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## Q. 36.4

What is the duty cycle of the output waveform in Fig. 36-13b?

## Verified Solution

Recall that the duty cycle is defined as the pulse width divided by the period. Duty cycle equals the conduction angle divided by 360°.

In Fig. 36-13b, the sine wave has a peak value of 10 V. Therefore, the input voltage is given by

$v_{in} = 10 \sin \theta$

The rectangular output switches states when the input voltage crosses +5 V. At this point, the foregoing equation becomes

$5 = 10 \sin \theta$

Now, we can solve for the angle $\theta$ where switching occurs:

$\sin \theta = 0.5$

or

$\theta$ = arcsin 0.5= 30° and 150°

The first solution, $\theta$ = 30°, is where the output switches from low to high. The second solution, $\theta$ = 150°, is where the output switches from high to low. The duty cycle is

$D=\frac{conduction angle}{360°}=\frac{150°-30°}{360°}= 0.333$

The duty cycle in Fig. 36-13b can be expressed as 33.3%.