FPGA Central

What is the meaning of high or low frequency for digital signal.( I
think this is not related to sampling frequency right?)

For example, we have an array and its size is 10. How can I understand
whether that 10 numbers represent high or low frequency?
Thanks.

On Mon, 24 Aug 2009 13:43:58 -0700, rotor cli wrote:

> What is the meaning of high or low frequency for digital signal.( I
> think this is not related to sampling frequency right?)
>
> For example, we have an array and its size is 10. How can I understand
> whether that 10 numbers represent high or low frequency? Thanks.

You mean "sampled signal?" A "digital signal" would be one that takes on
the values 0 or 1.

If you analyze the array for it's energy content vs. frequency, and most
of the energy is concentrated at low frequencies, then it's a low
frequency signal, and visa versa.

--
www.wescottdesign.com

On 25 AÄŸustos, 00:10, Tim Wescott <t...@seemywebsite.com> wrote:
> On Mon, 24 Aug 2009 13:43:58 -0700, rotor cli wrote:
> > What is the meaning of high or low frequency for digital signal.( I
> > think this is not related to sampling frequency right?)
>
> > For example, we have an array and its size is 10. How can I understand
> > whether that 10 numbers represent high or low frequency? Thanks.
>
> You mean "sampled signal?" Â*A "digital signal" would be one that takes on
> the values 0 or 1.
>
> If you analyze the array for it's energy content vs. frequency, and most
> of the energy is concentrated at low frequencies, then it's a low
> frequency signal, and visa versa.
>
> --www.wescottdesign.com

I meant discrete time signal that has 10 numbers.
I am just trying to understand frequency concept for discrete signal.
What makes these 10 numbers high or low frequency?

On Aug 24, 3:43*pm, rotor cli <rotor...@gmail.com> wrote:
> What is the meaning of high or low frequency for digital signal.( I
> think this is not related to sampling frequency right?)

Not right.

> For example, we have an array and its size is 10. How can I understand
> whether that 10 numbers represent high or low frequency?
> Thanks.

Ask youself what would happen if you sampled a 1-volt 1GHz sine wave
at sample rate of 1Gs/s, each time when wave is at its maximum (upper
hump), 10 samples. Your array would contain 10 values of 1. You would
get the exact same array of 10 values if you sampled a 1-volt 1Hz sine
wave at 1s/s. Which of these samples represent your high-frequency
signal and which represents the low-frequency signal?

It is not meaningful to digitize a signal without regard for what you
did when you digitized the signal.

-Le Chaud Lapin-

On Mon, 24 Aug 2009 14:55:03 -0700, rotor cli wrote:

> On 25 AÄŸustos, 00:10, Tim Wescott <t...@seemywebsite.com> wrote:
>> On Mon, 24 Aug 2009 13:43:58 -0700, rotor cli wrote:
>> > What is the meaning of high or low frequency for digital signal.( I
>> > think this is not related to sampling frequency right?)
>>
>> > For example, we have an array and its size is 10. How can I
>> > understand whether that 10 numbers represent high or low frequency?
>> > Thanks.
>>
>> You mean "sampled signal?" Â*A "digital signal" would be one that takes
>> on the values 0 or 1.
>>
>> If you analyze the array for it's energy content vs. frequency, and
>> most of the energy is concentrated at low frequencies, then it's a low
>> frequency signal, and visa versa.
>>
>> --www.wescottdesign.com
>
> I meant discrete time signal that has 10 numbers. I am just trying to
> understand frequency concept for discrete signal. What makes these 10
> numbers high or low frequency?

I told you.

But to give an example, if it goes 1,1,1,1,1,1,1,1,1,1 it has all of its
energy at frequency = 0, while if it goes 1,-1,1,-1,1,-1,1,-1,1,-1 then
it has all of its energy at frequency = 1/2 sample rate.

(Note that there's not too many possibilities with just ten samples --
the Nyquist rate is at 1/2 the sample rate and can't be changed, there's
only three well-defined frequency steps between that and DC).

--
www.wescottdesign.com

On Mon, 24 Aug 2009 14:55:03 -0700, rotor cli wrote:

> I meant discrete time signal that has 10 numbers. I am just trying to
> understand frequency concept for discrete signal. What makes these 10
> numbers high or low frequency?

This is a basic question, so you can start with looking at simple
discrete sine signals - which are the "single frequency" signals.

[The following are some musings which are not very strict but I hope
would be helpful.]

Roughly speaking, frequency - analog or discrete - means how fast a
periodic signal "repeats" itself (goes from one period to another). With
a discrete sine, its frequency also tells you *how fast* it may *change
its value* between two consecutive samples.

With analog signals, increasing a sine's frequency increases the maximal
value of its derivative [recall: the derivative of sin(w*t) is w*cos
(w*t)]. Here with discrete signals, it increases the maximal difference
of value a sample can have with a previous one [won't go into big maths
here]. So either way, the frequency essentialy tells you the pace of
change, whatever the mathematical details (derivative / difference).

A constant digital signal - a series of the same value, e.g. 0 - can be
interpreted (similarly as a constant analog signal) as a zero-frequency
(co)sine:

cos(0 * n) ,

where n is the sample number. It's constant, its value doesn't change, so
the frequency of its changes (oscillations) is zero.

As you look at discrete sines with an increasing frequency, you would get
a signal that switches from +1 to -1 going through zero (+1, 0, -1, 0,
+1, 0 and so on), which is really:

cos(pi/2 * n) ,

where n is the sample number. So, it's a (co)sine signal with the angular
frequency of pi/2. This means its "physical", temporal frequency is a
quarter of the sampling frequency (2*pi / 4 = pi/2, where 2*pi
corresponds to the sampling frequency).

The most rapidly changing discrete signal is +1, -1, +1, -1 and so on --
switching from the most positive value to the most negative value (and
back) every one sample. This is really:

cos(pi * n) ,

which means a half of the sampling frequency. This is the Nyquist
frequency mentioned in the sampling theorem. You just can't get more
rapid changes than in a series of alternating +/-1, or more generally, +/-
amplitude.

Now for more complicated signals, it's usually a matter of applying a
Discrete Fourier Transform to present the signal as a sum of separate
sine (or more generally, complex exponential) components - quite
similarly to continuous-time Fourier transforms, which you probably know.

Regards,
--
Krzysztof Lubański

On 25 Ağustos, 02:42, Krzysztof Lubański <lu...@nerdshack.com> wrote:
> On Mon, 24 Aug 2009 14:55:03 -0700, rotor cli wrote:
> > I meant discrete time signal that has 10 numbers. I am just trying to
> > understand frequency concept for discrete signal. What makes these 10
> > numbers high or low frequency?
>
> This is a basic question, so you can start with looking at simple
> discrete sine signals - which are the "single frequency" signals.
>
> [The following are some musings which are not very strict but I hope
> would be helpful.]
>
> Roughly speaking, frequency - analog or discrete - means how fast a
> periodic signal "repeats" itself (goes from one period to another). With
> a discrete sine, its frequency also tells you *how fast* it may *change
> its value* between two consecutive samples.
>
> With analog signals, increasing a sine's frequency increases the maximal
> value of its derivative [recall: the derivative of sin(w*t) is w*cos
> (w*t)]. Here with discrete signals, it increases the maximal difference
> of value a sample can have with a previous one [won't go into big maths
> here]. So either way, the frequency essentialy tells you the pace of
> change, whatever the mathematical details (derivative / difference).
>
> A constant digital signal - a series of the same value, e.g. 0 - can be
> interpreted (similarly as a constant analog signal) as a zero-frequency
> (co)sine:
>
> cos(0 * n) ,
>
> where n is the sample number. It's constant, its value doesn't change, so
> the frequency of its changes (oscillations) is zero.
>
> As you look at discrete sines with an increasing frequency, you would get
> a signal that switches from +1 to -1 going through zero (+1, 0, -1, 0,
> +1, 0 and so on), which is really:
>
> cos(pi/2 * n) ,
>
> where n is the sample number. So, it's a (co)sine signal with the angular
> frequency of pi/2. This means its "physical", temporal frequency is a
> quarter of the sampling frequency (2*pi / 4 = pi/2, where 2*pi
> corresponds to the sampling frequency).
>
> The most rapidly changing discrete signal is +1, -1, +1, -1 and so on --
> switching from the most positive value to the most negative value (and
> back) every one sample. This is really:
>
> cos(pi * n) ,
>
> which means a half of the sampling frequency. This is the Nyquist
> frequency mentioned in the sampling theorem. You just can't get more
> rapid changes than in a series of alternating +/-1, or more generally, +/-
> amplitude.
>
> Now for more complicated signals, it's usually a matter of applying a
> Discrete Fourier Transform to present the signal as a sum of separate
> sine (or more generally, complex exponential) components - quite
> similarly to continuous-time Fourier transforms, which you probably know.
>
> Regards,
> --
> Krzysztof Lubański

OK.
Thanks for all.

So; for example we have a picture and an array that represents its RGB
values of pixels.
Can we say how much RGB values of pixels are irrelevant it has a high
digital frequency and vice versa.

On 25 AÄŸustos, 12:34, rotor cli <rotor...@gmail.com> wrote:
> On 25 Ağustos, 02:42, Krzysztof Lubański <lu...@nerdshack.com> wrote:
>
>
>
> > On Mon, 24 Aug 2009 14:55:03 -0700, rotor cli wrote:
> > > I meant discrete time signal that has 10 numbers. I am just trying to
> > > understand frequency concept for discrete signal. What makes these 10
> > > numbers high or low frequency?
>
> > This is a basic question, so you can start with looking at simple
> > discrete sine signals - which are the "single frequency" signals.
>
> > [The following are some musings which are not very strict but I hope
> > would be helpful.]
>
> > Roughly speaking, frequency - analog or discrete - means how fast a
> > periodic signal "repeats" itself (goes from one period to another). With
> > a discrete sine, its frequency also tells you *how fast* it may *change
> > its value* between two consecutive samples.
>
> > With analog signals, increasing a sine's frequency increases the maximal
> > value of its derivative [recall: the derivative of sin(w*t) is w*cos
> > (w*t)]. Here with discrete signals, it increases the maximal difference
> > of value a sample can have with a previous one [won't go into big maths
> > here]. So either way, the frequency essentialy tells you the pace of
> > change, whatever the mathematical details (derivative / difference).
>
> > A constant digital signal - a series of the same value, e.g. 0 - can be
> > interpreted (similarly as a constant analog signal) as a zero-frequency
> > (co)sine:
>
> > cos(0 * n) ,
>
> > where n is the sample number. It's constant, its value doesn't change, so
> > the frequency of its changes (oscillations) is zero.
>
> > As you look at discrete sines with an increasing frequency, you would get
> > a signal that switches from +1 to -1 going through zero (+1, 0, -1, 0,
> > +1, 0 and so on), which is really:
>
> > cos(pi/2 * n) ,
>
> > where n is the sample number. So, it's a (co)sine signal with the angular
> > frequency of pi/2. This means its "physical", temporal frequency is a
> > quarter of the sampling frequency (2*pi / 4 = pi/2, where 2*pi
> > corresponds to the sampling frequency).
>
> > The most rapidly changing discrete signal is +1, -1, +1, -1 and so on --
> > switching from the most positive value to the most negative value (and
> > back) every one sample. This is really:
>
> > cos(pi * n) ,
>
> > which means a half of the sampling frequency. This is the Nyquist
> > frequency mentioned in the sampling theorem. You just can't get more
> > rapid changes than in a series of alternating +/-1, or more generally, +/-
> > amplitude.
>
> > Now for more complicated signals, it's usually a matter of applying a
> > Discrete Fourier Transform to present the signal as a sum of separate
> > sine (or more generally, complex exponential) components - quite
> > similarly to continuous-time Fourier transforms, which you probably know.
>
> > Regards,
> > --
> > Krzysztof Lubański
>
> OK.
> Thanks for all.
>
> So; for example we have a picture and an array that represents its RGB
> values of pixels.
> Can we say how much RGB values of pixels are irrelevant it has a high
> digital frequency and vice versa.

Are there 2 different digital frequency concept?

1-) In DSP books, they say:
omega = 2Pi*(fa/fs)
Here, there is a sampling so digital frequency is related with time
and sampling ratio.

2-)In image processing:
Frequency is about pixels values. How fast values are changed. For
example:
High frequency:
0 255 0 255 0 255 0 255
255 0 255 0 255 0 255 0
0 255 0 255 0 255 0 255
255 0 255 0 255 0 255 0
0 255 0 255 0 255 0 255
255 0 255 0 255 0 255 0

Low frequency:
255 255 255 255
255 255 255 255
255 255 255 255

Here, there is no time relation. But in first expression analog
frequency is involved.

I am confused.

>On 25 A=C4=9Fustos, 12:34, rotor cli <rotor...@gmail.com> wrote:
>
>Are there 2 different digital frequency concept?
>
>1-) In DSP books, they say:
>omega =3D 2Pi*(fa/fs)
>Here, there is a sampling so digital frequency is related with time
>and sampling ratio.
>
>2-)In image processing:
>Frequency is about pixels values. How fast values are changed. For
>example:
>High frequency:
>0 255 0 255 0 255 0 255
>255 0 255 0 255 0 255 0
>0 255 0 255 0 255 0 255
>255 0 255 0 255 0 255 0
>0 255 0 255 0 255 0 255
>255 0 255 0 255 0 255 0
>
>Low frequency:
>255 255 255 255
>255 255 255 255
>255 255 255 255
>
>Here, there is no time relation. But in first expression analog
>frequency is involved.
>
>I am confused.
>

There's a thing called "spatial frequency". The analogy is roughl
wavenumber (m^(-1)) is to frequency (s^(-1)=Hz) as wavelength (m) is t
period (s). Or you could use pixels instead of meters. It's all the sam
idea, so the two cases really only differ in physical interpretation (an
1d vs 2d).

Moreover, your image is still sampled into pixels: try resizing both o
those images (add more pixels for this experiment. If you take pixel
away, you'll alias in this case). You are looking at high and lo
frequency as relative concepts, compared to one another, or compared to th
Nyquist rate...either way. In the former case, you would still call thos
"high" and "low", but in the latter case (compared to Nyquist afte
doubling the image size), you might call them "medium" and "low".

On 25 AÄŸustos, 16:04, "Michael Plante" <michael.pla...@gmail.com>
wrote:
> >On 25 A=C4=9Fustos, 12:34, rotor cli <rotor...@gmail.com> wrote:
>
> >Are there 2 different digital frequency concept?
>
> >1-) In DSP books, they say:
> >omega =3D 2Pi*(fa/fs)
> >Here, there is a sampling so digital frequency is related with time
> >and sampling ratio.
>
> >2-)In image processing:
> >Frequency is about pixels values. How fast values are changed. For
> >example:
> >High frequency:
> >0 255 0 255 0 255 0 255
> >255 0 255 0 255 0 255 0
> >0 255 0 255 0 255 0 255
> >255 0 255 0 255 0 255 0
> >0 255 0 255 0 255 0 255
> >255 0 255 0 255 0 255 0
>
> >Low frequency:
> >255 255 255 255
> >255 255 255 255
> >255 255 255 255
>
> >Here, there is no time relation. But in first expression analog
> >frequency is involved.
>
> >I am confused.
>
> There's a thing called "spatial frequency". Â*The analogy is roughly
> wavenumber (m^(-1)) is to frequency (s^(-1)=Hz) as wavelength (m) is to
> period (s). Â*Or you could use pixels instead of meters. Â*It's all the same
> idea, so the two cases really only differ in physical interpretation (and
> 1d vs 2d).
>
> Moreover, your image is still sampled into pixels: Â*try resizing both of
> those images (add more pixels for this experiment. Â*If you take pixels
> away, you'll alias in this case). Â*You are looking at high and low
> frequency as relative concepts, compared to one another, or compared to the
> Nyquist rate...either way. Â*In the former case, you would still callthose
> "high" and "low", but in the latter case (compared to Nyquist after
> doubling the image size), you might call them "medium" and "low".

Hmm.
For you first paragraph:
Can you please make the analogy for fa and fs with pixels. For
"spatial domain"..
I mean how can fa and fs be related wit pixel example.
fa = pixel count? or fs = pixel count? or something else?

For second paragraph:
Books say: high frequencies are around pi and low frequencies are
around 0or(2Pi) in frequency domain.
What is that mean?

Yes, (as Nyquist)if fa = x and fs = 2x then 2Pi*(fa/fs) = pi. OK. But
if we increase sample ratio then pi increases, so why do books say low
frequencies are around 2Pi?

It is really hard to understand what DSP books are talking about....

>On 25 A=C4=9Fustos, 16:04, "Michael Plante" <michael.pla...@gmail.com>
>wrote:
>> >On 25 A=3DC4=3D9Fustos, 12:34, rotor cli <rotor...@gmail.com> wrote:
>>
>> >Are there 2 different digital frequency concept?
>>
>> >1-) In DSP books, they say:
>> >omega =3D3D 2Pi*(fa/fs)
>> >Here, there is a sampling so digital frequency is related with time
>> >and sampling ratio.
>>
>> >2-)In image processing:
>> >Frequency is about pixels values. How fast values are changed. For
>> >example:
>> >High frequency:
>> >0 255 0 255 0 255 0 255
>> >255 0 255 0 255 0 255 0
>> >0 255 0 255 0 255 0 255
>> >255 0 255 0 255 0 255 0
>> >0 255 0 255 0 255 0 255
>> >255 0 255 0 255 0 255 0
>>
>> >Low frequency:
>> >255 255 255 255
>> >255 255 255 255
>> >255 255 255 255
>>
>> >Here, there is no time relation. But in first expression analog
>> >frequency is involved.
>>
>> >I am confused.
>>
>> There's a thing called "spatial frequency". =C2=A0The analogy i
roughly
>> wavenumber (m^(-1)) is to frequency (s^(-1)=3DHz) as wavelength (m) i
to
>> period (s). =C2=A0Or you could use pixels instead of meters. =C2=A0It'
a=
>ll the same
>> idea, so the two cases really only differ in physical interpretatio
(and
>> 1d vs 2d).
>>
>> Moreover, your image is still sampled into pixels: =C2=A0try resizin
bot=
>h of
>> those images (add more pixels for this experiment. =C2=A0If you tak
pixe=
>ls
>> away, you'll alias in this case). =C2=A0You are looking at high an
low
>> frequency as relative concepts, compared to one another, or compared t
t=
>he
>> Nyquist rate...either way. =C2=A0In the former case, you would stil
call=
> those
>> "high" and "low", but in the latter case (compared to Nyquist after
>> doubling the image size), you might call them "medium" and "low".
>
>Hmm.
>For you first paragraph:
>Can you please make the analogy for fa and fs with pixels. For
>"spatial domain"..
>I mean how can fa and fs be related wit pixel example.
>fa =3D pixel count? or fs =3D pixel count? or something else?

I should not have analogized pixels to meters. Pixels are more lik
samples, and meters are more like seconds. Just like I could not tell yo
the sampling rate if you gave me a time array, I cannot give you "samplin
rate" for your image. This is a function of the camera and an
post-processing. It is often approximated with an instantaneous field o
view (IFOV), the angle subtended by that pixel (say in the horizonta
direction, e.g.). Of course, if you know the distance to your target an
the dot product with the target's normal vector, you may prefer to us
units of length along the target instead of angle across the target. Th
relation to the real world is not defined in your thought experiment. Onc
you know the relation of physical units to pixels, you can determine fs.

Much like with the analogous array of numbers representing a tim
sequence, all that can be given for an array of numbers representing a
image is fa/fs, unless you have more information about your image. Eac
scan line of the first image is at the highest representable frequency, an
the second is at DC.

>>On 25 A=C4=9Fustos, 16:04, "Michael Plante" <michael.pla...@gmail.com>
>>wrote:
>>> >On 25 A=3DC4=3D9Fustos, 12:34, rotor cli <rotor...@gmail.com> wrote:
>>>
>>> >Are there 2 different digital frequency concept?
>>>
>>> >1-) In DSP books, they say:
>>> >omega =3D3D 2Pi*(fa/fs)
>>> >Here, there is a sampling so digital frequency is related with time
>>> >and sampling ratio.
>>>
>>> >2-)In image processing:
>>> >Frequency is about pixels values. How fast values are changed. For
>>> >example:
>>> >High frequency:
>>> >0 255 0 255 0 255 0 255
>>> >255 0 255 0 255 0 255 0
>>> >0 255 0 255 0 255 0 255
>>> >255 0 255 0 255 0 255 0
>>> >0 255 0 255 0 255 0 255
>>> >255 0 255 0 255 0 255 0
>>>
>>> >Low frequency:
>>> >255 255 255 255
>>> >255 255 255 255
>>> >255 255 255 255
>>>
>>> >Here, there is no time relation. But in first expression analog
>>> >frequency is involved.
>>>
>>> >I am confused.
>>>
>>> There's a thing called "spatial frequency". =C2=A0The analogy is
>roughly
>>> wavenumber (m^(-1)) is to frequency (s^(-1)=3DHz) as wavelength (m
is
>to
>>> period (s). =C2=A0Or you could use pixels instead of meters
=C2=A0It's
>a=
>>ll the same
>>> idea, so the two cases really only differ in physical interpretation
>(and
>>> 1d vs 2d).
>>>
>>> Moreover, your image is still sampled into pixels: =C2=A0try resizing
>bot=
>>h of
>>> those images (add more pixels for this experiment. =C2=A0If you take
>pixe=
>>ls
>>> away, you'll alias in this case). =C2=A0You are looking at high and
>low
>>> frequency as relative concepts, compared to one another, or compare
to
>t=
>>he
>>> Nyquist rate...either way. =C2=A0In the former case, you would still
>call=
>> those
>>> "high" and "low", but in the latter case (compared to Nyquist after
>>> doubling the image size), you might call them "medium" and "low".
>>
>>Hmm.
>>For you first paragraph:
>>Can you please make the analogy for fa and fs with pixels. For
>>"spatial domain"..
>>I mean how can fa and fs be related wit pixel example.
>>fa =3D pixel count? or fs =3D pixel count? or something else?
>
>I should not have analogized pixels to meters. Pixels are more like
>samples, and meters are more like seconds. Just like I could not tel
you
>the sampling rate if you gave me a time array, I cannot give yo
"sampling
>rate" for your image. This is a function of the camera and any
>post-processing. It is often approximated with an instantaneous fiel
of
>view (IFOV), the angle subtended by that pixel (say in the horizontal
>direction, e.g.). Of course, if you know the distance to your targe
and
>the dot product with the target's normal vector, you may prefer to use
>units of length along the target instead of angle across the target.
The
>relation to the real world is not defined in your thought experiment.
Once
>you know the relation of physical units to pixels, you can determine fs.
>
>Much like with the analogous array of numbers representing a time
>sequence, all that can be given for an array of numbers representing an
>image is fa/fs, unless you have more information about your image. Each
>scan line of the first image is at the highest representable frequency
and
>the second is at DC.

To give firmer numbers for a camera, say the horizontal field of view i
60 degrees, and the vertical field of view is 45 degrees. We proceed t
capture the image at 320x240. Then the pixels are square and the IFOV i
each direction is 0.1875 degrees/pixel, or, equivalently, 16/
samples/degree. This is a sampling rate. (This assumes tan theta i
approximately theta, which is good enough for me, given all the spatia
distortion my camera introduces, and the fact I use it for closed-loo
control of the platform holding the camera. If I needed to do stuff usin
just one image, the computations would be more complex to better mode
reality. *Caveat to numbers given in this post*) Proceeding, if a featur
has a horizontally-varying intensity, and the period of that intensit
sinusoid is 3/8 of a degree, then I would obtain a scan line similar to on
in the image you gave. This might happen if you take a picture of a
nicely-oriented and pixel-aligned chessboard, where one square is 3/16 of a
degree (1 sample!) across, since 2 chessboard squares are one period of the
wave. From this numerical example, you could obtain fa and fs, rather than
just the ratio. If I doubled the sampling rate (fs) to 32/3 samples/degree
by capturing at 640x480, then I could now resolve periods of 3/16 of a
degree, instead of just 3/8, and the preceding high-frequency scan line
would now be at half Nyquist.




>Yes, (as Nyquist)if fa =3D x and fs =3D 2x then 2Pi*(fa/fs) =3D pi. OK.
>But if we increase sample ratio then pi increases, so why do books say
low
>frequencies are around 2Pi?

If you double the sample rate, the normalized frequency should be cut in
half, since you can now represent a higher real-world frequency. There are
certain weird wrap-around things people do with the normalized frequency,
and sometimes the range appears to be defined differently for real-valued
signals (which your image is). I couldn't give a hard answer about the low
frequency part, but, being primarily an analog person, I prefer to take DC
at 0. This would give a trivial answer to the low-frequency image: double
the image size, and 0 over 2 is still 0 (of course, if the "DC" image is
actually aliased, sampling at a higher rate would give a different answer
than resizing the already-sampled image).

On 25 AÄŸustos, 18:54, "Michael Plante" <michael.pla...@gmail.com>
wrote:
> >>On 25 A=C4=9Fustos, 16:04, "Michael Plante" <michael.pla...@gmail.com>
> >>wrote:
> >>> >On 25 A=3DC4=3D9Fustos, 12:34, rotor cli <rotor...@gmail.com> wrote:
>
> >>> >Are there 2 different digital frequency concept?
>
> >>> >1-) In DSP books, they say:
> >>> >omega =3D3D 2Pi*(fa/fs)
> >>> >Here, there is a sampling so digital frequency is related with time
> >>> >and sampling ratio.
>
> >>> >2-)In image processing:
> >>> >Frequency is about pixels values. How fast values are changed. For
> >>> >example:
> >>> >High frequency:
> >>> >0 255 0 255 0 255 0 255
> >>> >255 0 255 0 255 0 255 0
> >>> >0 255 0 255 0 255 0 255
> >>> >255 0 255 0 255 0 255 0
> >>> >0 255 0 255 0 255 0 255
> >>> >255 0 255 0 255 0 255 0
>
> >>> >Low frequency:
> >>> >255 255 255 255
> >>> >255 255 255 255
> >>> >255 255 255 255
>
> >>> >Here, there is no time relation. But in first expression analog
> >>> >frequency is involved.
>
> >>> >I am confused.
>
> >>> There's a thing called "spatial frequency". =C2=A0The analogy is
> >roughly
> >>> wavenumber (m^(-1)) is to frequency (s^(-1)=3DHz) as wavelength (m)
> is
> >to
> >>> period (s). =C2=A0Or you could use pixels instead of meters.
> =C2=A0It's
> >a=
> >>ll the same
> >>> idea, so the two cases really only differ in physical interpretation
> >(and
> >>> 1d vs 2d).
>
> >>> Moreover, your image is still sampled into pixels: =C2=A0try resizing
> >bot=
> >>h of
> >>> those images (add more pixels for this experiment. =C2=A0If you take
> >pixe=
> >>ls
> >>> away, you'll alias in this case). =C2=A0You are looking at high and
> >low
> >>> frequency as relative concepts, compared to one another, or compared
> to
> >t=
> >>he
> >>> Nyquist rate...either way. =C2=A0In the former case, you would still
> >call=
> >> those
> >>> "high" and "low", but in the latter case (compared to Nyquist after
> >>> doubling the image size), you might call them "medium" and "low".
>
> >>Hmm.
> >>For you first paragraph:
> >>Can you please make the analogy for fa and fs with pixels. For
> >>"spatial domain"..
> >>I mean how can fa and fs be related wit pixel example.
> >>fa =3D pixel count? or fs =3D pixel count? or something else?
>
> >I should not have analogized pixels to meters. Â*Pixels are more like
> >samples, and meters are more like seconds. Â*Just like I could not tell
> you
> >the sampling rate if you gave me a time array, I cannot give you
> "sampling
> >rate" for your image. Â*This is a function of the camera and any
> >post-processing. Â*It is often approximated with an instantaneous field
> of
> >view (IFOV), the angle subtended by that pixel (say in the horizontal
> >direction, e.g.). Â*Of course, if you know the distance to your target
> and
> >the dot product with the target's normal vector, you may prefer to use
> >units of length along the target instead of angle across the target.
> The
> >relation to the real world is not defined in your thought experiment.
> Once
> >you know the relation of physical units to pixels, you can determine fs.
>
> >Much like with the analogous array of numbers representing a time
> >sequence, all that can be given for an array of numbers representing an
> >image is fa/fs, unless you have more information about your image. Â*Each
> >scan line of the first image is at the highest representable frequency,
> and
> >the second is at DC.
>
> To give firmer numbers for a camera, say the horizontal field of view is
> 60 degrees, and the vertical field of view is 45 degrees. Â*We proceed to
> capture the image at 320x240. Â*Then the pixels are square and the IFOV in
> each direction is 0.1875 degrees/pixel, or, equivalently, 16/3
> samples/degree. Â*This is a sampling rate. Â*(This assumes tan theta is
> approximately theta, which is good enough for me, given all the spatial
> distortion my camera introduces, and the fact I use it for closed-loop
> control of the platform holding the camera. Â*If I needed to do stuffusing
> just one image, the computations would be more complex to better model
> reality. Â**Caveat to numbers given in this post*) Â*Proceeding, if a feature
> has a horizontally-varying intensity, and the period of that intensity
> sinusoid is 3/8 of a degree, then I would obtain a scan line similar to one
> in the image you gave. Â*This might happen if you take a picture of a
> nicely-oriented and pixel-aligned chessboard, where one square is 3/16 ofa
> degree (1 sample!) across, since 2 chessboard squares are one period of the
> wave. Â*From this numerical example, you could obtain fa and fs, rather than
> just the ratio. Â*If I doubled the sampling rate (fs) to 32/3 samples/degree
> by capturing at 640x480, then I could now resolve periods of 3/16 of a
> degree, instead of just 3/8, and the preceding high-frequency scan line
> would now be at half Nyquist.
>
>
>
> >Yes, (as Nyquist)if fa =3D x and fs =3D 2x then 2Pi*(fa/fs) =3D pi.. OK.
> >But if we increase sample ratio then pi increases, so why do books say
> low
> >frequencies are around 2Pi?
>
> If you double the sample rate, the normalized frequency should be cut in
> half, since you can now represent a higher real-world frequency. Â*There are
> certain weird wrap-around things people do with the normalized frequency,
> and sometimes the range appears to be defined differently for real-valued
> signals (which your image is). Â*I couldn't give a hard answer about the low
> frequency part, but, being primarily an analog person, I prefer to take DC
> at 0. Â*This would give a trivial answer to the low-frequency image: Â*double
> the image size, and 0 over 2 is still 0 (of course, if the "DC" image is
> actually aliased, sampling at a higher rate would give a different answer
> than resizing the already-sampled image).

Thanks Michael Plante

On 25 Aug, 15:49, rotor cli <rotor...@gmail.com> wrote:
> On 25 AÄŸustos, 16:04, "Michael Plante" <michael.pla...@gmail.com>
> wrote:
>
>
>
>
>
> > >On 25 A=C4=9Fustos, 12:34, rotor cli <rotor...@gmail.com> wrote:
>
> > >Are there 2 different digital frequency concept?
>
> > >1-) In DSP books, they say:
> > >omega =3D 2Pi*(fa/fs)
> > >Here, there is a sampling so digital frequency is related with time
> > >and sampling ratio.
>
> > >2-)In image processing:
> > >Frequency is about pixels values. How fast values are changed. For
> > >example:
> > >High frequency:
> > >0 255 0 255 0 255 0 255
> > >255 0 255 0 255 0 255 0
> > >0 255 0 255 0 255 0 255
> > >255 0 255 0 255 0 255 0
> > >0 255 0 255 0 255 0 255
> > >255 0 255 0 255 0 255 0
>
> > >Low frequency:
> > >255 255 255 255
> > >255 255 255 255
> > >255 255 255 255
>
> > >Here, there is no time relation. But in first expression analog
> > >frequency is involved.
>
> > >I am confused.
>
> > There's a thing called "spatial frequency". Â*The analogy is roughly
> > wavenumber (m^(-1)) is to frequency (s^(-1)=Hz) as wavelength (m) is to
> > period (s). Â*Or you could use pixels instead of meters. Â*It'sall the same
> > idea, so the two cases really only differ in physical interpretation (and
> > 1d vs 2d).
>
> > Moreover, your image is still sampled into pixels: Â*try resizing both of
> > those images (add more pixels for this experiment. Â*If you take pixels
> > away, you'll alias in this case). Â*You are looking at high and low
> > frequency as relative concepts, compared to one another, or compared tothe
> > Nyquist rate...either way. Â*In the former case, you would still call those
> > "high" and "low", but in the latter case (compared to Nyquist after
> > doubling the image size), you might call them "medium" and "low".
>
> Hmm.
> For you first paragraph:
> Can you please make the analogy for fa and fs with pixels. For
> "spatial domain"..
> I mean how can fa and fs be related wit pixel example.
> fa = pixel count? or fs = pixel count? or something else?
>
> For second paragraph:
> Books say: Â*high frequencies are around pi and low frequencies are
> around 0or(2Pi) in frequency domain.
> What is that mean?
>
> Yes, (as Nyquist)if fa = x and fs = 2x then 2Pi*(fa/fs) = pi. OK. But
> if we increase sample ratio then pi increases, so why do books say low
> frequencies are around 2Pi?
>
> It is really hard to understand what DSP books are talking about....- Hide quoted text -
>
> - Show quoted text -

If you have ever seen a film where a car is driving forward yet the
wheels are spinning in reverse, then you have experienced aliasing.
You need to draw some pictures with a pencil and paper to grasp what
is happening.