NPTEL Course : EMI / EMC and Signal Integrity | Answer: Week 1 : Assignment 1

 

Q1) Electromagnetic spectrum penetration through Earth’s atmosphere

Let’s evaluate each statement one by one.

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Statement I

“All electromagnetic waves can penetrate Earth’s atmosphere.”

❌ False
Many EM waves (like most gamma rays, X-rays, and UV) are absorbed by the atmosphere for life protection.

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Statement II

“Gamma rays and Microwaves can easily penetrate Earth’s atmosphere.”

❌ False

• Gamma rays → mostly absorbed by the atmosphere

• Microwaves → partially penetrate (some absorption by water vapor)
  Since gamma rays do not penetrate easily, this statement is wrong.

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Statement III

“Radio waves can penetrate Earth’s atmosphere.”

✅ True
Radio waves pass easily through the atmosphere and are widely used for communication and astronomy.

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Statement IV

“Visible light can penetrate Earth’s atmosphere.”

✅ True
Visible light passes through the atmosphere (this is why we can see sunlight).

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Correct combination

✔ Statement III is correct
✔ Statement IV is correct

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✅ Correct Answer: d) Both III and IV are correct

Q2) Identify the correct pairs (Emission & Susceptibility)

Concept

• Emission → Source is noisy

• Susceptibility → Victim is sensitive

• Radiated → Coupling through air

• Conducted → Coupling through cable

Given figures

• I → Noisy component radiating → Radiated Emission ✔

• II → Susceptible component affected via radiation → Radiated Susceptibility ✔

• III → Susceptible component affected via cable → Conducted Susceptibility ✔

• IV → Noisy component injecting noise via cable → Conducted Emission ✔

✅ Correct Answer: a) All are correct

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Q3) Difference between EMI and Noise

Concept

• EMI (Electromagnetic Interference)
  → Unwanted deterministic signals (clock harmonics, switching noise)

• Noise
  → Random and unpredictable (thermal noise, shot noise)

Option analysis

• a ❌ Incorrect

• b ✅ Correct definition

• c ❌ Incorrect

• d ❌ Incorrect

✅ Correct Answer: b

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Q4) 1st and 2nd harmonic of a 25% duty cycle pulse

Formula
For a pulse of amplitude A and duty cycle D:

$$|C_n| = \frac{A}{n\pi} \left|\sin(n\pi D)\right|$$

Given:

• A = 1 V

• D = 0.25

1st harmonic (n = 1)

$$|C_1| = \frac{1}{\pi}\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}\pi}$$

2nd harmonic (n = 2)

$$|C_2| = \frac{1}{2\pi}\sin\left(\frac{\pi}{2}\right) = \frac{1}{2\pi}$$

✅ Correct Answer: b

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Q5) Fourier coefficient C₀ (DC component)

Concept

$$C_0 = \text{Average value over one period}$$

From the waveform:

• Pulse height = 1

• ON time ≈ 2 ms

• Period ≈ 10 ms

$$C_0 = \frac{2}{10} = 0.2$$

✅ Correct Answer: a) 0.2

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Q6) Duty cycle from phase of 5th harmonic

Formula
For a pulse:

$$\angle C_n = -n\pi D$$

Given:

• n = 5

• Phase = −225° = −$\frac{5\pi}{4}$

$$5\pi D = \frac{5\pi}{4} \Rightarrow D = 0.25$$

Duty cycle = 25%

✅ Correct Answer: c

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Q7) Voltage at antenna base (dBµV)

Step 1: Cable loss

• 200 m ≈ 656 ft

• Loss = 8 dB / 100 ft

$$\text{Total loss} = 8 \times 6.56 \approx 52.5 \text{ dB}$$

Step 2: Power at antenna

$$-20 \text{ dBm} + 52.5 = +32.5 \text{ dBm}$$

Step 3: Convert dBm → dBµV (50 Ω)

$$V_{\text{dBµV}} = P_{\text{dBm}} + 107$$ $$32.5 + 107 = 139.5 \text{ dBµV}$$

✅ Correct Answer: a) 138 – 141 dBµV

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Q8) Electromagnetic Compatibility definition

EMC requires

• ✔ Does not interfere with others

• ✔ Does not interfere with itself

• ❌ Should NOT be susceptible to others

Thus:

• I ✔

• II ❌

• III ✔

✅ Correct Answer: d) I and III

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Q9) Output voltage of −40 dBm source

Step 1: Convert power

$$-40 \text{ dBm} = 0.1 \mu W$$

Step 2: Voltage across 75 Ω

$$V = \sqrt{PR} = \sqrt{0.1\times10^{-6} \times 75} \approx 2.7 \text{ mV}$$

✅ Correct Answer: a) 2 – 3 mV

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Q10) FCC harmonic testing above 1 GHz

FCC requirement

• Measurements up to 5th harmonic or 40 GHz, whichever is lower

✅ Correct Answer: b

Question	Correct Option
Q1                 d Q2	a
Q3	b
Q4	b
Q5	a
Q6	c
Q7	a
Q8	d
Q9	a
Q10	b