componentsdcpower
My task is to calculate participation (in percent) of DC component in total average power of signal from photo:
I represented this signal as complex Fourier series: $$u(t)=\frac{1}{2}+\sum_{n=-\infty,n\neq 0}^{n=\infty}\frac{1}{2n\pi}e^{j\frac{\pi}{2}}e^{jn*500t}.$$
Average power of signal is calculated using formula $$P=\lim_{T->\infty}\frac{1}{T}\int_{\tau}^{\tau+T}(f(t))^{2}dt$$ but how should I use it here? And how to calculate participation of DC component?
Thanks in advance.
I am confused on the difference between apparent, average and reactive power and how each is calculated
Average power is the real power you get billed for as kWh and, is the numerical multiplication of voltage and current waveforms, often done in an MCU these days but back in the old days a thing called a wattmeter did things magnetically then, along came analogue multiplier chips and solid state magnetic multipliers as patented by Landis & Gyr. This is all about four quadrant multiplication of the instantaneous signals and not to be confused with either of the following: -
Apparent power is RMS voltage x RMS current and only equals average power (above) on a purely resistive load. For pure sinewaves: -
\$Power_{Average} = Power_{Apparent}\cdot Power\space Factor\$
i.e. \$P=V\cdot I\cdot Cos{\phi}\$
Where power factor is the cosine of the phase angle between voltage and current - clearly when that angle is zero, power factor = 1.
Reactive power is \$\sqrt{Apparent^2 - Average^2}\$: -
See also this site for further details.
An integral is a sum like a + b + c... a, b,c are the values of the function in the part that you are integrating.
Where the value does not exist, it has the value 0. Adding 0 to a sum will not change that sum.
You already take that into account.
You say that the integral in the formula is \$\int_{\frac{-T}{2}}^{\frac{T}{2}}\$ and that \$T=\pi\$, which would lead to \$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\$, but actually the image shows that T = \$\frac{\pi}{2}\$, hence and the substitution is as you have written: \$\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\$. This adjusts the lower and upper bound so that the values that are summed up by the integral are those different from 0, because integrating those does not add anything to the integral.
That's why it is wrong to divide by two. In general: if you have a formula for something, use that formula, don't add more calculation to it, because that changes the formula and the result is something else.
Best Answer
This question is likely asking you to recall Parseval's identity, which is stated as
$$\sum_{n=-\infty}^\infty |c_n|^2 = \dfrac{1}{2\pi}\int_{-\pi}^\pi |f(x)|^2 \, dx$$
where \$c_n\$ are the coefficients of the Fourier series of a periodic function \$f(x)\$ with period \$2\pi\$.
You can see that the right-hand side of the identity is proportional to the average power of the signal. So, with some manipulation of the scaling factors, you can use this to assign a portion of the power of the single to each of the fourier components.
Note 1: I assume that your signal repeats on every interval of time T (that is, \$f(t+T)=f(t)\$). The way you've drawn it,it looks like the signal is 0 outside the region \$0 < t < T\$, but that would make it a non-periodic signal and it wouldn't make sense to talk about its Fourier series representation or average power.
Note 2: Notice that in your formula for the average power, you are have two different variables named T. One which you take to the limit of infinity, and one built in to the definition of the function f. Since f is periodic with period T, you don't need to take a limit and you can just use
$$ P = \dfrac{1}{T}\int_\tau^{\tau+T} f^2(t)\, dt$$
for any \$\tau\$ of your choice.