Fourier Analysis of Piecewise Function

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Question:Fourier Analysis of Piecewise Function

Posted: dareimert 35 Product: Maple

February 17 2018

How do I extract harmonics and plot a spectrum from a piecewise function?

The function is "Ia" shown below near the bottom in the "Plot Line Current" section.

(1)

Vmax = RmsMS phase to neutral voltage

The #1 diode is forward biased and switches on when Va>Vc (Va=Vc). The voltage VD will be equal to Va-Vb until diode #2 turns on when Vc< Vb. Then VD will equal Vc-Vc.

The switching angle for the diodes is given above. If phase control is used all commutation is delayed by the angle α.

Without phase delay the average (DC) voltage applied to the laod is the average value of the 6 commutation periods over one cycle:

Emax = RmsMS phase to neutral voltage

Eq1 := Vdo = 6*(int(Van-Vbn, theta = com1 .. com2))/(2*Pi)

Vdo = 3*3^(1/2)*2^(1/2)*Erms/Pi

(2)

Eq2 := Vdc = 6*(int(Van-Vbn, theta = com1+alpha .. com2+alpha))/(2*Pi)

Vdc = 3*2^(1/2)*Erms*3^(1/2)*cos(alpha)/Pi

(3)

Commutation

The switching between diodes can not occur instantaneously because the supply circuit is inductive. Current can not change instantaneously in and inductive circuit.
The transfer of load current from the out-going phase to the incomming phase is accomplshed by a period of symultaneous conduction of the two adjacent diodes the creation of a circulating current between these two phases that resembles short circuit between the two phases.

The circulating current produces equal voltage drop with opposid polarities in each phase. In the out-going phase the drop adds to the soruce voltage and in the incomming phase the drop subtracts from the source voltage. Only the drop in the incomming phase effects the DC output voltage.

During commutation intervil com1 to com2 phases a and c are shorted while Id is transfered from phase c to phase a. During this time voltage Va-Vb minus the voltage drop in a phase caused by the commutation. This voltage drop is 1/2 the Vac voltage.

(1/2)*2^(1/2)*Erms*sin(theta)-(1/2)*2^(1/2)*Erms*cos(theta+(1/6)*Pi)

(4)

and Va=Vc at π/6

`ΔE__Avg` = 6*(int(`Δe`, theta = (1/6)*Pi+alpha .. (1/6)*Pi+alpha+mu))/(2*Pi)

`ΔE__Avg` = 3*((1/2)*2^(1/2)*Erms*cos((1/6)*Pi+alpha)+(1/2)*2^(1/2)*Erms*sin((1/3)*Pi+alpha)-(1/2)*2^(1/2)*Erms*sin((1/3)*Pi+alpha+mu)-(1/2)*2^(1/2)*Erms*cos((1/6)*Pi+alpha+mu))/Pi

(5)

`ΔE__Avg` = (3/2)*2^(1/2)*Erms*(cos((1/6)*Pi+alpha)+sin((1/3)*Pi+alpha)-sin((1/3)*Pi+alpha+mu)-cos((1/6)*Pi+alpha+mu))/Pi

(6)

`ΔE__Avg` = (3/2)*2^(1/2)*Erms*3^(1/2)*cos(alpha)/Pi-(3/2)*Erms*2^(1/2)*3^(1/2)*cos(alpha)*cos(mu)/Pi+(3/2)*Erms*2^(1/2)*3^(1/2)*sin(alpha)*sin(mu)/Pi

(7)

`ΔE__Avg` = (3/2)*Erms*2^(1/2)*((-cos(mu)+1)*cos(alpha)+sin(alpha)*sin(mu))*3^(1/2)/Pi

(8)

`ΔE__Avg` = (1/2)*(-3*Erms*2^(1/2)*3^(1/2)*cos(alpha+mu)+3*2^(1/2)*Erms*3^(1/2)*cos(alpha))/Pi

(9)

`ΔE__Avg` = -(3/2)*3^(1/2)*2^(1/2)*Erms*(cos(alpha+mu)-cos(alpha))/Pi

(10)

DC output voltage

(11)

V__D = 3*2^(1/2)*Erms*3^(1/2)*cos(alpha)/Pi+(3/2)*3^(1/2)*2^(1/2)*Erms*(cos(alpha+mu)-cos(alpha))/Pi

(12)

V__D = (3/2)*2^(1/2)*Erms*3^(1/2)*(cos(alpha)+cos(alpha+mu))/Pi

(13)

Equating the intigral of the voltage drop between the phase A and C to the intigral of the voltage rise across the two phase inductances (thes intigrals represent eqivalent flux).

int(Van-Vcn, theta = (1/6)*Pi+alpha .. (1/6)*Pi+alpha+mu) = int(2*L*omega, theta = 0 .. Id)

2^(1/2)*Erms*cos((1/6)*Pi+alpha)+2^(1/2)*Erms*sin((1/3)*Pi+alpha)-2^(1/2)*Erms*cos((1/6)*Pi+alpha+mu)-2^(1/2)*Erms*sin((1/3)*Pi+alpha+mu) = 2*L*omega*Id