K:21) Find the steady state souti om respongPo GayadRdg + 19 = Vs (t)a,bwhereVs = loo Cos2t + 25 Sin 2t + 2ou Cos4t +25Sin4tand R= S0SL and C=0.04 F%3DSteady state soh9= A Sin 24 +B Cos Zt + c Sinut + D Cos4tVs vs t1 vsヒP lotuse ExceI,

Answered: Find the stFeady stete Sountio…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

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K:21)

Find the steady state souti om respong
Po Gayad
Rdg + 19 = Vs (t)
a,b
where
Vs = loo Cos2t + 25 Sin 2t + 2ou Cos4t +25Sin4t
and R= S0SL and C=0.04 F
%3D
Steady state soh
9= A Sin 24 +B Cos Zt + c Sinut + D Cos4t
Vs vs t
1 vsヒ
P lot
use ExceI,

Expert Solution

Step 1

The equation are given as

$R \frac{d q}{d t} + \frac{1}{C} q = V_{s} (t) 50 \frac{d q}{d t} + \frac{1}{0.04} q = 100 \cos 2 t + 25 \sin 2 t + 200 \cos 4 t + 25 \sin 4 t 50 \frac{d q}{d t} + 25 q = 100 \cos 2 t + 25 \sin 2 t + 200 \cos 4 t + 25 \sin 4 t \frac{d q}{d t} + 0.5 q = 2 \cos 2 t + 0.5 \sin 2 t + 4 \cos 4 t + 0.5 \sin 4 t$

Complementary function

$\frac{d q}{d t} + 0.5 q = 0 D + 0.5 = 0 D = - 0.5$

So,

$C F = C_{1} e^{- 0.5} t$

Step 2

Particular solution

$\begin{array}{rcl} P I & = & \frac{1}{D + 0.5} \times (2 \cos 2 t + 0.5 \sin 2 t + 4 \cos 4 t + 0.5 \sin 4 t) \\ = & \frac{D - 0.5}{D^{2} - 0.25} \times (2 \cos 2 t + 0.5 \sin 2 t + 4 \cos 4 t + 0.5 \sin 4 t) \\ = & \frac{(D - 0.5)}{(- 4 - 0.25)} (2 \cos 2 t) + \frac{(D - 0.5)}{(- 4 - 0.25)} (0.5 \sin 2 t) + \frac{(D - 0.5)}{(- 16 - 0.25)} (4 \cos 4 t) + \frac{(D - 0.5)}{(- 16 - 0.25)} (0.5 \sin 4 t) \\ = & - 0.47 (D - 0.5) (\cos 2 t) - 0.11 (D - 0.5) \sin 2 t - 0.25 (D - 0.5) \cos 4 t - 0.03 (D - 0.5) \sin 4 t \\ = & - 0.47 (- 2 \sin 2 t - 0.5 \cos 2 t) - 0.11 (2 \cos 2 t - 0.5 \sin 2 t) - 0.25 (- 4 \sin 4 t - 0.5 \cos 4 t) - 0.03 (4 \cos 4 t - 0.5 \sin 4 t) \\ = & 0.94 \sin 2 t + 0.235 \cos 2 t - 0.22 \cos 2 t + 0.055 \sin 2 t + \sin 4 t + 0.125 \cos 4 t - 0.12 \cos 4 t + 0.015 \sin 4 t \\ = & 0.995 \sin 2 t + 0.015 \cos 2 t + 1.015 \sin 4 t + 0.005 \cos 4 t \end{array}$

So, complete solution

$q =$ $C_{1} e^{- 0.5} t$ + $0 \begin{array}{rcl} . \end{array} \begin{array}{rcl} 995 \end{array} \begin{array}{rcl} s i n \end{array} \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} t \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} 0 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 015 \end{array} \begin{array}{rcl} c o s \end{array} \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} t \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} 1 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 015 \end{array} \begin{array}{rcl} s i n \end{array} \begin{array}{rcl} 4 \end{array} \begin{array}{rcl} t \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} 0 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 005 \end{array} \begin{array}{rcl} c o s \end{array} \begin{array}{rcl} 4 \end{array} \begin{array}{rcl} t \end{array}$

In steady state, exponential part become zero, so

$q_{s s} = 0 \begin{array}{rcl} . \end{array} \begin{array}{rcl} 995 \end{array} \begin{array}{rcl} s i n \end{array} \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} t \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} 0 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 015 \end{array} \begin{array}{rcl} c o s \end{array} \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} t \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} 1 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 015 \end{array} \begin{array}{rcl} s i n \end{array} \begin{array}{rcl} 4 \end{array} \begin{array}{rcl} t \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} 0 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 005 \end{array} \begin{array}{rcl} c o s \end{array} \begin{array}{rcl} 4 \end{array} \begin{array}{rcl} t \end{array}$

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