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. Applying the circuit model in Fig. 6.3 and making the weak coupling assumption,
(a.) IC = C dV/dt = (3 pF) (3.3 V)/(2 ns) = 5 mA. Multiplying by the parallel combination of the victim circuit impedances, Vcoupled = 10 mV. So, the coupled waveform is a positive or negative 2-mV pulse that lasts 2 ns that occurs with each signal transition. Note that the 5 pF load draws a current of C dV/dt = (5 pF) (2 mV)/(~1 ns) = 10 μA. This is much less than the 5 mA in the source resistance, so the load capacitance doesn't significantly affect the coupled voltage.
(b.) Terminating with a 50-Ω resistance would have little effect on the coupling in this case, because 50 Ω is still large compared to 2 Ω.
2. Moving the traces farther apart decreases C12 and slightly increases C11 and C22.
3. Moving the traces farther apart decreases L12.
4. The 50-Ω scope impedance and C12 form a voltage divider with 7.5 mV dropped across the 50 Ω. The current through 50-Ω scope resistance is 7.5 mV/50 Ω = 150 μA. Therefore, the impedance of C12 is approximately (1 V)/(150 μA) = 6.67 kΩ. At 1 MHz, this is a capacitance of C12 = 24 pF or 12 pF/m.
5. The current in the source circuit is (1 V)/(50 Ω) = 20 mA. The coupled voltage is ωM12I = .45 V. Therefore, M12 = 3.6 μH or 1.8 μH/m.
6. Below 100 kHz, the coupling would be very weak, and the coupled voltages would be hard to measure accurately. At frequencies much higher than 10 MHz, the 2-meter cable is no longer electrically short and doesn't have a constant voltage or current along its entire length.
7. A copper wire with a 1-mm diameter has a DC resistance of 21 mΩ/m. A 100-meter wire has a resistance of about 2.1 Ω. Carrying 8 A of current, the voltage drop across the wire would be about 17 volts. Since the sensor uses the same return wire, the sensor circuit would see this 17 volts whenever the motor started.
8. At these frequencies, the cable is electrically long. The maximum near-end crosstalk is 20 Log[C12/(C22+C12)] = -11 dB. At frequencies above 1 GHz, the wires will be lossy enough that the signal amplitudes and crosstalk will start to be attenuated.
9. Note that the end-to-end propagation delay for the cable is 53 ns. At 5 Mbps, the 10-meter cable is not electrically long. However, it is long relative to a transition time of 20 ns. Therefore, the coupled near-end voltage is a pulse with a maximum amplitude of (ΔV/2) x [C12/(C22+C12)] ≈ 170 mV. The pulse duration is 40 ns. Positive then negative pulses are coupled with every signal transition.
10. At 200 Mbps, the cable is electrically long. The maximum coupled voltage at the near end is still 170 mV but the coupled waveform has a shape similar to the signal waveform.
Chapter 7
1. With a 1-ns transition time, the 2nd knee frequency is 318 MHz (well above 35 MHz). Amplitude of first harmonic is 0.45 x 3.3 volts = 1.5 volts. Amplitude of 35th harmonics equals 1.5/35 = 42 mV. At the connector, the transition from unbalanced to balanced creates a common-mode voltage equal to half the signal voltage (i.e., 21 mV).
2. Without a change in the imbalance, there is no common-mode voltage generated.
3. h = C11/(C11+C22) = 0.4. Note that labeling the wider trace as conductor 1 would give a value of 0.6. Either value of h describes the same level of imbalance.
4. VCM = Δh x VDM = (0.4 - 0.5) 42 mV = -4.2 mV. (or +4.2 mV if the wider conductor is conductor 1). The polarity of VCM only becomes important when there are multiple imbalance changes in the same signal path. In that case, it's only necessary to be consistent with the definition of conductor 1 (keeping it on the same side of the signal).
5. Using ATLC2, h ≈ 0.33 (or, 0.67 depending on which conductor was labeled conductor 1).
6. Using the same procedure used in Example 7-4, the propagation velocity is 3 x 108 m/s, CDM is 24.5 pF/m. LDM is therefore 454 nH/m. k = 0.531. Applying Eq. 7.13 and 7.18, L11=L22 = 0.48 μH/m. L12 = kL11 = 0.26 μH/m.
7. Using the procedure used in Example 7-5, ZDM = 142 Ω and ZCM = 98 Ω. Z1 = 196 Ω and Z12 = 173 Ω.
8. The percentage of the differential-mode signal that drives the cable relative to the plane is zero. Both the differential-mode source and transmission lines are balanced, so there is no mode conversion. The percentage of the common-mode source voltage that drives the ribbon cable relative to the plane is close to 50%. The common-mode propagation on the board is very unbalanced (out on two traces and back on a wide plane). The common-mode propagation on the cable is nearly balanced (out on two wires and back on two wires). Approximately half of the common-mode voltage drives the cable, and the other half is converted to differential-mode.
9. Because common-mode and differential-mode propagation are independent, they can travel at different velocities. Single-ended signals have only one propagation mode, so they are not affected by this. Differential-mode signals only propagate in the differential-mode, so they are not directly impacted. However, if there is a common-mode component of the source voltage (e.g., as generated by pseudo-differential drivers), then the common-mode noise pulses travel with a different velocity.
10. Perhaps the main reason circuit designers don't take advantage of orthogonal propagation modes is the complexity involved in generating and receiving independent common-mode and differential-mode signals. Yes, it can be done, but at most practical transmission line lengths, the crosstalk is a relatively minor issue and doesn't warrant added cost or complexity in the source or receiving circuits. Another consideration is that maintaining perfect balance can be difficult, especially as signals propagate through connectors. And some crosstalk would occur in the driver and receiver circuitry. So, utilizing orthogonal propagation modes doesn't solve the crosstalk problem, it just redefines it.
Chapter 8
1. For each of the conductors commonly labeled ground listed below, indicate whether it is a safety ground, EMC ground, or current return.
(a.) Ground wire in building electrical distribution wiring (safety)
(b.) Frame ground in an electric oven (safety and EMC)
(c.) Frame ground in an automobile (EMC > 1 MHz, current return for <48 V and < 1 MHz, safety ground for >48 V in electric vehicles)
(d.) Frame ground in a large commercial aircraft (safety and EMC)
(e.) Frame ground in a cell phone (EMC)
(f.) Circuit board signal ground plane (current return)
(g.) Circuit board chassis ground plane (EMC)
(h.) Ground plane in an EMC test setup (EMC)
(i.) Coaxial cable shield (EMC and current return)
2. EMC grounds are mostly utilized at high frequencies (e.g., > 1 MHz). Intentional low-frequency currents in a product's frame ground generally don't create enough of a voltage to cause an EMC problem.
3. Common-impedance coupling can be a problem when a product's frame ground carries very large low-frequency currents. A good example of this is currents in aerospace vehicle frames induced by lightning or space charging. In these cases, circuits utilizing the frame as a current return path can be exposed to damaging levels of transient noise.
4. Cables that carry intentional high-frequency signal currents on their shield need to make a good high-frequency connection between the cable shield and the circuit board current-return plane. Also, shields that carry large unintentional high-frequency currents generated by common-mode sources on the board need to return those currents to the board's current-return plane.
5. Single-point grounds are designed to prevent common-impedance coupling. They are generally ineffective in preventing other types of EM coupling (and can often make other types of coupling worse).
6. In this example, the worst-case coupling between analog and digital components would (assuming a half-ounce copper plane) would be 3 mV. This is greater than the tolerance of the analog circuits, so Analog-GND and Digital-GND should be isolated if the switching occurs at a rate less than 100 kHz. The worst-case coupling between digital and power circuits would be 50 mV. This is far below the tolerance of the digital circuits, so Digital-GND and Power-GND should not be isolated. Note that even though the analog circuits require an isolated current return, the board should NOT have a split ground plane. Analog returns should be routed on traces or planes occupying a different layer in the stack-up.
7. The low-frequency resistance of #24 wire is 84 mΩ/m. If the maximum voltage we can drop across each wire is (5.0 - 4.5)/2 wires = 0.25 volts, then the maximum wire resistance is (0.25 V)/(2 A) = 0.125 Ω. This corresponds to a wire length of (0.125)/(0.084) = 1.5 meters.
8. If the heater used the cable shield as a current return and the camera uses the third wire, then they only share a power wire. This doubles the maximum length, (i.e., 3 meters.
9. If one side of the scope probe is grounded to the frame ground of the scope, that side also makes a low-frequency connection to the circuit ground. Connecting that probe across R2 puts a DC short in parallel with either the source or R1. If R1 is shorted, the voltage across R2 may be higher (at least the DC component). The impedance between the scope ground and the circuit ground is unknown at 1 MHz and its harmonics, so the effect on the waveform is unpredictable. If the source is shorted, it may stop working or put out a much lower voltage. Connecting Channel 2 across R1 is ok as long as the grounded side of the scope probe is connected to the grounded side of R1. The problem didn't specify whether the Channel input impedance was 50 Ω or high impedance. High-impedance inputs would attenuate the upper harmonics of the square wave. 50-Ω inputs could load the circuit and change the voltages being measured.
10. Using an ungrounded oscilloscope would eliminate the shorting problem when measuring R2 by itself. However, it would still not be possible to measure the voltages on R1 and R2 with Channel 1 and Channel 2 at the same time.
Chapter 9
1. The transition time is 2.2RC = 2.2 x (2 Ω) x (15 pF + 5 pF) = 88 ps. The bit width is 100 ns. To slow the transition time to 10 ns would require a series resistor with a value of 225 Ω.
2. This filter could be either a shunt capacitor or a series inductor. The shunt capacitor would need to have an impedance of about 0.2 Ω at 150 kHz (C = 5.3 μF). The series inductor would need to have an impedance of about 520 Ω at 150 kHz (L = 550 μH).
3. If the filter will have 20 dB of attenuation at 150 kHz, the cutoff frequency should be about 47 kHz. To make an LC filter, the capacitor in parallel with the load should have an impedance of 50 Ω at 47 kHz (C = 0.068 μF). To set the cutoff at 47 kHz, the value of L needs to be L = 170 μH.
4. If the filter will have 20 dB of attenuation at 150 kHz, the cutoff frequency should be 47 kHz. To make an LC filter, the inductor in series with the source should have an impedance of 2 Ω at 47 kHz (L = 6.8 μH). To set the cutoff at 47 kHz, the value of C needs to be C = 1.7 μF.
5. ...
