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16

Suppose you have a continuous time analog signal. It is continuous in both time and amplitude. Now when you sample it ,you get discrete samples every Ts seconds. Now you have discrete samples(discrete in time) each of which can take continuous value( in amplitude). This is normally referred to as discrete signal(discrete in time but continuous in amplitude). ...

9

Thanks to the plot in Olli Niemitalo's answer I got convinced that the formula given in the book has a sign error. The non-linearity used for fuzz or distortion is always some type of smoothed clipping function, which compresses the input signal. So small input amplitudes experience little change whereas high input amplitudes are (more or less) softly ...

9

A sinusoidal signal is represented as $$x(t) = \mathrm{cos}(\omega t) = \mathrm{cos}(2\pi f t)$$ $\omega$ is the angular frequency and $f$ is the frequency. See Frequency definition. Your signal \begin{align} x(t) &= 5 + 30\mathrm{cos}(2000\pi t) + 10\mathrm{cos}(6000\pi t)\\ &= 5\mathrm{cos}(2\pi \times 0 t) + 30\mathrm{cos}(2\pi \times 1000 t) + ...

8

I hope the plot below helps answer your question. Typically I have seen the "passband ripple" and "stopband attenuation" expressed in dB as shown in the picture translating the magnitude of the ripples to dB using $20log_{10}$ as shown. So the passband ripple is the amount of variation in the amplitude, within the designated passband of the filter, and stop ...

7

For no information to be lost on conversion back to continuous form, the signal would first need to be perfectly band-limited, and you would need an ideal reconstruction filter. A perfectly band-limited signal is infinite in extent. Since you want this arbitrary signal to be processed by a computer, your computer would need infinite memory. You would also ...

6

The sample rate needs to be GREATER than (NOT just equal to) twice the highest non-zero frequency content of the signal being sampled. Just a little bit greater might work, but the closer the sample rate is to twice the signal frequency, the longer in time you may need to sample to raise the signal above the noise and complex conjugate image in a DFT/FFT ...

6

The sampling theorem states that $f_\mathrm{S} \geq 2f_\mathrm{max}$, where $f_\mathrm{S}$ and $f_\mathrm{max}$ are the sampling and maximum signal freuqency, respectively. But there's an additional condition: The equal sign only holds if the signal spectrum does not contain a dirac impulse at $f_\mathrm{S}/2$ which is clearly the case in your example. ...

6

Here is a good article from National Instruments. In general, we ask on this site that you put in more effort into your question, e.g. links to articles that you've already looked at and things you haven't understood. Otherwise, you'll simply get an answer which is the first link you get from Google when typing in your question. In a nutshell, I/Q takes ...

6

The problem with your reasoning is that $\pi \ne \frac{22}{7}$; $\pi$ is an irrational number. There is no period $N$ for which $x[n] = x[n+N] \ \forall \ n \in \mathbb{Z}$. Hence, the sequence is not periodic.

6

A signal is indeed a function. Given a signal $f(x)$, according to whether continuous or discrete for both the variable $x$ and the function $f(x)$, there are four types of combinations: (1) $\mathbf{continuous}$ $x$ and $\mathbf{continuous}$ $f(x)$ This is the most common $\mathbf{analog}$ signal. (2) $\mathbf{continuous}$ $x$ and $\mathbf{discrete}$ $f(... 6 The optimal decision regions are the Voronoi Regions. I dont know, if this is what you are looking after. import numpy as np points = np.array([(1,1), (1,-1), (-1,1), (-1,-1), (3,3), (3,0), (3,-3), (0,-3), (-3,-3), (-3,0), (-3,3), (0,3), (5,0), (0,5), (-5,0), (0,-5)]) from scipy.spatial import Voronoi, voronoi_plot_2d vor = Voronoi(points) voronoi_plot_2d(... 6 As explained in Maximilian Matthé's answer, the exact computation of the symbol error probability of this constellation (ITU-T V.29 modem standard) is quite complex. However, you can quite easily compute an approximation which becomes very good for relatively large signal to noise ratios (SNRs). This approximation is based on the union bound. The symbol ... 5 |x| denotes the absolute value - the x / |x| bit of the formula is there to make sure that the sign of the input is preserved in the output. Regarding the implementation, yes, the steps you have listed are correct. 5 It's pretty simple, really. Basically, this relies on the fact that the images must be saved in an uncompressed file. If you think about how an image is stored, it's functionally a bunch of sequential numbers: (PX 0,0 - R:x, G:y, B:z) (PX 1,0 - R:x, G:y, B:z) (PX 2,0 - R:x, G:y, B:z) (PX 3,0 - R:x, G:y, B:z) .... (PX 639,0 - R:x, G:y, B:z) (PX 0,1 - R:x, ... 5 well i'm assuming you mean "conventional" DACs and not$\Sigma \Delta$DACs. in a conventional DAC (like an R-2R ladder or something), there are the micro errors that occur between neighboring DAC codes. e.g. non-monotonicity. i think the DSP solution to that is adding a teeny amount dither noise to the value that is output to the DAC. there is a more ... 5 You're misunderstanding what CDMA, TDMA and FDMA do: CDMA doesn't increase the channel capacity in any way. It's a MA, MA = Multiple Access mechanism. In other words, it's just a way of dividing the spectrum among multiple users. No matter what you do, you can't get more data through a channel than the physics and math allows – and Shannon's channel ... 5 From an abstract viewpoint, the space$\Omega$of all signals that are essentially band-limited to a bandwidth of$W$Hz (say from$f_0$Hz to$f_0+W$Hz) and essentially time-limited to duration$T$(say the epoch lasting from$t=0$to$t=T$) has approximately$2WT$orthonormal signals in it, that is, this set$\mathcal B$of$2WT$signals is a (orthonormal)... 4 A simple answer is that I/Q are the real and imaginary components of the complex-valued transmitted baseband signal. In communication systems "I/Q data" usually refers to the real (I) and imaginary (Q) samples of the constellation for the modulation type used. There are usually a lot of I/Q "samples" (rather than "data") that happen during interim ... 4 The S-transform allows you to deal with differential equations in an algebraic manner - so they become easier to solve. Since continuous/analog filters consist of integrators and differentiators the S-transform is therefore a natural way to deal with these systems. The z-transform provides an algebraic way of dealing with finite difference systems and ... 4 In the sampled digital realm, poles at the origin represent delay, which may be necessary to make a filter implementation strictly causal. This delay usually requires no additional arithmetic ops (as a pole elsewhere than zero would require). Sometimes when describing a filter where delay is irrelevant (offline processing, etc.), the filter is centered at ... 4 If you consider the transfer function of a causal IIR filter $$H(z)=\frac{B(z)}{A(z)}=\frac{\sum_{m=0}^M b_mz^{-m}}{\sum_{n=0}^N a_nz^{-n}},\quad a_0=1$$ then you always get the same number of poles and zeros, regardless of the choice of$M$and$N$(as already pointed out by Robert). However, what is meant by a system with "more zeros than poles", is a ... 4 Radix-2^3 is a special class of radix-2 algorithms where the basic decomposition is based on radix-8 and the 8-point DFTs are later on decomposed into radix-2, leading to an algorithm based on radix-2 butterflies. "Same" depends on what you mean and how you classify algorithms. In some sense, a valid way to implement a radix-8 butterfly is to decompose it ... 4 For now I will focus on the conceptual, pending feedback from you: What I'm interested in to know if it is possible to take two adjacent beams, and do another step of beamforming, The answer to this is yes, this is possible to do, and this is done in many beamforming applications. Think about it like this: All beamforming is really doing, is undoing ... 4 You are almost right: digital filters do deal with samples, but a sample can be any numerical representation of a given signal value at a given instant (so in general, they may accept zeros or ones). Moreover, a sample is usually represented by a binary word (e.g. 0001), so a digital filter actually deals with 0s and 1s. 4 A simple way to do it would be using a start bit like UART. When idle, don't transmit anything. When you want to transmit something, first transmit a start bit which is a '1' (ON). Then transmit a fixed number of bits known to both the transmitter and receiver. The receiver looks for the start bit and knows to begin demodulating after it receives the start ... 4 Yes, integration and differentiation can be linear filters. You can start from laplace properties that say:$ \int_{0}^{t} {x(t)dt} \longrightarrow \frac{X(s)}{s} \\ \frac{d}{dt}x(t) \longrightarrow sX(s) $So you can find transfer function of integration and differentiation:$ H_{INT}(s) = \frac{1}{s} \\ H_{DIFF}(s)=s $You can convert these transfer ... 3 Square waves are composed of sine waves at the odd harmonics- i.e. 1st, 3rd, 5th, etc. So yes, you can produce a sine wave from a square wave if you filter it with a low-pass filter whose bandwidth is at least as big as the first harmonic frequency, and whose cutoff frequency is lower than the third harmonic. 3 By definition, band-limited signals in the sense of the sampling theorem have finite energy. Sine waves are periodic and thus have infinite energy. So any dirac pulse in the Fourier transform is not permissible. To be more precise, the sampling theorems only applies to signals that can be represented as$$x(t)=\int_{-f_s/2}^{f_s/2} X(f)\,e^{2\pi i\,ft}\,df$...

3

I and Q signal concept is relatively complex topic to explain without signal background, you first need a basic knowledge about Passband and Baseband real and complex signals. if you are familiar whith these topics, you can jump to the last paragraph. In short signals are any representations of real life quantitative parameter value versus to another ...

3

Best book on the topic is probably http://www.amazon.com/DAFX-Digital-Udo-ouml-lzer/dp/0470665998. There is some freeware stuff (very mixed quality) here: http://www.musicdsp.org/ Many companies spend a lot of effort in making good sounding effects and well working tuners to make money with it, so access to "good" non-commercial sources if fairly limited.

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