Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 2 (Q.No. 6)
6.
A 12 kHz pulse wave-form is amplified by a circuit having an Upper cut-off frequency of 1 MHz. The minimum input pulse width that can be accurately reproduced is
Answer: Option
Explanation:
tr = 10% Pω = 
minimum Power(P) = 10 tr 10 x 0.35 μ sec = 3.5 μ sec.
Note: The I/P Pulse will be severely distorted if the rise time is more than 10% of Pulse width.
Discussion:
3 comments Page 1 of 1.
Jsree said:
7 years ago
Thank you @Kavya Sri.
Kavya sri said:
8 years ago
Rise time formula tr=0.35/bandwidth
What is bandwidth upper cut off frequency-lower cut off frequency it is approximately upper cut off frequency so rise time formula 0.35/fh.
0.35/10^6=0.3 msec.
You know that rise time initial to maximum 10% to 90% rise time so here asked minimum.
so 10% power p/10= tr.
p= 10tr so.ans is 3.5usec so if rise time is more than 10% input pulse is distorted so small value here option is 1 u sec.
What is bandwidth upper cut off frequency-lower cut off frequency it is approximately upper cut off frequency so rise time formula 0.35/fh.
0.35/10^6=0.3 msec.
You know that rise time initial to maximum 10% to 90% rise time so here asked minimum.
so 10% power p/10= tr.
p= 10tr so.ans is 3.5usec so if rise time is more than 10% input pulse is distorted so small value here option is 1 u sec.
Dolon said:
8 years ago
Can anyone please explain it?
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