# Frequency conversion

Discussion in 'Homework Help' started by marcodigio, Feb 25, 2011.

1. ### marcodigio Thread Starter New Member

Feb 25, 2011
3
0
Hello guys,
I am stuck with a task I have to do.
I know how to dao the task itself, find the currents in an impedance network.
All impedances are given in rectangular format ie. 3-j5 but the voltage source is given as V=15V f=100Hz
Before I start i need to convert the voltage source to rectangular form.
I know how to convert polar to rectangular but I have no experience for frequency in hertz.
I tried to do the following but i'm not sure if it is right.
w(omega)=2pf= 628.318 rad
next i converted to degrees deg=rad*180/p = 36000
when converted to rectangular it just comes back as 15+j0
This does not seem right to me, the imaginary part just doesn't exist?
Any help would be much appreciated!
thanks
marco

2. ### Kermit2 AAC Fanatic!

Feb 5, 2010
4,165
1,120
Correct if the source is assumed to be a perfect sinewave

3. ### marcodigio Thread Starter New Member

Feb 25, 2011
3
0
thanks for your reply!
sometimes the solution is the easy answer!

4. ### Heavydoody Active Member

Jul 31, 2009
140
11
While your answer is correct (possibly), you have come about it incorrectly. You have made a critical error in your units. ω is the angular frequency in radians/second, not the phase angle in radians. The voltage in the time domain would be expressed as a sine or cosine as:
$V_msin(\omega t+ \phi)$, where $\phi$ is the phase angle. Phase angles are like voltages in that they require a reference. This reference is typically the phase angle of the voltage source, with a phase angle of zero. Which is how you inadvertantly came up with the correct answer. The phasor representation of a sinusoid consists of the amplitude ($V_m$ in this case) and phase angle ($\phi$) in degrees for polar format.

5. ### marcodigio Thread Starter New Member

Feb 25, 2011
3
0
Hi Heavydoody...
I have tried to convert to time domain using $image=http://forum.allaboutcircuits.com/mimetex.cgi?V_msin(\omega+t++\phi)&hash=57a5e0a9c3a4d69e4c6f65582d45191d$, but missing t and the phase angle i got stuck. That's why I was trying to find another way to get to it.
Could you please clarify why in this case I haven't got a phase angle?
Does the 100Hz not come into play at all?
many thanks!
marco

6. ### Heavydoody Active Member

Jul 31, 2009
140
11
I am a little shaky on some of this myself, but let me try to explain. First, I was simply using the time domain representation so you could see the difference between omega and phi. You were correct in converting to the phasor (AKA frequency) domain, which is what you were doing by obtaining the rectangular form of the phasor representation of your voltage.

Phasors are not time dependent and, thus, contain no t. Furthermore, since only phasors of the same frequency can be manipulated together, omega is suppressed in their mathematical representation. So, no, omega does not come into play at this point, other than making sure that all signals being analyzed are of the same frequency. All you need is the magnitude (15) and the angle. Since no angle is given, we are assuming it to be zero, which, as I stated, is usually the case for the signal input, with all other angles referenced off of that. If you look here http://www.allaboutcircuits.com/vol_2/chpt_1/5.html it might help you understand this a bit better.

But, basically, in this case, your time domain representation would be:
$15sin(628t+0)$ and the frequency domain in polar would be $15 \angle 0$ while rectangular would be $15+j0$.