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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 ...


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

A real-valued system that doesn't distort the shape of the input signal must have the following input-output relation: $$y(t)=Ax(t-t_0)\tag{1}$$ with arbitrary real-valued constants $A>0$ and $t_0$. In the frequency domain, Eq. $(1)$ corresponds to $$Y(\omega)=Ae^{-j\omega t_0}X(\omega)\tag{2}$$ Consequently, the corresponding system is an LTI system ...


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

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

Consider a real, discrete-time signal $$ x[n] = A_1 \cos(\omega_1 n + \phi_1) + A_2 \cos(\omega_2 n + \phi_2) $$ Assuming that $\omega_1$ and $\omega_2$ are known, then in order to describe $x[n]$ completely and uniquely, you need both the amplitudes $A_1$, $A_2$, and also the phase angles, $\phi_1$ and $\phi_2$. That's what the various Fourier transforms ...


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

You have to remember that the actual signal you have is not an ideal complex exponential extending infinitely in time, but a rectangularly windowed exponential. You can compute the DFT analytically: $$X[k]=\sum_{n=0}^{N-1}e^{j2\pi f_0n}e^{j2\pi kn/N}\tag{1}$$ Using the formula for the geometric sum you'll end up with $$X[k]=e^{j\pi(N-1)(f_0+k/N)}\frac{\...


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.


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

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

For sinusoids that are not exactly integer periodic in the FFT length, an FFT measures the phase at a circular discontinuity. And that discontinuity flips direction as frequency changes from slightly below to slightly above an exact integer periodic-in-aperture frequency. This is part of the effect of the default rectangular windowing of any finite length ...


2

For a set of exact equations that describe how a pure real tone works in a DFT check out equations (23)-(25) in my blog article: DFT Bin Value Formulas for Pure Real Tones The phenomenon you have found can be seen clearer if you consider the complex definition of a real valued sinusiodal. $$ \cos( \theta ) = \frac{e^{i\theta}+e^{-i\theta}}{2} $$ So, when ...


1

You could try to Fourier transform, $F(k) = \int_{-\infty}^{\infty} x(t) e^{-2\pi i kt} dt$ which can be expressed in terms of magnitude and phase in the following way $F(k) = |F(k)|e^{i\Phi(k)} = a(k) + ib(k)$ with magnitude $|F(k)|$ and phase $\Phi(k) = \tan^{-1}(\frac{b}{a})$ for each frequency component $k$. Since it looks like you have a dominant $k$...


1

In QPSK, the bit information is embedded in the phase of the carrier signal. For example, for bits 00, the phase shift is 45degree. This means the carrier signal $x(t) = cos(2\pi f_c t + \phi) 0 \lt t \lt T$, the phase shift $\phi = \pi / 4$. So $x(t) = cos(2\pi f_c t + \pi / 4) = cos(2\pi f_c t) cos(pi/4) - sin(2\pi f_ct)sin(pi/4)$. The Inphase component ...


1

No, this doesn't make much sense. What you can do in the time domain is compute the analytic signal and derive the signal's instantaneous amplitude (envelope) and its instantaneous phase from it. Take as an example $$x[n]=A\sin(\omega_0n+\theta)\tag{1}$$ The corresponding analytic signal is $$x_a[n]=-jAe^{j(\omega_0n+\theta)}\tag{1}$$ Its instantaneous ...


1

This question has a simpler answer for the 2-D continuous-space Fourier transform but itsdDiscrete Fourier transform based verification requires some elaboration and careful implementation as @MarcusMüller has already mentioned. On the continuous-space it can be shown that rotating an image $f(x,y)$ by $\theta$ radians CCW (counter-clock wise) on the $xy$ ...


1

A DCT is identical to a DFT, twice as long, of the input data concatenated with its mirror image. Data concatenated with its mirror results in symmetric data. Since a symmetric vector is strictly even, there are no odd (sine or imaginary) components in the DFT result. Just cosine (or real or even) components. Thus you can reconstruct the original input ...


1

If you were to change the relative phase of some FFT result bins, the place where all the peaks would line up could change, thus representing a time domain shift of some peak. The peaks or transients would be moved to occur earlier or later in the FFT window. Sometimes, an FFT analysis cares about the shape of the time domain waveforms and what time (...


1

Suppose you have a sinusoidal that has a whole number of cycles ($k$) in your DFT frame containing $N$ sample points. It can be parameterized like this: $$ x[n] = A \cos \left( \left( k\frac{2\pi}{N}\right)n + \phi \right) $$ If you take the $1/N$ normalized DFT of this (FFT is a DFT that is computed efficiently), all the bins will be zero except for bins ...


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 ...


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