going from continuous to discrete fourier transform

sin(x)/(x) and its properties

polyphase filter banks

gaussian width after convolving 2 gaussians

Converting spectrum back to rms volts

Going from a continuous to discrete fourier transform: (top)

When going to the discrete, finite case of N samples we get:

t=>tThe FT becomes:_{s}* n n=0..N-1 since the samples are spaced by t_{s}.

f=> k/(t_{s}*N) k=0..N-1. N/2 samples are duplicated for real input samples.

X(k/(tIf we agree that X_{s}*N))= sum_{n}(x(n*t_{s})*exp(2*pi**(k/(ti_{s}*N))*n*t_{s}) * dt) n=0..N-1 or

X(k/(t_{s}*N))= sum_{n}(x(n*t_{s})*exp(2*pi**k*n/N) * dt) n=0..N-1 .i

X_{k}= sum_{n}(x_{n}*exp(2*pi**kn/N)) n=0..N-1, k=0,N-1i

sin(x)/x and its properties: (top)

To show that sin(x)/x is the FT of a boxcar window, we integrate the window from -T/2 to T/2 (to make it symmetric so the anit-symmetric imaginary part goes to zero):

X(f)=integral(exp(-2*pi*The imaginary part is odd so it integrates to 0. The real part integrate to:*f*t)dt) -T/2 <=t<=T/2 ..i

X(f)= 1/(2*pi*f) * [ 2*sin(2*pi*f*T/2)] = sin(pi*f*T)/(pi*f)= T* ( sin(pi*f*T)/(pi*f*T)If we change to the discrete format:

XThe zeros of sin(x)/x are at:_{k}=N * sin(pi*(k/N)*N)/(pi*k/N*N) = N*sin(pi*k)/(pi*k)

zeros:The extrema of sin(x)/x are found by differentiating and setting it to zero:sin(x) = 0 at x=(n+1)*pi (n=0..)

d/dx (sin(x)/x) = (cos(x)/x - sin(x)/x^2)=0Tan(x) goes to infinity at x=(2n+1)*pi/2. As x goes to infinity the extrema must approach (2n+1)pi/2 so for large x this is a pretty good approximation for the peaks. Looking at (sin(x)/x)^2 (the power) the difference between (2n+1)*pi/2 and the correct value is: (1st: .2 db, 2nd: .07db, 3rd: .03db,4th: .02db...)

x^2*cos(x)=x*sin(x) or x=tan(x)

The sin(x)/x
plot for (x>0) has a linear plot of the voltage on top and
a
db plot of the power (sinx/x)^2 below. Since both sin(x) and (x) are
anti
symmetric in (x), sin(x)/x is symmetric and x<0 is the same as
x>0.

The first sidelobe is -13.26 db down in power from
the peak. The values for the first 14 power sidelobes are printed on
the
bottom plot.

polyphase filter banks (top)

The simplest multi channel filter bank is created by fourier transforming (FT) the data. We take K input points and perform a FT. This uses a boxcar window in the time domain. The convolution theorem tells us that this will convolve the channels in the frequency domain with

The individual filters BN_{k} (k=0,K-1)
will
have a width that depends on how the filter bank was constructed. The
filter
BN_{k} will have some overlap with adjacent filters. When
a strong signal falls in filter BN_{j}, it will also appear
(almost
always) at a reduced level in BN_{k}. This overlap in response
is called the filter leakage. The polyphase filter technique allows you
to decrease this leakage at the expense of taking larger time
sequences.

*processing: x101/polyphase/polyphase.pro*

gaussian width after convolving 2 gaussians (top)

spc(f)
= (abs( 1/N*S_{t}(v(t)*exp(-2pi*f*t/N)))^2

the forward fft is getting divided by 1/fftlen in the idl
routine. The data was set so that the rms value was 10 counts. Below i refer to I and Q. These are the inphase (Real) and Quadrature (imaginary) inputs.

The plots show going from spectrum back to rms volts (.ps) (.pdf):

- Page 1 Sine wave:
- Top input voltages. Black is realI, red is imaginaryQ. Each has
an rms of 10 and an amplitude of 14.1 (sqrt(2)*rms).

- 2nd: power spectrum. All the power is in rmsI= avgSpectralPower*numspcChan)channel 16 (count from 0). The value 200= 2*rms^2. This comes from:
- pwr=rmsI^2 + rmsQ^2= 2*10^2.

- 3rd: input voltages with real rms of 10 and imaginary= rms of 0.
- 4th: power spectrum from 3. The power is now in 2 bins (+/-16).

- single channel=50
- total power=100 = rmsI^2
- Page 2 Noise:
- Top: raw voltage samples.
- Black is realI, red is imaginaryQ. Both have an rms of 10.
- 2nd: spectrum of I,Q data. I averaged 1000 spectrum to decrease
the noise on the spectrum.

- Mean spectral value: .05
- Total power= 200 = rmsI^2 + rmsQ^2
- 3rd: raw voltage samples for I, Q=0.
- bottom: spectrum for rmsI=10, rmsQ=0
- Mean spectral value: .02
- Total power=100 = rmsI^2 +rmsQ^2 = 100 + 0.

- TotalPower = rmsI^2 + rmsQ^2 = 2*rmsI^2 (if rmsI=rmsQ)
- rmsI or rmsQ = sqrt(TotalPower/Ndig) .. where Ndig is the number of digitizer (I,Q) with signal.
- TotalPower=AvgSpectralValue/numSpectralChannels.
- The important quantity is the rmsVoltage. If you have a sine wave then the rmsVolt=sqrt(2)*sinAmplitude.
- When using the forward transform, the spectra get divided by
1/spclen. This is done by the idl routine.