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Unformatted text preview: se of CIRCU125 diode elements, which have parameters that are set up to produce ideal
characteristics. The circuit in Problem Sketch 1 is modeled to be a simple half-wave rectifier with three elements:
• a voltage source • an ideal diode • a single load resistor In this case computing the Fourier series coefficients is a minor task. However, in Problem Sketch 2, a capacitor
is introduced into the circuit which causes a combination, sinusoidal-exponential output waveform, and hence
making the Fourier series calculation much more complex.
Definition: V LOAD( t ) = a0
2 ∑ n>0 2nπ
2nπ an cos ( T t ) + bn sin ( T t ) Problem Sketch One: Resistor, Diode and Voltage Source
ANSYS Verification Manual . ANSYS Release 9.0 . 002114 . © SAS IP, Inc. 1–523 VM226
We have: Vs(t) = Vsl*SIN(ωt)
Vsl = 135 Volts
ω = 2πf and f = (1)/T = 60 Hz
Fourier coefficients: 2
− a0 =
T T/2 ∫ Vs ( t ) dt = 2 * Vs1 0 π T/2 ∫ Vs ( t )c os(nωt ) dt 0 So we have: Vs1 1
1 n > 1, an = n + 1(1 − cos((n + 1)π)) − n − 1 (1 − cos((n − 1)π)
ωT a = 0
That is to say:...
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This note was uploaded on 12/09/2010 for the course DEPARTMENT E301 taught by Professor Kulasinghe during the Spring '09 term at University of Peradeniya.
- Spring '09
- The Land