020170
Full Member level 4

question for FFT
I've tested simple RC circuit through FFT
##########################################
Vdd vdd gnd 2.5
.option nomod fftout
.param fi = 1.0meg fs = 10meg fo = ' fs * (23/256)'
* Fin / Fs = Prime number / number of point
* input signal frequency = 1Meg, Sampling frequency = 10Meg
* output signal Frequency = Fs * ( 23 / 256 )
*va 1 gnd sin (0 1 fo)
va 1 gnd sin (0 1 fo)
* offset/amplitude/Frequency
R1 1 2 1K
C1 2 gnd 1p
* 100n * 256 = 25.6u
* 25.61u - 10n = 25.6u
* 10n?
.fft v(2) start=10n stop=25.61u np=256 format=norm dc=off
.op
.tran 10p 27u
.end
########################################
I didn't write some freq or window in FFT commmand.
is it right?
in my thought, simple RC circuit noise floor is determine like this
noise = kT/C
okay, noise is determined by C not R
in theory, when I increased the value of C, noise floor is decreased in FFT. right?
but I have the result that opposite of my thought
what is my problem?
I've tested simple RC circuit through FFT
##########################################
Vdd vdd gnd 2.5
.option nomod fftout
.param fi = 1.0meg fs = 10meg fo = ' fs * (23/256)'
* Fin / Fs = Prime number / number of point
* input signal frequency = 1Meg, Sampling frequency = 10Meg
* output signal Frequency = Fs * ( 23 / 256 )
*va 1 gnd sin (0 1 fo)
va 1 gnd sin (0 1 fo)
* offset/amplitude/Frequency
R1 1 2 1K
C1 2 gnd 1p
* 100n * 256 = 25.6u
* 25.61u - 10n = 25.6u
* 10n?
.fft v(2) start=10n stop=25.61u np=256 format=norm dc=off
.op
.tran 10p 27u
.end
########################################
I didn't write some freq or window in FFT commmand.
is it right?
in my thought, simple RC circuit noise floor is determine like this
noise = kT/C
okay, noise is determined by C not R
in theory, when I increased the value of C, noise floor is decreased in FFT. right?
but I have the result that opposite of my thought
what is my problem?