Monday, January 30, 2012
HW 2 Problem 19.1
There is some discussion in the 2010 posts about the sinusoidal approximation of v_s1(t) being multiplied by 1/2 because v_s(t) has a maximum value = V_g and minimum = 0, rather than the textbook example of +V_g and -V_g. A result of this scaling factor being that H(s) is also scaled. I don't think this is correct. The fundamental component of the Fourier series contains the amplitude term V_g ( (4*V_g/pi)*sin(w_s*t) ); that being the case, if V_g oscillates between V_g and 0 the sine wave will be accordingly 4*V_g/pi or 0. No need to scale. This will approximate the square wave. Does anyone else have some thoughts on this? Thanks.
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9 comments:
@Keith: I was thinking it would be the same (4*V_g/pi)*sin(w_s*t))+Vg/2 where the offset Vg/2 is required. I looked through the old blog as well, and frankly it's confusing. I'm working on part a, and it seems the ig(t) for 0<t<0.5Ts is simply a linear ramp with a negative peak current from inductor L. And interval 0.5Ts<t<0, ig(t)=0. Any 2 cents? How would one do sinusoidal approximation of such waveform?
Hi Khoa,
Thanks for the feedback. Regarding your question on i_g(t), I think that i_g(t) will be a square wave rather than a ramping waveform as you found it to be, for the following reasons: during subinterval t = 0 - T_s/2, Q1 is on, Q2 is off, and the supply current i_g(t) is flowing into capacitor C_b and the resonant tank. For t = T_s/2 - T_s, Q1 is off and Q2 is on, the supply is disconnected and i_g(t) = 0 (as you mentioned). The thing to keep in mind is that i_g(t) enters Cb before L. Cb is a DC blocking capacitor of large value and will therefore remove the average (DC) value from the i_g(t), resulting in an AC waveform with maximum +I_g/2 and minimum -I_g/2. This is the waveform that enters the resonant tank network. Since it is of square wave form, the sinusoidal approximation can be used. Hope that helps.
Hi Keith,
The current is(t) through Cb must be the same as L since they're in series. I agree the is(t) average is also zero. But for is(t) to be a square wave, implies that di/dt = 0 for the entire switching period. This also means the voltage across L is zero (vL=L*di/dt), which I don't think this is true. di/dt = V/L so the current must have a slope through L during intervals for vs(t)=Vg and 0.
For interval, 0<t<0.5Ts, the ig(t)=is(t)=current through L & Cb ramping down. And ig(t)=0 for 0.5Ts<t<Ts. This forms a right triangle for ig(t) for interval 0<t<0.5Ts. Then we need to compute the Fourier series for this signal???
Hi, guys. Regarding your discussion, I think:
1. For signal of square wave between -Vg and Vg, the magnitude is 2Vg, and therefore the fundamental harmonic has magnitude of 2*4/pi*Vg, or amplitude of 4/pi*Vg. However, for signal of square wave between 0 and Vg, the magnitude is Vg, and therefore the fundamental harmonic should have magnitude of 4/pi*Vg or amplitude of 2/pi*Vg.
2. I don't think i_s is square wave, nor is triangular wave. Since i_s=v_s/Zi, v_s is a square wave, and Zi is a 2nd order filter, it is difficult to say what exact waveform i_s really is. That is why we use the sinusoidal approximation of v_s to get an approximated sinusoidal i_s. regarding i_g, it is obvious that half cycle i_g=i_s, and the other half i_g=0. However, you don't really need to compute the fundamental harmonic of i_g, since it's on the DC side. You just need to take the average current, and model it as a DC current source.
All,
I may be doing it wrong, but I just did the Fourier Series of the square wave (ao + sum(ancos(wnt)) - sum(bn(sin(wnt)) where ao = 1/T int(f(x)dx,-T/2,T/2), an = 2/T int(f(x)cos(wnt)dx,-T/2,T/2) and bn = 2/T int(f(x)sin(wnt)dx,-T/2,T/2)
When I solve this guys (along with some trig identities), I get a DC offset of Vg/2, which also holds for the 1st harmonic approximation. If you use this sinusoidal approximation, you will get a kind of "rectified" sine wave as the current similar to Fig 19.8, with a DC offset
Just a thought, any comments?
Thanks to everyone for the discussion. After some more thought I am changing my mind as far as my original post goes. I think a scaling factor of V_g/2 is needed in v_s1(t) resulting in v_s1(t) = (2*V_g/pi)*sin(w_s*t). The reason for my change of thought is that I misinterpreted what the V_g term means. It is not v_g(t), a time dependent function that oscillates between the maximum V_g and 0 as shown in figure 19.53(b). It is the DC value of the v_g(t) waveform, which in this case is V_g/2.
Hi,
After reading all of your comments, I get somewhat confused. Here are my own thoughts.
Since this circuit makes use of a half bridge, the switching node will have a waveform that looks like a square wave between Vg and 0V. The DC value of that square wave is Vg/2. The blocking capacitor Cb filters the DC component out, and will therefore have a voltage Vg/2 across it. The signal presented to the LC tank therefore is a square wave with amplitudes +Vg/2 and -Vg/2. We can now apply the sinusoidal approximation to this signal. Since the LC tank is loaded with a rectifier, which acts as a resistive load, the output current is a sine wave current that is in phase with the output voltage. The input current is however a sine wave with a phase shift compared to the input voltage. Since the converter is operated above resonance, the current leads the voltage.
When Q1 is on, the input current ig is part of a sine wave around i=0 that leads the input voltage. When Q1 is off, the input current ig=0A.
Anybody agree/disagree?
Regards,
Toby
Toby,
I definitely agree with the comment about the input of the resonant tank, but I am not sure the output current and voltage are going to be in phase... wont Lf give you a phase shift?
Thanks
Hi Olga,
Well, if the load is purely resistive, the output current has to be be in phase with the voltage. The equivalent load of the rectifier plus filter network is given in eq. 19.26. Of course, if a complex impedance is used as a load, the current and voltage won't be in phase.
In this case, I would say output current and -voltage are in phase.
Regards,
Toby
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