Notes to myself
apr 2003
Notes to refresh what's left of my memory (or:
"an attempt to postpone the losing battle with entropy").
going
from continuous to discrete fourier transform
sin(x)/x
polyphase
filterbanks.
gaussian
width after convolving 2 gaussians.
Jon
Hagen's acf-> spectra memo (.pdf)
rms voltages needed for
2,3,4, and 5 bit sampling.
Radar
decoding.
clipping a sine wave
Creating
harmonics (how strong the various freq terms are)
Harmonics created when clipping a sine
wave.
Side band harmonics do not always
move farther out than at the fundamental.
Problems
using a 90 deg hybrid to convert linear to circular (lbw example)
properities of a gaussian
properteis of sin(x)/x
Computing Trcvr.
Telescope gain
Beamwidth,
near/far field transition
Using
idl
to process ao spectral line data
List of
cpu
compute servers at AO.
Disc
locations
by cpu
Backing up of
monitor data.
bitfield
storage format. converting big endian to little endian.
Ao Network
Info
- side band harmonics do not
always
move farther out than at the fundamental. (top)
Suppose you have a tone of Amplitude A and frequency
w and a side band of amplitude B and frequency (w+delta)
then:
(Aexp(iwt)+Bexp(i(w+delta)*t)2= AAexp(i2wt)+ ABexp(i(2wt
+delta))
+ BB*exp(i(2w+2delta)t)
If B << A then the AB term will be much larger than the BB term
so the +delta signal will be stronger than the 2*delta.
- Gaussian: (top)
A gaussian centered at x=0 can be defined as:
y1= 1./sqrt(2*pi)*e-.5*(x/sig)^2)
integral
normalized to unity
y2=
e-(x/sig)^2)
y2(x=0)=1.
y2(x=1)=e-1= .367879
Properties are:
| property |
y1 |
y2 |
| maxvalue y(0) |
y1(0)=.398942 |
y2(0)=1. |
| y(sig) |
y1(sig)=.2419707 |
y2(sig)=.367879 |
| integral -inf to inf |
1. |
sqrt(pi)=1.77245 |
FWHM=m*sig (see 1*)
sig=n*FWHM |
m=2.35482
n=.424661 |
m=1.66511
n=.60056118 |
1*
y1=A0/2=A0*e(-.5*x/sig)^2)
sqrt(2*ln(2))*sig=x(hwhm)
FWHM=2*x =2*sqrt(2*ln(2))
y2=1/2=e(-(x/sig)^2)
sqrt(ln(2))*sig=x(hwhm)
FWHM=2*x= 2*sqrt(ln(2))*sig
- sin(x)/x (top)
y=sin(x)/x . If x is defined in units of the first null, then use
sin(pi*x)/(pi*x)
| property |
|
| maxvalue : y(0) |
1. |
| minvalue :y |
y=-.2172336, x=4.4934100.. |
| zero crossings |
pi*n n=1.. |
| peaks,min |
when x=tan(x) (more
info)
for large x :close to x=(2n+1)pi/2 |
| x when y=.5 |
x=1.8954950 |
integral between 1st
nulls (-pi,pi) |
integral: 9.8696191 |
rms voltages needed for
2,3,4,
and 5 bit sampling. (top)
The optimum voltage
levels
for 2, 3, , and 5 bit sampling were computed using the threshold levels
from fred schwab. To compute the rmsVolts I just took
PktoPkVolts/Nlevels
* sigmaLevels. This ignores the problem of whether or not the levels
are
centered on 0 volts or not.
Rms Volts vs Nbits
| Nbits |
level threhsold
(in sigmas) |
Sigma (1./level)
(in levels) |
rms (Volts) assuming
2V PkToPk A/D |
rms (Volts) Assuming
5V PkToPk A/D (ri) |
| 2 |
.99568 |
1.004 |
0.502 |
1.255 |
| 3 |
.58601 |
1.706 |
0.427 |
1.067 |
| 4 |
.33520 |
2.983 |
0.373 |
0.932 |
| 5 |
.18814 |
5.315 |
0.332 |
0.830 |
Compute Trcvr. (top)
The receiver temperature is computed using a hot and
cold load at the input. let:
- T1 and T2 be the temperatures of these loads.
- gamma is the voltage reflection at the input to the horn. let
G2=(1-gamma^2).
It will be the fraction of the input power that gets into the device
- alpha is the loss in the omt, Tomt is the temperature of the omt.
(1-alpha)
of the entering temp will be passed through. (1-alpha)*Tomt will be
contributed
by the omt temp.
- Tamp is the amplifier temperature. Trcv is normally Tamp +
alpha*Tomt.
If we measure the output power ratio on load1 and load2:
Y=Pwr2/Pwr1
y=(Tamp + alpha*Tomt + (1-gamma^2)(1-alfa)T2)/(Tamp + alpha*Tomt + (1-gamma^2)(1-alpha)T1)
Solving for Tamp gives:
Tamp(1-Y)=Y(alpha*Tomt + (1-alpha)(1-gamma^2)*T1) - (alpha*Tomt + (1-alpha)(1-gamma^2)T2)
Tamp= (1-alpha)(1-gamma^2)*(T2-Y*T1)/(Y-1) - alpha*Tomt
If you assume that alpha,gamma are equal to zero, then you will get a
Tamp
that is higher than it really is.
Telescope gain (top)
- 1 K/Jy is 2762 m^2
- for a uniformilly illuminated circular appeture:
hpbw=1.02*lambda/D
- if plane wave with energy power density S hit telescope. The
telescope
extracts Pe energy then the effective area of the telescope is Ae=Pe/S
- The appeture efficiency is Na=Ae/Ag where Ag is the
geometric
area.
- Gmax=4*pi*Ae/lambda^2
- Ae*Omegaa=lambda^2 where OmegaA is the beam solid angle
(integrated
over
4pi steradian).
- OmegaMb is the main beam solid angle. Defined to be the solid
angle
that
has the same area as the mainbeam down to the first nulls.
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