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1

I think that's because of the DC offset of the signal. A time domain or frequency domain plot provided by you could have been helpful here for the confirmation of your doubt. Following are suggestions 1.try to filter the signals before FFT depending upon the signal band you are interested in 2.Subtract the signal by the mean of the entire signal (which is ...


0

I think there are some problems with your analysis structure. First of all, for being able to see from -0.5 to 0.5, without aliasing, you have to sample at 1. Secondly, if you want to down convert a signal, bring it to DC you need to multiply it with the corresponding cosine in time, or discrete time, or apply the convolution to its spectrum with the ...


0

So the original signal has a spectrum looking like this: So the 40Hz to 50Hz components are also shown at 50Hz to 60Hz. Assuming that the DAC just outputs the raw data without any interpolation (so it's just a zero-order hold between data points), then the spectrum of that signal will look like this: Note that all I've really done is sampled this at 10 $\...


1

Not 100% sure I understand the question but it seems to be just a confusion of terms. Discrete: A discrete time signal is one that is only defined at integer indices. Like you said, usually the notation for these indices are $n$. Discretize: A discretized signal is one that is made to be discrete time. For example, if you have a continuous time signal, $...


2

Signal $x[n]$ has $N=50$ samples and the filter $h[n]$ has $M=10$ samples. The output $y[n]$ (by linear conv) will have length $L = N+M-1=59$ samples. If you use time domain convolution, for each output sample (except the edges) you will be making $M=10$ real MACs. And this makes the number of total real MACs as $K = L \cdot M = 59 \times 10 = 590$. The ...


4

We know that $\displaystyle \frac{\sin(\theta (n+1))}{\sin(\theta)} u[n+1]$ has value $0$ for $n < -1$ since $u[n+1]=0$ for $n < -1$. At $n=-1$, $u[n+1]$ jumps to value $1$, but $\sin(\theta (n+1))\bigr|_{n=-1} = \sin(\theta (-1+1))$ has value $0$, and so $\displaystyle \frac{\sin(\theta (n+1))}{\sin(\theta)} u[n+1]$ just happens to equal $\...


0

@MBaz solution is neat. Another vision is to use the idea that discrete convolutions turn into polynomial products in the $z$-transform domain: $$[1,1]\ast [1,1] = [1,2,1]$$ is equivalent to $$(1+z^{-1})(1+z^{-1})=(1+2z^{-1}+z^{-2})$$ So separating a long filter into short convolved filters is equivalent to factorizing a polynomial. To ease notations, I'll ...


3

If the size $N$ of the DFT is even, only one "extremal" point (after fftshift) is Nyquist. If $N$ is even, you cannot have an arbitary $c_{-5}$ and $c_5$. They must add to be whatever your $c_{-5}$ term is. If the input to the DFT is purely real and if $N$ is even, consider the Nyquist point, at $c_{N/2}$ to be split in half. One half is the negative-...


1

First, we define $h_3[n] = h_2[n] \star h_2[n] = [1, 2, 1]$. From this result, we know that $h_1[n]$ must have 5 elements (so that $h[n]$ ends up with 7 elements). Let's define $h_1[n] = [g_1, g_2, g_3, g_4, g_5]$. We can find these as follows, using the definition of discrete convolution. First, we know that $g_1 h_3[0] = h[0] = 1$, so $g_1 = 1$. Then, ...


0

In the audio world, we have this notion of the Gerzon-Craven limit of bit reduction in quantization (this is not the same as lossy coding in a codec like MP3). And it just says that the Shannon result goes both ways (digital info over an analog channel and analog signal over a bit-limited digital channel). And they generalized the Shannon Channel Capacity ...


1

Ah okay thank you. I think what irritated me was that I didn’t consider that the sampled function is a periodic or periodically continued function so that the Fourier transform delivers amplitude values for the base frequency and multiples of it only (so the Fourier transform is discrete), which are indexed as n*(base frequency), so that it is obvious why F „...


0

This can be answered simply by considering the definition of the $N$-point DFT: $$X_N[n] = \sum_{k=0}^{N-1} x[k]e^{-j2\pi \frac {n}{N}k }$$ where it's easy to see that the DFT just compares your $N$-point input signal $x[k]$ to a sinusoid of frequency $\frac nN$. Thus, the lowest frequency is always $0$, and the resolution is always $\frac{f_\text{sample}...


0

Let us assume that you have a finite length discrete signal $x$, denoted by its samples $x_n$, $0\le n<N$; $x$ does not depend on $n$, but its is values are indexed by $n$. Once you index a signal with integers, it somehow "looses" its dependence to an "actual time" in seconds. In other words, one does not know how much time actually elapsed between $x_{...


0

Keep Signal Processing aside, if you just try to understand what is happening in Convolution and Correlation, both are very similar operations. The only difference is in Convolution, one of the variable is inverted(flipped) before performing the accumulation of the product. See that i am not using the word signal anywhere above. I am only talking in terms of ...


0

Marcus is correct, so this means something else is wrong. The left equality simply shows that, for continuous time, the unilateral PSD, $S_x(f)$, is two times the Fourier transform of the autocorrelation function, $R_x(\tau)$. See, e.g., A.B. Carlson, Introduction to Communication Systems, 2nd Ed., McGraw-Hill, NY ©1975, Chapter 2, equation 18a. Or see ...


2

Protocols like w-mbus or Zigbee typically use some form of frequency-spreading like DSSS. In DSSS we do not send symbols directly using some pulse shape and encoding the information in the sign but instead we use a rapidly alternating sequence of $\pm 1$ as a pulse. This leads to an increase in the occupied bandwidth ("spreading"). The receiver multiplies ...


20

If the ratio between your sampling frequency and the frequency of your signal is irrational, you will not have a periodic discrete signal. Assuming you have a 1-kHz sine wave and you sample at 3000*sqrt(2) Hz. You will have approximately 4.2 samples per period. However you will not be able to sample the sine wave exactly at the same place. Hence your ...


0

The most precise way to deal with them is to compare them using the algorythm used in the Wigner timer frequency distribution. It's a processing intensive way of comparing the real and imaginary results of the FFT, I am completely ignorant of the other types of FFT implementations. The wigner time frequency distribution uses both products to construct a ...


2

It depends on your spectrum analyzer. If you want to analyze a one channel real baseband signal, the mirrored frequencies are redundant. you should keep the first N/2+1 points. For communications signals, there can be an I and Q (two reals as a single complex), you want to keep them all. the frequencies will not be (in general) symmetric. If you want to ...


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