Harmonics in power system can be basicly introduced as the unintended frequency components, generally higher than system frequency, of current or voltage in the power system. The sources of current harmonics and the harm of them to the power system is stated in an article about Active Power Filters in the blog, thus; I will skip that part. Instead of the drowning in details, I will mention the case of two very closely located harmonics and their effect on harmonic spectrum. Almost everyone who is dealing with power system signal processing can know that the close harmonics tend to increase the magnitude of frequencies between them in FFT spectrum. However, the point that I will draw attention will be a different side of this situation, actually, the behaviors while they are approaching each other to almost 2 Hz.
Some loads in Metal Industry behaves very different than other widely known industrial load. For example, arc furnaces, laddle furnaces or induction furnaces for metal melting purposes have very distructive behaviours to the power system. For the introduction of this article, I will briefly mention some behaviours of induction furnaces. For one of the induction furnaces that I take measurements with my collegues in the field, We recognised that the load creates harmonic spectrum other than the characteristics of any other load in industry. The induction furnace creates variety of interharmonics whose frequency value is changing dynamically between 280 Hz and 550 Hz. In other words, there are not fixed frequency harmonics, there can be for example 312 Hz harmonic currents, while the harmonic was in 290 Hz 10 minutes before 312 Hz. The harmonic frequencies are changing during the melting process. I and my collegues are recognised that there can be two harmonics ,which are 2 Hz apart each other, at the same time. Let me know explain what happens at this case.
As you may know the whole current or voltage signal can be expressed as the sum of fundemental and harmonics frequency sine waves. Therefore, for our example ,for the sake of simplicity, we can consider the sum of 282 Hz and 280 Hz harmonic components whose magnitudes are A and B, respectively. Considering the fundemental Cosine Sum formulation in equation (1), we can apply the rule to our power system signal to get equation (2).
For the sake of simplicity, we can consider that the phases of our harmonics are equal to zero in order to reach equation (3). If we consider that the magnitudes A and B of the harmonic components are equal to each other as A, we can reach to equation (4).
When we apply our example to the equation (3) , by putting the frequency values 282 and 280 Hz, we can reach to the result of equation (5). Similarly if we consider that the magnitudes of the harmonics are the same at A, the equation (6) is reached. What is the interpretation of the equation (5) ? Equation (5) means a 282 Hz signal whose amplitude is modulated with a sine wave of 1 Hz. The physical meaning of this situation is that the 282 Hz harmonic component has high magnitude during 0.5 seconds and low magnitude in the following 0.5 seconds. Actually, if you have an active or passive filter for filtering this harmonic, you can hear an audible voice whose intensity is increasing and decreasing at every 0.5 seconds. What if the magnitudes are the same for two harmonics ? This causes the most severe result. The magnitude of the harmonic is increased up to double and decreased down to zero at a frequency of 1 Hz. Similarly, the physical meaning of this situation in active or passive filter case gives that the audible voice is extremely increased and totally ceased with a period of 0.5 seconds.
Now let me verify this situation using MATLAB.
MATLAB can be get from Amazon at link,
http://www.amazon.com/MATLAB-SIMULINK-Student-Version-R2007a/dp/0979223903/ref=sr_1_1?s=software&ie=UTF8&qid=1364394510&sr=1-1&keywords=matlab+software
Used MATLAB code for different magnitudes;
t = 0:1.5625e-4:1; % 6400 samples/second sampling frequency
A = 1; % magnitude of 282 Hz component
B = 0.3; % magnitude of 280 Hz component
X = A*cos(2*pi*282*t) + B*cos(2*pi*280*t); % Signal to be analysed
plot(X); % plotting the resultant waveform
Figure 1: Resultant Waveform with A = 1 and B = 0.3
Figure 2: Resultant Waveform(zoomed) with A = 1 and B = 0.3
Used MATLAB code for same magnitudes;
t = 0:1.5625e-4:1; % 6400 samples/second sampling frequency
A = 1; % magnitude of 282 Hz component
B = 1; % magnitude of 280 Hz component
X = A*cos(2*pi*282*t) + B*cos(2*pi*280*t); % Signal to be analysed
plot(X); % plotting the resultant waveform
Figure 3: Resultant Waveform with A = 1 and B = 1
Figure 4: Resultant Waveform with(zoomed) A = 1 and B = 1
