Q: Assume you sampled every 0.1ms (i.e. Delta =10^-4 sec) for N=1024 total samples,
so the total time sampling was about 0.1 sec, and then made a Fourier Transform (FFT).
What is the actual frequency (in Hz) of the first non-DC FFT component, 
i.e., the one Mathematica reports in Data[[2]].

1/(1024*10^-4 sec) APPROX 10 Hz


#40:
y=Table[0,{i, 20}]
x[n_]=UnitStep[n-6]

Do[
y[[n]] = ( x[n- 4] +  x[n- 3] +  x[n- 2] +  x[n- 1] +  x[n- 0])/5 ,{n,6,7}]

y[[7]]

2/5

A[w_]=(Exp[-I 4 w]+Exp[-I 3 w]+Exp[-I 2 w]+Exp[-I 1 w]+1)/5
Plot[Abs[A[f Pi]],{f,0,1}]

Nyquist=5000
f=2500 -> .5 Pi
Abs[A[.5 Pi]]
.2

y=Table[0,{i, 20}]
x[n_]=UnitStep[n-6]

Do[
y[[n]] = ( 1 x[n- 4] + 4 x[n- 3] + 6 x[n- 2] + 4 x[n- 1] + 1 x[n- 0])/68.94 +
-.05034 y[[n- 4]] + .3599 y[[n- 3]] -1.0459 y[[n- 2]] + 1.5042 y[[n- 1]],
{n,6,20}]
y[[7]]                                                                  

Out[4]= 0.0943458

ListPlot[y, Joined->True, PlotRange->All]

xx[k_]=Exp[I w k]
A[w_]=( 1 xx[-4] + 4 xx[-3] + 6 xx[-2] + 4 xx[-1] + 1 )/68.94/
(1-(-.05034 xx[-4] + .3599 xx[-3] -1.0459 xx[-2] + 1.5042 xx[-1]))
Abs[A[.5 Pi]]
Out[11]= 0.0507044

#41
a) 10,000 Hz
y=Table[0,{i, 20}]
y[[1]]=8/7
x[n_]=n

Do[
y[[n]] = 8/7 ( x[n] - x[n- 1] - y[[n- 1]]/8),
{n,2,20}]
ListPlot[y, Joined->True, PlotRange->All]
 {y[[1]],y[[2]],y[[3]]}                                                 

          8  48  344
Out[11]= {-, --, ---}
          7  49  343

quickly reaches & hold results =1
Note if y[[]]=1 then
  8/7 ( x[n] - x[n- 1] - y[[n- 1]]/8)= 8/7(1-1/8)=1

A[w_]=8/7( - xx[-1] + 1 )/(1+ xx[-1]/7)
Plot[Abs[A[f Pi]],{f,0,1}]

y=Table[0,{i, 20}]
x[n_]=UnitStep[n-6]

Do[
y[[n]] = 2 ( -x[n- 1] +  x[n- 0]-y[[n-1]]/2) ,{n,6,20}]

ListPlot[y, Joined->True, PlotRange->All]

NOT STABLE

A[w_]=( - xx[-1] + 1 )
Plot[Abs[A[f Pi]],{f,0,1}]
MORE BEND

#42

putting {x+I y} together makes this data be a complex number rotating in a particular
direction
datax=Re[data]
datay=Im[data]
data2=Transpose[{datax,datay}]
ListPlot[data2[[1;;50]],Joined->True,AspectRatio->Automatic]

you can perhaps see that {x,y} is rotating clockwise, i.e., like Exp[-I w t]
which Mathematica treats as a positive frequency (with no negative frequency component)

After RotateLeft, I see peaks at 86.5 and 102

freq= n/N f0 = 86.5/2048 20 & 102/2048 20
period=1/freq: 1.18382 year & 1.00392 year
                432.4 days    366.7  days

forcing function is not a monotone

#43

x=<<SunSpot2.m

y=Table[0,{i, 2052}];
Do[
y[[n]]=x[[n-4]]+4 x[[n-3]]+6 x[[n-2]]+4 x[[n-1]]+x[[n]]+
 (1.11894 ) y[[n-1]] - (0.884346) y[[n-2]] + (0.289792 ) y[[n-3]]- (0.0445722) y[[n-4]],{n,5,2052}]

ListPlot[y, Joined->True, PlotRange->All]
ListPlot[x, Joined->True, PlotRange->All]

the filter has gain...it amplifies

data=Drop[y,4]

Data=Fourier[data]
Data=RotateLeft[Data,1]
ListLinePlot[Abs[Data],PlotRange->All]

real data contains + and - frequencies with equal magnitudes (e.g., Cos[t]=(Exp[I t]+Exp[-I t])/2)
The negative frequencies are aliased up to N-n



ListLinePlot[Abs[Data],PlotRange->{{1,50},All}]

the peak is at n=16

freq = n/N f0 = 16/2048 12 (year^-1)
period= 1/freq = 10.7 years

The digital filter has attenuated frequencies above .2 *12 (year^-1) which corresponds
to n=.2*2048 approx 400

#59:

Nyquist =f0/2=5000
reading at x=.5 (half the Nyquist frequency)
@2500: Bessel=.5, Butterworth=.7, Chebyshev=.9

increasingly "sharp" (abrupt near-step change): Bessel, Butterworth, Chebyshev
Chebyshev is not flat in passed band...it has oscillations