Calculate the beam width between nulls of a 2-m paraboloid reflector used at 6GHz. Note: such reflectors are often used at that frequency as antennas outside broadcast
television microwave links.
Φ0= 2 x 70 λ D =140 x 0.05 2 Φ0=3.50
A CDMA mobile measures the signal strength from the base as -100 dBm. What should the mobile transmitter power be set to as a first approximation, considering - 76 dB at mobile power control?
PT = -76 dB – PR Where: PT = transmitted power in dBm PR = received power in dBm PT = -76 dB - PR = -76 dB – (-100dBm) = 24 dBm = 250 mW : A telephone signal takes 2ms to
reach its destination. Calculate the via net loss required for an
acceptable amount of echo.
VNL = 0.2t + 0.4 dB Where:
VNL = minimum required via net loss in dB t = time delay in ms for propagation one way along line
VNL = 0.2t + 0.4 dB = 0.2 X 2 + 0.4 = 0.8 dB
For a total capacity of 36 Mbps and 64-QAM of digital modulation, what is the ideal RF bandwidth?
For a 64-QAM, the total number of symbol combinations is 64 symbols and the number of bits to produce this is… n = log2 64 = 6 bits per symbol
Therefore the ideal RF bandwidth is… BW = fB / n = 36 Mbps / 6 bits
BW = 6 MHz What is the wavelength of a radio
signal travelling at a frequency of 220 MHz in a coaxial line having Teflon foam as its dielectric, εr = 2?
c = 3 x 108 m/sec = 300 x 106 m/sec velocity factor vf = 1 / sqrt (εr) = 1 / sqrt (2) = 0.707 = vP / c
Therefore the velocity of the radio signal in the coaxial line is …
vP = 0.707c = 0.707 (300 x 106) = 212,132,034.36 m/sec (or 212 x 106
m/sec approx.)
so that the wavelength of the signal is … λ = vP / f = 212 x 106 / 220 x 106 = 0.963636… m or 0.96 m approx.
What is the angle of refraction in a Teflon (εr2 = 2) medium of a radio wave from air (εr1 = 1.0) if its angle of incidence is 45°?
Using Snell’s Law…
Sqrt (εr1) sin θi = Sqrt (εr2) sin θr Sqrt (1) sin 45° = Sqrt (2) sin θr
θr = sin-1 [(Sqrt (1) sin 45°) / Sqrt (2)] = sin-1 (0.5) = 30°
What is the critical frequency of a layer if the maximum value of electron density is 2 x 106 per cm3?
Nmax = 2 x 106 per cm3 = 2 x 106 x 106 per m3 = 2 x 1012 per m3
fc = 9 x sqrt (Nmax) = 9 x sqrt (2 x 1012) fc = 12.73 MHz or 13 MHz approx.
A radio communication link is to be established via the ionosphere. The maximum virtual height of the layer is 110 km. at the midpoint of the path and the critical frequency is 4 MHz. If the distance between the radio stations is 500 km, what is the suitable value for the optimum working frequency? Use flat terrain analysis.
For flat terrain analysis, tan i = d/2hv
i = 1 [d/2hv] = 1 [500/2(110)] = tan-1 [2.273] = 66.25° angle of incidence MUF = fc x sec i = fc / cos i = 4 / cos 66.25° = 4 / 0.40275 = 9.932 MHz via Secant Law
OWF = 0.85 x MUF = 0.85 (9.932) = 8.4422 MHz or 8.4 MHz approx
The efficiency of an antenna is 82 %. Its radiation resistance is 30 Ω. What is the value of its loss
resistance? η = Rd/(Rd + Rloss) 0.82 = 30/(30 + Rloss) 0.82 (30 + Rloss) = 30 = 24.6 + 0.82 Rloss Rloss = (30 – 24.6) / 0.82 = 6.5854 Ω or 6.58 Ω approx.
A radio station operates at 11 meter wavelength. What is the designated band of station’s frequency?
c = 3 x 108 m/sec λ = 11 m = c / f
f = 3 x 108 / 11 = 27 MHz approx., which is within 3 – 30 MHz range designated as HF band
What is the total radio horizon
transmitting station and a 20 ft
receiving station? dRH (TOT) = sqrt (2 x 80) + sqrt (2 x 20) dRH (TOT) = 18.97 mi
A power density of 1.57 x 10-4 W/m2 is measured 50 meters from a test antenna whose directive gain is 2.15 dB. How much power was fed into the test antenna?
From Inverse Square Law, PD = PR/4πd2
PR = PD x 4πd2 = 1.57 x 10-4 x 4π(50)2 = 4.93 W approx. is the radiated power
GANT = log-1 [2.15 dB/10] = 1.64 PR = PFED x GANT
PFED = PR / GANT = 4.93 / 1.64 = 3.007 W or 3 W approx.
The output power of an FM transmitter is 5 kW. If the power loss in transmission line is around 10 W and the antenna has a power gain of 6 dB, what is the ERP of the station in kW?
GANT = 6 dB = 4
ERP = (5000 – 10) x 4 = 4990 x 4 = 19,960 W or 19.96 kW
Determine the visual carrier
frequency of TV channel 9. fNLB = 174 + (N – 7)6, using Arithmetic Progression
f9LB = 174 + (9 – 7)6 = 174 + (2)6 = 174 + 12 = 186 MHz, low band frequency of channel 9 fVC = 186 + 1.25 = 187.25 MHz, visual carrier frequency of channel 9
At NTSC standard, line frequency = 15734.264 Hz. What is the
equivalent line period
fH = 15734.264 Hz
H = 1 / fH = 1 / 15734.264
H = 63.556 x 10-6 = 63.556 usec Determine the radio horizon for a
transmit antenna that is 200 m high and a receiving antenna that is 100 m high
D = √ 17ht + √ 17hr
= √ 17(200) + √ 17(100) = 99.5 km
A group of filters has 1/3 octave of spacing. If the initial frequency is 25 Hz, what is the next frequency available for the filter?
f2 = 2x f1 where x = 1/3, fraction of an octave f2 = 21/3 x 25 = 31.5 Hz
+ 26 dBu is how many volts? V (dBu) = 20 log [ voltage in Volts / 0.775 Volts ] 26 = 20 log [ voltage in Volts / 0.775 V ]
26 / 20 = log [ voltage in Volts / 0.775 V ] = 1.3 Voltage = log-1 (1.3) x 0.775 V
Determine the sound power in Watts produced by the bank’s alarm if a by-stander heard the alarm at a sound pressure level of 100 dB-SPL. The by-stander is 100 ft away from the bank.
PWL = SPL + 20 log D(m) + 8 PWL = 100 + 20 log (100/3.28) + 8 = 137.68 dB-PWL 137.68 = 10 log W + 120 W = log-1 [(137.68 – 120)/10] W = 58.65 Watts A loudspeaker produces an SPL of 85dB-SPL at 1 meter distance and input electrical power of 1 W. How loud is the SPL at distance of 20 meters if this speaker is driven to 10 W of electrical power? SPL@1m/1W = 85 dB-SPL (given) SPL@20m/1W = 85 - 20 log (20m/1m) = 59 dB-SPL SPL@20m/10W = 59 + 10 log (10W/1W) = 69 dB-SPL
The moon orbits the earth with a period of approximately 28 days. How far away is it? Assume circular orbit. T =C v= 2 πR v v =2 πR T = 2 πR 28 days(where R=dkm+6400) v = 2 πR 28 days=
√
4 x 1011(
dkm+6400)
m s dkm=384,000 km What is the length of the path to ageostationary satellite from an Earth station if the angle of elevation is 300?
¿
√
(r+h)2−(r cosθ )2−r sinθ=
√
(6400+36000)2
−
(
6400 cos 300)
2−6400 sin 300=39000 kmFind the velocity of a satellite in a circular orbit 500 km above the earth’s surface. Formula : v 4 x1011 d6400
Where : v = velocity in meters per second d = distance above earth’s surface in km
v 4x1011
(5006400)
Find the orbital period of a satellite in a circular orbit 36,000 km above
the earth’s surface if the earth’s Formula :
C
radius is 6400 km. Where : T = orbital period C = circumference v = orbital velocity C = 2r = 2 ( 6400 + 36,000 ) = 266.4 x 106 m v 4 x1011 (36,0006400) 3.07km /s TC v 266.4x106m 3.07x103m /s24hrs or 1440 minutes Calculate the length of the path to a
geostationary satellite from an earth station where the angle of elevation is 30.
(earth’s radius = 6400 km, height of satellite above earth is 36 X 103 km)
Formula :
d (rh)2(rcos)2 rsin
d = distance to the satellite in km r = radius of the earth in km (6400 km) h = height of satellite above equator ( 36 x 103 km)
= angle of elevation to satellite at antenna site Solution :
d (640036x103)2(6400cos 30)2 6400sin 30 = 39,000 km
By how much should two antennas be separated for space diversity in
the 11 GHz band? m s x s m x f c d 5.5 10 11 10 3 200 200 200 9 8
What is the phase velocity of a rectangular waveguide with a wall separation of 3 cm and a desired frequency of operation of 6 GHz? fc= c 2 a= 3 x 108m s 2 (0.03 m)=5 GHz vp= c
√
1−(
fc f)
2= 3 x 108m s√
1−(
5 GHzc 6 GHz)
2=5.43 x 10 8m sA transmitter and a receiver is 45 km apart. Suppose that there is an obstacle midway between the
H1=17.32
√
d1d2fD =17.32
√
22.5 (22.5)
transmitter and receiver. By how much must the path between the towers clear the obstacle in order to avoid diffraction at a frequency of 11 GHz?
H1=10.51 m
How far from the transmitter could a signal be received if the transmitting and receiving antennas where 40 m and 20 m, respectively, above level terrain?
d=
√
17 ht+√
17 ht=√
17 ( 40)+√
17 (20 )=44.5 kmAn isolator has a forward loss of 0.7 dB and a return loss of 26 dB. A source provides 1 W to the isolator, and the load is resistive with an SWR of 3. How much power is dissipated in the load?
mW P P SWR SWR mW anti W P out LOAD out 35 . 638 1 5 . 0 1 1 14 . 851 10 7 . 0 log 1 2 An isolator has a forward loss of 0.7 dB and a return loss of 26 dB. A source provides 1 W to the isolator, and the load is resistive with an SWR of 3. How much power returns to the source?
W anti mW P mW P P mW P P SWR SWR mW anti W P returned out reflected out LOAD out 49 . 534 10 26 log 78 . 212 78 . 212 35 . 638 1 5 . 0 1 1 14 . 851 10 7 . 0 log 1 2 2Determine the Hamming distance
for the codewords ( 10101, 11001) 10101 XOR 11001 = 01100 ( presence of two 1) A multiplexer combines four
100-kbps channels using a time slot of 2 bits, determine the frame duration
The link carries 50,000 frames /sec The frame duration = 1/50,000 = 20s For the given parameters,
determine the energy per bit-to noise power density ratio
C = 10e-12 W fb = 60 kbps N = 1.2 x 10e-14`W B = 120 kHz Eb N0 =10 log
[
C N x B fb]
; Eb N0 =10 log[
10 −12 W 1.2 x 10−14W x 120 kHz 60 kbps]
A signal at the input to a mu-law compressor is positive with its voltage one-half the maximum value. What proportion of the maximum output voltage is produced? v =Vmax ln(1+ uVin Vmax) ln(1+u) v =Vmax ln(1+(255)(0.5 Vmax) Vmax ) ln(1+255) v =0.876 Vmax
A network with bandwidth of 10 Mbps can pass only an average of 12000 frames per minute with each frame carrying an average of 10000 bits. What is the throughput of this network?
Throughput=12000 x 10 000
60 = 2 Mbps
A telephone subscriber line must have anSNRdB above 40. What is the minimum number of bits?
SNRdB = 6.02nb + 1.76 = 40 n = 6.35
A mammography service examined 327 patients during the third
calendar quarter of 1996. 719 films were exposed during this period, eight of which were repeats. What is the repeat rate?
repeat rate=no . repeated films totalno . of films x 100
repeat rate= 8
719x 100
repeat rate=1.1
Determine the time taken for a signal to travel down a 10 m transmission line, if its velocity factor is 0.7.
T= L/ (Vf x Vc)
T= 10 m/ (0.7)(3x108 m/s) T= 47.6 ns
The maximum voltage standing wave of an RG-11/U foam coaxial cable is 52 V and its minimum voltage is 17 V. How many percent of the incident power is the reflected power? Г 2 = Pr/Pi Г = SWR-1 ; SWR = Vmax/ Vmin = 52/17 = 3.05 SWR+1 Г = 3.05 – 1 = 0.51 3.05 +1 Thus : Г 2 = (0.51)2 = 0.26
The reflected power is 26 % of the incident power
Determine the phase-shift
represented by a 75 ns delay of a 4 MHz signal to a 75-ft cable with a dielectric constant of 2.3.
Θ = 360 td / T
T =1/f = 1 / 4MHz = 250 ns Θ = 360 (75) / 250 = 108 degrees The modulated peak value of a
signal is 125V and the unmodulated carrier value is 85V. What is the modulation index?
m = Em/ Ec
Emax = Ec + Em ; Em = Emax – Ec = 125 – 85 = 40 V
m = 40V/ 85V = 0.47 The total transmitted power of an
AM broadcast transmitter with a carrier power of 50kW when modulated 80 percent is:
Pt = Pc ( 1 + m2/2) Pt = 50,000 ( 1 + 0.82 /2) Pt = 66,000 W
Pt = 66kW The input to an FM receiver has a
S/N of 2.8. What is the frequency deviation caused by the noise if the modulating frequency is 1.5KHz and the permitted deviation is 4KHz
= fm
= sin -1 N/S = sin -1 1/2.8 = 0.3652
= (0.3652)(1.5KHz) = 547.8 Hz Calculate the gain (relative to an
isotropic) of a parabolic antenna that has a diameter of 3 m, an efficiency of 60% and operates at a frequency of 4 GHz. G = 2 D2/ 2 = 3x108/ 4MHz = 0.075 m G = 0.62(3)2 0.0752 G= 9474.8 G = 10 log 9474.8 G = 39.8 dBi Determine the worst-case output
S/N for a narrowband FM receiver with deviation maximum of 10 kHz and a maximum intelligence
frequency of 3 kHz. The S/N input is 3:1. Φ = sin-1 (1/3) = 19.5 or 0.34 rad δ = Φ x fi = 0.34 x 3 kHz = 1 kHz The S/N will be 10 kHz/ 1 kHz = 10 The sensitivity of a radio receiver is
given in terms of dBm. If the
receiver is receiving 200 mW, what is the corresponding dBm of this signal?
dBm for 200 mW = 10 log (200mW/1mW) = 23 dBm
An antenna with a noise temperature of 75 Kelvin is
connected to a receiver input with a noise temperature of 300 K. Given
The total noise temperature of the antenna and receiver is NTtot = 75+ 300 = 375.
the reference temperature T0 = 290 K, find the noise figure of the system.
1
0
T NR
NT
where: NT – noise temp, NR = Noise ratio
Rearranging the above to find NR gives: NR = (375/290) +1 = 2.29
The Noise figure is the decibel equivalent of the noise ratio, such that:
NF = 10 log NR = 3.6 dB The modulation index of an AM
radio station is 0.75. If the carrier power is 500W, what is the total transmitted power?
641 W
An FM transmitter system is using a 1MHz crystal oscillator to generate a very stable 108MHz final carrier frequency. The output of the crystal oscillator is fed to a x36 multiplier circuit then mixed with a 34.5 MHz signal. The sum output of the mixer is filtered out, while the difference is fed to another multiplier in order to generate the final carrier frequency. What is the multiplier value?
Basic Block Diagram of the FM carrier generator: 1 MHz carrier > x 36 > Mixer > Multiplier > 108MHz carrier
First Multiplier output: 1MHz x 36 = 36 MHz Mixer output:
36MHz + 34.5 MHz = 70.5MHz (filtered output)
36MHz – 34.5 MHz = 1.5 MHz (difference fed to the next multiplier)
Multiplier stage: Output = 108MHz Output = 1.5MHz x n N = 108/1.5 = 72 Find the maximum dynamic range
for a linear PCM system using 16-bit quantizing.
DR = 1.76 + 6.02m; dB DR = 1.76 + 6.02(16) DR = 98.08 dB
Compute the baud rate for a 72000
bps 64-QAM signal. A 64-QAM signal means that there are 6 bits per signal element since 26 = 64. Therefore, baud rate = 72000/6 cos ƟFc
baud rate = 12000 baud A telephone signal takes 2ms to
reach its destination. Calculate the via net loss required for an
acceptable amount of echo
VNL = 0.2t + 0.4dB
VNL = 0.2 (2ms) + 0.4 dB VNL = 0.8 dB
We have an audio signal with a
bandwidth needed if we modulate the signal using AM? Ignore FCC regulations for now.
We have an audio signal with a bandwidth of 4 MHz. what is the bandwidth if we modulate the signal using FM? Ignore FCC regulations.
BW=10x4 = 40 MHz
For PCM system with the following parameters determine the minimum number of bits used.
Maximum analog frequency= 4 khz, maximum decoded voltage=±2.55, Minimum dynamic range= 199.5
n=log (199.5+1) log2
= 7.63 = 8 Find the maximum dynamic range
for linear PCM system using 8-bit quantizing
DR = 1.76 + 6.02(m) DR = 1.76 + 6.02(8) An attempt is made to transmit base
band frequency of 30khz using a digital audio system with a sampling rate of 44.1khz. What audible frequency would result?
fa = fs – fm
= 44.1KHz – 30KHz = 14.1 KHz
Calculate the capacity of a
telephone channel that has a S/N of 1023?
The telephone channel has a bandwidth of about 3KHz
C= BW log2 (1+ S/N) = 3 x 103 log2 (1+ 1023) =30000 b/s
A line of sight radio link operating at a frequency of 6GHz has a
separation of 40km between
antennas. An obstacle in the path is located 10km form the transmitting antenna. By how much must the beam clear the obstacle?
R=10.4
√
d 1 d 2 f (d 1+d 2)R=10.4
√
10 x 30 6(10+30) R = 11.6mDetermine the peak frequency deviation for a binary FSK signal with a mark frequency of 49 kHz, space frequency of 51kHz and an input bit rate of 2kbps.
∆ f =
|
fm−fs 2|
∆ f =
|
49 kHz−51 kHz2
|
∆ f =1 kHz
For a PCM system with a maximum analog frequency of 4kHz,
determine the minimum sample
rate. fs=2(4 kHz) fs=8 kHz
A dipole antenna has a radiation resistance of 67 ohms and a loss resistance of 5 ohms, measured at the feedpoint. Calculate the
efficiency
Rr / Rt = 67 / (67 + 5) = 93%
Find the binary equivalent of
x8+x3+x+1. x8+x7+x6+x5+x4+x3+x2+x+k 1 0 0 0 0 1 0 1 1 A telephone line has a bandwidth of
4.5 kHz and a signal-to-noise ratio of 40dB. A signal is transmitted down this line using a four level code. What is the maximum theoretical data rate?
C = 2Blog2M = 2(4.5x103)log2(4) = 18 kbps S/N = antilog (40/10) = 10000
C = Blog2(1+S/N) = (4.5x103)log2(1+10000) = 59.795 kbps
Since both results are maxima, we take the lesser of the two, 18 Kbps
The GSM radio system uses GMSK in a 200kHzchannel, with a channel data rate of 270.883 kbps.
Calculate the frequency shift between mark and space.
Fm-fs=0.5fb=0.5*270.833kHz=135.4165kHz
For Ethernet system with a length of 2.5km on coaxial cable with a velocity factor of 0.66. What is the propagation delay?
T=d/(VfC)=2.5km/(0.66*300000000m/s)=12.6 µs
A signal at the input to a µ-Law compressor is positive, with its volatage one-half the maximum value. What proportion of the maximum output voltage is produced?
Vo=Vo(ln(1+(255*0.5)))/ln(1+255)=0.876Vo
How long does it take to dial the number 784-3745 using pulse dialing with .5s inter digit time?
0.7 + 0.8 + 0.4 + 0.3 + 0.7 + 0.4 + 0.5 (6x0.5) = 6.8s
An optic fiber is made of glass with a refractive index of 1.55 and is clad with another glass with a refractive index of 1.51. Launching takes place from air. What numerical aperture does the fiber have?
∆=(n1-n2)/n1 =(1.55-1.51)/1.55 =0.0258
By equation (20.11), the numerical aperture is found to be:
NA=n1√(2∆)
=1.55√[(2)(0.0258)] = 0.352
a refractive index of 1.55 and is clad with another glass with a refractive index of 1.51. Launching takes place from air. What is the acceptance angle?
between the indexes is: ∆=(n1-n2)/n1
=(1.55-1.51)/1.55 =0.0258
By equation (20.11), the numerical aperture is found to be:
NA=n1√(2∆)
=1.55√[(2)(0.0258)] = 0.352
By equation (20.8), the acceptance angle is: Θ0(max)=sin-1NA= sin-10.352=20.6◦
For a single-mode optical cable with 0.25-dB/km loss, determine the optical power 100km from a 0.1-mW light source. P=0.1mW x 10-{[(0.25)(100)]/(10)} =1 x 10-4 x 10{[(0.25)(100)]/(10)] =(1 x 10-4)(1 x 10-25) =0.316µW P(dBm)= 10log (0.316 µW )0.001 = -35dBm
For an optical fiber 10km long with a pulse-spreading constant of 5ns/km, determine the maximum digital transmission rates for return-to-zero.
Fb=
1 5 ns
km x 10 km =20Mbps
For an electronic device operating at a temperature of 17°C with a bandwidth of 10 kHz, determine the thermal noise power in dBm .
N = KTB T(kelvin) = 17°C + 273° = 290 K B = 1 x 104 Hz = (1.38 x 10-23)(290)(1 x 104) = 4 x 10-17 W 10log (4 x 10-17) = -134 dBm 0.001
A 300Ω resistor is connected across the 300Ω antenna input of a
television receiver. The bandwidth of the receiver is 6MHz, and the resistor is at room temperature .Find the noise voltage applied to the receiver input.
VN =
√
4 kTBR= √ 4(1.38 x 10-23 J/K)(293 K)(6 x 106 Hz) (300 Ω)
= 5.4 x 10-6 V = 5.4 µV The signal power at the input to an
amplifier is 100µW and the noise power is 1 µW. At the output, the signal power is 1W and the noise power is 30mW. What is the amplifier noise figure, as a ratio?
(S/N)p = 100µW = 100 1µW (S/N)o = 1W = 33.3 0.03W NF (ratio) = 100 = 3 33.5
It is desired to operate a receiver with NF = 8dB at S/N = 15 dB over a 200-KHz bandwidth at ambient temperature. Calculate the receiver’s sensitivity.
S = Sensitivity = -74dBm + NF + 10log∆f + desired S/N
S = -174 + 8 + 10log(200,000) + 15 S = -98 dBm
A Satellite receiving system
includes a dish antenna ( Teq = 35 K) connected via a coupling
network (Teq= 40 K) to a microwave receiver (Teq=52 K referred to its input). What is the noise power to the receiver’s input over a 1-MHz frequency range?
PN = KT∆f
= 1.38 x 10-23 J/K (35 + 40 + 52)K (1MHz)
= 1.75 x 10-15 W
The noise voltage produced across a 50Ω is input resistance at a temperature of 302°C with a bandwidth of 6 MHz is ______ VN =