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4

Wouldn't it be much easier to just compare the left-hand side and the right-hand side of $$e^{-j2\pi ft_0}=|H(f)|e^{j\phi(f)}\tag{1}$$ to see that $|H(f)|=1$ and $\phi(f)=-2\pi ft_0$?


4

Or $2\pi$, or $3\pi$, or any integer multiple of $\pi$. Any odd multiple corresponds to -1 + 0i and any even multiple corresponds to 1 + 0i, aka -1 and 1. "Phase of a real number" is a little bit of a misleading label. What is required here is an understanding of the complex plane and what "phase" means in terms of a DFT bin value. Your question is ...


3

You need to make sure that the phase of continuous. The easiest way would be something like this. x(n) = cos(phase); phase = phase + 2*pi*current_frequency*sample_time; current_frequency= update_frequency(n); n = n + 1; Note, that this is very inefficient code and only for illustrative purposes.


3

The OP asked about a phase shift specifically but from the written details without seeing the actual demodulator implementation, I suspect he may possibly be asking how to implement the delay that is used for some simple FM demodulation approaches. I want to mention the possibility that this may actually be a delay element and not a phase shift in the sense ...


3

No problem at all. Inverse Fourier Transform is totally blind to phase wrapping. You can add any multiple integers of two pi to any phase and you will still get the exact same impulse response


3

the discrete function $$x_q[n]=\sin(\pi n)$$ is always zero for all of the integers $n$. the discrete function $$x_i[n]=\cos(\pi n)$$ is always $(-1)^n$ for integer $n$. so this general sinusoid at Nyquist that has a phase term: $$\begin{align} x[n] &= A \cos(\pi n + \theta) \\ &= A \big( \cos(\pi n) \cos(\theta) - \sin(\pi n) \sin(\theta) \...


2

The answers to this question explain the solution for the continuous-time case. In discrete time, the solution is completely analogous. The impulse response of a phase shifter with phase shift $\theta$ is given by $$h[n]=\cos(\theta)\delta[n]+\sin(\theta)g[n]\tag{1}$$ where $g[n]$ is the impulse response of an (ideal) discrete-time Hilbert transformer. Eq. ...


2

The phase shift is relative to the carrier of the analog FSK signal. If your FSK signal is a binary FSK (BFSK) with two frequencies $f_1$ and $f_2$; it makes the most sense to speak of the "carrier frequency" $f_0$ as being the average: $$ f_0 = \frac{f_1 + f_2}{2}. $$ The phase shift $\phi$ for a given time delay $\tau$ is: $$ \phi = 2 \pi f_0 \tau. $$ ...


2

I can't find any way to make the carrier recovery, I don't think you can or should really talk "carrier recovery" in OFDM systems; after all, OFDM is a multi-carrier system. So, you'd need to recover some 2 to a couple thousand carriers, depending on the OFDM system's number of subcarriers. but also the phase Uh-oh. Have you considered what this means? ...


2

This can be accomplished by changing the carrier frequency using a Numerically Controlled Oscillator (NCO) which maintains an accurate and continuous phase versus time trajectory via the phase accumulator. This is markedly different than changing the frequency with a classical PLL where we would typically break and reacquire lock to change frequency ...


2

There are a few things that can make this easier If you don't care about the phase response you can typically get the best result with calculating a minimum phase for your amplitude spectrum and using that as the complex target If you want to control the in-between and the behavior at the band edges you need to specify targets points there as well. If you ...


2

You're right that the transitions can be abrupt, but you can always choose to use pulse shaping, i.e., instead of using a rectangular pulse, use some smoother function $p(t)$. The corresponding baseband signal is then $$s(t)=\sum_kA_kp(t-kT)\tag{1}$$ where $A_k$ are the symbols, and $T$ is the symbol length. A smooth function $p(t)$ will avoid abrupt ...


2

This is easier to see graphically, on an s-plane, knowing that the phase is from the frequency response of the system and the frequency response is determined when we restrict s to be the $j\omega$ axis. Consider the simple ratio of a single pole and a single zero expressed as $$H(s) = \frac{s-z_1}{s-p_1}$$ Given s as any point on the s-plane, and $z_1$ ...


2

You can't design a filter that creates a phase shift that's constant with frequency for real valued input (if that's what you are trying to do). A Hilbert transformer appears to be doing this. However, the problem is, you can't implement a perfect Hilbert transformer since it's non causal with an infinite length impulse response. The tricky part is that ...


2

Yes, you are correct. If you take a DFT of square wave and only look at the amplitudes, doing an inverse DFT but using different or random phases for the sine components, it does not look like square wave in time domain any more. But it will have a matching spectra. Kind of like two racecars on a circular track, they might always have same velocity, but ...


2

how anyone can be confident in the result of an FFT when the signal being sampled has some frequency content that is at the frequency Fs/2. You can't. In practice you need a healthy margin between the highest frequency of interest and the Nyquist frequency. In audio for example the highest frequency is typically 20 kHz but you sample at 44.1 kHz or 48 ...


2

It is a similar question as the one I have asked a few weeks before, and I received a nice answer -> Shannon-Nyquist theorem reconstruct 1Hz sine wave from 2 samples I was trying to do the reverse: reconstruct a 1Hz sine wave from only 2 samples.


1

The problem is the definition of the phase. The command angle() computes the phase $\phi(\omega)$ according to $$H(e^{j\omega})=|H(e^{j\omega})|e^{j\phi(\omega)}\tag{1}$$ where $H(e^{j\omega})$ is the complex frequency response. At frequencies $\omega$ where the frequency response has zeros, the phase jumps by $\pi$. This is shown in the two left figures ...


1

You can get pretty good results with an approximate linear chirp spread out over the period: $$s(x) = \sum\limits_{n=1}^{N} \cos\left(2\pi \left(x n+ \frac{n^2}{2\times N}\right)\right), \quad N=10$$ $$m \approx 4.28$$ Group delay is the derivative of the phase with respect to frequency. For a linear (frequency as function of time) chirp, the phase should ...


1

The minimum phase shift in 16-QAM, 64-QAM, 256-QAM etc is 0 (same value regardless of whether you are using degrees or radians!) and occurs when the symbol transition is from one constellation point on some radius vector to another constellation point on the same radius vector. If we take the constellation points as having coordinates $(i,j)$ where $(i,j) \...


1

Symmetric zero-padding (in the center of an image around the N/2,N/2 sample) does not affect the FFT phase result. Or after an 2d fftshift before the 2DFFT, symmetric zero-padding around the edges (circularly around the 0,0 sample) does not add phase shift. An FFT phase measure the even to odd ratio around 0,0. This won't change with any padding that ...


1

Your understanding is correct. There is no difference between a phase of $\pi$ and a phase of $-\pi$. You can always add or subtract integer multiples of $2\pi$ to the phase without changing anything because $e^{j2\pi n}=1$ for $n\in\mathbb{Z}$. Clearly you have $e^{j\pi}=e^{-j\pi}=-1$. The probable reason why they used opposite signs in the phase plot for ...


1

There is more than one way to write the same thing. You can use $rect()$ functions or define it piece-wise, but in the end it is the same thing, since the $rect()$ is just a shorthand for a piece-wise definition. If you do it in sections, you don't need to use $rect()$ function at all, you could simply write is as $$X(j \omega) = \begin{cases} & j \...


1

I believe this algorithm isn’t a rotation (phase shift) by 90degrees but instead is a frequency shift of fsamp/4. In this algorithm each sample is rotated 90 degrees with respect to the last one causing a frequency shift: Samp1: no rotation Samp2: 90 degree rotation Samp3: 180 degree rotation Samp4: 270 degree rotation Samp5: 360 -> 0 degree rotation ...


1

Here is how I would do it: 1) Find the frequency of the clean signal, this should be trivial in the time domain for a clean square wave. 2) Heavily smooth the noisy signal with a symmetric filter. (IIR or FIR) 3) Calculate a single DFT bin for each signal using the length of a whole number of cycles for your frame. This will make the harmonics ...


1

If we use this definition of instantaneous frequency, you can construct a BFSK signal that is continuous in phase but is discontinuous in frequency. If we denote the two phase signals as $\phi_1(t)$ and $\phi_2(t)$, corresponding to frequencies $f_1,f_2$, and define them such that: $$ \phi_1(t) = 2\pi f_1t + k_1 \\ \phi_2(t) = 2\pi f_2t + k_2 $$ and denote ...


1

i am assuming this is a periodic (or quasi-periodic) function. also seems like the signal is a musical tone. periodic signals have a period, $P$, and the reciprocal of that period is the fundamental frequency; $f_0 \triangleq \frac{1}{P}$ or, as angular frequency; $\omega_0 = 2 \pi f_0 = \frac{2 \pi}{P}$ . the periodic signal is $x(t)$ where $$ x(t+P) = ...


1

Yes you are right. Let's practically implement it with Matlab / Octave. N = 8; h = [1,2,3,4,4,3,2,1]; % just an impulse response (FIR) H = fft(h,N); % DFT H[k] of h[n] as samples of H(w) at w = 2*pi*k/N Hm = abs(H); % extract the magnitude of H[k] Hp = angle(H); % extract the phase angle of H[k] Hr = Hm.*exp(j*Hp); ...


1

Regarding the phase: if you set all frequency components to have zero phase, then the signal becomes essentially a sum of sinusoids that tends to an impulse as the frequency increases. In other words, $$\delta(t) = \int_{-\infty}^\infty \cos(2\pi ft) df.$$ In your case, the maximum frequency is limited, which means that the signal will actually become a ...


1

The phase shift will be a factor of $e^{j2\pi k_0 m/M}$, where $m$ is the DFT bin number


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