[...]... same frequency to zero: C · j 1 · H1 ( 1 ) · ej 1 t + g1 · H1 ( 1 ) · ej 1 t + is · ej 1 t = 0 C · j 2 1 · H2 ( 1 , 1 ) · e j 3 1 t + g1 · H2 (ω, 11 ) · e (1. 60a) j 2 1 t 2 + g2 · H1 ( 1 ) · ej 2 1 t = 0 (1. 60b) where we have limited ourselves to the second-harmonic frequency From the first equation (eq (1. 60a)), we immediately have H1 ( 1 ) = − 1 g1 + j 1 C (1. 61) which is nothing but the solution... EXPANSION 19 When the voltage as in eq (1. 56) is replaced into Kirchhoff’s equation (eq (1. 57)), we get C · (j 1 · H1 ( 1 ) · ej 1 t + j 2 1 · H2 ( 1 , 1 ) · ej 2 1 t + · · ·) + g1 · (H1 ( 1 ) · ej 1 t + H2 ( 1 , 1 ) · ej 2 1 t + · · ·) + g2 · (H1 ( 1 ) · ej 1 t + H2 ( 1 , 1 ) · ej 2 1 t + · · ·)2 + is · ej 1 t = 0 (1. 58) Expanding the expressions in the parentheses, we get C · j 1 · H1 ( 1 ) ·... ( 1 ) · ej 1 t + C · j 2 1 · H2 ( 1 , ω2 ) · ej 2 1 t + g1 · H1 ( 1 ) · ej 1 t 2 + g1 · H2 ( 1 , 1 ) · ej 2 1 t + g2 · H1 ( 1 ) · ej 2 1 t + g2 · 2H1 ( 1 ) · H2 ( 1 , 1 ) · ej 3 1 t 2 + g2 · H2 ( 1 , 1 ) · ej 4 1 t + is · ej 1 t = 0 (1. 59) where the first and third rows are the linear terms, and the second row is the nonlinear term truncated to the second power We now try to split eq (1. 59) into... x(t) = A1 · cos( 1 t) + A2 · cos(ω2 t) (1. 52) The output signal is given by y(t) = y1 (t) + y2 (t) + y3 (t) + · · · (1. 53) y1 (t) = A1 · H1 ( 1 ) · cos( 1 t) + A2 · H1 (ω2 ) · cos(ω2 t) A2 A2 1 · H2 ( 1 , − 1 ) + 2 · H2 (ω2 , −ω2 ) 2 2 2 A A1 A2 · H2 ( 1 , ω2 ) · cos(( 1 + ω2 )t) + 1 · H2 ( 1 , 1 ) · cos(2 1 t) + 2 2 A2 + 2 · H2 (ω2 , ω2 ) · cos(2ω2 t) 2 3A3 3A3 1 2 · H3 ( 1 , 1 , − 1 ) · cos( 1 t)... (1. 63) 20 NONLINEAR ANALYSIS METHODS When the voltage in the form of eq (1. 63) is replaced into Kirchhoff’s eq (1. 57) we get C · j 1 · H1 ( 1 ) · ej 1 t + C · j ω2 · H1 (ω2 ) · ej ω2 t + C · j 2 1 · H2 ( 1 , 1 ) · ej 2 1 t + C · j ( 1 + ω2 ) · H2 ( 1 , ω2 ) · ej ( 1 +ω2 )t + C · j 2ω2 · H2 (ω2 , ω2 ) · ej 2ω2 t + g1 · H1 ( 1 ) · ej 1 t + g1 · H1 (ω2 ) · ej ω2 t + g1 · H2 ( 1 , 1 ) · ej 2 1 t + g1 ·... · cos(ω2 t) y3 (t) = 4 4 3A1 A2 3A2 A2 2 1 · H3 ( 1 , ω2 , −ω2 ) · cos( 1 t) + · H3 (ω2 , 1 , − 1 ) · cos(ω2 t) + 4 4 3A2 A2 1 + · H3 ( 1 , 1 , −ω2 ) · cos((2 1 − ω2 )t) 4 3A1 A2 2 + · H3 (ω2 , ω2 , − 1 ) · cos((2ω2 − 1 )t) 4 3A2 A2 3A3 1 1 · H3 ( 1 , 1 , 1 ) · cos(3 1 t) + · H3 ( 1 , 1 , ω2 ) · cos((2 1 + ω2 )t) + 4 4 A3 3A1 A2 2 · H3 ( 1 , ω2 , ω2 ) · cos((2ω2 − 1 )t) + 2 · H3 (ω2 , ω2 , ω2... H2 ( 1 , ω2 ) · H2 (ω2 , ω2 ) · ej ( 1 +3ω2 )t + g2 · 2 · H1 ( 1 ) · H2 ( 1 , 1 ) · ej 3 1 t + g2 · 2 · H1 ( 1 ) · H2 ( 1 , ω2 ) · ej (2 1 +ω2 )t + g2 · 2 · H1 ( 1 ) · H2 (ω2 , ω2 ) · ej ( 1 +2ω2 )t + g2 · 2 · H1 (ω2 ) · H2 ( 1 , 1 ) · ej (2 1 +ω2 )t + g2 · 2 · H1 (ω2 ) · H2 ( 1 , ω2 ) · ej ( 1 +2ω2 )t + g2 · 2 · H1 (ω2 ) · H2 (ω2 , ω2 ) · ej 3ω2 t + ej 1 t + ej ω2 t = 0 (1. 64) We see that eq (1. 64)... + g1 · H2 ( 1 , ω2 ) · ej ( 1 +ω2 )t + g1 · H2 (ω2 , ω2 ) · ej 2ω2 t 2 2 + g2 · H1 ( 1 ) · ej 2 1 t + g2 · H1 (ω2 ) · ej 2ω2 t + g2 · 2 · H1 ( 1 ) · H1 (ω2 ) · ej ( 1 +ω2 )t 2 2 2 + g2 · H2 ( 1 , 1 ) · ej 4 1 t + g2 · H2 ( 1 , ω2 ) · ej 2( 1 +ω2 )t + g2 · H2 (ω2 , ω2 ) · ej 4ω2 t + g2 · 2 · H2 ( 1 , 1 ) · H2 ( 1 , ω2 ) · ej (3 1 +ω2 )t + g2 · 2 · H2 ( 1 , 1 ) · H2 (ω2 , ω2 ) · ej (2 1 +2ω2 )t +... TIME-DOMAIN SOLUTION 0 .15 0 .15 0 .1 0 .1 0.05 0.05 0 0 −0.05 −0.05 −0 .1 −0 .1 −0 .15 −0 .15 −0.2 0 0.2 0.4 0.6 0.8 1 1.2 1. 4 1. 6 1. 8 t (s) 2 −0.2 0 0.2 0.4 0.6 0.8 10 −9 1 t (s) 1. 2 1. 4 1. 6 1. 8 2 10 −9 Figure 1. 4 Currents and voltages in the example circuit for two different amplitudes of a sinusoidal input current With the view to illustrate, the time-domain solution of our example circuit is given for a... · H2 ( 1 , ω2 ) · ej ( 1 +ω2 )t + g2 · 2 · H1 ( 1 ) · H1 (ω2 ) · ej ( 1 +ω2 )t = 0 from which we get H2 ( 1 , ω2 ) = 2g2 · H1 ( 1 ) · H1 (ω2 ) g1 + j ( 1 + ω2 )C (1. 66) (1. 67) The second-order nucleus is an explicit expression that includes only the first-order nucleus, already found from eq (1. 65) (or equivalently from eq (1. 61) ) By substitution we get H2 ( 1 , ω2 ) = 2g2 (g1 + j 1 C) · (g1 + j ω2 . 7 −0.2 −0 .15 −0 .1 −0.05 0 0 0.05 0 .1 0 .15 0.2 0.4 0.6 0.8 1 1.2 1. 4 1. 6 1. 8 2 10 −9 t (s) t (s) 10 −9 −0.2 −0 .15 −0 .1 −0.05 0 0 0.05 0 .1 0 .15 0.2 0.4 0.6 0.8 1 1.2 1. 4 1. 6 1. 8 2 Figure 1. 4 Currents. production. Contents Preface ix Chapter 1 Nonlinear Analysis Methods 1 1 .1 Introduction 1 1.2 Time-Domain Solution 4 1. 2 .1 General Formulation 4 1. 2.2 Steady State Analysis 7 1. 2.3 Convolution Methods 9 1. 3 Solution Through. Expansion 13 1. 3 .1 Volterra Series 13 1. 3.2 Fourier Series 22 1. 3.2 .1 Basic formulation (single tone) 23 1. 3.2.2 Multi-tone analysis 33 1. 3.2.3 Envelope analysis 43 1. 3.2.4 Additional remarks 45 1. 3.2.5