Hey everyone, There is a small mistake in y2(n). y2(n)={ 7, 2, -14, -12, -5, 10, 16, 32} And the final y(n) will be, y(n) = { 2, 2, 10, 8, 18, -13, -14, -12, -5, 10, 16, 32, 9, 9} Please note the correction and apologize for the inconvenience.
@@EnggClasses we've y1(n)={-12,-9,2,2,10,8,18,-13} after discarding first two numbers we are left with {2,2,10,8,18,-13......... But u have taken 12, -6 in final y(n) (comment section )
Bro because this is nothing but filter output right, we know tht linear convolution gives the filter output of length m+n-1, here m=12 and n=3,=> finally answer output length= 14, last 2 places bcz of X3...
Lenght of impulse wala =3, so we use 3-1=2 previous samples from input signal and we remove 2samples from output to remove points corrupted by aliasinsg
Yes Summaiya, i have updated the same in comment section. You have got your Maths correct, Well done.. Thanks for the concern. h(n) is of length 3 and x(n) is of length 12. Therefore final y(n) must be of length 12+3-1= 14. If you consider only y1(n) and y2(n), you will get y(n) of length 12, to get remaining 2 samples you have to consider y3(n) as well. Hope this helps.
Hey everyone,
There is a small mistake in y2(n).
y2(n)={ 7, 2, -14, -12, -5, 10, 16, 32}
And the final y(n) will be,
y(n) = { 2, 2, 10, 8, 18, -13, -14, -12, -5, 10, 16, 32, 9, 9}
Please note the correction and apologize for the inconvenience.
Sir, may I know why did u take 12,-6 instead of 2,2 in y(n) (in comment section)
12, -6 come from y1(n).
I'm not getting how do you get 2 and 2.
Plz let me know.
@@EnggClasses we've
y1(n)={-12,-9,2,2,10,8,18,-13}
after discarding first two numbers we are left with {2,2,10,8,18,-13.........
But u have taken 12, -6 in final y(n) (comment section )
You are right. Corrected.
Thank you so much.
Yes sir I got that ans 7,-14,10
Sir why u toook x1,x2,x3 when we can windup in x1,x2 by taking 6 coefficient each
Yes Sir same doubt
Could you please explain me what is the purpose of taking x3
same doubt anyone explain
Bro because this is nothing but filter output right, we know tht linear convolution gives the filter output of length m+n-1, here m=12 and n=3,=> finally answer output length= 14, last 2 places bcz of X3...
If the value of n was not given then how to calculate L ?
sir - why 2 samples from previous block ?= thank u sir
Because m-1 =2 if it is 3 then we have to take 3 previous samples..
Lenght of impulse wala =3, so we use 3-1=2 previous samples from input signal
and we remove 2samples from output to remove points corrupted by aliasinsg
🎉😊
Sir,y2[n] is wrong
It's y2[n]=[7 2 -14 -12 -5 10 16 32]
And y do we have to consider y3 [n]?
Yes Summaiya, i have updated the same in comment section. You have got your Maths correct, Well done..
Thanks for the concern.
h(n) is of length 3 and x(n) is of length 12. Therefore final y(n) must be of length 12+3-1= 14.
If you consider only y1(n) and y2(n), you will get y(n) of length 12, to get remaining 2 samples you have to consider y3(n) as well.
Hope this helps.
@@EnggClasses thank you sir
How can we know in advance the length of output if we consider only y1 and y2 ?m
Why elimate the fst two value there is any reason
This is overlap save method... We have to discard (M-1) values on the left hand side...