Electrical Engineering - Time Response of Reactive Circuits - Discussion
Discussion Forum : Time Response of Reactive Circuits - General Questions (Q.No. 22)
22.
What is the highest frequency contained in a pulse that has a rise and fall time equal to 10 microseconds (10 
s)?
s)?Discussion:
15 comments Page 1 of 2.
Prajakta said:
8 years ago
Rise Time/Fall Time. Rise time refers to the time it takes for the leading edge of a pulse (voltage or current) to rise from its minimum to its maximum value. Rise time is typically measured from 10% to 90% of the value.
So the extra time required to get 10% value can be any value.
If we consider it as say 5 us then also we get 40kHz so, this might be the reason that, maximum frequency form given options will be 35kHz.
So the extra time required to get 10% value can be any value.
If we consider it as say 5 us then also we get 40kHz so, this might be the reason that, maximum frequency form given options will be 35kHz.
Emmanuel wambua said:
5 years ago
Bandwidth= k/rise time.
Where k is a value between 0.35 and 0.45 depending on the shape of the oscilloscope frequency curve and pulse rise time response.
Oscilloscope with bandwidth< 1GHz typically have a value of 0.35, while oscilloscope with a bandwidth of >1GHz usually have a value between 0.40 and 0.45.
Where k is a value between 0.35 and 0.45 depending on the shape of the oscilloscope frequency curve and pulse rise time response.
Oscilloscope with bandwidth< 1GHz typically have a value of 0.35, while oscilloscope with a bandwidth of >1GHz usually have a value between 0.40 and 0.45.
Sewak said:
9 years ago
According to me,
Rise Time 10 micro Sec, Fall time 10 micro Sec.
So complete cycle time is 20 micro Sec.
Frequency = 1/Time span of 1 cycle.
Frequency = 1/20 micro Sec,
Frequency = 1/20 * 10^-6,
Frequency = 50 KHz.
Rise Time 10 micro Sec, Fall time 10 micro Sec.
So complete cycle time is 20 micro Sec.
Frequency = 1/Time span of 1 cycle.
Frequency = 1/20 micro Sec,
Frequency = 1/20 * 10^-6,
Frequency = 50 KHz.
Mkhan.q said:
4 years ago
Bandwidth can be calculated by using the formula.
Bandwidth= 0.35/rise time.
Here rise time is 10 microseconds.
Hence;
Bandwidth= 0.35/10x10^-6,
Bandwidth= 35k Hertz.
Bandwidth= 0.35/rise time.
Here rise time is 10 microseconds.
Hence;
Bandwidth= 0.35/10x10^-6,
Bandwidth= 35k Hertz.
(1)
Mohmedhusein Irfan Kapadia said:
8 years ago
35khz is right because fall time means fall up to 50% magnitude.
So total time is 10 +10+10 = 30us.
So 1/30u = 33.33khz.
Which is approx 35khz.
So total time is 10 +10+10 = 30us.
So 1/30u = 33.33khz.
Which is approx 35khz.
Rojohn Ogalinola said:
7 years ago
tr=10microsec,
tf=10microsec,
BW=Fh+Fl=Fh,
BW=.35/tr=Fh,
Therefore;
BW=.35/10x10-6=35khz=Fh.
tf=10microsec,
BW=Fh+Fl=Fh,
BW=.35/tr=Fh,
Therefore;
BW=.35/10x10-6=35khz=Fh.
BANU said:
10 years ago
1/t = 1/(10*10^-6) = 100,000 k.
Here, 1 khz = 1000 hz.
So 100 khz.
Here, 1 khz = 1000 hz.
So 100 khz.
Asif Baloch said:
8 years ago
Tr = 1/π f.
Using this formula we get 31.8KHz approx to 35kHz.
Using this formula we get 31.8KHz approx to 35kHz.
Sri said:
1 decade ago
Please explain how is the answer A?
Manoj said:
1 decade ago
1/10micro = 100k is the answer.
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