iecep-esat2-ps-12

IECEP FILES ESAT PROBLEM SOLVINGCompiled by : Mark Djeron C. Tumabao

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 x70 λD

=140 x0.05

2Φ0=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 – PRWhere:PT = transmitted power in dBmPR = 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 dBWhere:VNL = minimum required via net loss in dBt = 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 MHzWhat 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 = tan-1 [d/2hv] = tan-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 distance between an 80 ft transmitting station and a 20 ft receiving station?

dRH (TOT) = sqrt (2 x hT) + sqrt (2 x hR)

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

Voltage = 15.5 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

SPL@1m/1W = 85 dB-SPL (given)


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@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=Cv=2πR

v

v=2πRT

= 2 πR28days

(where R=dkm+6400)

v=2πR

28days=√ 4 x1011

(dkm+6400 )ms

dkm=384,000km

What is the length of the path to a geostationary satellite from an Earth station if the angle of elevation is 300?

¿√ (r+h )2− (r cosθ )2−r sinθ=√(6400+36000 )2−(6400 cos300)2−6400 sin 300=39000km

Find the velocity of a satellite in a circular orbit 500 km above the earth’s surface. Formula :

v=√ 4 x 1011

(d+6400 )Where : v = velocity in meters per second d = distance above earth’s surface in km

v=√ 4 x 1011

(500+6400)

Find the orbital period of a satellite in a circular orbit 36,000 km above the earth’s surface if the earth’s radius is 6400 km.

Formula : T=C

vWhere : T = orbital period C = circumference v = orbital velocity C = 2r = 2 ( 6400 + 36,000 ) = 266.4 x 106 m

v=√ 4 x 1011

(36 ,000+6400 )=3 . 07km / s

T=Cv=266 . 4 x 106 m

3 .07 x 103m /s=24hrs

or 1440 minutesCalculate 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=√(r+h )2−(r cosq )2−r sin q 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 siteSolution :

d=√(6400+36 x103)2−(6400 cos30 °)2−6400 sin 30°


= 39,000 km

By how much should two antennas be separated for space diversity in the 11 GHz band? d = 200 λ = 200 c

f=

200 (3 x 108 ms )

11 x 109

s

= 5.5 m

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?

f c=c

2a=

3 x108 ms

2 (0.03m )=5GHz

v p=c

√1−( f c

f )2=

3 x108 ms

√1−( 5GHzc

6GHz )2=5.43x 108 m

s

A transmitter and a receiver is 45 km apart. Suppose that there is an obstacle midway between the 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?

H 1=17.32√ d1 d2

fD=17.32√ 22.5 (22.5 )

11 (45 )=17.52m

H 1=10.51m

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=√17h t+√17ht=√17 (40 )+√17 (20 )=44.5km

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 is dissipated in the load?

Pout =1 W

anti log0 .710

= 851 .14 mW

Γ = SWR −1SWR + 1

= 0 .5

PLOAD = Pout(1 − Γ2 ) = 638 .35 mW

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?

Pout =1 W

anti log0 .710

= 851 .14 mW

Γ = SWR −1SWR + 1

= 0 .5

PLOAD = Pout(1 − Γ2 ) = 638 .35 mW

Preflected = Pout Γ2 = 212 .78 mW

Preturned = 212.78 mW

anti log2610

= 534 .49 μW

Determine the Hamming distance for 10101 XOR 11001 = 01100 ( presence of two 1)


the codewords ( 10101, 11001)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 ratioC = 10e-12 W fb = 60 kbpsN = 1.2 x 10e-14`W B = 120 kHz

Eb

N 0

=10 log [ CN

xBf b ] ;

Eb

N 0

=10 log [ 10−12W1.2 x10−14W

x120kHz60kbps ]

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=Vmaxln(1+ uVin

Vmax)

ln (1+u)

v=Vmaxln(1+

(255 )(0.5Vmax )Vmax

)

ln(1+255)v=0.876Vmax

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=12 000 x10 00060

= 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 filmstotal no .of films

x100 %

repeat rate= 8719

x100 %

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 nsΘ = 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/ EcEmax = Ec + Em ; Em = Emax – Ec = 125 – 85 = 40 Vm = 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 WPt = 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 mG = 0.62(3)2 0.0752G= 9474.8

G = 10 log 9474.8G = 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 kHzThe S/N will be10 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 reference temperature T0 = 290 K, find the noise figure of the system.

The total noise temperature of the antenna and receiver is NTtot = 75+ 300 = 375.The noise temperature is given by:

NT=T 0 ( NR−1 ) where: NT – noise temp, NR = Noise ratio


Rearranging the above to find NR gives: NR = (375/290) +1 = 2.29The 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 MHzMixer 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 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 + 0.4dBVNL = 0.2 (2ms) + 0.4 dBVNL = 0.8 dB

We have an audio signal with a bandwidth of 4KHz.What is the bandwidth needed if we modulate the signal using AM? Ignore FCC regulations for now.

BW = 2 x 4 KHz= 8 KHz

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.

n=log (199.5+1)

log2


Maximum analog frequency= 4 khz, maximum decoded voltage=±2.55, Minimum dynamic range= 199.5

= 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 3KHzC= 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 1d2f (d 1+d 2)

R=10.4√ 10 x306 (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=| f m−f s

2 |∆ f=|49kHz−51kHz

2 |∆ f=1kHz

For a PCM system with a maximum analog frequency of 4kHz, determine the minimum sample rate.

f s=2 f a

f s=2(4 kHz)f s=8kHz

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 kbpsS/N = antilog (40/10) = 10000C = Blog2(1+S/N) = (4.5x103)log2(1+10000) = 59.795 kbpsSince 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

Fm-fs=0.5fb=0.5*270.833kHz=135.4165kHz


frequency shift between mark and space.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 =0.0258By equation (20.11), the numerical aperture is found to be:NA=n1√(2∆) =1.55√[(2)(0.0258)] = 0.352

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 is the acceptance angle?

By equation (20.10), the fractional difference between the indexes is:∆=(n1-n2)/n1 =(1.55-1.51)/1.55 =0.0258By equation (20.11), the numerical aperture is found to be:NA=n1√(2∆) =1.55√[(2)(0.0258)] = 0.352By 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

5nskm

x10km=20Mbps

For an electronic device operating at a temperature of 17°C with a bandwidth of 10 kHz, determine the thermal noise

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


power in dBm . 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 = √4 KTBR = √4(1.38 x 10-23)(302)(6 x 106)(50) = 2.24µV

iecep-esat2-ps-12

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