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I am working on a lab in school with a common source JFET amplifier, and I needed to express \$v_{out}\$ in terms of time domain. If \$f\$ is the frequency and \$T\$ is the period, then I need $$v_{L}(t)=V_{m}\sin(2\pi f T \pm \theta)$$ Now this is just a concept question, when I plug in the numbers for the frequency and multiply that by 360, then multiply that by the period what is the next step? I think you evaluate; put into the calculator '\$\sin(x)\$' (whatever \$x\$ is) and then multiply that answer by \$V_{m}\$ or peak to peak of the system \$v_{out}\$. Don't worry about if I am at mid band or high frequency I just need to verify if what I am doing is right cause my numbers are iffy.

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  • \$\begingroup\$ What is "ifiy"? \$\endgroup\$ – diverger Nov 7 '14 at 5:50
  • \$\begingroup\$ frequency * period? \$\endgroup\$ – alex.forencich Nov 7 '14 at 7:42
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No, you are incorrect:

Time domain equation for a sin wave is:

$$v(t)= V_M \sin(2\pi f\:t+\theta) = V_M \sin\left(\frac{2\pi}{T}t+\theta\right)$$

Where:

  • \$V_M\$ is the amplitude of the sin wave, (relate to the peak to peak and RMS: \$V_M=\frac{V_{pk-pk}}{2}= \sqrt{2} V_{RMS}\$ )
  • \$f\$ is the frequency, in Hz
  • \$T\$ is the period in seconds (\$T=\frac{1}{f}\$)
  • \$\theta\$ is a phase shift in radians.
  • \$t\$ is the time in seconds.

So for example, if we pass Australia's mains grid power: 50Hz, \$V_{RMS}=240\text{V} \implies V_M=339.4\$V though a suitable RLC circuit that causes a phase shift of \$60^\circ=\frac{\pi}{3}\$ radians, the output would be: $$v(t)=339.4\sin(2\pi\times 50 \times t + \frac{\pi}{3})$$ This is a expression in terms of time -- the voltage is not constant, it goes up and down with time. You need to evaluate it at some fixed point in time to find some voltage.

Which if we wanted to know the voltage after 3 seconds: \$v(3)=339.4\sin(2\pi\times 50 \times 3 + \frac{\pi}{3})=293.9\$V.

The thing you had wrong was you were using the frequency form of the equation, but instead of putting in the time you wished to evaluate at, you put in the period.

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  • \$\begingroup\$ ok. when I do it your way it works but lets say I want to convert 2pi into 360 degrees, and then solve it. How would I do it that way or in degree mode as I think its called? \$\endgroup\$ – cloudhalo Nov 7 '14 at 19:27

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