Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 1 (Q.No. 4)
4.
If the input to the ideal comparator shown in the figure is a sinusoidal signal of 8 V (peak to peak) without any DC component, then the output of the comparator has a duty cycle of
Answer: Option
Explanation:
Discussion:
30 comments Page 2 of 3.
Mona said:
7 years ago
Why 5pi/6=?
Explanation is;
The total half cycle is of π.
2v is a reference so as per waveform projection will be at equal distance from 0 and 180°
now first projection is at π/6 and total is π
So,second projection =π-π/6 = 6π-π/6 = 5 π/6.
Explanation is;
The total half cycle is of π.
2v is a reference so as per waveform projection will be at equal distance from 0 and 180°
now first projection is at π/6 and total is π
So,second projection =π-π/6 = 6π-π/6 = 5 π/6.
Lakshmishree said:
7 years ago
Great explanations, thank you all.
Kanna said:
8 years ago
Please tell me how we got 5π/6?
Sangi said:
8 years ago
It's very helpful, thanks @Mani Kumar.
Dee said:
8 years ago
Thanks @Haritha and @Mani Kumar.
Lokesh said:
9 years ago
Is anyone tells the output of comparator irrespective of the duty cycle? I mean general comparator output.
Hima said:
1 decade ago
How that pi/6 will come in the formula?
Darshith S P said:
1 decade ago
As per me at 4V(+ve peak) the value of X-axis is PI/2, then at 2V the value of X-axis will become PI/4 & 3PI/4.
If so then the duty cycle will become 1/4. i.e. 25% duty cycle. Please anyone confirm with this doubt.
If so then the duty cycle will become 1/4. i.e. 25% duty cycle. Please anyone confirm with this doubt.
Umair said:
1 decade ago
Great explanation, thanks to all for the elaboration.
Mani kumar said:
1 decade ago
In general the form of sinusoidal signal is vmsin(x), where vm is peak voltage.
Given p-p is 8V, so vm=4v; so signal is 4sin(x) the period of sin wave is 2pi. equate 4sin(x) with 2 we get x=pi/6, 5pi/6; i.e sin(pi-pi/6) is also sin(pi/6), so on time is (pi-(2pi/6)).
Duty cycle = (4pi/6)/2pi = 1/3.
Given p-p is 8V, so vm=4v; so signal is 4sin(x) the period of sin wave is 2pi. equate 4sin(x) with 2 we get x=pi/6, 5pi/6; i.e sin(pi-pi/6) is also sin(pi/6), so on time is (pi-(2pi/6)).
Duty cycle = (4pi/6)/2pi = 1/3.
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