A low-noise amplifier senses a -80-dBm signal at 2.410GHz and two -20-dBm interf

Question

A low-noise amplifier senses a -80-dBm signal at 2.410GHz and two -20-dBm interferes at 2.420 GHz and 2.430 GHz. What IIP3 is required if the IM products must remain 20 dB below the signal? For simplicity, assume 50. Ohm interfaces at the input and output. Solution: Denoting the peak amplitudes of the signal and the interferes by Asig and Aint, respectively, we can write at the LNA output: 20 log | alpha 1asig| - 20 dB - 20 log |3 / 4 alpha 3 |. It follows that | alpha 1Asig| = |30 / 4 alpha |. In a 50- Ohm system, the -80 dBm and -20dBm levels respectively yield Asig = 31.6 mu Vp and Aint = 31.6 mVp. Thus, IIP3 = 4 / 3 | alpha 1 / alpha 3| = 3.65 Vp = + 15.2 dBm. Such an IP3 is extremely difficult to achieve, especially for a complete receiver chain.

Explanation / Answer

Voltage corresponding to -35dBm = 10^(-(35+10)/20 ) *Vp = 5.62m Vp

for -3dBm = 10^(-(3+10)/20) *Vp = 0.2234 Vp

IIP3 = sqrt[ (4/3) * (0.2234/(5.62 m) ) ] = 7.28 Vp

which is 20log(7.28) = +17.24 dBm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.