# Tag Info

65

The color burst is also an indicator that there is a color signal. This is for compatibility with black and white signals. No color burst means B&W signal, so only decode the luminance signal (no croma). No signal, no color burst, so the decoder falls back to B&W mode. Same idea goes to FM stereo/mono. If there is no 19 kHz subcarrier present, ...

26

In the absence of a valid color burst signal, the "color killer" circuit disables the color difference signals, otherwise you would indeed see colored noise. This is mainly intended for displaying weak signals in B/W without the colored noise. One step further is to mute the entire signal, substitute stable sync signals, and display a blue or black field ...

14

Analog and digital First off, you need to understand what an "analog" signal is, and what a "digital" signal is, how they are different, and how they are similar. The term "analog" comes from the old distinction between "analog" and "digital" computers. A "digital computer", even a very primitive one of decades ago, has always been more or less what we ...

12

Windowing is used because the DFT calculations operate on the infinite periodic extension of the input signal. Since many actual signals are either not periodic at all, or are sampled over an interval different from their actual period, this can produce false frequency components at the artificial 'edge' between repeated intervals, called leakage. By first ...

11

I think you are confusing two different operations. Windowing in the time domain is explained by @sam, so I won't repeat that. But windowing is not done to perform filtering. Filtering by multiplying the FFT of a signal by the filter frequency response is entirely reasonable in many situations, and is indeed done. The alternative for filtering is time-...

11

There have been several good answers to this question. However, I feel that one important point has not been made entirely clear. One part of the question was why we don't just multiply the FFT of a signal with the desired filter response. E.g., if we want to lowpass filter our signal, we could simply zero all frequency components higher than the desired cut-...

7

Windowing reduces spectral leakage. Say you start out with a $\sin(y) = \cos(\omega_0 t)$. The period is obviously $2 \pi/ \omega_0$. But if nobody told you that the period is $2 \pi/ \omega$ and you blindly choose the range $[0, 1.8 \pi/\omega_0]$ and take FFT of this truncated waveform, you will observe frequency components in other frequencies ...

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

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

If you don't use a non-rectangular window, then the FFT results will already be convolved with the transform of a default rectangular window (a periodic Sinc) before doing any frequency domain filtering. e.g. you will get two filters applied, one of which you probably don't want. By windowing in the time domain, before the FFT and frequency domain ...

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 $\mathbf{discrete}$ $f(... 5 Window in the time domain because we can guarantee zero at the edges of the window window functions have a nice analytic expression in spatial domain many window functions have a weird shaped spectrum that would be hard to approximate only a finite number of samples are needed (windowing can be done as the signal streams in) e.g. from wikipedia The hard ... 5 Film isn't absolutely "analog", as in continuous. Every individual silver halide molecule, after exposure and development, is either metalized or not; and there are a finite number of these molecules in every frame of film, thus quantizing the exposure measurement. However the density and location of the film grains and silver halide molecules is semi-... 5 To calculating the overall gain of a Bainter stage, would I simply work out the individual gains of the three op-amp sections. The overall gain would then be the product of the three individual gains? The short answer is: Yes, you can (probably) analyze them individually. When asking what happens when you cascade multiple analog filter stages, the ... 5 This sounds like a confusion in terminology. See http://en.wikipedia.org/wiki/Digitizing Digitization basically involves two steps: Discretization: Sampling the signal at discrete times Quantization: Turning the samples from (in theory) infinite resolution to finite resolution Look at all possible combinations you can have 4 different types of signal: ... 5 well i'm assuming you mean "conventional" DACs and not$\Sigma \Delta$DACs. in a conventional DAC (like an R-2R ladder or something), there are the micro errors that occur between neighboring DAC codes. e.g. non-monotonicity. i think the DSP solution to that is adding a teeny amount dither noise to the value that is output to the DAC. there is a more ... 4 Answer to "Why do we not (commonly, at least) use .... TDM to multiplex analog signals? If you have$N$analog (meaning continuous-time) signals to transmit over a common channel, then with TDM, each signal is transmitted for$(1/N)$-th of the time, e.g.$1000$signals are transmitted for one millisecond each in non-overlapping time slots during a$1$-... 4 There are two aspects to how this works. First, since the signal is oversampled there is a great deal of correlation between samples that we can take advantage of via the low-pass filter. The noise, on the other hand, has no correlation (assuming it is white noise), and thus will often destructively interfere with itself. Your question seems to be more ... 4 You have to clearly define what you mean by an "analog FIR filter". "No poles" is not correct because (discrete-time) FIR filters do have poles; they are just all at the origin of the$z$-plane (for causal FIR filters). Note that filters without poles do not exist. Take as an example the discrete-time transfer function $$H(z)=1-az^{-1}\tag{1}$$ with a zero ... 4 The main trick is knowing how typical lumped circuit element impedances are represented in the$s\$-domain. Recall the current-voltage relationship for a capacitor: $$i(t) = C \frac{dv(t)}{dt}$$ Transforming this to the Laplace domain yields: $$I(s) = CsV(s)$$ Note that voltage and current still have a proportional relationship in this domain, just ...

4

Try digitizing a 17GHz signal.. You will quickly find it difficult to procure a fast enough adc that does not cost a fortune and you will also need memory bandwidth of tens of Gb/s (depending on adc bit count) not to mention the processing power to process such amount of data in realtime. Analog circuits do not suffer from data storage problems and realtime ...

4

To elaborate on MM's answer just a little bit. Unless the receiver has some sort of expectation of the nature of the signal there is no way to tell. With an expectation (for instance a pure tone), only a best estimation can be made with the rest assumed to be noise. With something like serial communication a voltage past a threshold is considered a "1" ...

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 $$x(t)=\int_{-f_s/2}^{f_s/2} X(f)\,e^{2\pi i\,ft}\,df... 3 again, even with the Addendum, i think Lutz's answer misses the point. the point is (quoting Wikipedia): To illustrate the necessity of f_s \ > \ 2B, consider the family of sinusoids (depicted in Fig. 8) ) generated by different values of \theta in this formula:$$x(t) = \frac{\cos(2 \pi B t + \theta )}{\cos(\theta )}\ = \ \cos(2 \pi B t) - \sin(2 ...

3

For many applications, the technology often selected is the lowest cost one that meets the most common requirements (audio sample rates, etc.)

3

Another typical approach, that independently of my other answer works, is predistortion, for example with the look-up table mentioned by robert, or with a correction polynomial. If you can really pinpoint your nonlinearities to a simple digital-in/analog out curve, you can just find the inverse of that curve, and put it in a correcting mapping, and apply ...

3

My understanding of your question is that you have a small misunderstanding here. In the definition it is not said time and space, a signal can vary in time or space. Some signals vary with time, as your example human voice that varies over time in air pressure (or equivalently voltage), some vary with space, like image and some vary with both time and space ...

3

In PAL, the colour information (chrominance or chroma) is modulated onto the black and white (luminance or luma) baseband signal. The chroma is at ~4.4MHz offset from DC and is about 1.3 MHz wide. Assuming that your noise is centered around DC then, if it is less than ~3.5MHZ wide then it won't appear in the chroma spectrum and will only be in the luma. ...

3

It's often easier to design FIR filters for compensating group delay. At the same time they could also compensate the magnitude if necessary. The easiest method is to use a complex least squares method, which boils down to solving a system of linear equations for the filter coefficients. The difficult part is to choose an appropriate bulk delay in the ...

2

You should photograph a known calibration target with your system, (like this one or the one that I put in the thumbnail). This will allow you discovering the OECF of your system. You can either calculate it by yourself, given the technical data sheet or use third-party software like Imatest, ImageAnalyzer or others.

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