Multi Sine Power

I will show the superposition method. I insisted on doing the time-domain analysis as a double-check, as @Karan Chatrath has shown. There are three separate frequencies in the source voltage signal, so model three separate circuits independently; one for each frequency. Add the powers from all three circuits to get the total power. The resistance value remains the same at all frequencies, but the inductive reactance changes. Calculate the inductive reactances at the different frequencies:

$X_1 = \omega L \\ X_2 = 2 \omega L \\ X_3 = 3 \omega L$

Define the circuit impedances:

$Z_1 = R + j X_1 \\ Z_2 = R + j X_2 \\ Z_3 = R + j X_3$

Currents and voltages.

$V_1 = 10 + j0 \\ V_2 = 10 + j0 \\ V_3 = 10 + j0 \\ I_1 = \frac{V_1}{Z_1} \\ I_2 = \frac{V_2}{Z_2} \\ I_3 = \frac{V_3}{Z_3}$

Power values:

$P_1 = Re(V_1 I_1^*) = 50 \\ P_2 = Re(V_2 I_2^*) = 20 \\ P_3 = Re(V_3 I_3^*) = 10 \\ P_1 + P_2 + P_3 = 80$

The code below implements the steady state solution in the first part and the transient analysis in the second part. Both methods yield the same result.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87

import math

VSrms = 10.0

f = 60.0
w = 2.0*math.pi*f
T = 1.0/f

R = 1.0
L = 1.0/w

X1 = w*L
X2 = 2.0*X1
X3 = 3.0*X1

Z1 = complex(R,X1)
Z2 = complex(R,X2)
Z3 = complex(R,X3)

I1 = VSrms/Z1
I2 = VSrms/Z2
I3 = VSrms/Z3

I1conj = complex(I1.real,-I1.imag)
I2conj = complex(I2.real,-I2.imag)
I3conj = complex(I3.real,-I3.imag)

S1 = VSrms*I1conj
S2 = VSrms*I2conj
S3 = VSrms*I3conj

P1 = S1.real
P2 = S2.real
P3 = S3.real

print P1
print P2
print P3
print (P1+P2+P3)

print ""
print ""

#50.0
#20.0
#10.0
#80.0

###############################################################
###############################################################
###############################################################

VSmag = 10.0*math.sqrt(2.0)

dt = 10.0**(-6.0)

t = 0.0
count = 0

tf = 1000.0*T

I = 0.0
Id = 0.0

ES = 0.0

while t <= tf:

    I = I + Id*dt

    VS = VSmag*math.sin(w*t) + VSmag*math.sin(2.0*w*t) + VSmag*math.sin(3.0*w*t)

    Id = (VS - R*I)/L

    PS = VS*I

    ES = ES + PS*dt

    t = t + dt
    count = count + 1

    #if count % 100 == 0:
        #print (t/T),PS

print (ES/tf)

#80.0232187952

I took a numerical brute force approach to solve this problem. The circuit equation for this circuit is:

$L\dot{I} + IR = V_S$ $I(0) = 0$

IN this circuit, only the resistor is capable of dissipating power. The instantaneous power dissipated by the resistor is:

$P(t) = I(t)^2R$

This instantaneous power has been averaged out over a longer duration of time.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35

clear all
clc

% Parameters:
w    = 120*pi;
L    = 1/w;
R    = 1;

% Time initialisation:
dt   = 1e-6;
tf   = 50;
t    = 0:dt:tf;

% Initial condition:
I(1) = 0;

for k = 1:length(t)-1

    % Source Voltage:
    Vs = 10*sqrt(2)*(sin(w*t(k)) + sin(2*w*t(k)) + sin(3*w*t(k)));

    % Instantaneous power dissipated by resistor:
    P(k) = I(k)^2*R;

    % Kirchoff's voltage law applied and solving for current derivative:
    dI = (Vs - I(k)*R)/L;

    % Numerical integration:
    I(k+1) = I(k) + dt*dI;
end

% Average power dissipated:
Answer = mean(P)

% Answer = 80.0265 W