Amplitude spectrum of signal, power of signal etc

[etf]

Hi!
Here is my task:

1. Homework Statement

Periodic signal is input signal for circuit (photo).
a) Calculate and sketch amplitude spectrum of $$u(t+1ms)$$,
b) Calculate participation (in percent) of DC component of power of total mean power of signal $$u(t)$$,
c) Sketch power spectrum of signal $$u1(t)=u(t)-1V$$
d) Calculate power of third harmonic on output

Homework Equations

The Attempt at a Solution

[/B]
I represented $$u(t)$$ in terms of complex Fourier series as $$u(t)=\frac{E}{2}+\sum_{n=-\infty ,n\neq 0}^{n=\infty }F_ne^{jnw0t},$$ where $$F_n=\frac{E}{2n\pi }j,$$ $$T=4\pi *10^{-3}s,$$ $$w0=\frac{2\pi }{4\pi *10^{-3}},$$ $$E=1V.$$ Then I used time shift property of Fourier series to represent $$u(t+1ms)$$. I got $$u(t+1ms)=\frac{E}{2}+\sum_{n=-\infty,n\neq 0}^{n=\infty}F_ne^{-jnw0*0.001}e^{jnw0t}$$ ($$F_n$$ is same as for $$u(t)$$. Then I sketch it so task a) is completed. For task b) I calculated total mean power of signal as (if it's correct) $$P=\lim_{t->\infty}\frac{1}{T}\int_{\tau}^{\tau+T}(f(t))^{2}dt=\lim_{t->\infty}\frac{1}{T}\int_{\tau}^{\tau+T}(\frac{Et}{T})^{2}dt=...=\frac{E^{2}}{3}=\frac{1}{3}$$. What would be participation in percent of DC component? How to do other tasks?

 

[etf]

After substitution of E, T and w0 I get:
$u(t)=\frac{1}{2}+\sum_{n=-\infty,n\neq 0}^{n=\infty}\frac{1}{2n\pi}e^{j\frac{\pi}{2}}e^{jn*500t},$ $u(t+1ms)=\frac{1}{2}+\sum_{n=-\infty,n\neq 0}^{n=\infty}\frac{1}{2n\pi}e^{j(\frac{pi}{2}-\frac{n}{2})}e^{jn*500t}$

 

[lazyaditya]

You have calculated the Fourier series and also know the dc component present , calculate the power due to dc component . Contribution of dc power in total power will be their ratio, to express in percentage multiply it with 100.

 

[lazyaditya]

In order to calculate power of third harmonic on output, consider 3rd harmonic of your Fourier series as input signal and calculate output voltage considering transfer function of circuit ( reactance calculated with w=3w₀). Now you have output function you can calculate power.

 

[etf]

My third harmonic of input signals is $-\frac{1}{3\pi }\sin{(1500t)}$. It's representation in complex form is $\frac{1}{3\pi }e^{j*0}$, but I divided it with $sqrt(2)$ to get rms...
I put it in input and solved circuit using complex analysis. I got current $\underline{I}=\frac{1}{R-\frac{j}{wC}}\frac{1}{3\pi \sqrt{2}}=...=5.1941*10^{-5}+3.4628*10^{-5}j,$$\underline{Viz}=-\frac{j}{wC}\underline{I}=...=0.0231-0.0346j.$. Now I can calculate complex apparent power on output (on capacitor) as $\underline{S}=\underline{Viz}\underline{I^{*}}$ (I* is complex conjugate of I). Real part would be active and imaginary part reactive power?

 

[lazyaditya]

I have not understood your question correctly. Do you want the power supplied to capacitor or the power of the output signal due to third harmonic.

 

[lazyaditya]

The way you are trying to solve is calculating power supplied to capacitor . The power calculated will be volt ampere reactive power and thus 90 degree out of phase from real power across resistor.

Solve "VI*" and see that real part will cancel out only imaginary part would remain.

 

[lazyaditya]

And if you want to calculate the power in the output voltage signal you have to work as I explained before.

 

[etf]

I have not understood your question correctly. Do you want the power supplied to capacitor or the power of the output signal due to third harmonic.

Output on scheme, uiz(t), is voltage on capacitor, so by "Calculate power of third harmonic on output" they mean power on capacitor (output) with third harmonic in input?

 

[lazyaditya]

Yup that what I thought at 1st glance, since they haven't asked the power supplied to capacitor.

 

[etf]

Ok, I did it. Here is how I try to find power:
Average power of periodic signal can be calculated using relation $Ptotal=\sum_{n=-\infty}^{n=\infty}|Fn|^{2}$. Using Parseval's theorem, it is equal to $\frac{1}{T}\int_{\tau}^{\tau +T}(f(t))^{2}dt$. For my signal, $Ptotal=\frac{1}{T}\int_{\tau}^{\tau +T}(f(t))^{2}dt=\frac{1}{T}\int_{0}^{T}(\frac{Et}{T})^{2}dt=\frac{E}{T^{3}}\frac{T^{3}}{3}=\frac{E}{3} (E=1)=\frac{1}{3}.$ For DC component I have:$\frac{1}{T}Pdc=\int_{0}^{T}(\frac{1}{2})^{2}dt=\frac{1}{4}.$ Is it ok?

 

[etf]

I made typo, It should be Pdc=(1/T)*integral... But it doesn't affect solution :)

 

[lazyaditya]

Ok, I did it. Here is how I try to find power:
Average power of periodic signal can be calculated using relation $Ptotal=\sum_{n=-\infty}^{n=\infty}|Fn|^{2}$. Using Parseval's theorem, it is equal to $\frac{1}{T}\int_{\tau}^{\tau +T}(f(t))^{2}dt$. For my signal, $Ptotal=\frac{1}{T}\int_{\tau}^{\tau +T}(f(t))^{2}dt=\frac{1}{T}\int_{0}^{T}(\frac{Et}{T})^{2}dt=\frac{E}{T^{3}}\frac{T^{3}}{3}=\frac{E}{3} (E=1)=\frac{1}{3}.$ For DC component I have:$\frac{1}{T}Pdc=\int_{0}^{T}(\frac{1}{2})^{2}dt=\frac{1}{4}.$ Is it ok?

I got the same values.