Version 3e 1.0
January 2022
Solutions to
Homework Problems
,Last Revised 1/26/2022 at 5:32 PM Version 3e 1.0, January 2022
Chapter 1
1. Compute the range R corresponding to echo delays t0 of 1 ns, 1 s, 1 ms, and 1 second.
Equation (1.1): R = ct0/2. With c = 3×108 m/s, the results are:
Delay t0 1 ns 1 s 1 ms 1s
Range 0.15 m 150 m 150 km 1.5×108 m
2. Compute the time delays for two-way propagation to targets at distances of 100 km, 100
statute miles, and 100 feet.
Equation (1.1) rearranged gives t0 = 2R/c. With the conversion factors of 1 statute mile =
1609.34 m and 1 foot = 0.3048 m, the results are:
100 statute
Range 100 km 100 feet
miles
Delay t0 666.6 s 1.073 ms 203.2 ns
3. Radar is routinely used as one means of measuring the distance to objects in space. For
example, it has been used to calculate the orbital parameters and rate of rotation of the
planet Jupiter. The distance from Earth to Jupiter varies from 588.5×10 6 km to 968.1×106
km. What are the minimum and maximum time delays in minutes from the time a pulse is
transmitted in the direction of Jupiter until the time the echo is received? If pulses are
transmitted at a rate of 100 pulses per second, how many pulses are in flight, either on their
way to Jupiter or back again, at any given instant?
Minimum delay 2 Rmin c 2 5.885 1011 m 3 108 3923 s = 65.4 minutes
Minimum delay 2 Rmax c 2 9.681 10 m
11
3 108 6454 s = 107.6 minutes
Number of pulses in flight = trip time 100 pps 392,333 pulses (min)
645,400 pulses (max)
4. Table 1.1 defines the millimeter wave (MMW) band to extend from 30 to 300 GHz. Only
certain frequencies in this band are widely used for radar. This is partly due to frequency
allocation rules (which frequencies are allotted to which services), but also due to
atmospheric propagation. Based on Fig. 1.3, list two frequencies in the MMW band that
might be preferable for radar use, and two that would not be suitable. Explain.
Chapter 1 problems and solutions 1
,Last Revised 1/26/2022 at 5:32 PM Version 3e 1.0, January 2022
Figure 1.3 is repeated here:
Generally, frequencies with less atmospheric attenuation are more favorable than those
with higher attenuation because the received echo power will be greater for a given
target and range. Therefore, four favorable MMW frequency bands are, roughly, 30 to 40
GHz, 80 to 100 GHz, 140 to 150 GHz, and 230 to 270 GHz. Unfavorable frequencies
occur in the vicinity of about 60, 120, and 190 GHz.
5. Consider two equal-power transmitters, one at 10 GHz and the other at 30 GHz. The power
received from each is measured at a distance of 10 km and compared. Based on Figs. 1.3
and 1.4, estimate how many decibels weaker than the 10 GHz signal will the 30 GHz signal
be in sea level clear air conditions. Repeat for a medium rainfall of 2.5 mm/h and a tropical
downpour of 50 mm/h.
Figure 1.3 gives the attenuation in clear air at sea level. By inspection, the attenuation at
10 GHz is about 0.02 dB/km, and at 30 GHz is about 0.1 dB/km. The difference is 0.08
dB/km. Over 10 km, the difference in atmospheric attenuation will be (10)(0.08) = 0.8
dB.
Figure 1.4 is used in the same way to estimate the difference in loss for the two rainfall
cases. For a 2.5 mm/hr rain rate, the attenuation is about 0.02 dB/km at 10 GHz and 0.5
at 30 GHz, a difference of 0.48 dB/km. Therefore the 30 GHz signal is expected to be
about 4.8 dB weaker than the 10 GHz signal.
For the tropical downpour the attenuations are about 1 and 10 dB/km, giving rise at 10
km to a difference of 9 dB.
6. Compute the bandwidth needed to achieve range resolutions of 1 m, 1 km, and 100 km.
What is the length of a rectangular pulse having this Rayleigh bandwidth (peak-to-first null
width of the Fourier transform) for each value of resolution?
Chapter 1 problems and solutions 2
, Last Revised 1/26/2022 at 5:32 PM Version 3e 1.0, January 2022
Equation (1.2) states that R = c/2, so = c/(2∙R). The Rayleigh bandwidth of a pulse
of length seconds is 1/Hz (see App. B). The results are:
Resolution R 1m 1 km 100 km
Bandwidth 150 MHz 150 kHz 1.5 kHz
Pulse length 66.6 ps 66.6 ns 666.6 s
7. Derive Eq. (1.10) from Eq. (1.8).
From Eq. (1.8), E sin Dy sin Dy sin . The peak occurs when =
0, which gives us 0/0; but applying L’Hôpital’s rule shows us that E(0) = 1. The first null
of E() occurs at the first nonzero value of the argument of the numerator that makes the
numerator equal zero, i.e. when the argument of the sine function in the numerator
equals . So
Dy sin arcsin Dy
The peak-to-null (Rayleigh) width of the mainlobe is the distance between these two
values of , so it is just arcsin Dy .
Applying the small angle approximation sin() ≈ gives
Dy sin Dy Dy ,
which is the desired result of Eq. (1.10). However, another and equivalent way to arrive
at this is to consider the Taylor (or Maclaurin) series for the arcsin, namely
arcsin x x 1 6 x3 . Applying this in the expression arcsin Dy and
keeping only the first term gives arcsin Dy Dy and also leads directly to Eq.
(1.10). However, this version is slightly more convenient for the next problem …
8. Using the results from Prob. 7, find the value of Dy t for which the error in the small-
angle approximation of the Rayleigh beamwidth is 10 percent. What are the exact and
approximate Rayleigh beamwidths for that value of Dy t ?
The second term in the series for the arcsin is a rough estimate of the error in computing
Dy due to the small angle approximation. We want the error to be less than 10% of
the actual value, i.e. 1 6 x3 x 10 , where x = Dy . This leads to x 6 10 =
0.775 radians; this is an acceptable answer for this problem. However, at this value of
Chapter 1 problems and solutions 3
January 2022
Solutions to
Homework Problems
,Last Revised 1/26/2022 at 5:32 PM Version 3e 1.0, January 2022
Chapter 1
1. Compute the range R corresponding to echo delays t0 of 1 ns, 1 s, 1 ms, and 1 second.
Equation (1.1): R = ct0/2. With c = 3×108 m/s, the results are:
Delay t0 1 ns 1 s 1 ms 1s
Range 0.15 m 150 m 150 km 1.5×108 m
2. Compute the time delays for two-way propagation to targets at distances of 100 km, 100
statute miles, and 100 feet.
Equation (1.1) rearranged gives t0 = 2R/c. With the conversion factors of 1 statute mile =
1609.34 m and 1 foot = 0.3048 m, the results are:
100 statute
Range 100 km 100 feet
miles
Delay t0 666.6 s 1.073 ms 203.2 ns
3. Radar is routinely used as one means of measuring the distance to objects in space. For
example, it has been used to calculate the orbital parameters and rate of rotation of the
planet Jupiter. The distance from Earth to Jupiter varies from 588.5×10 6 km to 968.1×106
km. What are the minimum and maximum time delays in minutes from the time a pulse is
transmitted in the direction of Jupiter until the time the echo is received? If pulses are
transmitted at a rate of 100 pulses per second, how many pulses are in flight, either on their
way to Jupiter or back again, at any given instant?
Minimum delay 2 Rmin c 2 5.885 1011 m 3 108 3923 s = 65.4 minutes
Minimum delay 2 Rmax c 2 9.681 10 m
11
3 108 6454 s = 107.6 minutes
Number of pulses in flight = trip time 100 pps 392,333 pulses (min)
645,400 pulses (max)
4. Table 1.1 defines the millimeter wave (MMW) band to extend from 30 to 300 GHz. Only
certain frequencies in this band are widely used for radar. This is partly due to frequency
allocation rules (which frequencies are allotted to which services), but also due to
atmospheric propagation. Based on Fig. 1.3, list two frequencies in the MMW band that
might be preferable for radar use, and two that would not be suitable. Explain.
Chapter 1 problems and solutions 1
,Last Revised 1/26/2022 at 5:32 PM Version 3e 1.0, January 2022
Figure 1.3 is repeated here:
Generally, frequencies with less atmospheric attenuation are more favorable than those
with higher attenuation because the received echo power will be greater for a given
target and range. Therefore, four favorable MMW frequency bands are, roughly, 30 to 40
GHz, 80 to 100 GHz, 140 to 150 GHz, and 230 to 270 GHz. Unfavorable frequencies
occur in the vicinity of about 60, 120, and 190 GHz.
5. Consider two equal-power transmitters, one at 10 GHz and the other at 30 GHz. The power
received from each is measured at a distance of 10 km and compared. Based on Figs. 1.3
and 1.4, estimate how many decibels weaker than the 10 GHz signal will the 30 GHz signal
be in sea level clear air conditions. Repeat for a medium rainfall of 2.5 mm/h and a tropical
downpour of 50 mm/h.
Figure 1.3 gives the attenuation in clear air at sea level. By inspection, the attenuation at
10 GHz is about 0.02 dB/km, and at 30 GHz is about 0.1 dB/km. The difference is 0.08
dB/km. Over 10 km, the difference in atmospheric attenuation will be (10)(0.08) = 0.8
dB.
Figure 1.4 is used in the same way to estimate the difference in loss for the two rainfall
cases. For a 2.5 mm/hr rain rate, the attenuation is about 0.02 dB/km at 10 GHz and 0.5
at 30 GHz, a difference of 0.48 dB/km. Therefore the 30 GHz signal is expected to be
about 4.8 dB weaker than the 10 GHz signal.
For the tropical downpour the attenuations are about 1 and 10 dB/km, giving rise at 10
km to a difference of 9 dB.
6. Compute the bandwidth needed to achieve range resolutions of 1 m, 1 km, and 100 km.
What is the length of a rectangular pulse having this Rayleigh bandwidth (peak-to-first null
width of the Fourier transform) for each value of resolution?
Chapter 1 problems and solutions 2
, Last Revised 1/26/2022 at 5:32 PM Version 3e 1.0, January 2022
Equation (1.2) states that R = c/2, so = c/(2∙R). The Rayleigh bandwidth of a pulse
of length seconds is 1/Hz (see App. B). The results are:
Resolution R 1m 1 km 100 km
Bandwidth 150 MHz 150 kHz 1.5 kHz
Pulse length 66.6 ps 66.6 ns 666.6 s
7. Derive Eq. (1.10) from Eq. (1.8).
From Eq. (1.8), E sin Dy sin Dy sin . The peak occurs when =
0, which gives us 0/0; but applying L’Hôpital’s rule shows us that E(0) = 1. The first null
of E() occurs at the first nonzero value of the argument of the numerator that makes the
numerator equal zero, i.e. when the argument of the sine function in the numerator
equals . So
Dy sin arcsin Dy
The peak-to-null (Rayleigh) width of the mainlobe is the distance between these two
values of , so it is just arcsin Dy .
Applying the small angle approximation sin() ≈ gives
Dy sin Dy Dy ,
which is the desired result of Eq. (1.10). However, another and equivalent way to arrive
at this is to consider the Taylor (or Maclaurin) series for the arcsin, namely
arcsin x x 1 6 x3 . Applying this in the expression arcsin Dy and
keeping only the first term gives arcsin Dy Dy and also leads directly to Eq.
(1.10). However, this version is slightly more convenient for the next problem …
8. Using the results from Prob. 7, find the value of Dy t for which the error in the small-
angle approximation of the Rayleigh beamwidth is 10 percent. What are the exact and
approximate Rayleigh beamwidths for that value of Dy t ?
The second term in the series for the arcsin is a rough estimate of the error in computing
Dy due to the small angle approximation. We want the error to be less than 10% of
the actual value, i.e. 1 6 x3 x 10 , where x = Dy . This leads to x 6 10 =
0.775 radians; this is an acceptable answer for this problem. However, at this value of
Chapter 1 problems and solutions 3
