IFT 372 MIDTERM COMPREHENSIVE GUIDE
QUESTIONS 2026 TRUSTED ANSWERS
READY
⫸ What logarithmic base is used in digital communications? Answer:
Two
⫸ How do you find a log2 value on a calculator that only has log10
functionality? (Hint: what is the formula?) Answer: log2(a) =
[log10(a) / log10(2)]
⫸ What is the definition of a decibel? Answer: A decibel is a
mathematical representation of a power level on a logarithmic scale.
a = 10lob(b), where a is the decibel value and b is the decimal value.
⫸ You cannot mix decimal and decibel values in an equation (i.e.,
you either have to work with decimal values or decibel values)
Answer: True
⫸ There can be more than one power value in a decibel (dB)
equation. Answer: False
⫸ When you add two decibel values, you are multiplying those
values in decimal (I.e. 3 dB + 6 dB = 9 dB which is equivalent to 2 *
4 = 8 in decimal) Answer: True
,⫸ __________ is a decibel value referenced to watts and ______ is a
decibel value referenced to milliwatts. Answer: dBW & dBm
⫸ How do you convert dBW to dBm? Answer: Add 30 dB to dBW to
obtain dBm value
⫸ The net gain or (loss) of a transmission system is a radio between
the _________ __________ and ____________ ____________ where
the ___________ is the numerator and the ___________
____________ is the denominator. Answer: Output power & Input
power
⫸ What parameter of the intelligent signal causes or determines the
instantaneous rate of carrier frequency? Answer: Amplitude of the
modulating signal
⫸ What parameter of the intelligent signal causes frequency deviation
of the carrier? Answer: Amplitude of the modulating signal
⫸ A modulated carrier occupies a single frequency. Answer: False
⫸ Determine the noise power (in dBm) at 27° C in a 500 kHz
bandwidth. Answer: PN = 10 log (1.38*10^-23*(27+273)*500kHz)
= -146.8 dBw
=> -116.8 dBm
,⫸ Determine the noise power (in dBm) at 27° C in a 20 MHz
bandwidth system. Answer: Pn=kTB; k=1.38*(10^-23)W°K-Hz;
T(°K)=27°C+273=300°K;
B=20*10^6Hz
=10log(1.38*(10^-23)*300*20*10^6)
=10log(8.28*(10^-14))
=-130.82 dBW
=-130.82 dBW +30 dB
=-100.8 dBm
⫸ Determine the noise power (in dBm) at 23°C in a 500 kHz
bandwidth. Answer: k = 1.38 * 10^23
T = 27 + 273 = 300
B = 5e5
PN = 10log(kTB)
= 10log(1.38 * 10^23 * 300 * 5e5)
= -146.8
= -146.8 + 30 dB
= -116.8 dBm
⫸ For a 12-bit linear sign-magnitude PCM code with a resolution of
0.02 V, determine the voltage range that would be converted to the
PCM code 110000000000 Answer: 1024 * .02 => 20.47V to 20.49V
, ⫸ An FM signal has a deviation of 75 kHz and a modulating
frequency of 15 kHz. The carrier frequency is 1600 MHz. Using
Carson's rule, what is the required bandwidth for this system?
Answer: BSCR = 2 * (75kHz + 15kHz) = 180kHz
⫸ Which of the following are strengths of OFDM? Answer:
Frequency selective fading does not adversely affect all subcarriers &
effectively uses the wireless spectrum.
⫸ Given a three stage amplifier system with stage 1 having a gain of
16 dB and a NF=3.5 dB, Stage 2 gain of 19 dB and a NF = 4.8 dB,
Stage 3 having a gain of 20 dB and a NF = 6.2 dB, calculate the
system noise figure. Answer: 16 dB => 40;
3.5 dB => 2.24;
19 dB => 80;
4.8 dB => 3.02;
20 dB => 100;
6.2 dB => 4.17
F = 2.24 + ((3.02-1)/40) + ((4.17-1)/(40*80))
= 2.29 => 3.6 dB
⫸ Consider an OFDM implementation that uses 15 kHz subcarriers
and use an OFDM symbol of 2048 subcarriers. The nominal cyclic
prefix accounts for a 7% guard time. The extended cyclic prefix can
use up to 25%. 1600 subcarriers can be used for data transmission.
The rest are needed for pilot and null subcarriers. The transmission
bandwidth is 25 MHz using 8 QAM modulation.
QUESTIONS 2026 TRUSTED ANSWERS
READY
⫸ What logarithmic base is used in digital communications? Answer:
Two
⫸ How do you find a log2 value on a calculator that only has log10
functionality? (Hint: what is the formula?) Answer: log2(a) =
[log10(a) / log10(2)]
⫸ What is the definition of a decibel? Answer: A decibel is a
mathematical representation of a power level on a logarithmic scale.
a = 10lob(b), where a is the decibel value and b is the decimal value.
⫸ You cannot mix decimal and decibel values in an equation (i.e.,
you either have to work with decimal values or decibel values)
Answer: True
⫸ There can be more than one power value in a decibel (dB)
equation. Answer: False
⫸ When you add two decibel values, you are multiplying those
values in decimal (I.e. 3 dB + 6 dB = 9 dB which is equivalent to 2 *
4 = 8 in decimal) Answer: True
,⫸ __________ is a decibel value referenced to watts and ______ is a
decibel value referenced to milliwatts. Answer: dBW & dBm
⫸ How do you convert dBW to dBm? Answer: Add 30 dB to dBW to
obtain dBm value
⫸ The net gain or (loss) of a transmission system is a radio between
the _________ __________ and ____________ ____________ where
the ___________ is the numerator and the ___________
____________ is the denominator. Answer: Output power & Input
power
⫸ What parameter of the intelligent signal causes or determines the
instantaneous rate of carrier frequency? Answer: Amplitude of the
modulating signal
⫸ What parameter of the intelligent signal causes frequency deviation
of the carrier? Answer: Amplitude of the modulating signal
⫸ A modulated carrier occupies a single frequency. Answer: False
⫸ Determine the noise power (in dBm) at 27° C in a 500 kHz
bandwidth. Answer: PN = 10 log (1.38*10^-23*(27+273)*500kHz)
= -146.8 dBw
=> -116.8 dBm
,⫸ Determine the noise power (in dBm) at 27° C in a 20 MHz
bandwidth system. Answer: Pn=kTB; k=1.38*(10^-23)W°K-Hz;
T(°K)=27°C+273=300°K;
B=20*10^6Hz
=10log(1.38*(10^-23)*300*20*10^6)
=10log(8.28*(10^-14))
=-130.82 dBW
=-130.82 dBW +30 dB
=-100.8 dBm
⫸ Determine the noise power (in dBm) at 23°C in a 500 kHz
bandwidth. Answer: k = 1.38 * 10^23
T = 27 + 273 = 300
B = 5e5
PN = 10log(kTB)
= 10log(1.38 * 10^23 * 300 * 5e5)
= -146.8
= -146.8 + 30 dB
= -116.8 dBm
⫸ For a 12-bit linear sign-magnitude PCM code with a resolution of
0.02 V, determine the voltage range that would be converted to the
PCM code 110000000000 Answer: 1024 * .02 => 20.47V to 20.49V
, ⫸ An FM signal has a deviation of 75 kHz and a modulating
frequency of 15 kHz. The carrier frequency is 1600 MHz. Using
Carson's rule, what is the required bandwidth for this system?
Answer: BSCR = 2 * (75kHz + 15kHz) = 180kHz
⫸ Which of the following are strengths of OFDM? Answer:
Frequency selective fading does not adversely affect all subcarriers &
effectively uses the wireless spectrum.
⫸ Given a three stage amplifier system with stage 1 having a gain of
16 dB and a NF=3.5 dB, Stage 2 gain of 19 dB and a NF = 4.8 dB,
Stage 3 having a gain of 20 dB and a NF = 6.2 dB, calculate the
system noise figure. Answer: 16 dB => 40;
3.5 dB => 2.24;
19 dB => 80;
4.8 dB => 3.02;
20 dB => 100;
6.2 dB => 4.17
F = 2.24 + ((3.02-1)/40) + ((4.17-1)/(40*80))
= 2.29 => 3.6 dB
⫸ Consider an OFDM implementation that uses 15 kHz subcarriers
and use an OFDM symbol of 2048 subcarriers. The nominal cyclic
prefix accounts for a 7% guard time. The extended cyclic prefix can
use up to 25%. 1600 subcarriers can be used for data transmission.
The rest are needed for pilot and null subcarriers. The transmission
bandwidth is 25 MHz using 8 QAM modulation.
