design the output for a low pass filter for a given signal

[sameer7]

Hi All,

I faced the below scenario , so I want to take your suggestions after I put forward my approach.

we have a signal and impulse ,𝑥(𝑡)=2⋅sin150𝜋 +sin250𝜋 and 𝑔(𝑡)=𝑥(𝑡)sin250𝜋 . We pass the signal 𝑔(𝑡) through an ideal lowpass filter with cutoff frequency 300𝜋 and passband gain of 3. what is the output of the Low Pass filter in this case?

As a solution approach, I multiplied x(t) with g(t) to get the sine product(and remove those values of frequencies above cutoff in the derived formula since it is a low pass filter) but I am not sure what I should do with passband gain? Could anyone suggest the possible approach if my assumptions for the above problem looks fine?

Kind regards
Sameer

 

[barry]

Your post doesn’t make sense.

1) you say you have a “signal and impulse“, but then define two signals. I don’t see an impulse.
2) why do you multiply x(t) by g(t)? You’ve already defined g(t) as a function of x(t).

i think you need to restate your problem. also, I think your missing ‘t’ factors in your signal definitions.

 

[sameer7]

Your post doesn’t make sense.

1) you say you have a “signal and impulse“, but then define two signals. I don’t see an impulse.
2) why do you multiply x(t) by g(t)? You’ve already defined g(t) as a function of x(t).

i think you need to restate your problem. also, I think your missing ‘t’ factors in your signal definitions.

you are right.My bad ,I am sorry,it was a misplacement and a typo .Yes g(t) is related to x(t) and we need to evaluate g(t) by multiplying x(t) and sin250 pi with a consideration of the cut off frequency 300pi and passband gain of 3.


I am not sure what to do with passgain value in the final product formula...please advice ...

 

[kaz1]

you are right.My bad ,I am sorry,it was a misplacement and a typo .Yes g(t) is related to x(t) and we need to evaluate g(t) by multiplying x(t) and sin250 pi with a consideration of the cut off frequency 300pi and passband gain of 3.


I am not sure what to do with passgain value in the final product formula...please advice ...

If I run:
x(t) = 2*sin(150*t*pi) + sin(250*t*pi); blue plot below, it doesn't look that nice, very low values.
g(t) = x(t)*sin(250*t*pi); green plot, looks worse, very very low values.

As mentioned above you need to redefine your input before going to filter gain issue.

 

[barry]

If I run:
x(t) = 2*sin(150*t*pi) + sin(250*t*pi); blue plot below, it doesn't look that nice, very low values.
g(t) = x(t)*sin(250*t*pi); green plot, looks worse, very very low values.

View attachment 173607

As mentioned above you need to redefine your input before going to filter gain issue.

Your plots are very wrong.

 

[kaz1]

Your plots are very wrong.

True but I used arbitrary t of 0:1023.
if I use t = 0:1 for 1024 samples I get better result. The post does not define time:

 

[sameer7]

True but I used arbitrary t of 0:1023.
if I use t = 0:1 for 1024 samples I get better result. The post does not define time:

View attachment 173609

Actually the time duration has to be taken any arbitrary value as taken above by you..

In the above case scenario based on your signal sample time t 0:1024,how does the final output in look like if cutoff frequency for low pass is defined as 300pi and passgain as 3?

I am little confused that will be require to multiply pass gain as a factor and then evaluate the cut off?

 

[kaz1]

Hi All,

I faced the below scenario , so I want to take your suggestions after I put forward my approach.

we have a signal and impulse ,𝑥(𝑡)=2⋅sin150𝜋 +sin250𝜋 and 𝑔(𝑡)=𝑥(𝑡)sin250𝜋 . We pass the signal 𝑔(𝑡) through an ideal lowpass filter with cutoff frequency 300𝜋 and passband gain of 3. what is the output of the Low Pass filter in this case?

As a solution approach, I multiplied x(t) with g(t) to get the sine product(and remove those values of frequencies above cutoff in the derived formula since it is a low pass filter) but I am not sure what I should do with passband gain? Could anyone suggest the possible approach if my assumptions for the above problem looks fine?

Kind regards
Sameer

Your cutoff of 300pi is not meaningful to me and passband gain of 3 possibly means 3dB.
Cutoff ratio must be relative to either sampling rate (1 or 2pi) or relative to Nyquist (0.5 or pi). I am not clear about your equations using 150 pi, 250 pi and so on.

 

[barry]

OP has a problem with writing "t", but:
300pi*t=> 150 Hz.

 

[barry]

Also, there's nothing to indicate this is a digital filter. Not sure why you're bringing sample rate into the discussion.

 

[andre_luis]

there's nothing to indicate this is a digital filter.

Respectfully, I must remind that OP posted in the Digital Signal Processing section, not at the Analog Circuit Design section.