Download CMA43

CMA46
Questions & Answers
Q1
A system attenuates the input signal by 32
dB. If the input level is 10mW what is the
output level?
• Attenuation =-32db
• Pout-Pin=Gain  Pout – (10mW) = -32 dB
• Pout and Pin needs to be measured in dBm 
• Covert to dBm  10mW=10log(10*10-3)=10log(0.01)=-20.
 Pin =-20 dBm.
• Therefore Pout-(-20) = -32  Pout in dBm = -52
• Convert it to Watt  =10log(P)=-52 
• P=10-5.2=6.3 W
Q2
What is 45 dBm expressed in W?
• Q2 Option D
• 45 dBm is a power level 45 dB above 1
mW, thus the power ratio of 45dB needs to
be calculated.
• 45=10 log(Power Ratio)  4.5= log Power
ratio)  104.5=Power Ratio = 31622.7 mW
• Thus=45 dBm is a power of 31.6 W.
A sine wave of 1.2mV is input to an amplifier. The
output is undistorted and has an amplitude of 1.5V.
Q3
If the gain is expressed in dB what is the
assumption that is often made?
• Q3 Option H.
• dB are defined as a power ratio and a
voltage attenuation can only be defined
when the impedance levels are the same.
A The power gain is expressed in nepers
E The same formula applies to Vand power gain
B The output is voltage generator.
F The amplifier acts like a current generator
C Amplifiers attenuate signals
G The input resistance is high
D The maximum gain is 20 dB
H The input resistance and output load are the same value
A sine wave of 1.2mV is input to an amplifier. The output is
undistorted and has an amplitude of 1.5V
Q4
Determine the gain of the amplifier in dB
• Q4 Option C
• Note Power ratio =10log (Po/Pi) = 20
log(Vo/Vi)
• The gain is given be 1500/1.2 = 1250
• and dB 20 Log 10 (1250) = 61.9 dB.
A copper cable transmits electrical signals at
frequencies where the attenuation is modelled on a
f approximation. A 100 kHz signal transmitted with
an input amplitude of 5V is found to have an
amplitude of 300mV after a distance of 35 km.
Q5
Determine the attenuation of the cable in dB per km.
• Q5 F
• The attenation after 35 km is 20log (5000/300) =
24.4 dB
• For 35 km, this represents 24.4/35=0.7 dB/km.
Q6
If the signal frequency is reduced to 70 kHz, but
with the same amplitude, what would the signal at
the same distance?
• Q6 Option D
• The frequency has reduced by a factor of 0.7
(70/100) and thus the attenuation will reduce by a
factor of 0.7 = 0.837. The attenuation will be
24.4 * 0.837 = 20.65 dB.
• 20.65 db=20 log (5/V2)  log(5/V2)=1.03  is a
voltage ratio of 101.03 = 10.77  V2=5/10.77
• The signal level at the distance of 35 at 70 kHz
will be 5/10.77 = 0.46 V
Q7
A radio system suffers a fourth-power law
attenuation with distance. At a distance of 3 km from
the transmitter the Carrier/ Noise ratio (C/N) is 60
db. Determine the C/N ratio approximately at a
distance of 12 km from the transmitter. The noise
level can be assumed constant.
• Q7 Option D
• The distance has increased by a factor of 4
so the attenuation will increase by a factor
of 44 (i.e. 10log256=24.1 db.)
• Thus the C/N is now only (60-24.1)=35.9 dB
Determine the mean value <x>.
• Q8 Option B
• The distribution is symmetrical about x =
0.5 and thus this is the mean value.
Q9
What is the maximum
value of the probability
density function?
• Q9 Option E
• The area under the curve has to be unity
otherwise it is not a pdf.
• Area = 2 *[(2.5 – 0.5) + 0.5(5.0 – 2.5)] * p MAX = 6.5 * PMAX
• Thus pMAX = 0.154 (=area/6.5=1/6.5)
Q10
What is the probability
of x being between 1
and 5?
• Q10 Option D
• The area required = 0.154 * ( 1.5 + 0.5 *
2.5) = 0.4325. Thus the probability is 0.44
approx
Q11-A digital system uses comparator to distinguish
between the received signal states 0V and 4V. The noise in
the system has a zero mean and has a Gaussian probability
density. 0’s and 1’s in the received data are equally likely.
Determine the maximum noise level if the probability of error
is not to exceed 3 in 10000.
• Q11 Option F
• The probability of a bit error can be determined from
Q(V/2) where V is the separation between logic levels
and  is the r.m.s. noise level.
• From the Figure 6.15 in the reference Book.V/2 equal
to 3.35 (approx) is necessary for the error rate to be less
than Qz ( 3 * 10-4.)
• V/2 < 3.35  4/2 =3.35   < 4/(2*3.35) = 0.59 V
Q12
A signal at a level of –15 dBm has band-limited
noise accompanying it in the range 40 to 60 kHz.
The power density of the noise is 12 pW/ Hz.
Estimate the Signal-to-Noise ratio as a ratio.
• Q12 Option F
• -15 dBm converts to a Signal power level of 10–1.5 mW =
0.0316 mW
• The noise power is 12 *10-12 * (60-40) * 103 = 0.0024 mW
• And thus the SNR = 10 log(0.0316/0.0024)=21.2 dB
Q13
What value is the maximum burst size, in the
ATM leaky bucket operation, given that Ts =
5s T = 3s and s = 20 s.
• Q13 Option A
• The expression for MBS (Maximum Burst
Size) is given on page 117 of Block 6.
 S 
 20 
MBS  1  
 11cells
  1 

5  3
 TS  T 
Q14-What is the minimum frame length that
can be detected on a fast-Ethernet segment,
if the segment has a maximum length of
200m and is a continuous cable. Assume a
propagation speed of 200*106 m/s.
• Q14 Option C
• Fast Ethernet is 100 Mbits/sec. A 400 m
round trip takes 400/(200 * 106) = 2s.
• In 2s. 100* 106* 2* 10-6 =200 bits are transmitted
thus the minimum frame length is 200 bits
Q15
What is ‘ ’, the cell duration for an ATM cell,
if the physical layer uses an SDH STM-16
link?
• Q15 Option G
• The bit rate for STM-16 frames is 2488.32
Mbits/s (Reference Book Chapter 7 table 7.4)
• and there 53 * 8 bits per cell. So  = 53 * 8/(
2488.32 * 106) = 0.17 s