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Hello there I am trying to understand this proof: foo. I dont understand why $$\cos{(2 \omega t +\phi)} $$ vanishes when it's integrated between 0 and T. When I try to do this alone I get: $$V_{rms}=\sqrt{\frac{V_{pk}^2}{2}-\frac{V_{pk}^2}{(2T)(2\omega)}(\sin{(2\omega T+ 2\phi)-\sin{(2\phi))}}}$$ Can I get some help to understand this please?

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  • \$\begingroup\$ Can we assume \$T=2\pi/\omega\$? \$\endgroup\$ – The Photon Apr 23 '18 at 16:09
  • \$\begingroup\$ yes sir, that is right! \$\endgroup\$ – Zacky Apr 23 '18 at 16:12
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I dont understand why $$\cos{(2 \omega t +\phi)} $$ vanishes when it's integrated between 0 and T.

Because regardless of what \$\phi\$ is, if \$T\$ is the period of the sine wave (or an integer multiple of the period), then within the time \$T\$ the sine wave spends half it's time below 0 and half the time above 0 (and the shape is indeed symmetric).

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  • \$\begingroup\$ Thank you sir, makes sense! Mind If I ask how the integral arises to evaluate this? \$\endgroup\$ – Zacky Apr 23 '18 at 16:17
  • \$\begingroup\$ @Zacky, it should be easily evaluated using the usual rules for integrals. (i.e., antideriative of cosine is negative sine, and the symmetry will give you two opposite values of the sine at 0 and T). \$\endgroup\$ – The Photon Apr 23 '18 at 16:23
  • \$\begingroup\$ Yes I know to evaluate the integral, I am curious how it appears here. \$\endgroup\$ – Zacky Apr 23 '18 at 16:27
  • \$\begingroup\$ @Zacky, there is an identity \$\sin^2 \theta \equiv \frac{1-\cos 2\theta}{2}\$. \$\endgroup\$ – The Photon Apr 23 '18 at 16:32
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Even though its intuitive why it's zero. We can just integrate to find that result.

Solution:

$$\int_0^Tcos(2\omega t+\phi)$$ $$=\int_0^Tcos(2.\frac{2\pi}{T} t+\phi)$$ $$=\int_0^Tcos(\frac{4\pi}{T} t+\phi)$$ $$=\left[\frac{sin(\frac{4\pi}{T}t+\phi)}{4\pi/T}\right]^T_0$$ $$=\frac{sin(4\pi+\phi)-sin\phi}{\frac{4\pi}{T}}$$ $$=\frac{sin4\pi.cos\phi+sin\phi.cos4\pi-sin\phi}{\frac{4\pi}{T}}$$ $$= \frac{0.cos\phi+sin\phi.1-sin\phi}{\frac{4\pi}{T}}$$ $$= \frac{sin\phi-sin\phi}{\frac{4\pi}{T}} $$ $$=0$$

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  • \$\begingroup\$ nice, I didnt thought to use the alternate version for $\omega$. Thank you alot! \$\endgroup\$ – Zacky Apr 23 '18 at 16:29

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