IIR Filter Design

I was given a continous signal
 
x(t) =
1.272*cos(2*pi*50*t)−0.424*cos(2*p*150*t)+0.255*cos(2*pi*250*t)
and I generated discrete signal {x(n)}, 0 <=n <=256. Sampling frequnecy is
fs = 1000 Hz.

Now I need to design a Discrete low pass filter (Butterworth ili
Chebishev)that will  attenuate 150 Hz signal componentat most 1dB and
attenuate 250 Hz signal component not more than 40 dB.

I appreciate your help,

Marie

On 7 Mar, 13:39, "jerseygirl" <odmenet...@yahoo.com> wrote:
> I was given a continous signal
>
> x(t) =
> 1.272*cos(2*pi*50*t)-0.424*cos(2*p*150*t)+0.255*cos(2*pi*250*t)
> and I generated discrete signal {x(n)}, 0 <=n <=256. Sampling frequnecy is
> fs = 1000 Hz.
>
> Now I need to design a Discrete low pass filter (Butterworth ili
> Chebishev)that will  attenuate 150 Hz signal componentat most 1dB and
> attenuate 250 Hz signal component not more than 40 dB.
>
> I appreciate your help,

Homework? There ought to be a recipe for this in
your textbook. The general recipe is as follows:

1) Set up the specification (attenuation or ripple,
   corner frequencies) in discrete time domain.
2) Pre-warp these to analog domain
3) Determine the filter order from the analog domain
   specification
4) Compute the poles and zeros in analog domain
5) Use the bilinear transform on each 2nd or 1st
   order section to transfer the filter back to
   discrete time domain
6) Use a frequency transform to get your final
   filter.

Rune

On Mar 7, 7:39 am, "jerseygirl" <odmenet...@yahoo.com> wrote:
> I was given a continous signal
>
> x(t) =3D
> 1.272*cos(2*pi*50*t)=E2=88=920.424*cos(2*p*150*t)+0.255*cos(2*pi*250*t)
> and I generated discrete signal {x(n)}, 0 <=3Dn <=3D256. Sampling frequne=
cy is
> fs =3D 1000 Hz.
>
> Now I need to design a Discrete low pass filter (Butterworth ili
> Chebishev)that will  attenuate 150 Hz signal componentat most 1dB and
> attenuate 250 Hz signal component not more than 40 dB.
>
> I appreciate your help,
>
> Marie

Those are strange filter specifications. Usually, a Butterworth filter
is specified in terms of its 3-dB point (the passband edge); the 1-dB
value you give sounds like something that you would use for a type-I
Chebyshev filter (maximum passband ripple specification). And, I think
you probably mean "attenuate a 250 Hz signal component by at least 40
dB", as the stopband behavior is usually given in a "minimum
attenuation" number. In any event, as Rune said, this should be
covered in most textbooks out there. If you care, there's a
particularly good treatment (IMHO) of this topic in "A Course in
Digital Signal Processing" by Porat.

Jason

On 7 Mar, 14:19, cincy...@gmail.com wrote:
> On Mar 7, 7:39 am, "jerseygirl" <odmenet...@yahoo.com> wrote:
>
> > I was given a continous signal
>
> > x(t) =3D
> > 1.272*cos(2*pi*50*t)=E2=88=920.424*cos(2*p*150*t)+0.255*cos(2*pi*250*t)
> > and I generated discrete signal {x(n)}, 0 <=3Dn <=3D256. Sampling frequ=
necy is
> > fs =3D 1000 Hz.
>
> > Now I need to design a Discrete low pass filter (Butterworth ili
> > Chebishev)that will =C2=A0attenuate 150 Hz signal componentat most 1dB =
and
> > attenuate 250 Hz signal component not more than 40 dB.
>
> > I appreciate your help,
>
> > Marie
>
> Those are strange filter specifications. Usually, a Butterworth filter
> is specified in terms of its 3-dB point (the passband edge); the 1-dB
> value you give sounds like something that you would use for a type-I
> Chebyshev filter (maximum passband ripple specification).

It is perfectly possible to design a Butterworth filter
with this sort of ripple. It only means that the 3 dB
point is somewhere in the transition band, not on the
passband corner frequency.

> And, I think
> you probably mean "attenuate a 250 Hz signal component by at least 40
> dB", as the stopband behavior is usually given in a "minimum
> attenuation" number.

You are right, but provided the OP cites the spec
correctly, it is a perfectly possible -- if maybe
a bit unreasonable -- specification. The one filter
that meets the spec is the one that attenuates
exactly 40 dB on the passband corner. Which might
be the kind of devious twist I might give in a
homework excercise...

Rune