Thread Subject: ifft

i have a base band frequency domain impulse response that i
need to transform to the time domain.
it is in the form of two vectors one containing the real
part and the other the imaginary one.
the frequency span is from -11MHz to +11MHz and the sampling
frequency is 88Mhz. and i have 500 samples so the frequency
separation is of 500/22MHz.
i cannot seam to catch how to use ifft in order to perform this.
any help would be appreciated.
thank u in advance

On Jul 23, 1:53 pm, "hani tok" <aht...@ndu.edu.lb> wrote:
> i have a base band frequency domain impulse response that i
> need to transform to the time domain.
> it is in the form of two vectors one containing the real
> part and the other the imaginary one.
> the frequency span is from -11MHz to +11MHz and the sampling
> frequency is 88Mhz. and i have 500 samples so the frequency
> separation is of 500/22MHz.

Something is not right.

N = 500 samples at a sampling frequency Fs = 88e6 yields
a spectrum with a frequency spacing of df = Fs/N = 176e3
that has a "b"ipolar domain of fb = -fQ:df:fP where

fQ = (Q-1)*df = 44e6
fP = (P-1)*df = 43.824e6
Q = ceil((N+1)/2) = 251
P = floo((N+1)/2) = 250

On the other hand, a bipolar domain of fb ~[-11e6,11e6]
with 500 samples implies Fs ~ 22e6 and df ~ 22e6/500
= 44e3 and which are 4 times smaller.

Please resolve this. Is your signal band-limited to 11MHz?

> i cannot seam to catch how to use ifft in order to perform this.
> any help would be appreciated.

x = ifft(fftshift(Re+i*Im));

t = dt*(0:N-1);
dt = 1/Fs

However, it is not clear if Fs is 88MHz or 22MHz.

Hope this helps.

Greg

Hello Greg,
first of all thank you for your help.

in fact one of my biggest problems is that i have no control
what so ever on the data, it is given to me in a text file
format with the description as stated previously. that is
why i am having problems, nevertheless i am sure that the
length of each vector is 500, so i have 500 samples per
impulse response. i know that the response is for a wifi
channel so it has a total bandwidth of 22MHz, and i know
that the data i have is for baseband (donno how it got there
lol). so what i wanted to know is: would a straight forward
IFFT as the one u wrote at the end of your reply would be
good enough for me to get the effective time domain response
???? (what are the capabilities and limits of ifft ???)

thank you once again for your help


Greg Heath <heath@alumni.brown.edu> wrote in message
<459109e2-844d-4908-8e4a-980b178d8ff4@u36g2000pro.googlegroups.com>...
> On Jul 23, 1:53 pm, "hani tok" <aht...@ndu.edu.lb> wrote:
> > i have a base band frequency domain impulse response that i
> > need to transform to the time domain.
> > it is in the form of two vectors one containing the real
> > part and the other the imaginary one.
> > the frequency span is from -11MHz to +11MHz and the sampling
> > frequency is 88Mhz. and i have 500 samples so the frequency
> > separation is of 500/22MHz.
>
> Something is not right.
>
> N = 500 samples at a sampling frequency Fs = 88e6 yields
> a spectrum with a frequency spacing of df = Fs/N = 176e3
> that has a "b"ipolar domain of fb = -fQ:df:fP where
>
> fQ = (Q-1)*df = 44e6
> fP = (P-1)*df = 43.824e6
> Q = ceil((N+1)/2) = 251
> P = floo((N+1)/2) = 250
>
> On the other hand, a bipolar domain of fb ~[-11e6,11e6]
> with 500 samples implies Fs ~ 22e6 and df ~ 22e6/500
> = 44e3 and which are 4 times smaller.
>
> Please resolve this. Is your signal band-limited to 11MHz?
>
> > i cannot seam to catch how to use ifft in order to
perform this.
> > any help would be appreciated.
>
> x = ifft(fftshift(Re+i*Im));
>
> t = dt*(0:N-1);
> dt = 1/Fs
>
> However, it is not clear if Fs is 88MHz or 22MHz.
>
> Hope this helps.
>
> Greg
>

CORRECTED FOR THE UNPARDONABLE SIN OF TOP_POSTIN

On Jul 24, 12:03=A0pm, "hani tok" <aht...@ndu.edu.lb> wrote:
> Greg Heath <he...@alumni.brown.edu> wrote in message
> <459109e2-844d-4908-8e4a-980b178d8...@u36g2000pro.googlegroups.com>...
> > On Jul 23, 1:53 pm, "hani tok" <aht...@ndu.edu.lb> wrote:
> > > i have a base band frequency domain impulse response that i
> > > need to transform to the time domain.
> > > it is in the form of two vectors one containing the real
> > > part and the other the imaginary one.
> > > the frequency span is from -11MHz to +11MHz and the sampling
> > > frequency is 88Mhz. and i have 500 samples so the frequency
> > > separation is of 500/22MHz.
>
> > Something is not right.
>
> > N =3D 500 samples at a sampling frequency Fs =3D 88e6 yields
> > a spectrum with a frequency spacing of df =3D Fs/N =3D 176e3
> > that has a "b"ipolar domain of fb =3D -fQ:df:fP where
>
> > fQ =A0=3D =A0(Q-1)*df =A0 =A0 =A0 =A0 =A0=3D =A044e6
> > fP =A0=3D =A0(P-1)*df =A0 =A0 =A0 =A0 =A0=3D =A043.824e6
> > Q =A0 =3D =A0ceil((N+1)/2) =A0 =3D =A0251
> > P =A0 =3D =A0floo((N+1)/2) =A0 =3D =A0250
>
> > On the other hand, a bipolar domain of fb ~[-11e6,11e6]
> > with 500 samples implies =A0Fs ~ 22e6 and df ~ 22e6/500
> > =3D 44e3 and which are 4 times smaller.
>
> > Please resolve this. =A0 Is your signal band-limited to 11MHz?
>
> > > i cannot seam to catch how to use ifft in order to
> perform this.
> > > any help would be appreciated.
>
> > x =3D ifft(fftshift(Re+i*Im));
>
> > t =3D dt*(0:N-1);
> > dt =3D 1/Fs
>
> > However, it is not clear if Fs is 88MHz or 22MHz.
>
> Hello Greg,
> first of all thank you for your help.
>
> in fact one of my biggest problems is that i have no control
> what so ever on the data, it is given to me in a text file
> format with the description as stated previously. that is
> why i am having problems, nevertheless i am sure that the
> length of each vector is 500, so i have 500 samples per
> impulse response. i know that the response is for a wifi
> channel so it has a total bandwidth of 22MHz,

So why did you say Fs =3D 88MHz?

>and i know
> that the data i have is for baseband (donno how it got there
> lol). so what i wanted to know is: would a straight forward
> IFFT as the one u wrote at the end of your reply would be
> good enough for me to get the effective time domain response
> ???? (what are the capabilities and limits of ifft ???)

Check a text. Look for auxiliary topics like

aliasing
Gibbs Phenomenon
windowing
zero-padding

> thank you once again for your help

Hope this helps.

Greg

Greg Heath <heath@alumni.brown.edu> wrote in message
<bd75cc62-1df1-4224-9591-fb5c43271e11@z11g2000prl.googlegroups.com>...
> CORRECTED FOR THE UNPARDONABLE SIN OF TOP_POSTIN
>
> On Jul 24, 12:03=A0pm, "hani tok" <aht...@ndu.edu.lb> wrote:
> > Greg Heath <he...@alumni.brown.edu> wrote in message
> >
<459109e2-844d-4908-8e4a-980b178d8...@u36g2000pro.googlegroups.com>...
> > > On Jul 23, 1:53 pm, "hani tok" <aht...@ndu.edu.lb> wrote:
> > > > i have a base band frequency domain impulse response
that i
> > > > need to transform to the time domain.
> > > > it is in the form of two vectors one containing the real
> > > > part and the other the imaginary one.
> > > > the frequency span is from -11MHz to +11MHz and the
sampling
> > > > frequency is 88Mhz. and i have 500 samples so the
frequency
> > > > separation is of 500/22MHz.
> >
> > > Something is not right.
> >
> > > N =3D 500 samples at a sampling frequency Fs =3D 88e6
yields
> > > a spectrum with a frequency spacing of df =3D Fs/N =3D
176e3
> > > that has a "b"ipolar domain of fb =3D -fQ:df:fP where
> >
> > > fQ =A0=3D =A0(Q-1)*df =A0 =A0 =A0 =A0 =A0=3D =A044e6
> > > fP =A0=3D =A0(P-1)*df =A0 =A0 =A0 =A0 =A0=3D =A043.824e6
> > > Q =A0 =3D =A0ceil((N+1)/2) =A0 =3D =A0251
> > > P =A0 =3D =A0floo((N+1)/2) =A0 =3D =A0250
> >
> > > On the other hand, a bipolar domain of fb ~[-11e6,11e6]
> > > with 500 samples implies =A0Fs ~ 22e6 and df ~ 22e6/500
> > > =3D 44e3 and which are 4 times smaller.
> >
> > > Please resolve this. =A0 Is your signal band-limited
to 11MHz?
> >
> > > > i cannot seam to catch how to use ifft in order to
> > perform this.
> > > > any help would be appreciated.
> >
> > > x =3D ifft(fftshift(Re+i*Im));
> >
> > > t =3D dt*(0:N-1);
> > > dt =3D 1/Fs
> >
> > > However, it is not clear if Fs is 88MHz or 22MHz.
> >
> > Hello Greg,
> > first of all thank you for your help.
> >
> > in fact one of my biggest problems is that i have no control
> > what so ever on the data, it is given to me in a text file
> > format with the description as stated previously. that is
> > why i am having problems, nevertheless i am sure that the
> > length of each vector is 500, so i have 500 samples per
> > impulse response. i know that the response is for a wifi
> > channel so it has a total bandwidth of 22MHz,
>
> So why did you say Fs =3D 88MHz?
>
> >and i know
> > that the data i have is for baseband (donno how it got there
> > lol). so what i wanted to know is: would a straight forward
> > IFFT as the one u wrote at the end of your reply would be
> > good enough for me to get the effective time domain response
> > ???? (what are the capabilities and limits of ifft ???)
>
> Check a text. Look for auxiliary topics like
>
> aliasing
> Gibbs Phenomenon
> windowing
> zero-padding
>
> > thank you once again for your help
>
> Hope this helps.
>
> Greg


hello Greg,
thanks once again for the help.
i am looking at such topics.

On Jul 23, 1:53 pm, "hani tok" <aht...@ndu.edu.lb> wrote:
> i have a base band frequency domain impulse response that i
> need to transform to the time domain.
> it is in the form of two vectors one containing the real
> part and the other the imaginary one.
> the frequency span is from -11MHz to +11MHz and the sampling
> frequency is 88Mhz. and i have 500 samples so the frequency
> separation is of 500/22MHz.

Something is not right.

N = 500 samples at a sampling frequency Fs = 88e6 yields
a spectrum with a frequency spacing of df = Fs/N = 176e3
that has a "b"ipolar domain of fb = -fQ:df:fP where

fQ = (Q-1)*df = 44e6
fP = (P-1)*df = 43.824e6
Q = ceil((N+1)/2) = 251
P = floo((N+1)/2) = 250

On the other hand, a bipolar domain of fb ~[-11e6,11e6]
with 500 samples implies Fs ~ 22e6 and df ~ 22e6/500
= 44e3 and which are 4 times smaller.

Please resolve this. Is your signal band-limited to 11MHz?

> i cannot seam to catch how to use ifft in order to perform this.
> any help would be appreciated.

x = ifft(fftshift(Re+i*Im));

t = dt*(0:N-1);
dt = 1/Fs

However, it is not clear if Fs is 88MHz or 22MHz.

Hope this helps.

Greg

Hello Greg,
first of all thank you for your help.

in fact one of my biggest problems is that i have no control
what so ever on the data, it is given to me in a text file
format with the description as stated previously. that is
why i am having problems, nevertheless i am sure that the
length of each vector is 500, so i have 500 samples per
impulse response. i know that the response is for a wifi
channel so it has a total bandwidth of 22MHz, and i know
that the data i have is for baseband (donno how it got there
lol). so what i wanted to know is: would a straight forward
IFFT as the one u wrote at the end of your reply would be
good enough for me to get the effective time domain response
???? (what are the capabilities and limits of ifft ???)

thank you once again for your help


Greg Heath <heath@alumni.brown.edu> wrote in message
<459109e2-844d-4908-8e4a-980b178d8ff4@u36g2000pro.googlegroups.com>...
> On Jul 23, 1:53 pm, "hani tok" <aht...@ndu.edu.lb> wrote:
> > i have a base band frequency domain impulse response that i
> > need to transform to the time domain.
> > it is in the form of two vectors one containing the real
> > part and the other the imaginary one.
> > the frequency span is from -11MHz to +11MHz and the sampling
> > frequency is 88Mhz. and i have 500 samples so the frequency
> > separation is of 500/22MHz.
>
> Something is not right.
>
> N = 500 samples at a sampling frequency Fs = 88e6 yields
> a spectrum with a frequency spacing of df = Fs/N = 176e3
> that has a "b"ipolar domain of fb = -fQ:df:fP where
>
> fQ = (Q-1)*df = 44e6
> fP = (P-1)*df = 43.824e6
> Q = ceil((N+1)/2) = 251
> P = floo((N+1)/2) = 250
>
> On the other hand, a bipolar domain of fb ~[-11e6,11e6]
> with 500 samples implies Fs ~ 22e6 and df ~ 22e6/500
> = 44e3 and which are 4 times smaller.
>
> Please resolve this. Is your signal band-limited to 11MHz?
>
> > i cannot seam to catch how to use ifft in order to
perform this.
> > any help would be appreciated.
>
> x = ifft(fftshift(Re+i*Im));
>
> t = dt*(0:N-1);
> dt = 1/Fs
>
> However, it is not clear if Fs is 88MHz or 22MHz.
>
> Hope this helps.
>
> Greg
>

CORRECTED FOR THE UNPARDONABLE SIN OF TOP_POSTIN

On Jul 24, 12:03=A0pm, "hani tok" <aht...@ndu.edu.lb> wrote:
> Greg Heath <he...@alumni.brown.edu> wrote in message
> <459109e2-844d-4908-8e4a-980b178d8...@u36g2000pro.googlegroups.com>...
> > On Jul 23, 1:53 pm, "hani tok" <aht...@ndu.edu.lb> wrote:
> > > i have a base band frequency domain impulse response that i
> > > need to transform to the time domain.
> > > it is in the form of two vectors one containing the real
> > > part and the other the imaginary one.
> > > the frequency span is from -11MHz to +11MHz and the sampling
> > > frequency is 88Mhz. and i have 500 samples so the frequency
> > > separation is of 500/22MHz.
>
> > Something is not right.
>
> > N =3D 500 samples at a sampling frequency Fs =3D 88e6 yields
> > a spectrum with a frequency spacing of df =3D Fs/N =3D 176e3
> > that has a "b"ipolar domain of fb =3D -fQ:df:fP where
>
> > fQ =A0=3D =A0(Q-1)*df =A0 =A0 =A0 =A0 =A0=3D =A044e6
> > fP =A0=3D =A0(P-1)*df =A0 =A0 =A0 =A0 =A0=3D =A043.824e6
> > Q =A0 =3D =A0ceil((N+1)/2) =A0 =3D =A0251
> > P =A0 =3D =A0floo((N+1)/2) =A0 =3D =A0250
>
> > On the other hand, a bipolar domain of fb ~[-11e6,11e6]
> > with 500 samples implies =A0Fs ~ 22e6 and df ~ 22e6/500
> > =3D 44e3 and which are 4 times smaller.
>
> > Please resolve this. =A0 Is your signal band-limited to 11MHz?
>
> > > i cannot seam to catch how to use ifft in order to
> perform this.
> > > any help would be appreciated.
>
> > x =3D ifft(fftshift(Re+i*Im));
>
> > t =3D dt*(0:N-1);
> > dt =3D 1/Fs
>
> > However, it is not clear if Fs is 88MHz or 22MHz.
>
> Hello Greg,
> first of all thank you for your help.
>
> in fact one of my biggest problems is that i have no control
> what so ever on the data, it is given to me in a text file
> format with the description as stated previously. that is
> why i am having problems, nevertheless i am sure that the
> length of each vector is 500, so i have 500 samples per
> impulse response. i know that the response is for a wifi
> channel so it has a total bandwidth of 22MHz,

So why did you say Fs =3D 88MHz?

>and i know
> that the data i have is for baseband (donno how it got there
> lol). so what i wanted to know is: would a straight forward
> IFFT as the one u wrote at the end of your reply would be
> good enough for me to get the effective time domain response
> ???? (what are the capabilities and limits of ifft ???)

Check a text. Look for auxiliary topics like

aliasing
Gibbs Phenomenon
windowing
zero-padding

> thank you once again for your help

Hope this helps.

Greg

Greg Heath <heath@alumni.brown.edu> wrote in message
<bd75cc62-1df1-4224-9591-fb5c43271e11@z11g2000prl.googlegroups.com>...
> CORRECTED FOR THE UNPARDONABLE SIN OF TOP_POSTIN
>
> On Jul 24, 12:03=A0pm, "hani tok" <aht...@ndu.edu.lb> wrote:
> > Greg Heath <he...@alumni.brown.edu> wrote in message
> >
<459109e2-844d-4908-8e4a-980b178d8...@u36g2000pro.googlegroups.com>...
> > > On Jul 23, 1:53 pm, "hani tok" <aht...@ndu.edu.lb> wrote:
> > > > i have a base band frequency domain impulse response
that i
> > > > need to transform to the time domain.
> > > > it is in the form of two vectors one containing the real
> > > > part and the other the imaginary one.
> > > > the frequency span is from -11MHz to +11MHz and the
sampling
> > > > frequency is 88Mhz. and i have 500 samples so the
frequency
> > > > separation is of 500/22MHz.
> >
> > > Something is not right.
> >
> > > N =3D 500 samples at a sampling frequency Fs =3D 88e6
yields
> > > a spectrum with a frequency spacing of df =3D Fs/N =3D
176e3
> > > that has a "b"ipolar domain of fb =3D -fQ:df:fP where
> >
> > > fQ =A0=3D =A0(Q-1)*df =A0 =A0 =A0 =A0 =A0=3D =A044e6
> > > fP =A0=3D =A0(P-1)*df =A0 =A0 =A0 =A0 =A0=3D =A043.824e6
> > > Q =A0 =3D =A0ceil((N+1)/2) =A0 =3D =A0251
> > > P =A0 =3D =A0floo((N+1)/2) =A0 =3D =A0250
> >
> > > On the other hand, a bipolar domain of fb ~[-11e6,11e6]
> > > with 500 samples implies =A0Fs ~ 22e6 and df ~ 22e6/500
> > > =3D 44e3 and which are 4 times smaller.
> >
> > > Please resolve this. =A0 Is your signal band-limited
to 11MHz?
> >
> > > > i cannot seam to catch how to use ifft in order to
> > perform this.
> > > > any help would be appreciated.
> >
> > > x =3D ifft(fftshift(Re+i*Im));
> >
> > > t =3D dt*(0:N-1);
> > > dt =3D 1/Fs
> >
> > > However, it is not clear if Fs is 88MHz or 22MHz.
> >
> > Hello Greg,
> > first of all thank you for your help.
> >
> > in fact one of my biggest problems is that i have no control
> > what so ever on the data, it is given to me in a text file
> > format with the description as stated previously. that is
> > why i am having problems, nevertheless i am sure that the
> > length of each vector is 500, so i have 500 samples per
> > impulse response. i know that the response is for a wifi
> > channel so it has a total bandwidth of 22MHz,
>
> So why did you say Fs =3D 88MHz?
>
> >and i know
> > that the data i have is for baseband (donno how it got there
> > lol). so what i wanted to know is: would a straight forward
> > IFFT as the one u wrote at the end of your reply would be
> > good enough for me to get the effective time domain response
> > ???? (what are the capabilities and limits of ifft ???)
>
> Check a text. Look for auxiliary topics like
>
> aliasing
> Gibbs Phenomenon
> windowing
> zero-padding
>
> > thank you once again for your help
>
> Hope this helps.
>
> Greg


hello Greg,
thanks once again for the help.
i am looking at such topics.