Post on 16-Oct-2021
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Homework H5 Solution 1. Turn in: A flute tone is played for a duration of 2 seconds at a frequency of π=440 Hz (middle A on the keyboard). Note that π=2 ππ. a. As a percentage of the frequency π, what is the uncertainty βπ in the pitch being played? Now someone hits a drum for 1 millisecond, making a sound centered at π =440 Hz (a βtuned drum like they use in orchestras). b. As a percentage of the frequency π, now what is the uncertainty βπ in the pitch being played? c. Based on your finding above, which of these two instruments would be more useful for playing a melody? d. If you sped up your playing and you could play a note on the flute every 1 millisecond, is it still possible in principle to tell the difference between an A and A# (440 Hz and 466.1 Hz)? Can one play melody on a flute that fast as a matter of basic principle, no matter how nimble? Solution: a. As it is played for 2 seconds, Ξπ‘ = 2π . You learnt that the Fourier conjugate variables π
and t are related as
Ξπ Ξπ‘ =14π
Therefore, Ξπ = !!!= 0.03978 π»π§.
β΄Ξππβ 100 % = 9.04 β 10!! %
b. Now, Ξπ‘ = 10!!π . So, Ξπ = !!!β!"!!
= 79.57 π»π§. β΄ !!
!β 100 % = 18.08 %
c. Thus, for playing a melody, the flute would be more useful if the time duration is high
like in part a, since the frequency, or the note played would be more precise. d. If you speed up playing the flute so that Ξπ‘ = 10!! π , the uncertainty in the
frequency would again be 79.57 Hz. So in principle it would be impossible to distinguish the notes A and A#, which have a difference of 26.1Hz (< 79.57 Hz !)
2. The inverse Fourier transform is given by π(t) = (1/2π) β« dπ π (π) exp[-Ββiπt]. Itβs just like the Fourier transform, with a minus sign in front of the βiβ. The Fourier transform gets you from π (t) to π (π), and the inverse transform gets you from π (π) back to π (t) . Comparing the two,
π (π) = 1 . β« dt π (t) exp[+iπt]
and
π (t) = (1/2π) β« dπ π (π) exp[-Ββiπt]. Itβs shown in the βT1 Readingβ handout on the course schedule.
Use this knowledge to prove, using integration by parts of the inverse Fourier transform like what we did in lecture with the Fourier transform, that β/βπ β +it.
Solution:
Begin similarly as how you learnt in class about proving !!" βΆ βππ.
β±π!! !" !
!"= !
!!ππ !" !
!" π!!"#!
!! [ Integrating by parts] = !
!!π π π!!"#|!!! + ππ π π ππ‘ π!!"#!
!! = !"
!!ππ π π!
!! π!!"# [lim!βΆΒ±! π π = 0 ] = ππ‘ β±π!![π π ] Thus,
!!"
β +ππ‘