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In the singlepulse width modulation method, the output voltage waveform is symmetrical about __________
The waveform is a positive in the first half cycle and symmetrical about π/2 in the first half.
The shape of the output voltage waveform in a single PWM is
Positive and the negative half cycles of the output voltage are symmetrical about π/2 and 3π/2 respectively. The shape of the waveform obtained is called as quasisquare wave.
In the singlepulse width modulation method, the Fourier coefficient an is given by
As the positive and the negative half cycles are identical the coefficient a_{n} = 0.
In case of a singlepulse width modulation with the pulse width = 2d, the peak value of the fundamental component of voltage is given by the expression
For the fundamental component put n = 1.
V_{o} = (4Vs/π) sin (d) sin (ωt)
Hence the peak value is (4Vs/π) sin d.
Find the peak value of the fundamental component of voltage with a pulse width of 2d = 90 and Vs = 240 V for singlepulse modulation in a full wave bridge inverter.
The peak value of the fundamental component of voltage is given by (4Vs/π) sin d.
In the singlepulse width modulation method, the output voltage waveform is symmetrical about ____________ in the negative half cycle.
In the negative half the wave is symmetrical about 3π/2.
In the singlepulse width modulation method, the Fourier coefficient b_{n} is given by
The Fourier analysis is as under:
b_{n} = (2/π) ∫ Vs sin nωt .d(ωt) , Where the integration would run from (π/2 + d) to (π/2 – d)
2d is the width of the pulse.
In the singlepulse width modulation method, when the pulse width of 2d is equal to its maximum value of π radians, then the fundamental component of output voltage is given by
The Fourier representation of the output voltage is given by
Put 2d = π & n = 1.
In case of a singlepulse width modulation with the pulse width = 2d, to eliminate the nth harmonic from the output voltage
To eliminate, the nth harmonic, nd is made equal to π radians, or d = π/n.
From the below expression,
when nd = π. sin nd = 0 hence, that output voltage harmonic is eliminated.
In case of a singlepulse width modulation with the pulse width = 2d, to eliminate the 3rd harmonic from the output voltage waveform, the value of the pulse width (2d) must be
To eliminate the nth harmonic, nd = π.
Therefore, d = π/n = π/3 = 60°
Hence, 2d = 120°.
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