1. CMA46
Questions & Answers

2. Q1 A system attenuates the input signal by 32 dB
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)=  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

3. 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 = mW
Thus=45 dBm is a power of 31.6 W.

4. A sine wave of 1. 2mV is input to an amplifier
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

5. Note Power ratio =10log (Po/Pi) = 20 log(Vo/Vi)
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) = dB.

6. 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.

7. 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 = The attenuation will be 24.4 * = dB.
20.65 db=20 log (5/V2)  log(5/V2)=1.03  is a voltage ratio of =  V2=5/10.77
The signal level at the distance of 35 at 70 kHz will be 5/10.77 = 0.46 V

8. Thus the C/N is now only (60-24.1)=35.9 dB
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 ( )=35.9 dB

9. Determine the mean value <x>.
Q8 Option B
The distribution is symmetrical about x = 0.5 and thus this is the mean value.

10. The area under the curve has to be unity otherwise it is not a pdf.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 = (=area/6.5=1/6.5)

11. Q10
What is the probability of x being between 1 and 5?
Q10 Option D
The area required = * ( * 2.5) = Thus the probability is 0.44 approx

12. 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
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

13. 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 = mW
The noise power is 12 *10-12 * (60-40) * 103 = mW
And thus the SNR = 10 log(0.0316/0.0024)=21.2 dB

14. 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.

15. 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

16. 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 Mbits/s (Reference Book Chapter 7 table 7.4)
and there 53 * 8 bits per cell. So  = 53 * 8/( * 106) = 0.17 s