and voltages would be sinusoids of the same frequency as that of the sources

And voltages would be sinusoids of the same frequency

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and voltages would be sinusoids of the same frequency as that of the sources. Since the equations to solve for the circuit voltages and currents are always linear in this case, these are of the form a 1 x 1 ( t ) + a 2 x 2 ( t ) + a 3 d dt x 3 ( t ) = a 4 x 4 ( t ) . (1.4) If x 1 ...... 3 are all sinusoids (or co-sinusoids), it is easy to show that x 4 is definitely a sinusoid. So, if x 1 ( t ) = X 1 cos( ωt + φ 1 ) = ℜ{ X 1 e j ( ωt + φ 1 ) } = ℜ{ ( X 1 e 1 ) e jωt } = ℜ{ ( X 1 ) e jωt } x 2 ( t ) = X 2 cos( ωt + φ 2 ) = ℜ{ X 2 e j ( ωt + φ 2 ) } = ℜ{ ( X 2 e 2 ) e jωt } = ℜ{ ( X 2 ) e jωt } x 3 ( t ) = X 3 cos( ωt + φ 3 ) = ℜ{ X 3 e j ( ωt + φ 3 ) } = ℜ{ ( X 3 e 3 ) e jωt } = ℜ{ ( X 3 ) e jωt } (1.5) with X 1 , X 2 , X 3 and φ 1 , φ 2 , φ 3 all constants, then, x 4 ( t ) = 1 a 4 bracketleftbigg a 1 ℜ{ X 1 e j ( ωt + φ 1 ) } + a 2 ℜ{ X 2 e j ( ωt + φ 2 ) } + a 3 d dt parenleftBig ℜ{ X 3 e j ( ωt + φ 3 ) } parenrightBig bracketrightbigg = 1 a 4 bracketleftbigg a 1 ℜ{ X 1 e jωt } + a 2 ℜ{ X 2 e jωt } + a 3 parenleftbigg d dt { X 3 e ( 3 + ωt ) } parenrightbiggbracketrightbigg = 1 a 4 bracketleftBig ℜ{ ( a 1 X 1 ) e jωt } + ℜ{ ( a 2 X 2 ) e jωt } + a 3 parenleftBig ℜ{ ( jωX 3 e 3 ) e jωt } parenrightBigbracketrightBig = 1 a 4 bracketleftbig ℜ{ ( a 1 X 1 ) e jωt } + ℜ{ ( a 2 X 2 ) e jωt } + ℜ{ ( jωa 3 X 3 ) e jωt } bracketrightbig = 1 a 4 bracketleftbig ℜ{ ( a 1 X 1 + a 2 X 2 + jωa 3 X 3 ) e jωt } bracketrightbig (1.6) which shows that x 4 is a sinusoid of frequency ω . Hence, expressing x 4 as x 4 ( t ) = X 4 cos( ωt + φ 4 ) = ℜ{ ( X 4 ) e jωt } 2 c circlecopyrt Dr. P. Sensarma, Department of Electrical Engg, IIT-Kanpur, India.
1.2. PHASORS it can be easily seen that ℜ{ ( X 4 ) e jωt } = bracketleftbigg { 1 a 4 ( a 1 X 1 + a 2 X 2 + jωa 3 X 3 ) } e jωt bracketrightbigg (1.7) Thus a complete and exact solution of the circuit equation (1.4) is obtained if the quantity X 4 is determined and the common frequency ω is known. Comparing both sides of equa- tion (1.7), it is obvious that X 4 is equal to the term inside the chain brackets on the right.

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