Smallest duration pulse frequency

  • Thread starter Polkadot
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Homework Statement


An electronic device is designed to amplify signals in the frequency range 80 MHz to 111 MHz. what is the smallest duration pulse in this frequency range that can be amplified without distortion?

Homework Equations


T=1/f

The Attempt at a Solution


the smallest duration would be 1/1.11*10^9Hz and this is 9.0*10^-9s. this is wrong and I don't understand why.
 

Answers and Replies

  • #2
Simon Bridge
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Explain your reasoning,
 
  • #3
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So period is how long for a repetition to occur and frequency is how many repetitions per time unit. if units are Hz for frequency and seconds for period. Then you can say that T=1/f so that equation should give you the smallest duration (or period) of the pulse when you plug in the largest frequency that the device can amplify.
 
  • #4
vela
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One obvious mistake is that mega means 106, not 109.

What class is this for? I'm guessing this problem is a bit more involved than you think.
 
  • #5
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Your mistake may come from the assumption that smallest duration = period. A pulse is not a sine-like function ;)
 
  • #6
Simon Bridge
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I'm guessing your course has covered representing a pulse as a sum of sine waves ... is this correct?
You will need to use your understanding of this and how it relates to distortion to work out the answer.

But you may want to consider a pulse like ##g(t)=A\sin \omega t : 0<t<\pi / \omega, 0##: else.
 
  • #7
haruspex
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One obvious mistake is that mega means 106, not 109.
Polkadot did not take it as 109. There appears to be a typo implying MHz interpreted as 105, but as the next number, 9 10-9s, makes sense, I think it was just a typo in the post.
 
  • #8
haruspex
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Your mistake may come from the assumption that smallest duration = period. A pulse is not a sine-like function ;)
It could be sine-like, but as you say, the pulse duration would not be the period of the sine wave.
 
  • #9
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It could be sine-like, but as you say, the pulse duration would not be the period of the sine wave.
I'm not sure what else the duration of the pulse could be. I read somewhere that it is the time it takes from the the moment the pulse reaches half of its amplitude to the time it drops to the same level, but I'm not sure how that applies here or how I can use that information to solve the question.
 
  • #10
rcgldr
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It's not clear to me what is meant by pulse here. I would consider the displacement or force versus time of an idealized pulse as being half of a square wave, in which case it could not be amplified without infinite frequency. If instead the displacement or force versus time of a pulse is considered to be half a sine wave, then the minimum duration would be related to the maximum frequency.

Using a morse code key as another example, in order to reduce bandwidth, the ARRL reccomends that off/on or on/off transitions each take about 5 ms to prevent "key clicks". In this case the shortest duration full amplitude pulse would be 10 ms.
 
  • #11
haruspex
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I'm not sure what else the duration of the pulse could be. I read somewhere that it is the time it takes from the the moment the pulse reaches half of its amplitude to the time it drops to the same level, but I'm not sure how that applies here or how I can use that information to solve the question.
That might be right, or maybe from being zero to next being zero.
Either way, that is well short of a complete period of 2π.
If we take your version, at what angle x is sin(x) half its maximum?
 
  • #12
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So I interpreted the formula wrong. The formula for a range of frequencies is Δt=1/Δf and not t=1/f. So the answer becomes Δt=1/(111MHz-80MHz) which then gives the answer 3.22*10^-8s when MHz are converted to Hz.
 

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