3cos50t+4sin50t by adding sinx to cosx how does phase shift?
- Thread starter eee
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If the above is the exact text of the exercise, then I think it's no wonder you're confused! :shock:eee said:
Please consult with your instructor regarding a corrected statement of the exercise, including what the actual question is (what you're supposed to do) and how "x" and "t" are related. Thank you!
Eliz.
The question is i(t)=3cos50t + 4sin50t amps, then the question is what is the phaser representation. The answer is a phaser written as 5 at an angle of -53.13 degrees in amps. In the s domain, it is plus or minus j50rad/s. I don't see where the -53.13 degrees phase shift comes from in the phaser answer. This is a sinusoidal electrical current being transformed from a time function to a phaser. I see from pythagoreans theorem that the squre root of (3 squared + 4 squared) = 5, but still can't get the phase shift angle of -53.13 degrees. I see the phase shift easily when I plot it and add the sine to the cosine on the graph, but I can't see how it is arrived at mathematically.
By solving this problem in the reverse direction going from 5<-53 degrees (the phasor) to 5e**-53t to 5cos53.13 degrees + 5jsin53.13 degrees to 3 + j4, I found that the arctangent of 4/3 is 53.13 degrees and this is the correct answer. Why that this problem does not care about the frequency or the fact that a sine wave is added to a cosine wave is weird and understandable to me. Atleast I see how it is solved mathematically, just don't understand the mathematical theory.
D
Deleted member 4993
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If you are given:
\(\displaystyle A\cdot \cos(\theta) \, + \, B\cdot \sin(\theta)\)
then define:
\(\displaystyle \cos(\phi) \, = \, \frac{A}{\sqrt{A^2 \, + \, B^2}}\)
and
\(\displaystyle \sin(\phi) \, = \, \frac{B}{\sqrt{A^2 \, + \, B^2}}\)
and
\(\displaystyle \tan(\phi) \, = \, \frac{B}{A}\)
then
\(\displaystyle A\cdot \cos(\theta) \, + \, B\cdot \sin(\theta) \,\)
\(\displaystyle = \, \sqrt{A^2 \, + \, B^2}[\frac{A}{\sqrt{A^2 \, + \, B^2}}\cdot \cos(\theta) \, + \, \frac{B}{\sqrt{A^2 \, + \, B^2}}\cdot \sin(\theta)]\)
\(\displaystyle = \, \sqrt{A^2 \, + \, B^2}[\cos(\phi)\cdot \cos(\theta) \, + \, \sin(\phi)\cdot \sin(\theta)]\)
\(\displaystyle = \, \sqrt{A^2 \, + \, B^2}[\cos(\theta \, + \, \phi)]\)
\(\displaystyle A\cdot \cos(\theta) \, + \, B\cdot \sin(\theta)\)
then define:
\(\displaystyle \cos(\phi) \, = \, \frac{A}{\sqrt{A^2 \, + \, B^2}}\)
and
\(\displaystyle \sin(\phi) \, = \, \frac{B}{\sqrt{A^2 \, + \, B^2}}\)
and
\(\displaystyle \tan(\phi) \, = \, \frac{B}{A}\)
then
\(\displaystyle A\cdot \cos(\theta) \, + \, B\cdot \sin(\theta) \,\)
\(\displaystyle = \, \sqrt{A^2 \, + \, B^2}[\frac{A}{\sqrt{A^2 \, + \, B^2}}\cdot \cos(\theta) \, + \, \frac{B}{\sqrt{A^2 \, + \, B^2}}\cdot \sin(\theta)]\)
\(\displaystyle = \, \sqrt{A^2 \, + \, B^2}[\cos(\phi)\cdot \cos(\theta) \, + \, \sin(\phi)\cdot \sin(\theta)]\)
\(\displaystyle = \, \sqrt{A^2 \, + \, B^2}[\cos(\theta \, + \, \phi)]\)