# Tag Info

1

Usually, there are two types of noise that you can eliminate by using the spectrum. One type is a noise that is in a different frequency band than the signal (it can be a high-frequency noise). In such a case, there is a clear separation between the signal content and the noise in the spectrum and you have to do nothing. The second case is when the noise ...

4

Start with what we know: The DFT works perfectly for a single cycle waveform input. For a single cycle waveform, all harmonics have an integer number of cycles (by the definition of “harmonic”). “Spectral leakage” occurs when the input is not a single cycle waveform. As such, at least one frequency component is not represented by an integer number of cycles....

2

IMHO, leakage is a poor term for it. Best representation is a much more descriptive though wordy description. Imagine a stereo with one of those displays with the bars that correspond to frequency. They usually bounce around a lot, but they do give a representative view of the frequency content. Now, let's take the stereo into the lab and feed it a sweep ...

2

Imagine that a continuous signal has a frequency component of amplitude $A$ at exactly 20 Hz (you can imagine it is alone, a single perfect wave). With a FFT, or any discrete Fourier transform, you could hope that, if you have acquired the signal correctly (above Nyquist), the discrete spectrum will give you a clear peak with amplitude $A$ at 20 Hz. This is ...

0

$\displaystyle \hbox{ DTFT}(\alpha x_1 + \beta x_2) = \alpha\cdot \hbox{ DTFT}(x_1) + \beta \cdot\hbox{ DTFT}(x_2)$ Linearity property cannot applied here.

0

If I have an output from band-pass filter and want to convert it back to time domain. Is it also the correct approach to cyclic shift to left and pad zeros in the middle before applying windowing and convert back to time-domain?

2

You're right, the ROC of the Laplace transform of a two-sided signal is a strip in the complex plane. In your case, the imaginary axis is inside the ROC, and the ROC is limited by the poles in the right and left half-planes. If the ROC were the right half-plane, the signal would be right-sided, which is clearly not the case.

2

OK, so your signal is described in one period, $T_0$, as $$x_{T_0}(t) = \left\{\begin{array}{ll} 0, & -T_0/2 < t \leq 0 \\ 1, & 0 < t \leq T_0/2 \end{array}\right.$$ Let's keep it that way (without any other shortcuts like $\prod$ or $\mathrm{rect(\cdot)}$ etc). One way to find the Fourier coefficients of the even and the odd part of a signal ...

1

Assuming that you want the Fourier Transform of this signal, you can easily obtain it by using a very useful property and a well known FT pair in continuous time. The pair is $$x(t) = e^{-at}u(t), \: \: a>0 \longleftrightarrow X(f) = \frac{1}{a+j2\pi f}$$ and the so-called duality property says that if $$x(t) \longleftrightarrow X(f)$$ is an FT pair, ...

0

I came back to this and tried deriving the discrete version which helped make things more sense: Somehow $f_k t_n = f(n, k, N)$ $f_k = \frac{f_s}{N}k$ and $t_n = \frac{T}{N}n$ $f_s = \frac{N}{T}$ So $f_k t_n = \frac{f_s}{N}k\frac{T}{N}n = \frac{N}{TN}k\frac{T}{N}n = \frac{kn}{N}$ Done!

1

The second image you find online as you mentioned is magnitude and the first image must be either real part or the imaginary part (but not both). If you calculate complex value magnitude for each point, i.e. $Mag_{i,j} = R_{i,j}^2 + I_{i,j}^2$ you get something similar to the following image. I = imread('cameraman.tif'); F = fft2(I); F_Centered = fftshift(F)...

2

Let's go through a few ways to solve this: Fourier transform: an ideal integrator is an LTI system, so its response to a sinusoidal input signal is a sinusoid with the amplitude and phase changed according to the frequency response evaluated at the input frequency (if it exists). For the ideal integrator we have $$H(\omega)=\pi\delta(\omega)+\frac{1}{j\... 2 HINT: Note that the given h(t) can be written as$$h(t)=g(t)-g(t-T_0)\tag{1}$$with some g(t) the Fourier transform G(f) of which you know. So from (1) you then get$$H(f)=G(f)\left(1-e^{-j2\pi fT_0}\right)\tag{2}$$2 With feedback systems such as the one given it's often easy to define an additional signal at the output of the adder. This gives the following equations:$$U(f)= X(f)-H_2(f)Y(f)\tag{1}$$and$$Y(f)=U(f)H_1(f)\tag{2}$$Now you can solve Eqs (1) and (2) to get the frequency response H(f)=Y(f)/X(f). 3 If I understand correctly, you want to verify the energy calculation in the frequency domain by computing the energy as$$E_x=\int_{-\infty}^{\infty}|X(f)|^2df\tag{1}$$with$$X(f)=\mathcal{F}\big\{x(t)\big\}=\frac{A}{A+i2\pi f}\tag{2}$$From (2) we get$$|X(f)|^2=\frac{A^2}{A^2+(2\pi f)^2}=\frac{1}{1+\left(\frac{2\pi f}{A}\right)^2}\tag{3}$$With (... 0 Two conjugate poles: (z-(0.51+0.7i))(z-(0.51-0.7i)) = z^{2} - 1.02z +0.76 Two conjugate zeros: (z-(0.57+0.78i))(z-(0.57-0.78i)) = z^{2} - 1.14z +0.94 2 Your math looks correct, thanks for including what you have done. Also for such a block diagram of a linear system, you can rearrange each of the three blocks in any order at the points where the nodes come together (can't break loops) - For example, you can move the derivative to the end without changing the overall result. This may make it even more ... 0 Bottom Line$$(\theta_2-\theta_1) = 2\pi f(T_2-T_1)n -(\phi_2[n]-\phi_1[n]) \tag{1}$$f: frequency in Hz of two tones of the same frequency and fixed phase offset (\theta_2-\theta_1): phase difference in radians of tones being sampled T_1: period of sampling clock 1 with sampling rate f_{s1} in seconds T_2: period of sampling clock 2 with ... 1 ifft works on the full spectrum as pointed by Hilmar. In the discrete description of Fourier Transform theory, it means that each sample supplied as the argument to ifft is a frequency bin in the range [0,1] in cycles per sample (see normalized frequency if you do not know what that means), which corresponds to [0,2\pi] radians per sample or [0, F_s] ... 1 First of all, note that the discrete-time Fourier transform (DTFT) of a sequence is always periodic with period 2\pi:$$X(\omega)=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\tag{1}$$This is obvious because e^{-jn\omega} is 2\pi-periodic in \omega. Next, consider the inverse DTFT:$$x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(\omega)e^{jn\omega}d\omega\...

2

Your problem is at this step. You're doing half of a shortcut, but not the other half: $$x_k = \frac{2A}{T} \int_{0}^{T/2} \frac{2t-T}{T} e^{-i2 \pi k f_0’ t }$$ Start with the Fourier series definition (with the notation tidied up): $$x_k = \frac{A}{T_0} \int_{-T_0/2}^{T_0/2} \frac{2t-T_0}{T_0} e^{-i2 \pi \frac{k}{T_0} t } dt$$ What you did with this ...

4

This is one of the best problems to demonstrate Fourier Series properties, and specifically the time derivative property: $$\frac{d}{dt}x(t) \overset{FS}\longleftrightarrow j2\pi kf_0 X_k$$ Instead of computing the integral $$X_k=\frac{1}{T_0}\int_{T_0}x(t)e^{-j2\pi kf_0t}dt$$ which is time consuming, you can take the first derivative of your signal and ...

1

The typical practice with decibels is to append the unit. Unfortunately, it's informal, so you'll see things like "dBm" in RF circuits which refers to "dB with a reference of 1mW", not "dB with a reference of 1 meter". you could: Put a note in your graph that $0\mathrm{dB} = \mathrm{1 m/s^2}$ Use "dBg" (and use 1g as your reference) Just keep the y-axis ...

3

Not a full solution but a few hints: Note that your function has an odd symmetry. Hence the even Fourier coefficients should be zero. They don't seem to be in your case and I think that's a consequence of how you integrate: you integrate from 0 to $T/2$ instead of $-T/2$ to $T/2$. If you did both halves, they would turn out with opposing signs and then ...

0

You have to realize that all sequences in DFT relations are inherently periodic. This is usually understood, but it can be made explicit, for instance by that subscript. So both sources are correct, but one explicitly notes the periodicity of the sequence. If you write down the DFT relation $$\delta[n]=\frac{1}{N}\sum_{k=0}^{N-1}e^{j2\pi nk/N}\tag{1}$$ ...

0

I suppose you've learned about the Hilbert transform relationship between the real part and the imaginary part of the Fourier transform of a real-valued and causal signal. So you can obtain the real part from the imaginary part using this relationship. Since the imaginary part is just composed of sinusoids, the real part can be obtained very easily (note ...

1

Assume $x[n]$ and $y[n]$ represent two jointly WSS (wide-sense stationary) discrete-time random processes. Further assume that they have zero means, without losing the generality. If $x[n]$ and $y[n]$ are independent, then they are also uncorrelated, and since they have zero means, they are also orthogonal; i.e. $$E\{ x[n]y^*[n-k]\} = 0 \tag{1}$$ Let $$... 4 I just finished writing an expository note on this topic recently. I hope you would find it useful. The link between Fourier transform (FT), Fourier series (FS), Discrete-Time Fourier Transform (DTFT) and Discrete Fourier Transform (DFT) is actually hidden in the Poisson summation formula$$ \sum_{n=-\infty}^{\infty} f(x+nT) = \frac{1}{T} \sum_{n=-\infty}^{\...

1

If you are interested one more step into transform theory, I suggest you to go through Exponential Fourier series where the basis functions are complex exponentials, unlike sine and cosine as in Trigonometric Fourier series. I will make an attempt to find the Fourier spectrum of $x(t) = A \cos (2\pi f_0 t)$ using Exponential FS and its properties(let me ...

3

Sketching the signal always helps in such cases. Let's consider a cosine of period $T_0$, that is, of fundamental frequency equal to $f_0=1/T_0$. That is the signal on the left. As you can see on the right, $|A\cos(2\pi f_0 t)|$ does not share the same period with $A\cos(2\pi f_0 t)$. The period of the signal you are looking for is $T_0^{\prime}=T_0/2$ and ...

1

I think a better definition of the power spectrum is the following: The power spectrum of $x(t)$ is the Fourier transform of the autocorrelation function of $x(t)$, where $x(t)$ can be either a deterministic power signal, or a wide-sense stationary (WSS) random process. The definition of the autocorrelation function depends on the model for $x(t)$. If $x(... 1 "I totally understand the concept of Fourier transform" Lucky you if you really do. Some of us (me, in first place) don't (in totality). The Fourier transform (and its avatars) is a prototype for duality. Duality here means that you can represent a signal on some primal domain (time) onto a dual domain (here frequency). This transformation is meant to ... 0 In the discrete case, as when feeding a digitized signal to a DFT or FFT, each output point in the FFT spectrum is the magnitude output from a simple bandpass filter (a Goertzel filter result, or equivalent complex FIR filter). For a rectangular windowed FFT, each of those filters has a Sinc (or Dirichlet) shaped filter response. The more input signal ... 0 The magnitude of$X(\omega)$for a given$\omega$signifies how much does the signal$e^{j\omega t}$exist in the signal$x(t)$; indicated as an inner product between$x(t)$and$e^{j\omega t}$. 0 The magnitude of each bin is the magnitude of that frequency component for that waveform in the time-domain, specifically when the time domain waveform is expressed as a sum of complex exponential frequencies. In the frequency domain that includes positive and negative frequencies, each impulse for the Fourier Transform of an arbitrary waveform is the ... 1 There are various flavors physical interpretations of the continuous Fourier Transform. Here is the one that works best for me: The amplitude of the Fourier Transform is a metric of spectral density. If we assume that the unit's of the original time signal$x(t)$are Volts than the units of it's Fourier Transform$X(\omega)$will be Volts/Hertz or$V/Hz\$. ...

Top 50 recent answers are included