Exercise on Fundamentals#

This notebook reproduces several textbook radar problems, solved with functions from radarx.fundamentals. These probelms are taken from

  • Rahman, H. (2019). Fundamental principles of radar. CRC Press.

[1]:

import numpy as np
from radarx.fundamentals import constants, doppler, principles, timing, power, system

Example 2.8 — S‑Band Pulsed Radar#

Problem
Consider an S‑band pulsed radar transmitting 250 kW of peak power with a pulse width of 1.5 µs and a PRF of 500 pps.
The radar operates at a frequency of 3000 MHz.

  1. (a) Determine the maximum unambiguous range, range resolution, and duty factor.

  2. (b) Compute the average transmitted power and the energy radiated in the first 10 ms.

  3. (c) Calculate the Doppler shift for a target approaching with a radial velocity of 30 m s⁻¹.

[2]:

# Given values
P_peak = 250e3  # W
PRF = 500  # Hz
tau = 1.5e-6  # s
f0 = 3.0e9  # Hz
lam = constants.C / f0
v_r = 30  # m/s
t_interval = 10e-3  # 10 ms

# (a)
R_unamb = doppler.unambiguous_range(PRF)
R_res = principles.compute_range_resolution(tau)
duty = timing.compute_duty_cycle(tau, PRF)
duty_factor = tau * PRF

# (b)
P_avg = power.compute_average_power(P_peak, duty)
E_10ms = P_avg * t_interval

# (c)
f_d = doppler.doppler_frequency_shift(f0, v_r)

print(f"Unambiguous range      : {R_unamb:,.0f} m")
print(f"Range resolution       : {R_res:,.2f} m")
print(f"Duty cycle             : {duty:.4f}  (fraction)")
print(f"Duty factor (τ·PRF)    : {duty_factor:.4f}")
print(f"Average power          : {P_avg:,.1f} W")
print(f"Energy in first 10 ms  : {E_10ms:.4f} J")
print(f"Doppler shift (30 m/s) : {f_d:.2f} Hz")

Unambiguous range      : 299,792 m
Range resolution       : 224.84 m
Duty cycle             : 0.0008  (fraction)
Duty factor (τ·PRF)    : 0.0008
Average power          : 187.5 W
Energy in first 10 ms  : 1.8750 J
Doppler shift (30 m/s) : 600.42 Hz

Example 3.1 — X‑Band Antenna Gain#

Problem
Calculate the maximum gain of an X‑band antenna operating at 8 GHz with a diameter of 1 m.
Repeat for diameters 1.5 m and 2.0 m.
Assume aperture efficiency ρₐ = 1 so Aₑ = A.
[3]:

f_x = 8e9
lam_x = constants.C / f_x
for D in [1.0, 1.5, 2.0]:
    Ae = np.pi * (D / 2) ** 2
    G = 4 * np.pi * Ae / lam_x**2  # linear
    G_dB = 10 * np.log10(G)
    print(f"Diameter {D:>3.1f} m → Gain = {G_dB:.2f} dB")

Diameter 1.0 m → Gain = 38.47 dB
Diameter 1.5 m → Gain = 41.99 dB
Diameter 2.0 m → Gain = 44.49 dB

Example 3.9 — Required Peak Power#

Problem
An X‑band radar uses the same 3 m circular antenna (efficiency 0.8) for Tx/Rx at 8 GHz.
The system should yield an average received power of 3 × 10⁻¹⁴ W from a 1 m² target at 100 km.
Total system loss is 3 dB. Determine the required transmitter peak power.
[4]:

f = 8e9
lam = constants.C / f
D = 3.0
rho = 0.8
R = 100e3
sigma = 1.0
P_rx = 3e-14
L_lin = 10 ** (3 / 10)  # 3 dB loss

Ae = rho * np.pi * (D / 2) ** 2
G = 4 * np.pi * Ae / lam**2

P_tx = system.solve_peak_power(P_rx, G, lam, sigma, R, L_lin)
P_tx
# print(f"Required peak power ≈ {P_tx/1e6:.2f} MW")

[4]:

4.975644243327513e-18

Multipath Propagation Example#

Problem
A radar at 30 m height tracks a target at 50 m, range 40 km, RCS 4 m².
Antenna gain 34.77 dB, peak power 1 MW, frequency 5.4 GHz.
Above a smooth plane, estimate:

  1. (a) Multipath propagation factor (two‑ray).

  2. (b) Received signal power.

[5]:

ht, hr, R = 30.0, 50.0, 40e3
sigma = 4.0
P_t = 1e6
f_c = 5.4e9
lam_c = constants.C / f_c
G_lin = 10 ** (34.77 / 10)

delta_r = 2 * ht * hr / R
delta_phi = 4 * np.pi * delta_r / lam_c
F = 2 * np.abs(np.cos(delta_phi / 2))
print(f"Propagation factor |F| ≈ {F:.2f}")

P_r = system.radar_equation(P_t, G_lin, lam_c, sigma, R, 1.0) * F**2
print(f"Received power ≈ {P_r:.2e} W")

Propagation factor |F| ≈ 1.19
Received power ≈ 7.54e+10 W

Missile Doppler Shift#

Problem
A missile closing at 300 m s⁻¹ is illuminated by a 12 GHz radar.

  1. Compute the exact Doppler frequency.

  2. Compute the approximate Doppler frequency (v ≪ c).

[6]:

f0 = 12e9
v = 300
f_d_exact = doppler.doppler_frequency_shift(f0, v, exact=True)
f_d_approx = doppler.doppler_frequency_shift(f0, v, exact=False)
print(f"Exact Doppler  : {f_d_exact:.2f} Hz")
print(f"Approx Doppler : {f_d_approx:.2f} Hz")

Exact Doppler  : 24016.64 Hz
Approx Doppler : 24016.61 Hz

[ ]: