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25

To demodulate a phase-shift keyed signal, of which BPSK is the simplest, you have to recover the carrier frequency, phase, and symbol timing. Bursty Signals Some signals are bursty and provide a known data sequence called a preamble or mid-amble (depending on whether it shows up at the beginning or middle of the burst). Demodulators can use a matched ...


18

You aren't doing anything wrong, but you also aren't thinking carefully about what you should expect to see, which is why you're surprised at the result. For question 1, your conjecture is close, but you actually have things backwards; it's numerical noise that's plaguing your second one, not your first one. Pictures may help. Here's plots of the magnitude ...


18

Suppose that a linear filter has impulse response $h(t)$ and frequency response/transfer function $H(f) = \mathcal F [h(t)]$, where $H(f)$ has the property that $H(-f) = H^*(f)$ (conjugacy constraint). Now, the response of this filter to complex exponential input $x(t) = e^{j2\pi f t}$ is $$y(t) = H(f)e^{j2\pi f t} = |H(f)|e^{j(2\pi f t + \angle H(f))}$$ ...


16

They don't both measure how much a sinusoid is delayed. Phase delay measures exactly that. Group delay is a little more complicated. Picture a short sine wave with an amplitude envelope applied to it so that it fades in and fades out, say, a gaussian multiplied by a sinusoid. This envelope has a shape to it, and in particular, it has a peak that represents ...


14

If you want the shifted output of the IFFT to be real, the phase twist/rotation in the frequency domain has to be conjugate symmetric, as well as the data. This can be accomplished by adding an appropriate offset to your complex exp()'s exponent, for the given phase slope, so that the phase of the upper (or negative) half, modulo 2 Pi, mirrors the lower ...


10

One-dimensional version The one-dimensional version that you list won't work. When there is a large enough shift in images (more than one or two pixels in real-world images), there will be nothing relating the column pixels. For an example of this, try: I5 = rand(100,100)*255; I6 = zeros(100,100); I6(11:100,22:100) = I5(1:90,1:79); So that we have I5: ...


9

For those who still cannot chalk the difference here is an simple example Take long transmission line with simple sine signal with an amplitude envelope, $v(t)$, at its input $$v(t) \cdot \sin(\omega t)$$ If you measure this signal at the transmission line end, it might come somewhere like this: $$ v(t-\tau_g) \cdot \sin(\omega t + \phi)\\ = v(t-\tau_g)...


9

Phase Noise and Frequency Noise are not two different noise sources, they are artifacts of the same noise, it is just a matter of what units you want to use. Frequency and Phase are directly related as frequency is phase changing with time, so if you have one you will always have the other; frequency and phase are related by derivatives and integrals: the ...


9

Compare $y= sin(\omega t)$ to $y=sin(\omega t-\phi)$ as shown in the plot below. $\phi$ is the additional phase term in radians where $\omega$ represents the frequency in radians per second. Thus the phase term shifts a sinusoid along the horizontal axis. So at a given frequency this will result in a time delay for that frequency, although to be noted that a ...


8

Nice question! It uses one of my favorite trig identities (which can also be used to show that quadrature modulation is actually simultaneous amplitude and phase modulation). The impulse response of the system described above is given by: Block diagram:


8

Phase (or carrier) Recovery for BPSK can be done over the entire sequence using the information from every sample. Here are common approaches to doing Carrier Recovery: Frequency Doubling (squaring): If you square a BPSK signal (multiply it by itself) a strong tone will be created at twice your carrier frequency. The Squaring operation strips the data ...


8

The book's formula is right. Let $$H(w) = 1 - r e^{j(\theta - w)} = [1-r \cos(\theta - w)] + j [-r \sin(\theta - w)]$$ Since the group delay $\tau$ is the negative of the derivative of the phase of $H(w)$, we first define the phase as: $$\phi(w) = \tan^{-1}\left( \frac{-r \sin(\theta - w)}{1-r \cos(\theta - w)} \right)$$ Using the derivative rule for the ...


8

What you need is carrier phase synchronization. This is a complicated topic with many different approaches. The approach that you'll choose could depend on things like: Data-aided versus blind: Does the underlying sequence contain any known data (e.g. a training or sync sequence of some kind) that you can use to divine the phase offset? Or, do you have to ...


7

You can use the cross-correlation function to determine the lag between the two signals.


7

Answers about using cross-correlation are correct. But if your input and output signals are sinusoid you can use more simple and faster method. input - $y1=a*\sin(\omega*t)$ output - $y2=b*\sin(\omega*t+\phi)$ multiply input and output: $y1*y2=a*b*\sin(\omega*t)*\sin(\omega*t+\phi)=a*b*1/2*(\cos(\phi) - \cos(2*\omega*t+\phi))$ eliminate high-frequency ...


7

Your question is tricky because it is hard to define what "discarding the phase information" means without specifying a practical way of doing it; but then we encounter problems/artifacts which are specific to this particular method. Since you mentioned STFT, let us assume phase information is suppressed by doing a STFT, keeping the magnitudes, setting to 0 ...


7

The group delay of a filter is defined as minus the change in the phase response with respect to frequency. If the phase response of a filter is $\Phi(\omega)$, the corresponding group delay $\tau_g$ is given by: \begin{equation} \tau_g = -\frac{d\Phi(\omega)}{d\omega} \end{equation} In Matlab code, the group delay of a 4th order Butterworth filter can be ...


6

There are a couple of interesting aspects of "reconstruction to unity". First, there are two ways of combining two filters: parallel and in series. For a parallel topology it is ALWAYS possible to find a complimentary filter so that the pairs add to unity. It's easy enough, actually. Simply do $\tilde{H}(\omega) = 1-H(\omega)$. In the time domain that means ...


6

I think wikipedia can more than adequately answer your question. http://en.wikipedia.org/wiki/Phase_(waves) However, in brief, the phase term describes the relationship between a waveform and a fixed reference point in time. For example the sinusoid $sin(\omega t)$ is zero at $t = 0$ whereas $sin(\omega t - \phi)$ is zero at $t = \phi$. $\phi$ could be ...


6

There are filters that cause a ,,linear'' phase shift, that is, constant delay. It is not possible to filter anything at all (causally) without causing any delay.


6

If you have a signal $$f[n]=\cos(\Omega_0n)$$ and you apply a time shift of $n_0$ you get $$f[n+n_0]=\cos(\Omega_0(n+n_0))=\cos(\Omega_0n+\Omega_0n_0)=\cos(\Omega_0n+\phi)$$ where $\phi=\Omega_0n_0$ is the phase shift. The other way around, if you have a phase shift of $\phi$, this is not always equivalent to a time shift of the original signal: $$g[n]=...


6

This is a slightly tedious but nevertheless straightforward exercise in computing the derivative of a function: $$\begin{align}\tau(\omega)&=-\frac{d\phi(\omega)}{d\omega}=-\frac{d}{d\omega}\arctan(f(\omega))\tag{1}\end{align}$$ with $$f(\omega)=\frac{r\sin(\omega-\theta)}{1-r\cos(\omega-\theta)}\tag{2}$$ From $(1)$ we have $$\tau(\omega)=-\frac{f'(\...


6

There are several reasons why the two results don't match: the coefficients of the FIR Hilbert transformer are wrong the FIR Hilbert transformer is too short to even come close to the performance of the FFT-based implementation the frequency of the input signal is too low for the FIR Hilbert transformer to perform properly. A FIR Hilbert transformer always ...


5

There is nothing inherently 'magical' about performing a power of 2 DFT, other than the fact that performing a power of 2 DFT allows one to perform the DFT in $O(Nlog(N))$ instead of $O(N^2)$. So the power of 2 DFT, (The algorithm that does this is known as the FFT), allows you to simply speed up your DFT computation by a huge factor. I apply the fft ...


5

Modern FFT libraries, such as FFTW and Apple's Accelerate framework can do non-power-of-2 FFTs very efficiently, as long as all the prime divisors of the composite length are fairly small (2,3,5,etc.) A power of 2 makes it simpler (about 1 page of source code) if you have to code your own FFT for some reason, or are otherwise constrained as to max program ...


5

This is indeed correct. There is a few things you could do to make it a better graph: Label the X-axis Label the Y-axis Use real physical units if it's applicable. Phase is measured either in radians or in degrees. These are VERY different things, so proper labeling with units helps to clarify what you are using. Similar if f is a frequency in Hz (or kHz or ...


5

You're probably better off skipping the $5u(t)$ idea and going straight to a periodic input, probably a sinusoid. A phase locked loop is for signals that have a phase to lock onto, whereas $u(t)$ doesn't have a phase. The Stanford reference looks pretty good. Start by comparing that to your feedback system: Note that the phase detector is a ...


5

The z domain transfer function of the system is the z transform of the system impulse response, so start by taking the Z transform of h[n] ... $$H[z] = -z^1 + 1 + 2z^{-1} + 2z^{-2} + z^{-3} -z^{-4}$$ You may be able to message this into a nicer form, but that isn't necessary. Next, to get the the frequency response, replace z with $e^{jw}$ So this ...


5

If you are looking for a frequency-independent delay applied to any given input signal by the filter (apart from amplifying and attenuating certain frequency components), then you won't be able to find it because there is no such delay. As you can see in your plots, group delay and phase delay are generally frequency dependent. Furthermore, for general input ...


5

I'm not sure if I understand your problem, but I'll give it a try. If you have a complex frequency response $$H(\omega)=|H(\omega)|e^{j\phi(\omega)}\tag{1}$$ where $\phi(\omega)$ is the phase, then you probably know that for a given frequency $\omega_0$, the frequency response might as well be written as $$H(\omega_0)=|H(\omega_0)|e^{j(\phi(\omega_0)+2k\...


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