Mixing
polynomial long division quick question about the algorithm
I was making use of polynomial long division in inverse Z transform and I got stuck in a brainfart in one stage of the polynomial long division.
I posted the original question into digital signal processing stack exchange, but nobody answered it so I thought about sharing the link to math stack exchange.
So here is the link to the my earlier post containing the question and my concerns
https://dsp.stackexchange.com/questions/53426/inverse-z-transform-confused-about-polynomial-long-division-lti-and-causal
if you don't want to look at the link then I can put the same photo here and show the direct question also
I don't know what I should do here, I have two terms (17/2−5z). Which one do you divide by (2z^2) and why is that?
z-transform
asked Nov 20 '18 at 5:11
Late347
366
add a comment |
I was making use of polynomial long division in inverse Z transform and I got stuck in a brainfart in one stage of the polynomial long division.
I posted the original question into digital signal processing stack exchange, but nobody answered it so I thought about sharing the link to math stack exchange.
So here is the link to the my earlier post containing the question and my concerns
https://dsp.stackexchange.com/questions/53426/inverse-z-transform-confused-about-polynomial-long-division-lti-and-causal
if you don't want to look at the link then I can put the same photo here and show the direct question also
I don't know what I should do here, I have two terms (17/2−5z). Which one do you divide by (2z^2) and why is that?
z-transform
asked Nov 20 '18 at 5:11
Late347
366
add a comment |
I was making use of polynomial long division in inverse Z transform and I got stuck in a brainfart in one stage of the polynomial long division.
I posted the original question into digital signal processing stack exchange, but nobody answered it so I thought about sharing the link to math stack exchange.
So here is the link to the my earlier post containing the question and my concerns
https://dsp.stackexchange.com/questions/53426/inverse-z-transform-confused-about-polynomial-long-division-lti-and-causal
if you don't want to look at the link then I can put the same photo here and show the direct question also
I don't know what I should do here, I have two terms (17/2−5z). Which one do you divide by (2z^2) and why is that?
z-transform
asked Nov 20 '18 at 5:11
Late347
366
I was making use of polynomial long division in inverse Z transform and I got stuck in a brainfart in one stage of the polynomial long division.
I posted the original question into digital signal processing stack exchange, but nobody answered it so I thought about sharing the link to math stack exchange.
So here is the link to the my earlier post containing the question and my concerns
https://dsp.stackexchange.com/questions/53426/inverse-z-transform-confused-about-polynomial-long-division-lti-and-causal
if you don't want to look at the link then I can put the same photo here and show the direct question also
I don't know what I should do here, I have two terms (17/2−5z). Which one do you divide by (2z^2) and why is that?
z-transform
z-transform
asked Nov 20 '18 at 5:11
Late347
366
asked Nov 20 '18 at 5:11
Late347
366
asked Nov 20 '18 at 5:11
Late347
366
asked Nov 20 '18 at 5:11
Late347
366
asked Nov 20 '18 at 5:11
Late347
366
366
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You've gone too far. After the first step, when you have $5z-4$, you should stop because the exponent in $z^2$ is too high for it to go into $5z$. You simply have a remainder of $5z-4$, thus:
$$frac{4z^2-5z}{2z^2-5z+2} = 2 + frac{5z-4}{2z^2-5z+2}$$
Technically, you could keep dividing forever to get an infinite series, as you started to do, but usually the idea is to stop as soon as the degree of the current expression is less than that of the divisor. This is like normal division with numbers, stopping at the decimal point rather than getting the infinite decimal expansion.
answered Nov 20 '18 at 5:20
Théophile
19.5k12946
so, the logic is that you have: 17/2 * z^{0} - 5*z^{-1}, so that then z^0 is the highest power term, so that is being divided? I think the idea was to do exactly that to receive infinite series, or a couple of the first terms. Then you can do the inverse Z transform (if you can see the explicit form from the terms of the series, that you obtained from long division)
– Late347
Nov 20 '18 at 5:45
1
@Late347 If you do want to get the terms of an infinite series, then yes, at each step, look at the highest power term, like $17/2$ in $17/2 - 5/z$.
– Théophile
Nov 20 '18 at 23:58
add a comment |
The long division algorithm for dividing $a(z)$ by $b(z)$ is meant to find
the unique polynomials $q(z)$ and $r(z)$ such that
$$a(x) = q(z)b(z) + r(z)$$ with the degree of $r(z)$ strictly less than the degree of $b(z)$.
In your case after one step you already have this situation for $q(z)=2$ (constant) and $r(z) = 5z-4$ which has degree 1 < degree $b(z)$ which is $2$. So you're done after one step.
answered Nov 20 '18 at 6:10
Henno Brandsma
105k346113
add a comment |
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2 Answers
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2 Answers
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You've gone too far. After the first step, when you have $5z-4$, you should stop because the exponent in $z^2$ is too high for it to go into $5z$. You simply have a remainder of $5z-4$, thus:
$$frac{4z^2-5z}{2z^2-5z+2} = 2 + frac{5z-4}{2z^2-5z+2}$$
Technically, you could keep dividing forever to get an infinite series, as you started to do, but usually the idea is to stop as soon as the degree of the current expression is less than that of the divisor. This is like normal division with numbers, stopping at the decimal point rather than getting the infinite decimal expansion.
answered Nov 20 '18 at 5:20
Théophile
19.5k12946
so, the logic is that you have: 17/2 * z^{0} - 5*z^{-1}, so that then z^0 is the highest power term, so that is being divided? I think the idea was to do exactly that to receive infinite series, or a couple of the first terms. Then you can do the inverse Z transform (if you can see the explicit form from the terms of the series, that you obtained from long division)
– Late347
Nov 20 '18 at 5:45
1
@Late347 If you do want to get the terms of an infinite series, then yes, at each step, look at the highest power term, like $17/2$ in $17/2 - 5/z$.
– Théophile
Nov 20 '18 at 23:58
add a comment |
You've gone too far. After the first step, when you have $5z-4$, you should stop because the exponent in $z^2$ is too high for it to go into $5z$. You simply have a remainder of $5z-4$, thus:
$$frac{4z^2-5z}{2z^2-5z+2} = 2 + frac{5z-4}{2z^2-5z+2}$$
Technically, you could keep dividing forever to get an infinite series, as you started to do, but usually the idea is to stop as soon as the degree of the current expression is less than that of the divisor. This is like normal division with numbers, stopping at the decimal point rather than getting the infinite decimal expansion.
answered Nov 20 '18 at 5:20
Théophile
19.5k12946
so, the logic is that you have: 17/2 * z^{0} - 5*z^{-1}, so that then z^0 is the highest power term, so that is being divided? I think the idea was to do exactly that to receive infinite series, or a couple of the first terms. Then you can do the inverse Z transform (if you can see the explicit form from the terms of the series, that you obtained from long division)
– Late347
Nov 20 '18 at 5:45
1
@Late347 If you do want to get the terms of an infinite series, then yes, at each step, look at the highest power term, like $17/2$ in $17/2 - 5/z$.
– Théophile
Nov 20 '18 at 23:58
add a comment |
You've gone too far. After the first step, when you have $5z-4$, you should stop because the exponent in $z^2$ is too high for it to go into $5z$. You simply have a remainder of $5z-4$, thus:
$$frac{4z^2-5z}{2z^2-5z+2} = 2 + frac{5z-4}{2z^2-5z+2}$$
Technically, you could keep dividing forever to get an infinite series, as you started to do, but usually the idea is to stop as soon as the degree of the current expression is less than that of the divisor. This is like normal division with numbers, stopping at the decimal point rather than getting the infinite decimal expansion.
answered Nov 20 '18 at 5:20
Théophile
19.5k12946
You've gone too far. After the first step, when you have $5z-4$, you should stop because the exponent in $z^2$ is too high for it to go into $5z$. You simply have a remainder of $5z-4$, thus:
$$frac{4z^2-5z}{2z^2-5z+2} = 2 + frac{5z-4}{2z^2-5z+2}$$
Technically, you could keep dividing forever to get an infinite series, as you started to do, but usually the idea is to stop as soon as the degree of the current expression is less than that of the divisor. This is like normal division with numbers, stopping at the decimal point rather than getting the infinite decimal expansion.
answered Nov 20 '18 at 5:20
Théophile
19.5k12946
edited Nov 20 '18 at 5:25
answered Nov 20 '18 at 5:20
Théophile
19.5k12946
answered Nov 20 '18 at 5:20
Théophile
19.5k12946
answered Nov 20 '18 at 5:20
Théophile
19.5k12946
19.5k12946
so, the logic is that you have: 17/2 * z^{0} - 5*z^{-1}, so that then z^0 is the highest power term, so that is being divided? I think the idea was to do exactly that to receive infinite series, or a couple of the first terms. Then you can do the inverse Z transform (if you can see the explicit form from the terms of the series, that you obtained from long division)
– Late347
Nov 20 '18 at 5:45
1
@Late347 If you do want to get the terms of an infinite series, then yes, at each step, look at the highest power term, like $17/2$ in $17/2 - 5/z$.
– Théophile
Nov 20 '18 at 23:58
add a comment |
so, the logic is that you have: 17/2 * z^{0} - 5*z^{-1}, so that then z^0 is the highest power term, so that is being divided? I think the idea was to do exactly that to receive infinite series, or a couple of the first terms. Then you can do the inverse Z transform (if you can see the explicit form from the terms of the series, that you obtained from long division)
– Late347
Nov 20 '18 at 5:45
1
@Late347 If you do want to get the terms of an infinite series, then yes, at each step, look at the highest power term, like $17/2$ in $17/2 - 5/z$.
– Théophile
Nov 20 '18 at 23:58
so, the logic is that you have: 17/2 * z^{0} - 5*z^{-1}, so that then z^0 is the highest power term, so that is being divided? I think the idea was to do exactly that to receive infinite series, or a couple of the first terms. Then you can do the inverse Z transform (if you can see the explicit form from the terms of the series, that you obtained from long division)
– Late347
Nov 20 '18 at 5:45
so, the logic is that you have: 17/2 * z^{0} - 5*z^{-1}, so that then z^0 is the highest power term, so that is being divided? I think the idea was to do exactly that to receive infinite series, or a couple of the first terms. Then you can do the inverse Z transform (if you can see the explicit form from the terms of the series, that you obtained from long division)
– Late347
Nov 20 '18 at 5:45
1
1
@Late347 If you do want to get the terms of an infinite series, then yes, at each step, look at the highest power term, like $17/2$ in $17/2 - 5/z$.
– Théophile
Nov 20 '18 at 23:58
@Late347 If you do want to get the terms of an infinite series, then yes, at each step, look at the highest power term, like $17/2$ in $17/2 - 5/z$.
– Théophile
Nov 20 '18 at 23:58
add a comment |
The long division algorithm for dividing $a(z)$ by $b(z)$ is meant to find
the unique polynomials $q(z)$ and $r(z)$ such that
$$a(x) = q(z)b(z) + r(z)$$ with the degree of $r(z)$ strictly less than the degree of $b(z)$.
In your case after one step you already have this situation for $q(z)=2$ (constant) and $r(z) = 5z-4$ which has degree 1 < degree $b(z)$ which is $2$. So you're done after one step.
answered Nov 20 '18 at 6:10
Henno Brandsma
105k346113
add a comment |
The long division algorithm for dividing $a(z)$ by $b(z)$ is meant to find
the unique polynomials $q(z)$ and $r(z)$ such that
$$a(x) = q(z)b(z) + r(z)$$ with the degree of $r(z)$ strictly less than the degree of $b(z)$.
In your case after one step you already have this situation for $q(z)=2$ (constant) and $r(z) = 5z-4$ which has degree 1 < degree $b(z)$ which is $2$. So you're done after one step.
answered Nov 20 '18 at 6:10
Henno Brandsma
105k346113
add a comment |
The long division algorithm for dividing $a(z)$ by $b(z)$ is meant to find
the unique polynomials $q(z)$ and $r(z)$ such that
$$a(x) = q(z)b(z) + r(z)$$ with the degree of $r(z)$ strictly less than the degree of $b(z)$.
In your case after one step you already have this situation for $q(z)=2$ (constant) and $r(z) = 5z-4$ which has degree 1 < degree $b(z)$ which is $2$. So you're done after one step.
answered Nov 20 '18 at 6:10
Henno Brandsma
105k346113
The long division algorithm for dividing $a(z)$ by $b(z)$ is meant to find
the unique polynomials $q(z)$ and $r(z)$ such that
$$a(x) = q(z)b(z) + r(z)$$ with the degree of $r(z)$ strictly less than the degree of $b(z)$.
In your case after one step you already have this situation for $q(z)=2$ (constant) and $r(z) = 5z-4$ which has degree 1 < degree $b(z)$ which is $2$. So you're done after one step.
answered Nov 20 '18 at 6:10
Henno Brandsma
105k346113
answered Nov 20 '18 at 6:10
Henno Brandsma
105k346113
answered Nov 20 '18 at 6:10
Henno Brandsma
105k346113
answered Nov 20 '18 at 6:10
Henno Brandsma
105k346113
105k346113
add a comment |
add a comment |
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