Complex Periodic Waves

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Homework Statement



A 100Ω resistor is connected in series with a 2 uF capacitor and a 20 mH inductor. The voltage 1 sin 5000t + 0.5 sin 1000t V is applied across the circuit.

Determine:

a) The effective (rms) voltage

b) The effective (rms) current

c) The true power

d) The power factor

Homework Equations



For question (a)

Vrms = √Vo^2 +(V1m^2 + V2m^2 / 2)

Vrms = √0+ (1^2) + (0.5^2) / 2 = √1+ 0.25 /2 = √1.25/2 = √0.625 = 0.79

Am i correct so far?

The Attempt at a Solution

 

Answers and Replies

  • #2
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Could you check my first steps?
 
  • #3
gneill
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Looks alright to me. You could also have computed the rms voltage from the definition, re:

$$V_{rms} = \sqrt{\frac{1}{2 \pi}\int_0^{2 \pi}\left(sin(5\theta) + \frac{1}{2}sin(\theta)\right)^2 d\theta}$$

or used this to check your result.
 
  • #4
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For question (b) IRMS

Peak Value Current (Io) = (VRMS x √3) / R

Peak Value Current (Io) = (0.79 X 1.73205) / 100

Peak Value Current (Io) = 1.3683195 / 100

Peak Value Current (Io) = 0.01368 A


IRMS = Io / √2

IRMS = 0.01368 / 1.4142

IRMS = 0.0098 A RMS

Can someone please let me know if I am heading in the right direction with this question?
 
  • #5
gneill
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I don't think your approach is right. The voltage waveform is composed of two different frequencies, so not a true sinewave. Thus the usual conversions for peak / average / rms are not going to work for this waveform as a whole.

Effectively you have two different circuits since the impedance is different for each frequency. If you compute the currents for each frequency separately and then add them in quadrature (square root of sum of squares), you should be okay.

I'd begin by computing the two impedances; one for each frequency. Then if you let E1 = 1V and E2 = (1/2)V be the two peak voltages, compute I1 = E1/Z1 and I2 = E2/Z2. These will be complex currents. That is, they contain magnitude and phase information for each. You can take their magnitudes for the peak current values. They are individually both sinewaves, so you can convert to rms by dividing the peak values by root 2. So, then add the rms values in quadrature:

##I_{rms} = \sqrt{\left|\frac{I1}{\sqrt{2}}\right|^2 + \left|\frac{I2}{\sqrt{2}}\right|^2}##.

I think that should give you good results.
 
  • #6
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OK. Thank you very much for the response. I just need to first of all know if I am going down the right route by using the following equation to then work out the two impedances?

Z = R2 + (XL - XC)2

XL = ωL

XC = 1 / ωC
 
  • #7
gneill
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That will give you the reactance, which is the magnitude of the impedance. I'd use the complex form for impedance:
$$Z = R + j\omega L + \frac{1}{j \omega C}$$
a complex value which gives you the magnitude and phase angle.
 
  • #8
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Once again thanks for the support. This question seems to be getting rather deep. Do, I then use the following equations to include in the previous equation you provided?

ω = 2 ∏ f

Then, the equation for (f) is:

f = 1 / (4∏2 LC)
 
  • #9
gneill
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You can read the ω's right off the voltage expressions. The form is V(t) = k sin(ωt).
 
  • #10
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Can you help me with the form you provided? I am genuinely stuck. I have researched into this area and still no further forward. Thanks!
 
  • #11
gneill
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In a sinewave (or cosinewave) as a function of time, a voltage may be written in the form
$$V(t) = V_m cos(ω t + \phi)$$
Where ##V_m## is the peak voltage, ω is the angular frequency, and ##\phi## is the phase angle. You are given a voltage function that is the sum of two such expressions with different values for ω. Thus the 'signal' has two component frequencies.

You can read the values for these ω's directly from the expressions you were given. Call them ω1 and ω2.

To each of these frequency components of the signal the circuit will look slightly different. This is because the impedances of reactive components (inductors, capacitors) depends upon frequency. Effectively there's one circuit for each of the frequency components of the driving signal.

Start by computing the net impedance of the circuit for both of the given frequencies. Then you can calculate individual currents, powers, etc. for each frequency component of the source.
 
  • #12
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Could you please help me with net impedance, i cant find equation for it? And im stuck!


Z= (R+ jwL) + (1/ ( jwC)) how can i calculate j for this equation?
 
  • #13
gneill
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Could you please help me with net impedance, i cant find equation for it? And im stuck!


Z= (R+ jwL) + (1/ ( jwC)) how can i calculate j for this equation?
j is the square root of -1, as usual. Impedance is a complex number.
 
  • #14
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Can you help me more, i have 2 hours to solve it!
 
  • #15
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I think there is an error in task condition

u(t)=1.sin(5000t)+0.5sin(1000t)V

It should be u(t)=1.sin(5000t)+0.5sin(10000t)V

Intead of 10000t is written 1000t which is not a second harmonic, because ω=1000rad/s, hence 2ω=10000rad/s
 
Last edited:
  • #16
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rms-voltage:

U=[itex]\sqrt{\frac{1}{2}(U^{2}_{m1}+U^{2}_{m2})}[/itex]
 
  • #17
gneill
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I think there is an error in task condition

u(t)=1.sin(5000t)+0.5sin(1000t)V

It should be u(t)=1.sin(5000t)+0.5sin(10000t)V

Intead of 10000t is written 1000t which is not a second harmonic, because ω=1000rad/s, hence 2ω=10000rad/s
There's nothing in the problem statement to suggest that the frequencies (or periods) of the two voltage components have to be related harmonically. Any two frequencies might have been used. The fact that one is 5x the other is a nice happenstance though.
 
  • #18
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Ok, but when you calculate rms-current.Don't you have to calculate impedance of first harmonic, then impedance of second harmonic?Is it wrong to assume that they are harmonically connected?

Z1=R+j(ωL-1/ωC),Z1=100Ω and [itex]\varphi[/itex]1=0
Z2=R+j(2ωL-1/2ωC) Z2=180.3Ω and [itex]\varphi[/itex]2=56.31°

Hence

Im1=Um1/z1 and Im2=Um2/z2

then rms-current is
I=[itex]\sqrt{\frac{1}{2}(I^{2}_{m1}+I^{2}_{m2})}[/itex]
 
Last edited:
  • #19
gneill
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Ok, but when you calculate rms-current.Don't you have to calculate impedance of first harmonic, then impedance of second harmonic?Is it wrong to assume that they are harmonically connected?
The thing is, they are not necessarily harmonics. They are simply two voltage sources in series. They could have been specified as relatively prime numbers and it wouldn't matter. Just call them frequency components.

Z1=R+j(ωL-1/ωC),Z1=100Ω and [itex]\varphi[/itex]1=0
Z2=R+j(2ωL-1/2ωC) Z2=180.3Ω and [itex]\varphi[/itex]2=56.31°

Hence

Im1=Um1/z1 and Im2=Um2/z2

then rms-current is
I=[itex]\sqrt{\frac{1}{2}(I^{2}_{m1}+I^{2}_{m2})}[/itex]
Sure. That should work.
 

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