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برسی و آموزش منطقه فرنل Fresnel Zone ونکته هایی که باید در مورد آن بدانیم

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  • برسی و آموزش منطقه فرنل Fresnel Zone ونکته هایی که باید در مورد آن بدانیم

    دوستان گلم از اونجایی که میبینم خیلی از دوستان عموما در درک فرنل مشکل دارن و این امر باعث میشه نتونن مشکلات خودشون رو عیب یابی کنن تصمیم گرفتم که توضیح جامع و کاملی همراه با مثال در مورد منطقه فرنل بدم

    تئوری منطقه فرنل ابتدا بر پایه یک خط دید مستقیم(که آنرا LOS مینامیم "Line Of Sight") مابین 2 دستگاه رادیو(که در اینجا نقطه A,Bنام گذاری خواهد شد) است و سپس فضای 3 بعدی پیرامون این خط که خطوط مستقیمی را از نطقه A به B ایجاد میکند

    توجه داشته باشید که منطقه فرنل زیادی وجود دارد اما ما فقط با LOSیعنی منطقه 1فرنل کار داریم
    برای اندازه گیری ارتفاع نصب تجهیزات باید ما مقدار شعاع سیگنال رو حساب کنیم که اون رو با r نشون میدیدم
    ((r = 17.31 * sqrt(N(d1*d2)/(f*d
    r=شعاع سیگنال به متر
    n=منطقه مورد محاسبه
    d1,d2=مسافت از مانع تا هریک از رادیو ها به متر
    d=تمام مسافت لینک به متر
    f=فرکانس به مگاهرتز
    مثال:
    یک لینک به طول 2 کیلومتر در فرکانس 5180 چه مقدار ارتفاع نیاز دارد (بدون مانع)
    زمانی که هیچ مانعی وجود ندارد Pدقیقا در مرکز است.p شعاع سیگنال در محل تلاقی d1,d2است

    (r=17.31*sqrt(1*1000*1000/5180*2000
    (r=17.31*sqrt(1000000/10360000
    r=5.37 متر
    و یا با فرمولی دیگر
    (r=8.657*sqrt(D/f
    D=مسافت کل به کیلومتر
    f=فرکانس به گیگا هرتز
    5.37=(r=8.657*sqrt(2/5.18
    در این ارتباط در صورتی که شما هیچ مانعی نداشته باشید نصب آنتن در ارتفاع 6 متری باعث میشود که شما 60% از سیگنالتون رو به طور صحیح دریافت کنید.
    آیا این ارتفاع برای ارتباط فول داپلکس مناسبه؟جواب خیر بدلیل اصل هوگن و رعایت اون در الگوریتم ارتباطی رادیوها، شرکتهای سازنده برای جولوگیری از تصادم و کاهش تداخل از الگوی تغییر فاز در ارتباط داپلکس استفاده میکنن که در این صورت شما به 100% منطقه پاک احتیاج دارید

    نکته
    زمانی که شما شعاع سیگنال رو حساب کردید طبیعتا با برسی موانع تون اقدام به نتیجه گیری برای ارتفاع مورد نیاز میکنید اما باید توجه داشته باشید که زمین گرد هست و در لینکهای طولانی باید منحنی زمین رو هم محاسبه و حاصل رو با ارتفاع مانع تون جمع کنید با فرمول زیر

    H=1000*D2/8Er
    H=ارتفاع قوس زمین در وسط مابین 2 نقطه
    D=کل مسافت به کیلومتر
    Er=شعاع موثر زمین بر حسب کیلومتر که 8504 کیلومتردرنظرگرفته میشود


    فایل اکسل محاسبات بالا(با تشکر از parmid)

  • #2
    شروین جان دست گلت درد نکه آموزش خوبی بود
    مشاوره و اجرای m.programer@gmail.com
    LAN - WAN - WIRELESS -VOIP - CCTV Architecture

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    • #3
      با تشکر از آقا شروین فقط لطف میکنی بگی sqrt و N منطقه مورد محاسبه دقیقا چیه؟ اینم بی زحمت توضیح بدین دیگه بحث کامل میشه.

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      • #4
        نوشته شده توسط IliyaBlack مشاهده پست
        با تشکر از آقا شروین فقط لطف میکنی بگی sqrt و N منطقه مورد محاسبه دقیقا چیه؟ اینم بی زحمت توضیح بدین دیگه بحث کامل میشه.
        فکر نمیکردم کسی باشه که ندونه اینا چی هست چون 2تا مطلب ریاضی هست که همه باید بلد باشن sqrt یعنی رادیکال و N تعداد

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        • #5
          مرسی خوب البته شاید ذهن ادم اولش نره سمت ریاضیات
          دیگه خیلی از مدرسمون گذشته

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          • #6
            سلام
            من این فایل اکسل رو بر اساس گفته های آقای شروین درست کردم برای راحتی کار امیدوارم به دردتون بخوره
            دانلود

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            • #7
              دو تا عکس اضافه کردم که مطلب ملموس تر باشه برای دوستان

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              • #8
                نوشته شده توسط parmid مشاهده پست
                سلام
                من این فایل اکسل رو بر اساس گفته های آقای شروین درست کردم برای راحتی کار امیدوارم به دردتون بخوره
                دانلود
                احسنت آفرین چه کار جالبی انجام دادی ببینم بجز اکسل برنامه نویسی هم بلدی که یه نرم افزار محاسبه کامل طراحی کنیم بزرایم برای همه دانلود کنن؟

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                • #9
                  با c# راحت میشه نوشت

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                  • #10
                    Fresnel Zone

                    The exact theory of Fresnel (pronounced “Fray-nell”) zones is quite complicated. However, the concept is quite easy to understand: we know from the Huygens principle that at each point of a wavefront new circular waves start, We know that microwave beams widen. We know that waves of one frequency can interfere with each other. Fresnel zone theory simply looks at a line from A to B, and then at the space around that line that contributes to what is arriving at point B. Some waves travel directly from A to B, while others travel on paths off axis. Consequently, their path is longer, introducing a phase shift between the direct and indirect beam. Whenever the phase shift is one full wavelength, you get constructive interference: the signals add up optimally. Taking this approach and calculating accordingly, you find there are ring zones around the direct line A to B which contribute to the signal arriving at point B.
                    Figure 2.10: The Fresnel zone is partially blocked on this link, although the visual line of sight appears clear..
                    Note that there are many possible Fresnel zones, but we are chiefly concerned with zone 1. If this area were blocked by an obstruction, e.g. a tree or a building, the signal arriving at the far end would be diminished. When building wireless links, we therefore need to be sure that these zones be kept free of obstructions. Of course, nothing is ever perfect, so usually in wireless networking we should check that that the area containing about 60 percent of the first Fresnel zone should be kept free.
                    Here is one formula for calculating the first Fresnel zone:
                    r = 17.31 * sqrt(N(d1*d2)/(f*d))
                    ...where r is the radius of the zone in meters, N is the zone to calculate, d1 and d2 are distances from obstacle to the link end points in meters, d is the total link distance in meters, and f is the frequency in MHz. Note that this gives you the radius of the zone. To calculate the height above ground, you need to subtract the result from a line drawn directly between the tops of the two towers.
                    For example, let's calculate the size if the first Fresnel zone in the middle of a 2km link, transmitting at 2.437GHz (802.11b channel 6):
                    r = 17.31 sqrt(1 * (1000 * 1000) / (2437 * 2000))
                    r = 17.31 sqrt(1000000 / 4874000)
                    r = 7.84 meters
                    Assuming both of our towers were ten meters tall, the first Fresnel zone would pass just 2.16 meters above ground level in the middle of the link. But how tall could a structure at that point be to clear 60% of the first zone?
                    r = 17.31 sqrt(0.6 *(1000 * 1000) / (2437 * 2000))
                    r = 17.31 sqrt(600000 / 4874000)
                    r = 6.07 meters
                    Subtracting the result from 10 meters, we can see that a structure 3.93 meters tall at the center of the link would block up to 60% of the first Fresnel zone. To improve the situation, we would need to position our antennas higher up, or change the direction of the link to avoid the obstacle.



                    کد:
                    http://www.vias.org/wirelessnetw/wndw_04_08b.html

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                    • #11
                      Fresnel zone calculation


                      A particularly common problem with radio systems is communication errors. The occurrence of communication errors changes significantly depending on the method of placement of the radio equipment. With radio communication, it is important to ensure "line of sight" between the transceivers.
                      When we can see the receiving antenna we use the expression "line of sight" without much thought, but with radio waves, it can be said that "line of sight" is only achieved when the Fresnel zone indicated below is established.
                      With radio systems used at about 1.5 m above ground, if a Fresnel zone is not established, multipath interference will occur, so that the true electric field intensity cannot be obtained. If operation of the equipment is unstable, it is necessary to consider whether the fixed station can be set up in a higher place.


                      Radio system errors
                      Radio system errors occur when the bit strings of 0's and 1's that make up the data cannot be identified by the electronic circuits inside the receiver, due to corruption of the transmission data.

                      Viewed on a temporal basis, assessment of the receive data is performed in the middle position of the receive bit width, and in terms of level, the bit is determined to be 1 or 0 by whether the received voltage of the median value (threshold value) is bigger or smaller. This identified signal is the signal after FSK demodulation. Because the signal and noise are a composite waveform to begin with, if the difference between the signal level and noise level (inside the receiver) is low, with certain sampling timing, since the decision is made with a waveform in which the threshold value is exceeded due to noise, an error results. Therefore in general, the signal level is required to be about 20 dB (100 times the voltage ratio) greater than the level of noise.

                      FSK radio equipment is immune to the influence of amplitude noise, but if there is an obstruction between the antennas (the ground, buildings, natural objects and so on), the radio waves will be reflected resulting in multipath interference. At the receiving point, the composite wave incorporating the delayed radio waves is distorted in amplitude, resulting in an error. In addition, the ratio of the noise component of signals that enter the demodulation circuit in a location with weak electric field intensity is high in relation to the signal, and this causes errors. Of course, signals generated by the antenna due to external noise will also have an impact on amplitude and frequency.

                      Fresnel zone
                      In order for radio waves emitted from the transmitter to reach the receiver without attenuation of power, a certain amount of space is required. The energy cannot reach the receiver via one straight line in space. It is easy to understand for example that the waves will not get there through a hole the size of a needle in a concrete wall.
                      The space required is a spheroid with its center along the shortest distance between antennas, and this is called the Fresnel zone. In fact this space expands indefinitely, but the part that principally contributes to communicating the energy is called the 1st Fresnel zone.
                      If there are obstacles inside the Fresnel zone, insufficient energy is transmitted so that received field intensity becomes weak. If the received field intensity is weak, the probability that errors will occur becomes gradually higher. The receive sensitivity of the receiver is absolute, and propagation loss which depends on the distance traveled by the radio waves cannot be avoided. Therefore in order to prevent errors from occurring, it is important to ensure that the received radio waves are as close as possible to the theoretical value.

                      The 1st Fresnel zone is a spheroid space formed within the trajectory of the path when the path difference when radio wave energy reaches the receiver by the shortest distance, and when it gets there by another route, is within λ/2. In this case, λ is the wave length of the radio wave (wave length = speed of light / frequency) which at 400 MHz is 0.75 m.

                      When positioning radio systems
                      The distance between the Fresnel zone boundary and a straight line running the shortest distance between the antennas is called the Fresnel radius, and if there are no obstacles in the space forming 60% of this distance, propagation characteristics are said to be the same as in free space.

                      < Click to move to the calculation window >
                      Fresnel zone formula



                      Instructions



                      1. Click the "Figure / Graph" button to display the calculation screen.
                      2. Enter the required items in the yellow input fields for entering parameters. To calculate the Fresnel zone, after setting the parameters in the input fields, push the Enter key on your PC, or click the "Calculate" button.
                      3. Set the distance d in the pink input field.
                      4. To find the Fresnel radius, decide the position to find (distance on the X axis), and left-click with your mouse on the graph.

                      * Right panel
                      The values shown in the panel on the right are the Fresnel radius and diameter at the midpoint.

                      آخرین ویرایش توسط nimadrayaneh; در تاریخ/ساعت 2013/12/05, 12:37 AM.

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                      • #12
                        اینم یک امورش خوب
                        http://s5.picofile.com/file/81027769..._Zone.pdf.html

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                        • #13
                          سلام .
                          يه سئوال از كارشناسان محترم دارم ، من خيلي جاها در مورد فرنل زون خوندم . اما يه موضوع مهم واسم پيش اومده ،

                          تا اون جايي كه من ميدونم هرچي گين آنتن بيشتر باشه تمركز امواج آنتن بيشتر ميشه پس در نتيجه شعاع فضاي فرنل بايد باريكتر بشه ، سئوال من اينه كه چرا هيچ جايي توي محاسبات ناحيه فرنل از گين آنتن صحبت نشده ؟؟؟ در صورتي كه وقتي گين آنتن بالا باشه تمركز امواج بيشتر و شعاع ناحيه فرنل كمتر بايد باشه ..


                          ممنون ميشم جوابم روبدين ..

                          ياعلى ..

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