IFT 372 Final Study Guide
1. Given that a repeater is 25 km from a transmitter sending data at a frequency of
1800 MHz, what is the path loss between the transmitter and repeater?
a. 125.5 dB b. 132.0 dB
c. 130.5 dB d. None of these
LdB=20∗log ( f )+ 20∗log ( d ) −147.56 dB
20∗log ( 1800∗106 ) + 20 log ( 25∗103 ) −147.56 dB ≈ 125.5 dB
2. Short term fading is caused mainly by terrain configurations and the built
environment.
True False
3. What is the power level entering the first active stage of the receiver at 1900 MHz
being fed by a 25 meter coax with a loss of 20 dB/km, given that the Isotropic
Receive Level is -134 dBW and the receiving station has a 1 meter radius dish? (3
points)
RSL=IRL+ antenna gain−lossdB
Gain = 20 log(1) + 20 log(1.9) + 17.8 dB = 23.375
RSL=(−134 dbW )+antenna gain−( 10 log ( 2.5 )) dB
How to calculate gain?
4. Given the following phasor diagram, what is the M (level) value for this system?
16
1
, 5. A certain transmission channel has a bandwidth of 2 MHz, neglecting the effects
of noise, what is the channel capacity for 32 level encoding?
C=2 B log 2 (M ) C = 2 * B * log2 (M) = 2 * 2 * 106 * log2 (32)
6
C = 2 * (2 * 10 Hz) * log2 (32)
C = 20 * 106 bps = 20 MHz
6. A certain transmission channel has a bandwidth of 2 MHz, what is the channel
capacity if the S/N is 27 dB?
I =B log 2 (1+ SNR)
I =2∗log 2 (1+ 27 dB)
I =2∗log 2 (1+ 501.19)
I =2∗log 2 (502.19)
I =2∗8.97=17.94
≈ 18 MHz
7. If a signal were to be sent across a channel with 1 MHz bandwidth, 27 dB of SNR
and 64-level encoding, what would be the limiting factor to the maximum data
rate?
Use Shannon Hartley (both equations) and whichever is the smaller value is the
limiting factor.
a. SNR
b. 64-Level Coding
c. Either, there are no limitations
d. Both, they are the same
e. I don’t like any of these choices, so I am selecting this answer
8. The receiver on a digital line of sight microwave link has a noise figure of 6 dB.
What is the N0 for this receiver?
N0 = K * 290 * F = -228.6 dB + (10*log(290)) + 6 dB = -198 dB
OR
N0 = K * 290 * F = ( 1.38∗10−23 ) * 290 * 3.98 = 1.592796 * 10-20 ≈ -198 dB
9. Based on Nyquist’s Sampling Theorem, if a signal is sampled at a rate of 32 kHz,
what would be the highest frequency component recovered?
2
1. Given that a repeater is 25 km from a transmitter sending data at a frequency of
1800 MHz, what is the path loss between the transmitter and repeater?
a. 125.5 dB b. 132.0 dB
c. 130.5 dB d. None of these
LdB=20∗log ( f )+ 20∗log ( d ) −147.56 dB
20∗log ( 1800∗106 ) + 20 log ( 25∗103 ) −147.56 dB ≈ 125.5 dB
2. Short term fading is caused mainly by terrain configurations and the built
environment.
True False
3. What is the power level entering the first active stage of the receiver at 1900 MHz
being fed by a 25 meter coax with a loss of 20 dB/km, given that the Isotropic
Receive Level is -134 dBW and the receiving station has a 1 meter radius dish? (3
points)
RSL=IRL+ antenna gain−lossdB
Gain = 20 log(1) + 20 log(1.9) + 17.8 dB = 23.375
RSL=(−134 dbW )+antenna gain−( 10 log ( 2.5 )) dB
How to calculate gain?
4. Given the following phasor diagram, what is the M (level) value for this system?
16
1
, 5. A certain transmission channel has a bandwidth of 2 MHz, neglecting the effects
of noise, what is the channel capacity for 32 level encoding?
C=2 B log 2 (M ) C = 2 * B * log2 (M) = 2 * 2 * 106 * log2 (32)
6
C = 2 * (2 * 10 Hz) * log2 (32)
C = 20 * 106 bps = 20 MHz
6. A certain transmission channel has a bandwidth of 2 MHz, what is the channel
capacity if the S/N is 27 dB?
I =B log 2 (1+ SNR)
I =2∗log 2 (1+ 27 dB)
I =2∗log 2 (1+ 501.19)
I =2∗log 2 (502.19)
I =2∗8.97=17.94
≈ 18 MHz
7. If a signal were to be sent across a channel with 1 MHz bandwidth, 27 dB of SNR
and 64-level encoding, what would be the limiting factor to the maximum data
rate?
Use Shannon Hartley (both equations) and whichever is the smaller value is the
limiting factor.
a. SNR
b. 64-Level Coding
c. Either, there are no limitations
d. Both, they are the same
e. I don’t like any of these choices, so I am selecting this answer
8. The receiver on a digital line of sight microwave link has a noise figure of 6 dB.
What is the N0 for this receiver?
N0 = K * 290 * F = -228.6 dB + (10*log(290)) + 6 dB = -198 dB
OR
N0 = K * 290 * F = ( 1.38∗10−23 ) * 290 * 3.98 = 1.592796 * 10-20 ≈ -198 dB
9. Based on Nyquist’s Sampling Theorem, if a signal is sampled at a rate of 32 kHz,
what would be the highest frequency component recovered?
2
