By Nicolaas Govert de Bruijn

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25E-6). / Ikn Figure 5. 1 - 6 Pole Butterworth (fco = 1 6 Hz) capacitance). 42E-6. Z I 3 is, of course, 0, 0, 0 because there is no mutual impedance between these loops (that is, the current flowing in loop 3 will introduce no voltage into loop I , so we multiply 13 by zero in the loop I equation. Similarly with Z14. 082E-6. 082E-6 is the series sum ofthe two capacitors in this loop! 94E-6 and Z24 is, again, 0, 0, o. Okay, you know how to input the rest of the data. [Note that we didn't have to input Z2 1 because it's identical to Z 1 2.

_ "; F; , PR I NT FREQUENCY 1 0 1 30 PR I NT U S I N G "##. ####_ "; F ( 1 , I ) * RTERM; , ? REAL COMPONENT 1 0 1 40 J S = "+j " : I F SGN ( F ( 2 , I » < 0 T HE N J S = " - j " 1 0 1 50 PR I NT " "; JS; 1 0 1 60 PR I NT US I NG "#. ####_ "; ABS( F ( 2 , I )*RTERM ) ; ' PR I NT I MAG I NARY 1 0 1 70 MAG = SQR ( F ( 1 , I ) A 2 + F ( 2 , I ) A 2 ) , F I ND MAGN I TUDE 1 0 1 80 I F MAG = 0 THEN PR I NT " 0 . 0000 10 . 00" : GOTO 1 0240 1 0 190 PR I NT US I NG "#. ####_ # "; RTERM * MAG; : PR I NT " I " ; 1 0200 I F F ( 1 , I ) = 0 AND F ( 2 , I ) >= 0 T HEN B T A = P I I 2 : GOTO 1 0230 1 02 1 0 I F F ( 1 , I ) = 0 AND F ( 2 , I ) < 0 T HEN BTA = - P I I 2: GOTO 1 0230 1 0220 BTA = ATN( F ( 2 , 1 )1 F ( 1 , I » : I F F ( 1 , 1 ) <0 T HEN BTA = BTA+P I 1 0230 PR I NT USING "### .

6. 1 - Transient/Steady State paramount. 1 BAND LIMITED SIGNALS We know, via Fourier analysis, that we may decompose signals into their harmonic components. , the width of the band offrequencies that make up any given signal), is referred as its bandwidth. Theoretically, discontinuous waveforms contain an infinite number of components. A perfect square wave, for example, requires an infinite number of harmonics, and this implies an infinite bandwidth. When working with real square waves, howeve� we will always disregard the harmonics above some arbitrarily high value after all, the harmonics are becoming smaller and smaller with increasing frequency.