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Re: A problem using IDL? [message #60079]

Fri, 02 May 2008 09:20

R.G. Stockwell
Messages: 363
Registered: July 1999

Senior Member

a bit better code. The top plot is the aliased signal.
The bottom shows the well sampled (not aliased) signal, which
is sampled much more finely. The thick squares show the aliased
signal from the top plot.

window,xsize=800,ysize=400
min=0
max=100
n=205
x=fltarr(n+1)
for i=0,n do x[i]=min+double(i)*(max-min)/n
;print,x
y=sin(2*3.141593*x)

!P.multi=[0,1,2]

plot,x,y,psym=-4,xr=[0,10]
x1 = x
y1 = y

n=205*8
x=fltarr(n+1)
for i=0,n do x[i]=min+double(i)*(max-min)/n
;print,x
y=sin(2*3.141593*x)

plot,x,y,psym=-4,xr=[0,10]
oplot,x1,y1,psym=6,thick=4

end

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[Message index]

		Re: A problem using IDL? By: R.G. Stockwell on Fri, 02 May 2008 09:20
		Re: A problem using IDL? By: R.G. Stockwell on Fri, 02 May 2008 09:16
		Re: A problem using IDL? By: Vince Hradil on Thu, 01 May 2008 08:41
		Re: A problem using IDL? By: pgrigis on Thu, 01 May 2008 07:49

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