# Tag Info

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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 ...

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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 ...

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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 $$y(t) = H(f)e^{j2\pi f t} = |H(f)|e^{j(2\pi f t + \angle H(f))}$$ ...

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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 ...

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For those who still cannot chalk the difference here is an simple example Take long transmission line with simple quasi-sinusoidal signal with an amplitude envelope, $a(t)$, at its input $$x(t) = a(t) \cdot \sin(\omega t)$$ If you measure this signal at the transmission line end, $y(t)$, it might come somewhere like this: \begin{align} y(t) &= a(... 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: ... 10 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: 10 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 You can use the cross-correlation function to determine the lag between the two signals. 9 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 ... 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 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. LetH(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)$$Using the derivative rule for the ... 8 What you need is carrier phase synchronization. This is a complicated topic with many different approaches. The approach that you'll choose could depend on things like: Data-aided versus blind: Does the underlying sequence contain any known data (e.g. a training or sync sequence of some kind) that you can use to divine the phase offset? Or, do you have to ... 8 Creating the Frequency Vector The arrangement of the output of fft() depends on whether you use an odd or even number of points for your fft. I think this post nicely summarises how the frequencies are arranged. Have a look at it. Since you are using an even number of points, the Nyquist frequency, F_N = F_s/2, is present in the output of your fft, and is ... 8 The frequency response of a real-valued discrete-time system with linear phase has the form$$H(e^{j\omega})=A(\omega)e^{-j\omega\tau},\qquad\omega\in [-\pi,\pi]\tag{1}$$where A(\omega) is either a real-valued even function or a purely imaginary odd function, and \tau is some real-valued parameter (the delay). If A(\omega) is purely imaginary, then ... 7 The group delay of a filter is defined as minus the change in the phase response with respect to frequency. If the phase response of a filter is \Phi(\omega), the corresponding group delay \tau_g is given by: $$\tau_g = -\frac{d\Phi(\omega)}{d\omega}$$ In Matlab code, the group delay of a 4th order Butterworth filter can be ... 7 I think wikipedia can more than adequately answer your question. http://en.wikipedia.org/wiki/Phase_(waves) However, in brief, the phase term describes the relationship between a waveform and a fixed reference point in time. For example the sinusoid sin(\omega t) is zero at t = 0 whereas sin(\omega t - \phi) is zero at t = \phi. \phi could be ... 6 I know this is a pretty old question, but I've been looking for a derivation of the expressions for group delay and phase delay on the internet. Not many such derivations exist on the net so I thought I'd share what I found. Also, note that this answer is more of a mathematical description than an intuitive one. For intuitive descriptions, please refer to ... 6 There are a couple of interesting aspects of "reconstruction to unity". First, there are two ways of combining two filters: parallel and in series. For a parallel topology it is ALWAYS possible to find a complimentary filter so that the pairs add to unity. It's easy enough, actually. Simply do \tilde{H}(\omega) = 1-H(\omega). In the time domain that means ... 6 There are filters that cause a ,,linear'' phase shift, that is, constant delay. It is not possible to filter anything at all (causally) without causing any delay. 6 If you have a signal$$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]=...

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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 ...

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Your integrator is a perfect accumulator, which is an unstable (marginally stable) system with a pole at $z=1$: $$H(z)=\frac{1}{1-z^{-1}}\tag{1}$$ This means that the value of the frequency response at $\omega=0$ (corresponding to the pole at $z=1$) is undefined. Consequently, the value of the group delay at $\omega=0$ is also undefined. Note that you can ...

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Note that a constant group delay is not sufficient for a band-limited signal to exhibit no dispersion. It is the phase delay that needs to be constant. If the phase is affine, i.e., if we have $$\phi(\omega)=a+b\omega,\qquad \omega>0\tag{1}$$ the group delay is constant $$\tau_g(\omega)=-\frac{d\phi(\omega)}{d\omega}=-b\tag{2}$$ but the phase delay is ...

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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'(\...

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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 ...

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Answer : No, any causal LTI system with frequency response $H(f)$ cannot produce the output $y(t)$ in advance. And, the answer lies in the causality of input signal $x(t)$ being applied to $h(t)$. Any causal input $x(t)$ which has an identifiable beginning cannot truly be Narrow-Band or Band-Limited. It will have non-zero frequency content at all frequencies....

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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 ...

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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 ...

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