Mobilne Telekomunikacije Ispit (Test 1)
Pitanja sa papira pod rednim brojevima 3, 4, 5, 6 i 7 nisu i ovom kvizu.
Questions and Answers
- 1.
Ako je predajnik snage 10W, snaga izrazena u dBm je:
- A.
40dBm
- B.
30dBm
- C.
10dBm
Correct Answer
A. 40dBmExplanation
The correct answer is 40dBm because the power in dBm is calculated by taking the logarithm (base 10) of the power in milliwatts and multiplying it by 10. In this case, since the power is given as 10W, we need to convert it to milliwatts by multiplying it by 1000 (1W = 1000mW). Taking the logarithm of 10,000 (10W * 1000mW/W) gives us 4, and multiplying it by 10 gives us 40dBm.Rate this question:
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- 2.
Ako je snaga predajnika 10W, dobitak predajne antene 10dBi koliki je EIRP?
- A.
100dBW
- B.
20dBW
- C.
50dBW
Correct Answer
B. 20dBWExplanation
The EIRP (Effective Isotropic Radiated Power) is calculated by adding the transmit power (in dBW) to the gain of the transmitting antenna (in dBi). In this case, the transmit power is 10W (or 10dBW) and the gain of the transmitting antenna is 10dBi. Adding these values together, we get 10dBW + 10dBi = 20dBW. Therefore, the EIRP is 20dBW.Rate this question:
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- 3.
Izabrati tip antene koji ima najmanju direktivnost:
- A.
Yagi antena
- B.
Monopol antena
- C.
Dipol antena
Correct Answer
C. Dipol antenaExplanation
The dipole antenna is the type of antenna that has the least directivity. This means that it radiates and receives signals equally in all directions. Unlike the yagi and monopol antennas, which are more directional and focus their energy in a specific direction, the dipole antenna is omnidirectional. It is commonly used in applications where a wide coverage area is desired, such as in broadcast radio and television.Rate this question:
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- 4.
Ako je radna frekvencija antene 3GHz odrediti talasnu duzinu:
- A.
0,3m
- B.
0,1m
- C.
1m
Correct Answer
B. 0,1mExplanation
The wavelength of an antenna can be calculated by dividing the speed of light by the frequency of the antenna. In this case, the frequency is given as 3GHz (3 billion Hz). By using the formula wavelength = speed of light / frequency, we can find the wavelength. The speed of light is approximately 3 x 10^8 meters per second. Dividing this by 3 x 10^9 Hz gives us a wavelength of 0.1 meters, which is equivalent to 0.1m. Therefore, the correct answer is 0.1m.Rate this question:
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- 5.
Ako je snaga predajnika 10W, dobitak predajne snage 10dBi koliki je ERP u slucaju polutalasnog dipola
- A.
18.75
- B.
17,85
- C.
15,75
Correct Answer
B. 17,85Explanation
The question asks for the Effective Radiated Power (ERP) in the case of a half-wave dipole antenna, given that the transmitter power is 10W and the gain of the transmitting antenna is 10dBi. The ERP can be calculated by adding the transmitter power to the gain of the antenna. In this case, the gain is given in decibels relative to an isotropic radiator (dBi), so we need to convert it to a linear scale. 10dBi is equivalent to a gain of 10^(10/10) = 10 times. Therefore, the ERP is 10W + 10W = 20W. However, the answer choices provided do not match this result, so the correct answer cannot be determined.Rate this question:
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- 6.
Koja je maksimalna frekvencija koju prima mobilni koji se krece brzinom od 30m/s, ako je ucestanost nosioca fc=6GHz?
- A.
Fc+600Hz
- B.
Fc+300Hz
- C.
Fc+127Hz
Correct Answer
A. Fc+600HzExplanation
As the mobile is moving with a velocity of 30m/s, there will be a Doppler shift in the frequency of the carrier wave. The Doppler shift formula is given by Δf/fc = v/c, where Δf is the change in frequency, fc is the carrier frequency, v is the velocity of the mobile, and c is the speed of light. Substituting the given values, we can calculate the change in frequency to be 600Hz. Therefore, the maximum frequency that the mobile receives is fc + 600Hz.Rate this question:
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- 7.
Doplerov pomeraj je jednak nuli kada:
- A.
Vozilo se krece direktno u pravcu predajnika
- B.
Vozilo se krece direktno od predajnika
- C.
Vozilo se krece upravno na pravac dolaska emitovanog talasa
Correct Answer
C. Vozilo se krece upravno na pravac dolaska emitovanog talasaExplanation
When a vehicle is moving perpendicular to the direction of the emitted wave, the Doppler shift is equal to zero. This means that there is no change in the frequency of the wave as observed by the receiver on the vehicle. This occurs because the vehicle's motion is neither towards nor away from the transmitter, resulting in no apparent change in frequency.Rate this question:
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- 8.
Frekvencijski selektivan feding nastaje kada je frekvencijski opseg signala:
- A.
Manji od koheretnog opsega kanala
- B.
Ne zavisi od opsega kanala
- C.
Veci od koheretnog opsega kanala
Correct Answer
C. Veci od koheretnog opsega kanalaExplanation
Frequency selective fading occurs when the frequency range of the signal is larger than the coherent bandwidth of the channel. In other words, the signal experiences different levels of attenuation or distortion across different frequencies within the channel. This can result in variations in signal strength and quality, leading to potential communication errors or disruptions.Rate this question:
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- 9.
Ravan feding nastaje kada:
- A.
Se korisnik krece veoma brzo
- B.
Kada je trajanje simbola veliko u odnosu na standardnu devijaciju kasnjenja visestrukih putanja signala
- C.
Kada se korisnik krece sporo
Correct Answer
B. Kada je trajanje simbola veliko u odnosu na standardnu devijaciju kasnjenja visestrukih putanja signalaExplanation
Ravan fading occurs when the duration of symbols is large compared to the standard deviation of the delay of multiple signal paths. This means that if the symbols are transmitted for a longer duration and there is a significant difference in the delay of the multiple signal paths, ravan fading will occur. The speed of the user's movement is not a factor in ravan fading.Rate this question:
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- 10.
Brz feding nastaje kada je trajanje simbola:
- A.
Vece od koherentnog vremenskog intervala kanala
- B.
Mnogo manje od keheretnog vremenskog intervala kanala
- C.
Ne zavisi od trajanja simbola
Correct Answer
A. Vece od koherentnog vremenskog intervala kanalaExplanation
Brz feding nastaje kada je trajanje simbola veće od koherentnog vremenskog intervala kanala. This means that fast fading occurs when the duration of symbols is longer than the coherent time interval of the channel. In fast fading, the channel conditions change rapidly over time, causing significant fluctuations in the received signal strength. This can result in errors and degraded performance in wireless communication systems.Rate this question:
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- 11.
Spor feding nastaje kada je trajanje simbola:
- A.
Mnogo manje od koherentnog vremenskog intervala kanala
- B.
Mnogo vece od koherentnog vremenskog intervala
- C.
Ne zavisi od trajanja simbola
Correct Answer
A. Mnogo manje od koherentnog vremenskog intervala kanalaExplanation
Sports fading occurs when the duration of symbols is much shorter than the coherent time interval of the channel. This means that the symbols transmitted in the channel are not able to maintain their coherence due to the rapid changes in the channel. As a result, the received signal experiences fading, causing fluctuations in the signal strength. This phenomenon is particularly common in mobile communication systems where the channel conditions can vary rapidly.Rate this question:
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- 12.
Zaokruziti netacno tvrdjenje:
- A.
Koheretni opseg predstavlja opseg frekvencija u kome je uticaj kanala na dve spektralne komponente signala priblizno jednak.
- B.
Dve sinusoide sa frekvencijskim razmakom manjim od Vs (koheretnog opsega kanala) podlezu razlicitom uticaju kanala.
- C.
Ako je frekvencijski opseg signala mnogo manji od koheretnog opsega kanala tada nema ili je minimalna distorzija signala usled propagacije po vise putanja.
Correct Answer
B. Dve sinusoide sa frekvencijskim razmakom manjim od Vs (koheretnog opsega kanala) podlezu razlicitom uticaju kanala.Explanation
If the frequency separation between two sinusoids is smaller than the coherent bandwidth of the channel (Vs), they will experience different channel effects. In other words, the channel will introduce different distortions or variations in amplitude and phase for each sinusoid. This is because the channel's influence on the signal depends on its frequency content, and if the sinusoids have different frequencies, they will be affected differently. Therefore, the statement that "Dve sinusoide sa frekvencijskim razmakom manjim od Vs (koheretnog opsega kanala) podlezu razlicitom uticaju kanala" is incorrect.Rate this question:
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- 13.
Zaokruziti netacno tvrdjenje:
- A.
Koheretni vremenski interval predstavlja vremenski interval u kome je impulsni odziv kanala uglavnom nepromenljiv.
- B.
Koheretni vremenski interval je reciprocan Doplerovom sirenju spektra.
- C.
Ako je trajanje simbola mnogo manje od koheretnog vremenskog intervala kanala impulsni odziv kanala se brzo menja u odnosu na promene signala u osnovnom opsegu.
Correct Answer
C. Ako je trajanje simbola mnogo manje od koheretnog vremenskog intervala kanala impulsni odziv kanala se brzo menja u odnosu na promene signala u osnovnom opsegu.Explanation
The correct answer is "Ako je trajanje simbola mnogo manje od koheretnog vremenskog intervala kanala impulsni odziv kanala se brzo menja u odnosu na promene signala u osnovnom opsegu." This statement is incorrect because if the symbol duration is much smaller than the coherent time interval of the channel, the channel's impulse response changes slowly compared to the signal variations in the baseband.Rate this question:
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- 14.
Ako je trajanje koheretnog vremenskog intervala 0,001s, i ako je trajanje simbola 10-6s. Kanal se ponasa kao kanal sa:
- A.
Sporim fedingom
- B.
Brzim fedingom
- C.
Ravnim fedingom
Correct Answer
A. Sporim fedingomExplanation
The correct answer is "sporim fedingom" because the given information states that the duration of the coherent time interval is 0.001s, and the duration of the symbol is 10-6s. Slow fading refers to a situation where the channel conditions remain relatively constant over the duration of the coherent time interval, but may change significantly over longer periods of time. In this case, since the symbol duration is much smaller than the coherent time interval, it indicates that the channel conditions are expected to remain relatively constant over the symbol duration, thus exhibiting slow fading.Rate this question:
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- 15.
Ako je najmanje trajanje simbola 1*10-6s, kakva je brzina signaliziranja?
- A.
Iznosi najvise 1Mbps
- B.
Iznosi najvise 10-6 Mbps
- C.
Iznosi najvise 10 Mbps
Correct Answer
A. Iznosi najvise 1MbpsExplanation
The correct answer is "Iznosi najvise 1Mbps" because if the minimum duration of a symbol is 1*10-6s, it means that the maximum number of symbols that can be transmitted in one second is 1/1*10-6, which is equal to 1,000,000 symbols. Since the symbol rate is equal to the data rate in this case, the maximum data rate that can be achieved is 1Mbps.Rate this question:
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- 16.
Brz feding se obicno javlja za:
- A.
Veoma niske brzine prenosa
- B.
Veoma visoke brzine prenosa
- C.
Optimalne brine prenosa
Correct Answer
A. Veoma niske brzine prenosaExplanation
Buffer underrun typically occurs at very low transfer speeds. When the transfer speed is too slow, the buffer in the device being written to cannot be filled fast enough, causing the writing process to pause and resulting in a buffer underrun error. This error can be prevented by ensuring that the transfer speed is sufficient to keep the buffer consistently filled during the writing process.Rate this question:
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- 17.
Zaokruziti netacno tvrdjenje:
- A.
Povecanjem broja celija u klasteru smanjuje se kapacitet sistema
- B.
Povecanjem borja celija u klasteru povecava se odnos signal interferencija
- C.
Broj celija u klasteru ne utce na kapacotet sistema ni na odnos signal interferencija
Correct Answer
C. Broj celija u klasteru ne utce na kapacotet sistema ni na odnos signal interferencijaExplanation
The correct answer is that the number of cells in the cluster does not affect the system capacity or the signal interference ratio. This means that increasing the number of cells in the cluster does not decrease the system capacity nor increase the signal interference ratio.Rate this question:
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- 18.
Istokanalna interferencija (co-channel interference) raste:
- A.
Ako broj celija u klasteru stagnira
- B.
Ako broj celija u klasteru opada
- C.
Ako broj celija u klasteru raste
Correct Answer
B. Ako broj celija u klasteru opadaExplanation
Co-channel interference refers to the interference caused by multiple cells using the same frequency channel. When the number of cells in a cluster decreases, it means that there are fewer cells using the same frequency channel. As a result, the interference caused by co-channel interference also decreases. Therefore, the correct answer is "Ako broj celija u klasteru opada" which translates to "If the number of cells in the cluster decreases."Rate this question:
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- 19.
Kapacitet sistema se povecava:
- A.
Ako broj celija u klasteru opada
- B.
Ako broj celija u klasteru ostane isti
- C.
Ako broj celija u klasteru ne prelazi 25
Correct Answer
A. Ako broj celija u klasteru opadaExplanation
The correct answer is "if the number of cells in the cluster decreases." This means that as the number of cells in the cluster decreases, the system capacity increases. This can be explained by the fact that with fewer cells in the cluster, there is less interference and congestion, allowing for better utilization of resources and increased capacity.Rate this question:
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- 20.
Sektorizacijom se:
- A.
Umanjuje istokanalna interferencija
- B.
Smanjuje kapacitet sistema
- C.
Povecava efikasnost trunking tehnike
Correct Answer
A. Umanjuje istokanalna interferencijaExplanation
Sektorizacija se koristi kako bi se smanjila istokanalna interferencija. To znači da se delovi frekvencijskog spektra koji se koriste za prenos signala preklapaju u manjoj meri, što smanjuje smetnje i poboljšava kvalitet prenosa. Umanjenje istokanalne interferencije omogućava bolju komunikaciju i smanjuje mogućnost grešaka u prenosu podataka.Rate this question:
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- 21.
Kontrolom snage se:
- A.
Minimizuje snaga koju emituju korisnici
- B.
Maksimizuje snaga koju emituju korisnici
- C.
Minimizuje snaga koju emituju uredjajii
Correct Answer
A. Minimizuje snaga koju emituju korisniciExplanation
By controlling the power, the goal is to minimize the power emitted by the users. This means that the aim is to reduce the amount of power being transmitted by the users in order to achieve a more efficient and optimal use of resources. By minimizing the power emitted by the users, it helps in reducing interference, conserving energy, and improving overall system performance.Rate this question:
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- 22.
Polu-dupleks komunikacija podrazumeva:
- A.
Predaju i prijem po istom kanalu, ali ne u isto vreme
- B.
Predaju i prijem po istom kanalu u isto vreme
- C.
Predaju i prijem u dva kanala, alli ne u isto vreme
Correct Answer
A. Predaju i prijem po istom kanalu, ali ne u isto vremeExplanation
Polu-dupleks komunikacija podrazumeva da se predaja i prijem podataka vrše preko istog kanala, ali ne istovremeno. To znači da se jedan uređaj može koristiti za slanje podataka dok drugi uređaj istovremeno prima podatke, ali ne mogu istovremeno oba uređaja slati i primati podatke.Rate this question:
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- 23.
Za dinamicku dodelu kanala generalno vazi:
- A.
Smanjuje verovatnocu blokiranja poziva
- B.
Povecava verovatnocu blokiranja poziva
- C.
Blokira pozive
Correct Answer
A. Smanjuje verovatnocu blokiranja pozivaExplanation
For dynamic channel allocation, it generally reduces the probability of call blocking. This means that by dynamically assigning channels, there is a higher chance that calls will be successfully connected without being blocked.Rate this question:
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- 24.
Kakav je oblik Gausove funkcije f(x) i gde se nalazi njen maksimum?
- A.
Zvonasti oblik, sa maksimumom za x jednako srednjoj vrednosti
- B.
Kruzni oblik, sa maksimumom za x jednako srednjoj vrednosti
- C.
Zvonasti oblik, sa maksimumom za x jednako maksimlanoj vrednosti
Correct Answer
A. Zvonasti oblik, sa maksimumom za x jednako srednjoj vrednostiExplanation
The correct answer is "zvonasti oblik, sa maksimumom za x jednako srednjoj vrednosti". This means that the shape of the Gaussian function is bell-shaped, and the maximum value of the function is located at the mean value of x.Rate this question:
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- 25.
Uniformna raspodela na intervalu graficki ima oblik:
- A.
Pravougaonika
- B.
Kvadrata
- C.
Sinusoide
Correct Answer
A. PravougaonikaExplanation
The correct answer is "pravougaonika" (rectangle). A uniform distribution on an interval means that all values within the interval have an equal probability of occurring. Graphically, this can be represented by a rectangle, where the height of the rectangle represents the probability density function and the width represents the interval. Since all values within the interval have the same probability, the height of the rectangle remains constant. Therefore, the shape of a uniform distribution on an interval is that of a rectangle.Rate this question:
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- 26.
Amplitudski spektar kosinusne i sinusne funkcije iste frekvencije i amplitude:
- A.
Izgleda isto
- B.
Izgleda razlicito
- C.
Izgleda slicno
Correct Answer
A. Izgleda istoExplanation
The amplitude spectrum of a cosine and sine function with the same frequency and amplitude will look the same because the only difference between the two functions is a phase shift of 90 degrees. The amplitude spectrum represents the magnitude of each frequency component in the signal, and since the frequency and amplitude are the same for both functions, their amplitude spectra will be identical.Rate this question:
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- 27.
Furijeov red se koristi za spektralnu analizu:
- A.
Kontinualnih i preiodicnih signala
- B.
Kontinualnih i aperiodicnih signala
- C.
Diskretnih signala
Correct Answer
A. Kontinualnih i preiodicnih signalaExplanation
The correct answer is "kontinualnih i preiodicnih signala" which means "continuous and periodic signals" in English. The Fourier series is used to analyze signals that are continuous and periodic in nature. It breaks down these signals into their individual frequency components, allowing for a better understanding of their spectral characteristics. This analysis is commonly used in fields such as telecommunications, audio processing, and image analysis.Rate this question:
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- 28.
Furijeova transformacija se koristi za spektralnu analizu:
- A.
Kontinualnih i aperiodicnih signala
- B.
Kontinualnih i periodicnih signala
- C.
Diskretnih signala
Correct Answer
A. Kontinualnih i aperiodicnih signalaExplanation
The correct answer is "kontinualnih i aperiodicnih signala" which translates to "continuous and aperiodic signals" in English. The Fourier transform is used to analyze signals in the frequency domain, and it is applicable to continuous signals (those that exist over a continuous range of time) as well as aperiodic signals (those that do not repeat in a regular pattern). This transform allows us to decompose a signal into its individual frequency components, providing valuable information about its spectral content.Rate this question:
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- 29.
Diskretna Furijeova transformacija se koristi za spektralnu analizu:
- A.
Diskretnih signala
- B.
Kontinualnih i periodicnih signala
- C.
Kontinualnih i aperiodicnih sgnala
Correct Answer
A. Diskretnih signalaExplanation
The correct answer is "diskretnih signala". The Discrete Fourier Transform (DFT) is used to analyze the frequency components of discrete signals. It transforms a sequence of discrete data points into its equivalent representation in the frequency domain. This allows for the identification of specific frequencies present in the signal and is commonly used in fields such as telecommunications, audio processing, and image analysis.Rate this question:
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- 30.
Kako se naziva klasa algoritama velike efikasnosti za izracunavanje deiskretne Furijeove transformacije?
Correct Answer
FFT,fftExplanation
The class of algorithms that are highly efficient for calculating discrete Fourier transform is called Fast Fourier Transform (FFT). This algorithm is commonly denoted as FFT or fft. It is widely used in various fields such as signal processing, image processing, and data compression due to its ability to quickly compute the Fourier transform of a sequence or signal.Rate this question:
- 31.
Povecanjem broja clanova konacnog Furijeovog reda aproksimacija originalnog signala je:
Correct Answer
bolja,BOLJAExplanation
Increasing the number of terms in the finite Fourier series improves the approximation of the original signal.Rate this question:
- 32.
Raspodela sume velikog broja slucajnih promenljivih tezi:
Correct Answer
Gausovoj raspodeli,gausovoj raspodeli,GAUSOVOJ RASPODELIExplanation
The explanation for the given correct answer is that the distribution of a large number of random variables is being described as Gaussian, also known as the normal distribution. The repetition of "Gausovoj raspodeli, gausovoj raspodeli, GAUSOVOJ RASPODELI" emphasizes that the random variables are following a Gaussian distribution.Rate this question:
- 33.
Kojom raspodelom se moze staticki opisati signal koji se prenosi kroz multipath fading kanal?
Correct Answer
Rejlijevom raspodelom,rejlijevom raspodelom,REJLIJEVOM RASPODELOMExplanation
The correct answer is "Rejlijevom raspodelom". The Rayleigh distribution is commonly used to describe the amplitude of a signal that is transmitted through a multipath fading channel. In this type of channel, the signal takes multiple paths to reach the receiver, resulting in constructive and destructive interference. The Rayleigh distribution models the random variations in the amplitude of the received signal, which is caused by the different path lengths and phase shifts.Rate this question:
- 34.
Sum u fizickom sistemu se cesto karakterise kao:
Correct Answer
Gausov sum,GAUSOV SUM,gausov sumExplanation
The sum in a physical system is often characterized as a Gaussov sum, which refers to the Gaussian distribution or the normal distribution. This distribution is commonly used to describe the probability distribution of a continuous random variable. It is symmetric and bell-shaped, with the majority of values clustering around the mean. The term "Gausov sum" is likely a variation of the term "Gauss sum," which is used to describe the summation of a series of numbers following a specific pattern.Rate this question:
- 35.
Na slici je prikazan profil snage. Odrediti srednje vreme kasnjenja.
- A.
1/1.1µs
- B.
1.1µs
- C.
1/0.1µs
Correct Answer
A. 1/1.1µsExplanation
The correct answer is 1/1.1µs. The average delay time can be determined by taking the reciprocal of the given value, which in this case is 1/1.1µs. This means that on average, there is a delay of 1.1µs between the input and the output.Rate this question:
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- 36.
Ako klaster sadrzi 100 kanala i koristi se faktor ponovnog koriscenja frekvencija 4, tada svaka celija sadrzi kanala:
- A.
25
- B.
4
- C.
100
Correct Answer
A. 25Explanation
If a cluster contains 100 channels and uses a frequency reuse factor of 4, then each cell contains 25 channels. This is because with a frequency reuse factor of 4, the available channels are divided into groups of 4 and each group is assigned to a different cell. Therefore, each cell will have 1/4th of the total number of channels, which is 25.Rate this question:
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- 37.
Ako je faktor visestrukog ponavljanja frekvencija 13 koliko je najblizih istokanalnih celija:
- A.
6
- B.
4
- C.
13
Correct Answer
A. 6Explanation
If the factor of multiple repetition of frequency is 13, it means that the frequency is repeated 13 times. In order to have the closest neighboring cells, we need to consider the cells that have the same frequency. Since the factor is 13, the closest neighboring cells would be the 6 cells that have the same frequency, as they are the closest in terms of the repetition factor. Therefore, the correct answer is 6.Rate this question:
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