AMI paradox
DOCX · 75.8 KB
Open DOCX file
Typed calculation note by Phil dated 3.24.13, in the support folder of his spectral theory book. It compares the Fourier spectrum of a square wave with his AMI result for general p. Setting a=(1-2p)→-1, he shows the tan² factor becomes a periodic comb of delta functions by integrating near the singularity. The result agrees with the square-wave spectrum except for a factor of 2. Equation text is partly garbled.
AI-written summary; may contain errors. This description is approximate.
Extracted text (machine-read; may contain errors)
The AMI Paradox PhL 3.24.13
If p = 1, then AMI is just a square wave of amplitude V and period T1. Finally after much mucking around, I show this in Section 25 : [ the T2 appearing here is half T1]
X(ω) = !Syntax Error, I sinc(m ) 2πδ(ω - mω2/2) = !Syntax Error, I sinc(m ) 2πδ(ω - mω1) (25.12)
I have to multiply by 2V to get the amplitude right, so
X(ω) = 2V !Syntax Error, I sinc(m ) 2πδ(ω - mω1) = 2V T1 !Syntax Error, I sinc(m ) 2πδ(ωT1 - mω1/T1)
= V T1 Σm_odd sinc(m ) 2πδ(ωT1 - mω1/T1)
= V T1 Σm sinc(m ) 2πδ(ωT1 - mω1/T1)
Now I can manually square
|X(ω)|2 = V2T12 Σm sinc(m ) 2π δ(ωT1 - mω1/T1) Σm' sinc(m' ) 2π δ(ωT1 - mω1/T1)
= V2T12 Σm sinc2(m ) 2π δ(ωT1 - mω1/T1) 2πδ(0)
so that
|X(ω)|2/ [2πδ(0)] = V2T12 Σm sinc2(m ) 2π δ(ωT1 - mω1/T1)
|X(ω)|2/ [2πδ(0)] = V2T12 2 !Syntax Error, I sinc2(m ) 2π δ(ωT1 - mω1/T1)
|X(ω)|2/ [2πδ(0)] = V2T1 2 !Syntax Error, I sinc2(m ) 2π δ(ω - mω1)
I can express this in terms of T2 as follows
|X(ω)|2/ [2πδ(0)] = V2 2T2 2 !Syntax Error, I sinc2(m ) 2πδ(ω - mω2/2)
In my AMI world I have T1 set to the pulse width, so my AMI limit I expect to be this
|X(ω)|2/ [2πδ(0)] = (2V)2T1 !Syntax Error, I sinc2(m ) 2πδ(ω - mω1/2) ***********
Meanwhile, here is my AMI result for general p
= |Xpulse(ω)|2 a ≡ (1-2p) AMI
Xpulse(ω) = (VT1) sinc(ωT1/2)
If I take p→ 1, then a = -1 and we have
= |Xpulse(ω)|2
= |Xpulse(ω)|2
= |Xpulse(ω)|2
= |Xpulse(ω)|2
= |Xpulse(ω)|2 (1/4) (1-a2) tan2(ωT1/2)
= |Xpulse(ω)|2 (1/4) (1-a2) tan2(π )
Footnote: I found a PDF with the spectrum stated this way
If I take p = 1 in this result, the leading factor vanishes, the denominator is 1 + 1 + 2 cos(2πfT) so this result seems to have the same problem, it is no doubt the same result. The PDF plots the thing
Go back now to my result
f(ω) = |Xpulse(ω)|2 (1/4) (1-a2) tan2(π )
The tangent blows up at π = π/2. We can see f(ω) is periodic.
π = π + 2π
= + 2
ω + Δω = ω + 2ω1 Δω = 2ω1
So whatever happens repeats with period Δω = 2ω1 . So we need only examine the region near π/2. Surely I can show a delta function there. So go back to
f(ω) ≡
Change variable to x = ωT1/2 so we have
f(ω(x)) =
Now let's look near the singular point
x = π/2 + δx
sin(x) = sin(π/2+δx) = cos(δx) ≈ 1
cos(2x) = cos(π + 2δx) = - cos(2δx)
So near the problem we have
f(ω(x)) =
Now expand
cos(2δx) = 1 - (2δx)2/2 = 1 - 2 δx2
Then denominator is
1 + a2 + 2a(1-2δx2) = 1 + a2 + 2a - 4aδx2 = 1 + 1 + 2 - 4δx2 = 4(1-δx2)
Plan B: I want to show this
lima→0 = k δ(x-π/2) x = ωT1/2 = π ω/ω1
Change variables to y = x-π/2
sin(x) = sin(y+π/2) = cos(y)
cos(2x) = cos(2y + π) = -cos(2y)
So I want to show this
lima→0 = k δ(y)
The numerator I feel plays no role, so we really want to show that
lima→0 = k δ(y)
NOW integrate both sides from -ε to ε
lima→0!Syntax Error, Idy = k
lima→0 { (1-a2) !Syntax Error, Idy [ 1 + a2+2acos(2y)]-1 }
= lima→0 { (1-a2) 2 (a2-1)-1 tan-1( tanε (a-1)/(a+1)) }
= lima→0 { - 2 tan-1( tanε (-2)/(a+1)) }
Now as a→-1, the arg of tan-1 approaches -∞, so tan-1 = -π/2 and we seem then to get
= lima→0 { -2 -π/2 } = π
So I have shown then that
lima→0 = πδ(y)
But since periodic, this is really
lima→0 = πδ(y)
But LHS has period π, so we have
lima→0 = πΣm=-∞∞δ(y-mπ)
This is the main idea. Now we can to back to x
lima→0 = πΣm=-∞∞ δ(x+π/2-mπ)
Here is my AMI result with x = ωT1/2.
= |Xpulse(ω)|2 a ≡ (1-2p) AMI
So I have
= |Xpulse(ω)|2 π Σm=-∞∞ δ(ωT1/2+π/2-mπ)
= |Xpulse(ω)|2 π Σm=-∞∞(2/T1) δ(ω+π/T1-2mπ/T1)
= |Xpulse(ω)|2 π Σm=-∞∞(2/T1) δ(ω -(2m-1)ω1/2)
Let k = 2m-1 so sum ranges
m = -2,-1,0,1,2
k = -5,-3,-1,1,3,5
= |Xpulse(ω)|2 π Σk odd (2/T1) δ(ω -kω1/2)
= (2/T1)|Xpulse(ω)|2 2π Σk odd+ δ(ω -kω1/2)
= (2/T1)|Xpulse(ω)|2 !Syntax Error, I 2π δ(ω -mω1/2)
and this is looking a lot like what I want. For my AMI case I want
Xpulse(ω) = (VT1) sinc(ωT1/2)
|Xpulse(ω)|2 = (VT1)2 sinc2(ωT1/2)
So continue the AMI limit to get
= (2/T1) (VT1)2 sinc2(ωT1/2) !Syntax Error, I 2π δ(ω -mω1/2)
= (2/T1) (VT1)2 sinc2(π ω/ω1) !Syntax Error, I 2π δ(ω -mω1/2)
= (2/T1) (VT1)2 !Syntax Error, I sinc2(π ω/ω1) 2π δ(ω -mω1/2)
= (2/T1) (VT1)2 !Syntax Error, I sinc2(π m/2) 2π δ(ω -mω1/2)
= 2 V2 T1 !Syntax Error, I sinc2(π m/2) 2π δ(ω -mω1/2)
This agrees but off by factor of 2.