Amplitude Response ?

Hi guys,

I'm new to this stuff. I want to calculate the amplitude response so I can 
graph it, I'm using this example:

y(n)=x(n)+x(n-1)

Amplitude Response: G(w)=2.0*cos(Pi*f*T);

According to the tutorial I'm using I think:

f=frequency;
w=2.0*Pi*f;
T=1.0/sample_rate;

...but I'm not really sure because it's all so confusing.

How do I get meaningful values out of that, meaning the gain going from 2.0 
down to 0.0 with frequencies from 0.0 to 22050.0 (half the sample rate) ?

Am I on the right track here ?

Thanks.

PS - I'm doing this all in C(++) so please don't give any MATLAB 
instructions or anything like that.

Robert A. skrev:
> Hi guys,
>
> I'm new to this stuff. I want to calculate the amplitude response so I can
> graph it, I'm using this example:
>
> y(n)=x(n)+x(n-1)
>
> Amplitude Response: G(w)=2.0*cos(Pi*f*T);
>
> According to the tutorial I'm using I think:
>
> f=frequency;
> w=2.0*Pi*f;
> T=1.0/sample_rate;
>
> ...but I'm not really sure because it's all so confusing.
>
> How do I get meaningful values out of that, meaning the gain going from 2.0
> down to 0.0 with frequencies from 0.0 to 22050.0 (half the sample rate) ?
>
> Am I on the right track here ?

One reason that might have caused your confusion, is that lots of
DSP stuff is explained in "normalized angular frequency", that is,
the frequency is expressed as

w = 2pi f = [0,...,2*pi]

where

f = [0,...,1]

while the useful stuff relates to "physical frequency" F. Now, the
key to unravel this is the sampling frequency (or sampling rate) Fs,

Fs = 1/T.

Normalizing from physical frequency F to normalized frequency
is simple, 

f = F/Fs = F/(1/T) = FT.

Did that help?

Rune

Robert A. wrote:

> Hi guys,
> 
> I'm new to this stuff. I want to calculate the amplitude response so I can 
> graph it, I'm using this example:
> 
> y(n)=x(n)+x(n-1)
> 
> Amplitude Response: G(w)=2.0*cos(Pi*f*T);
> 
> According to the tutorial I'm using I think:
> 
> f=frequency;
> w=2.0*Pi*f;
> T=1.0/sample_rate;
> 
> ...but I'm not really sure because it's all so confusing.
> 
> How do I get meaningful values out of that, meaning the gain going from 2.0 
> down to 0.0 with frequencies from 0.0 to 22050.0 (half the sample rate) ?
> 
> Am I on the right track here ?
> 
> Thanks.
> 
> PS - I'm doing this all in C(++) so please don't give any MATLAB 
> instructions or anything like that.
> 
> 
This is a job for the z transform.  Learning the transform itself is a 
bit of a pain, but once you know it many things come clear.

http://www.wescottdesign.com/articles/zTransform/z-transforms.html

No Matlab!

Robert Lyon's book "Understanding Digital Signal Processing" has a 
chapter on the z transform, as well as lots of other useful DSP stuff -- 
here's a link to buy it: 
http://www.powells.com/partner/30696/biblio/0-13-108989-7.  My book, 
"Applied Control Theory for Embedded Systems" also has a chapter (based 
on the above article), but mine is really more about feedback control 
than DSP.  Should you be interested anyway, here's a link to buy _it_: 
http://www.powells.com/partner/30696/biblio/0750678399.

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google?  See http://cfaj.freeshell.org/google/

"Robert A." wrote:
> 
> Hi guys,
> 
> I'm new to this stuff. I want to calculate the amplitude response so I can
> graph it, I'm using this example:
> 
> y(n)=x(n)+x(n-1)
> 
> Amplitude Response: G(w)=2.0*cos(Pi*f*T);
> 
> According to the tutorial I'm using I think:
> 
> f=frequency;
> w=2.0*Pi*f;
> T=1.0/sample_rate;
> 

Yes this looks correct.

> ...but I'm not really sure because it's all so confusing.
> 
> How do I get meaningful values out of that, meaning the gain going from 2.0
> down to 0.0 with frequencies from 0.0 to 22050.0 (half the sample rate) ?



 Plugging in 22050 for f and 1/44100 for T you get 0 for the gain at
fs/2. Plugging in  0 for f gives you a gain of 2 at DC. 
Where is the problem?

-jim

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On Sun, 16 Apr 2006 14:05:47 -0700, Tim Wescott <tim@seemywebsite.com>
wrote:

>Robert A. wrote:
>
>> Hi guys,
>> 
>> I'm new to this stuff. I want to calculate the amplitude response so I can 
>> graph it, I'm using this example:
>> 
>> y(n)=x(n)+x(n-1)
>> 
>> Amplitude Response: G(w)=2.0*cos(Pi*f*T);
>> 
>> According to the tutorial I'm using I think:
>> 
>> f=frequency;
>> w=2.0*Pi*f;
>> T=1.0/sample_rate;
>> 
>> ...but I'm not really sure because it's all so confusing.
>> 
>> How do I get meaningful values out of that, meaning the gain going from 2.0 
>> down to 0.0 with frequencies from 0.0 to 22050.0 (half the sample rate) ?
>> 
>> Am I on the right track here ?
>> 
>> Thanks.
>> 
>> PS - I'm doing this all in C(++) so please don't give any MATLAB 
>> instructions or anything like that.
>> 
>> 
>This is a job for the z transform.  Learning the transform itself is a 
>bit of a pain, but once you know it many things come clear.
>
  (snipped)

Hi Tim,
   you're not kiddin' that the z-transform 
makes things clear.  If Robert gets serious about learning 
DSP, he can learn how the z-transform allows him 
to start with his "y(n)=x(n)+x(n-1)" difference equation 
and derive the G(w) freq response equation himself.

When I was first exposed to the z-transform, I thought 
that this transform was merely a way for university 
professors to make our lives miserable.  Not so!
Knowing how to use the transform as a tool allows us 
to understand and analyze discrete networks 
in a very thorough way.

Thanks for "preachin' the z-transform gospel" Tim.

[-Rick-]
Quote of the Day:
   "Research is what I'm doing when I don't know 
    what I'm doing."   -Wernher von Braun

Rick Lyons skrev:

> Quote of the Day:
>    "Research is what I'm doing when I don't know
>     what I'm doing."   -Wernher von Braun

So...  a "researcher" doesn't know what he does?

Rune

On Sun, 16 Apr 2006 16:49:17 -0400, "Robert A." <invalid@invalid.org>
wrote:

>Hi guys,
>
>I'm new to this stuff. I want to calculate the amplitude response so I can 
>graph it, I'm using this example:
>
>y(n)=x(n)+x(n-1)
>
>Amplitude Response: G(w)=2.0*cos(Pi*f*T);
>
>According to the tutorial I'm using I think:
>
>f=frequency;
>w=2.0*Pi*f;
>T=1.0/sample_rate;
>
>...but I'm not really sure because it's all so confusing.
>
>How do I get meaningful values out of that, meaning the gain going from 2.0 
>down to 0.0 with frequencies from 0.0 to 22050.0 (half the sample rate) ?
>
>Am I on the right track here ?
>
>Thanks.

Hi,
  Here's my two cents:
I think you're on the right track.

Think, for a moment about your
y(n)=x(n)+x(n-1) equation.  It says the 
current output sample is equal to the current 
input sample plus the previous input sample.

OK, now think about a zero-hertz (DC) input such as 
a long sequence of samples whose values were all seven (7).
In this case your output sequence would be a long sequence 
of samples whose values are all 14.  So the system's 
gain at zero hertz is output/input = 14/7 = 2.

Now think about an input sinusoid (with peak value 
of 7) at the highest 
frequency that we have in our world of DSP.
That frequency is half the sample rate.  So an 
input sequence of +7, -7, +7, -7, etc.
would be a discrete sinewave whose freq is half the 
sample rate.  With such an input sequence, what would be 
the output of your system?

That's right, the output would be a sequence 
of all zeros.  So the system's gain at half the sample 
rate is output/input = 0/7 = 0.

Hope that made some sense.

Tim 
Good Luck,
[-Rick-]

Rune Allnor wrote:
> Rick Lyons skrev:
> 
> 
>>Quote of the Day:
>>   "Research is what I'm doing when I don't know
>>    what I'm doing."   -Wernher von Braun
> 
> 
> So...  a "researcher" doesn't know what he does?

Herr Braun was too stark. Research is what he's doing when he doesn't 
know what the result will be. "

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Jerry Avins skrev:
> Rune Allnor wrote:
> > Rick Lyons skrev:
> >
> >
> >>Quote of the Day:
> >>   "Research is what I'm doing when I don't know
> >>    what I'm doing."   -Wernher von Braun
> >
> >
> > So...  a "researcher" doesn't know what he does?
>
> Herr Braun was too stark. Research is what he's doing when he doesn't
> know what the result will be. "

Wouldn't somebody who knows his trade but is doing new stuff
related to his trade be a scientist?

What would one call somebody who does stuff that is "trivial"
to some profession but new or unknown to himself, and gets
paid for it?

All right, you have seen enough of my previous rants on the
subject, so I'll leave it there.

Rune

Tim Wescott wrote:

> Robert A. wrote:
>
> > Hi guys,
> >
> > I'm new to this stuff. I want to calculate the amplitude response so I can
> > graph it, I'm using this example:
> >
> > y(n)=x(n)+x(n-1)                                           (1)
...
> This is a job for the z transform.  Learning the transform itself is a
> bit of a pain, but once you know it many things come clear.

There really is no need to invoke the whole z-transform machinery for
such a trivial example. All you need to know is that equation (1)
defines an LTI system, which means that a sinusoidal input results in a
sinusoidal output with scaled amplitude and shifted phase, ie.

L x(n) := x(n) + x(n-1)
=> L cos(w n) = A(w) cos(w n + phi(w) ), for any 0<= w < pi.

Now you can compute A(w) and  phi(w) using trigonometric identies:

L cos(w n)
= cos(w n) + cos(w (n-1))
= cos(w n) + cos(w n) cos(w) + sin(w n) sin(w)
= cos(w n) (1+ cos(w) ) + sin(w n) sin(w)
= 2 cos(w/2) cos(w n - w/2)

Regards,
Andor