We know ID =1/2 kn (VGS + vgs – Vt)2. Let the signal vgs be a sine wave with amplitude Vgs, and substitute vgs = Vgs sin ω t in Eq.(5.43). Using the trigonometric identity show that the ratio of the signal at frequency 2ω to that at frequency ω , expressed as a percentage (known as the second-harmonic distortion) is

Correct Answer: 1/4Vgs/Vov x 100%
n: