Electromagnetic Compatibility Course Notes - Supplement
This page is a repository for supplemental information related to the book Electromagnetic Compatibility Course Notes. It will be updated with error corrections, content updates, problem solutions, and links to information related to the material presented in the book.
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Problem Solutions
Chapter 1
1. For each of the EMC problems below, identify the probable source,
coupling path and receptor.
a.) Source: Microwave Oven, Coupling Path: Radiated Field, Receptor: Computer Display
b.) Source: Starter Motor, Coupling Path: Conducted (through power wiring), Receptor: Car Radio
c.) Source: Lightning, Coupling Path: Conducted (Power or Communication Wiring), Receptor: Modem
d.) Source: Electric Blanket, Coupling Path: Radiated (likely) or Conducted through Power Wiring, Receptor: WiFi Router or Device
e.) Source: Airport Radar or Transmitter, Coupling Path: Radiated Field, Receptor: Tire Pressure Monitoring System
2. What are alternative names for conducted, inductive, and capacitive coupling?
Conducted Coupling ⇒ Common Impedance Coupling
Inductive Coupling ⇒ Magnetic-Field Coupling
Capacitive Coupling ⇒ Electric-Field Coupling
3.
4. 150 mV = 43.5 dB(mV) = 104 dB(μV) = –3.5 dBm
5. 6 dB (same as ratio of rms amplitudes)
6. +12 dB - 6 dB = +6 dB (or a factor of 2 in voltage), so output amplitude is 2 volts.
7. The term 50% gain is ambiguous. Does it mean 50% of the original amplitude (factor of 0.5) or 50% more (factor of 1.5)?
By the first interpretation, answers are 0.5 V, 1 V, 2V and 5V. By the second interpretation, answers are 1.5 V, 2 V, 3 V and 6V. These answers would all be different, if we interpreted the 50% gain to be a power gain instead of a voltage gain. Note that a 100% gain is generally interpreted as a doubling of the amplitude. However, a 200% gain is also generally considered a doubling of the amplitude. Signal gains, losses, or error margins should usually be expressed in dB in order to avoid misinterpretation.
8. 46 dB(μV/m) - 40 dB(μV/m) = 6 dB. Note that it is NOT 6 dB(μV/m), which is a signal amplitude totally unrelated to the correct answer.
9. 200 μV/m = 46 dB(μV/m). 100 μV/m = 40 dB(μV/m). 46 dB(μV/m) - 40 dB(μV/m) = 6 dB. Note that we could also say that the measured field is 200 μV/m - 100 μV/m = 100 μV/m over the limit. In most cases, however, it will be more helpful to know the ratio in dB rather than the difference in the field strengths.
10. Bulk Current Injection is an immunity test.
11. Radiated Immunity and Bulk Current Injection testing should be done in a shielded room to contain the fields and prevent them from interfering with the test equipment and other nearby electronic devices. Some radiated emissions tests also specify an absorber-lined shielded room, although useful emissions measurements could still be made outside the room. A shielded room is not required for conducted emissions and transient immunity measurements.
12. Conducted Emissions, Bulk Current Injection, Lightning Immunity and Electrical Fast Transient Immunity tests can only be done on products with an attached cable.
Chapter 2
1. @ DC: R ≈ 170 mΩ/m. @ 1 MHz: R ≈ 330 mΩ/m. @ 1 GHz: R ≈ 10 Ω/m.
2. C ≈ 52 pF/m
3. L ≈ 52 nH/m
4. At DC, we'll assume that the plane resistance is negligible relative to the trace resistance. The trace resistance is 4.0 Ω/m or 40 mΩ/cm.
At 10 MHz, the plane current is focused under the trace, so we'll assume that the plane resistance is approximately equal to the resistance of the trace. Skin depth is 21 μm, which is slightly thicker than the half-ounce copper. Resistance of the trace is still about 40 mΩ/cm. Resistance of the trace and plane is about 80 mΩ/cm or a little less.
5. C/ε = G/σ therefore G = C(σ/ε) = 10-8 F/m x 5.4 x 10-4 S/m/(4.3 x 8.854 x 10-12 F/m) = 0.14 S/m (or 1.4 x 10-3 S/cm).
6. Inductance of each loop is 0.146 μH. Leakage inductance is approximately the inductance per unit length of the wire pair (edge-to-edge spacing is 0.5 mm) times the circumference of the loop. This is 0.52 μH/m x 0.16 m = 0.083 μH. So mutual inductance is approximately 0.146 - 0.08 = 0.066 μH. Note that his is a coupling coefficient of only k = 66/146 = 0.45 (about 45%). Even though the loops share a lot of area, most of the flux is near the wire and the wire-to-wire spacing is too large to get efficient coupling. Using the same method, reducing the wire insulation thickness to 0.025 mm would increase the coupling coefficient to 80%.
7. Due to the symmetry of the wire cross-section, the mutual inductance must be half the self-inductance of the circuit using wires 1 and 3. The self-inductance of that circuit is 0.92 μH/m x 0.5 m = 0.46 μH. The mutual inductance is therefore 0.23 μH.
8. The impedance of a 1-μF capacitor is less than 5 Ω at frequencies above 1/2πRC = 31.8 kHz. A 5-nH inductor has an impedance less than 5 Ω at frequencies below R/2πL = 159 MHz. Therefore, the range of frequencies where the capacitor's impedance is less than 5 Ω is about 32 kHz to 160 MHz.
9. The impedance of a 2.2-μF capacitor is less than 5 Ω at frequencies above 1/2πRC = 14.5 kHz. A 5-nH inductor has an impedance less than 5 Ω at frequencies below R/2πL = 159 MHz. Therefore, the range of frequencies where the capacitor's impedance is less than 5 Ω is about 14.5 kHz to 160 MHz.
10. At DC, the ground strap impedance is its resistance: R ≈ 0.35 mΩ. At 10 kHz, the resistance is about the same and the inductance (from Equation 2.32) is about 460 nH, which is an impedance of about 29 mΩ. The inductance dominates, so the ground strap impedance is 29 mΩ. At 10 MHz, the inductance is still 460 nH, so the ground strap impedance is about 29 Ω.
Chapter 3
1. At DC, the resistance of the 20-meter return wire is 1.68 Ω. The source current is VS1 / 100 Ω, so the voltage dropped across the return conductor is VS1 x 1.68/ 100 Ω = 0.0168 VS1.. The portion dropped across the victim circuit load is (100/110) x 0.0168 VS1., so the ratio VS2/VS1 is 0.0153 or -36 dB.
At 1 MHz, the resistance of the 20-meter return wire is 3.25 Ω. All other values are the same as they were at DC. Therefore, the coupled voltage is a factor of 3.25/1.68 higher. This is 0.030 or -31 dB.
At 3 MHz, the resistance is 5.63 Ω. The coupled voltage is a factor of 5.63/1.68 higher than at DC. This is 0.051 or -26 dB.
2. The mutual capacitance is 52 pF/m x 20 m = 1.04 nF. At DC, there is no capacitive crosstalk. Using Equation 3.11, the crosstalk at 1 MHz is -23.7 dB. At 3 MHz, it is a factor of 3 higher or -14 dB.
3. The mutual inductance is 200 nH/m x 20 m = 4.0 μH. At DC, there is no inductive crosstalk. Using Equation 3.40, the calculated crosstalk at 1 MHz is -12 dB. At 3 MHz, simply assuming the crosstalk is a factor of 3 higher yields a crosstalk value of -3 dB. This value suggests the weak coupling assumption has been violated. We do not have enough information to calculate the crosstalk precisely; however, the self-inductance of the source loop must be greater than 4.0 μH. At 3 MHz, this is 75 Ω and is nearly has high as the load resistance. In other words, the signal is significantly distorted, and the precise value of the crosstalk is probably irrelevant.
4. Changing the receiver impedance from 100 Ω to 1 MΩ drops the current in the source circuit to near zero. The common-impedance coupling is virtually eliminated (-115 dB @DC, -110 dB @1 MHz, -100 dB @3 MHz). The inductive coupling is also virtually eliminated (-92 dB @1 MHz, -82 dB @3 MHz).
The only significant coupling is due to capacitive coupling, which is non-existent at DC, but is still -23.7 dB @1 MHz and -14 dB @3 MHz.
5. The resonant dipole input resistance is approximately 72 Ω. Source voltage is 1 mVrms. Therefore, the power delivered to the dipole and radiated is approximately 14 nW.
6. Power density in the direction of maximum radiation is Prad (D0/4πR2) where the dipole directivity, D0, is 1.6. So, the maximum power density is 200 pW/m2. This corresponds to a field strength in free space of about 270 μV/m. In a semi-anechoic environment, the field strength will be doubled in places where the source and its image create fields that add in phase. This means the maximum field strength in a semi-anechoic environment will be 540 μV/m.
7. The differential-mode current is (1 v) / (102 Ω) = 9.8 mA. Using Equation 3.62, the radiated field strength at 3 meters is 43 μV/m.
8. Here's one example. Length is 1.3 meters. Length of a half-wave dipole at 20 MHz is 7.5 meters.
9. Since log-periodic arrays have elements that are approximately a half wavelength long, these antennas would be very large. They would be essentially unusable at heights of 1-meter above a ground plane as required for most radiated emissions measurements.
10. Due to the skin-effect, thin wire antennas start to become pretty lossy at GHz frequencies. The smallest elements of a log-periodic array need to be fairly closely (and precisely) spaced. This does not work well with thick elements. Horn and patch antennas don't have thin wires and tend to work better at GHz frequencies.
Chapter 4
1. a.) In peak detection mode, the spectrum analyzer measures the peak rms value of the signal, which is 1.41 volts or 16 dBm.
b.) The signal is continuous with constant amplitude, so the quasi-peak and peak values are the same, 16 dBm.
c.) The signal is continuous with constant amplitude, so the average and peak values are the same, 16 dBm.
2. The first harmonic of a square wave has an rms amplitude that is 0.45 times the peak-to-peak value. In this case, 0.9 volts or 12 dBm @ 1 MHz. The third harmonic is one-third of this value, 0.3 or 2.55 dBm. The 15th harmonic is 60 mV or -11.4 dBm. The 35th harmonic is 26 mV or -19 dBm.
3. A 40-ns transition time places the second knee frequency at 8 MHz. The amplitudes of the harmonics at 1 MHz and 3 MHz are not changed much. The amplitudes of each harmonic calculated using Equation 4.16 to get the peak amplitude and then multiplying by 0.707 to get the rms amplitude are:
1 MHz — 0.9 volts or 12 dBm (same as square wave)
3 MHz — 0.29 volts or 2.3 dBm (-0.35 dB compared to square wave)
15 MHz — 0.03 volts or -17 dBm (-5.6 dB compared to square wave)
35 MHz — 0.0056 volts or -32 dBm (-13 dB compared to square wave)
4 and 5. Plot shown here.
6. Plot shown here.
7. 1.4 mV or -88 dBm.
8. Noise power is proportional to resolution bandwidth. Reducing bandwidth by 9/120 reduces noise floor by 10 Log (9/120) = -11 dB. So new noise floor is at -91 dBm.
9. (a.) FFT BW = 1/fs. Displayed bandwidth (one-sided) BWmax < 1/2 x 109) = 500 MHz.
(b.) Δfres = 1/T = 125 kHz.
(c.) To get better frequency resolution, it's necessary to sample more of the input signal.
10. As indicated in Figure 4.13, the quasi-peak attack time constant is 1 ms and the decay time constant is 160 ms. So any signal that is present for 2.2 ms and doesn't go away for substantially longer than 2.2 ms will not have a quasi-peak value significantly lower than the peak value. Therefore, pulses occurring at a rate of 15 kHz, 100 kHz, and 1 MHz will have quasi-peak values approximately equal to their peak values. Pulses occurring with a repetition rate of 60 Hz (period of 17 ms) would be expected to have a quasi-peak value less than the peak value.
Chapter 5
1. The velocity of propagation is 3x108/sqrt(4.3) = 1.45x106 m/s. The propagation delay is length/velocity or about 690 ps.
2. 
3. LC = με, so L = με/C = 4.8 nH/m. Z0 = sqrt(L/C) = 0.69 Ω. [This is not a very realistic value. The problem statement was supposed to say 100 pF/m.]
4. The reflection coefficient for the forward-traveling wave only requires a load end. If the transmission line is matched at the load end, there are no reflections.
5. Increasing the width of a circuit board trace means,
(a.) Characteristic impedance decreases.
(b.) Inductance per unit lengths decreases.
(c.) Capacitance per unit length increases.
(d.) Propagation delay stays the same for a stripline, decreases slightly for a microstrip line.
(e.) Resistance per unit length decreases.
(f.) Conductance per unit length increases.
6. Cable properties
(a.) Inductance per unit length: 630 nH/m
(b.) Capacitance per unit length: 51 pF/m
(c.) Resistance per unit length: 325 mΩ/m
(d.) Characteristic impedance: 111 Ω
(e.) Propagation velocity: 1.76 x 108 m/s
(f.) Cable attenuation at 1 MHz: (from Eqs. 5.31 and 5.32) 12.7 dB/km.
7. For the transmission line in the problem statement,
(a.) Γ = 0.41
(b.) VSWR = 2.4
(c.) At 250 MHz, the cable wavelength is 80 cm. βl = 1.25π. tan(βl) = 1.0. Zin = 50 (120 + j50)/(50 +j120) = 35.5 + j35.2 Ω.
(d.) Magnitude of current in = 3.3/|40.5 + j35.2| = 0.061 amps. Power in is Pin = I2Rin = 134 mW. Line is lossless, so all of Pin is delivered to the load.
(e.) V = sqrt(Pin x R) = sqrt(0.134 x 120) = 4.0 Vrms. (Note that the load voltage is higher than the open-circuit source voltage.)
8. From Eq. 5.50, Leq = 5.1 nH.
9. The fields above and below the trace must travel together. Their velocity is slower than a plane wave traveling in air, but faster than a plane wave traveling entirely in the dielectric.
10. The only meaningful voltages in an RLCG model are the differential voltages between the two conductors at any position along the length. Any voltage differences between points at two different locations along the length of the transmission line are artifacts of the calculation. They do not correspond to meaningful voltage differences that could be measured on a real transmission line. In fact, voltage differences along the length of the line are not uniquely defined.
Models that put resistances and inductances on both sides of the RLCG equivalent, are unnecessarily complex. But worse, they imply that there is a unique and calculable voltage difference between the reference on one end of the line and the reference on the other end.
Chapter 6
1. For ...
