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FORMATION OF AN IMAGE on the screen:
GENERAL PRINCIPLES
(Compilation, continued)
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(C) Tertius Gaudens
2.4 The number of scan lines
Z oredelyaetsya based on the spatial frequency response of
the system. Found that at distances up to 5 frame height enough
Z = 660. At other angular size enough Z within [720 2400], ie a
further increase in Z does not improve the playback quality.
For interlaced scanning in European TV systems adopted
And oh And oh
Z = Z + Z 4 0 4 0 = 2Zp 4 0 = 2 (Zp + Zp) = 625 =
= 575 + 50 = 2 * 312.5 = 2 (288 + 24.5)
A
where Z 5 0 - the number of active (reproduced on the screen)
lines of the frame;
oh
Z - number of rows flyback (OH);
Zp - the number of rows in the field.
Similarly, for the NTSC
Z = 525 = 452 + 73 = 2 * 262.5 = 2 (226 + 36.5)
Accordingly, in progressive scan on the basis of the
European TV standard
Z = 312 = 287 + 25
For NTSC
Z = 262 = 225 + 37
Frequency sweep lines in standard 50/625 is
F 0 = F 4c 4r 0 * Z = 25 Hz * 625 = 15.625 kHz
In this case, the allowable standard deviation of the
horizontal frequency is
F 4s 0 = 15625 + / - July 2003 Hz
We can obtain the tolerance framerate.
Accordingly, the duration of the line is
Tc 4 0 = 1/Fc 4 0 = 64 ms
for the duration of its active part
A oh
Tc = Tc - Tc = 64 - 12 = 52 ms
oh
4c, where T 0 - the duration of the GO line. A
In NTSC Fc 4 0 = 29.97 * 525 = 15.734 kHz
Tc 4 0 = 63.55 ms, Tc 5 0 = 52.6 ms.
With progressive scan requirements to speed increases:
Ave CR
F. = 2Fk
Hence it is necessary to either
Ave CR
Fc = 2Fs,
or
Ave CR
Z = Z / 2
Vlyanie Z on the quality perception is evaluated based on
the actual detail of the image, which is
A 2
N = (F * Z)
Given the logarithmic nature of the perception of the eye,
can be
A
calculate that a decrease in Z 5 0vdvoe reduces the number
of perceived by the elementary parts (the elements of
disintegration image pixel, ELP), only 13%.
From this it follows that the quality perception f. 4 0 is th
A
Section of greater importance than Z. That is why when
dealing with display is more suitable progressive scan on NTSC
and not a European TV standard.
2.5 Bandwidth
Required sampling frequency video fd 4
0zavisit from the cutoff frequency df, defined as the maximum
the frequency spectrum relative to the signal, Fourier series.
In this case, according to Theorem Kotelnikov need to fdmin
= 2df. To guarantee the security token being a signal when it is
reverse conversion into analog form fd 4 0vybirayut,
from the condition
fd 4 0> fdmin = 2df
fd 4 0 = F * Fc
where F 7 0 - number of harmonics for which decomposes the
signal.
From here you can get a general expression for the
decomposition of the image:
2
TC = F * Z / (2df * hr)
A A
hr = (Z / Z) * (Tc / Tc)
where hr - coefficient of performance scan.
By the way, substituting the reference values can be
obtained
625 525
hr = 0,75; hr = 0,71
In PAL accepted df = 5,5, in NTSC - 5.01 in SECAM - 6,5
MHz. For all TV systems accepted fd 4 0 = 13.5 MHz. It is a
harmonic frequency of 2.25 MHz - the least common multiple of
systems with expansion to 525 (F 7 0 = 143) and 625 (D 7 0 =
144) lines.
Hence, for PAL / SECAM number of readings - on line 4 0 = Nx
A
864, in the active part strokm Nx 5 0 = 702, dlitelnlst
reference td 4 0 = 74.1 ns.
A
For NTSC Nx = 858, and Nx = 710.
Since the number of lines, progressive scan twice
Meunier, the value of Nx 4 0tozhe reduced accordingly: for
PAL / SECAM fd 4 0 = 6.75 MHz; td 4 0 = 148.2 (approximately 7
0150) ns;
A
Nx 4 0 = 432; Nx = 351; df = 3,375 MHz.
Compare these data with the requirements for audio
sampling. The range of perceived ear - 2 Hz ... 20 kHz. Based
on the This sampling frequency of sound is assumed to be
star star
2df 5 0
