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I hope someone here could clarify for me. With the waveform function f()=(A)sin(wt + theta)

where

A=magnitude of the wave,

sin= type of wave form,

wt=frequency rads/sec,

theta = phase shift

...isn't wt + theta just giving us a different frequency once we replace wt and theta with values?
If I have a frequency of 6000 Rad/s and I add a phase shift of 45 then I have a frequency of 6045 Rad/s which is not the same as 6000 Rad/s yet the theory I'm told says it is, only offset by 45 degrees. What am I not getting?

Chef Flambe
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2 Answers2

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Just think about it like a transformation of a function $f(x)$: the function $f(ax+b)$ has been compressed or stretched horizontally by a factor of $a$, and shifted horizontally by $b/a$, either left or right depending on the sign of $b$. In your example, the horizontal compression/stretch factor $w$ gives you the frequency, and the phase factor $\theta$ tells you how much the function is shifted horizontally. The shift doesn't change the stretch though, so the frequency stays the same.

To further convince yourself, try graphing $\sin(2t)$, $\sin(t-\pi)$ and $\sin(2t-\pi)$.

icurays1
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Seems likely that what you really want to write is $f(t)=A \sin (\omega t +\theta)$. This is a function of $t$. The frequency is $\omega$, not $omega t$. The extra angle $\theta$ changes the precise angle at the beginning of the process, not the frequency. It "shifts" your sine function to the left in the $t$ axis. To understand the function better, you could use matlab, mapple, or mathematica, to plot the function you are interested (with perhaps a smaller value of $\omega$) and change the angle $\theta$ in different plots.

F. Solis
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