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I'm reading through a chapter on sinusoidal alternating waveforms and I'm having some difficulty in the section on Phase Relations. The generic expression is mentioned below for a waveform that has been shifted. Following that I posted a list of expressions that show their geometric relationship. Can someone explain the list of expressions in your own context or understanding.

I do not understand these angles and/or functions. Their meaning as well as their context in phase shifting is confusing to me. I need a better foundation of understanding to have a better intuitive approach towards problem solving. I've had Calculus 1 and I'm currently taking Calculus 2. Therefore, I'm familiar with trig functions but I just can't bring it all together.

I posted this question in the electronics forum for a better engineering standpoint. However, I feel that I also need a mathematical point of view also.

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A problem that is associated with this is as follows......

What is the phase relationship between the sinusoidal waveforms?

NOTE:

(w) is Angular Velocity

(t) is time

(i) is a instantaneous value of current

(v) is a instantaneous value of voltage

Also numbers inside parenthesis are in degrees

i = -2cos(wt-60)

v = 3sin(wt-150)

Their solution and answer is as follows. This is the part I'm not understanding at all.

SOLUTION:

i = -2cos(wt-60) = 2cos(wt-60-180) = 2cos(wt-240)

2cos(wt-240) = 2sin(wt-240+90) = 2sin(wt-150)

ANSWER:

v and i are in phase

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I find it curious that they went through as many steps as they did. You have $$i=-2\cos(wt-60^\circ) \\ v=3\sin(wt-150^\circ)$$ Now use $\cos \alpha=\sin (\alpha+90^\circ)$ in the first to get $$i=-2\sin(wt+30^\circ)$$ and $\sin \alpha=-\sin (\alpha-180^\circ)$ in this to get $$i=2\sin(wt-150^\circ)$$ Because both $v$ and $i$ have a sine wave with a phase of $-150^\circ$ they are in phase.

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  • $\begingroup$ Can you perhaps elabrate briefly on why you used sin(a+90) to do this? $\endgroup$ – Shane Yost May 8 '13 at 0:02
  • $\begingroup$ @ShaneYost: because I wanted to make the trig functions match. Originally they were one $\cos$ and one $\sin$, this made them both $\sin$. I could have done it in one step using $-\cos \alpha = \sin (\alpha-90^\circ)$ $\endgroup$ – Ross Millikan May 8 '13 at 0:12

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