frac{1}{\sqrt{\Delta t}}$. The PSD is then defined as  ...

1

I am assuming $X(t)$ is a passband signal having finite support (non zero value for only a finite range of frequencies within spectrum analyzers range). In your equation (1) what you have missed is $f_{LO}$ to $f_{IF}$ conversion. So $W_{\Delta t}$ should be $W_{IF}(t) = e^{j2\pi (f_{LO}-f_{IF})t}W_{\Delta t}(t)$. When you convolve $X(t)$ and $W_{IF}(t)$, ...

0

What is the link between them? Power should not be function of frequency (it is the integral on all frequencies of power spectral density) As far as a Spectrum Analyzer is concerned, to display the power spectrum in dBm, what it does it to take the FFT and compute square of magnitude $P_n = 20 \times log_{10}(|X(f_n)|^2*1000/100)$ with $f_n$ being the ...

0

In addition to the resolution bandwidth fact, i think power spectral density has something to do with stochastic signals. If your signal is predetermined, you have to work with power instead of power spectral density. One good example is the very sine wave you mentioned. The mean power for that signal in its main frequency can be defined using the integral ...

1

@Amelia. Hi. I tried to run your MATLAB code, but it contains several errors regarding vector lengths and produces error messages. But that's not the main issue here. The answer to your question is: If you compute the DFT of an N-point x(n) sequence you produce an N-point complex-valued xDFT(m) frequency-domain sequence. You can only reconstruct the original ...

0

At the risk of sounding stupid, I am going to give a fresh perspective (different from the link referenced above) to this problem. I will list my reasoning and finally give the code I tried. When the signal length is 1000, auto-correlation will have 1999 length. Hence FFT of PSD should have at least 1999 length. In my example I used 2000 for simplicity. I ...

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