Expand $X(z) = frac{1}{1+az}$ into a causal sequence
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For a HW problem, I'm told to expand $frac{1}{1+az}$ into a causal and noncausal sequence. I found the noncausal sequence by long division (the result is $1-az+(az)^2-dots$) and found the region of convergence, but I'm not sure how to find the causal sequence. The full question is :
sequences-and-series convergence z-transform
edited Jan 14 at 23:46
Arnaud D.
16.2k52445
asked Jan 14 at 23:32
Kevin SilkenKevin Silken
31
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add a comment |
$begingroup$
For a HW problem, I'm told to expand $frac{1}{1+az}$ into a causal and noncausal sequence. I found the noncausal sequence by long division (the result is $1-az+(az)^2-dots$) and found the region of convergence, but I'm not sure how to find the causal sequence. The full question is :
sequences-and-series convergence z-transform
edited Jan 14 at 23:46
Arnaud D.
16.2k52445
asked Jan 14 at 23:32
Kevin SilkenKevin Silken
31
$endgroup$
add a comment |
$begingroup$
For a HW problem, I'm told to expand $frac{1}{1+az}$ into a causal and noncausal sequence. I found the noncausal sequence by long division (the result is $1-az+(az)^2-dots$) and found the region of convergence, but I'm not sure how to find the causal sequence. The full question is :
sequences-and-series convergence z-transform
edited Jan 14 at 23:46
Arnaud D.
16.2k52445
asked Jan 14 at 23:32
Kevin SilkenKevin Silken
31
$endgroup$
For a HW problem, I'm told to expand $frac{1}{1+az}$ into a causal and noncausal sequence. I found the noncausal sequence by long division (the result is $1-az+(az)^2-dots$) and found the region of convergence, but I'm not sure how to find the causal sequence. The full question is :
sequences-and-series convergence z-transform
sequences-and-series convergence z-transform
edited Jan 14 at 23:46
Arnaud D.
16.2k52445
asked Jan 14 at 23:32
Kevin SilkenKevin Silken
31
edited Jan 14 at 23:46
Arnaud D.
16.2k52445
asked Jan 14 at 23:32
Kevin SilkenKevin Silken
31
edited Jan 14 at 23:46
Arnaud D.
16.2k52445
edited Jan 14 at 23:46
Arnaud D.
16.2k52445
edited Jan 14 at 23:46
Arnaud D.
16.2k52445
16.2k52445
asked Jan 14 at 23:32
Kevin SilkenKevin Silken
31
asked Jan 14 at 23:32
Kevin SilkenKevin Silken
31
asked Jan 14 at 23:32
Kevin SilkenKevin Silken
31
31
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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If $frac{1}{1+az}=sum b_n z^{-n}$, then plugging in $w=1/z$, we get $frac{1}{1+a/w}=sum b_n w^n$. So let us rewrite $frac{1}{1+a/w}=frac{1}{frac{w+a}{w}}=frac{w}{a+w}=(w/a)frac{1}{1-(-w/a)}=(w/a)sum (-w/a)^n$. Now just expand and plug back in for $z$.
answered Jan 15 at 0:20
AaronAaron
16.1k22855
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add a comment |
$begingroup$
HINT
You have
$$
frac{1}{1+az}
= frac{1}{az} left[ frac{1}{1 + 1/(az)} right]
$$
answered Jan 15 at 0:16
gt6989bgt6989b
36k22557
$endgroup$
$begingroup$
So do partial fraction decomposition on this? And it would chain into a series of these terms?
$endgroup$
– Kevin Silken
Jan 15 at 0:19
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@KevinSilken no, the bracket is a geometric series, like the one you did, except in $1/z$...
$endgroup$
– gt6989b
Jan 15 at 0:20
add a comment |
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2 Answers
2
active
oldest
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2 Answers
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active
oldest
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$begingroup$
If $frac{1}{1+az}=sum b_n z^{-n}$, then plugging in $w=1/z$, we get $frac{1}{1+a/w}=sum b_n w^n$. So let us rewrite $frac{1}{1+a/w}=frac{1}{frac{w+a}{w}}=frac{w}{a+w}=(w/a)frac{1}{1-(-w/a)}=(w/a)sum (-w/a)^n$. Now just expand and plug back in for $z$.
answered Jan 15 at 0:20
AaronAaron
16.1k22855
$endgroup$
add a comment |
$begingroup$
If $frac{1}{1+az}=sum b_n z^{-n}$, then plugging in $w=1/z$, we get $frac{1}{1+a/w}=sum b_n w^n$. So let us rewrite $frac{1}{1+a/w}=frac{1}{frac{w+a}{w}}=frac{w}{a+w}=(w/a)frac{1}{1-(-w/a)}=(w/a)sum (-w/a)^n$. Now just expand and plug back in for $z$.
answered Jan 15 at 0:20
AaronAaron
16.1k22855
$endgroup$
add a comment |
$begingroup$
If $frac{1}{1+az}=sum b_n z^{-n}$, then plugging in $w=1/z$, we get $frac{1}{1+a/w}=sum b_n w^n$. So let us rewrite $frac{1}{1+a/w}=frac{1}{frac{w+a}{w}}=frac{w}{a+w}=(w/a)frac{1}{1-(-w/a)}=(w/a)sum (-w/a)^n$. Now just expand and plug back in for $z$.
answered Jan 15 at 0:20
AaronAaron
16.1k22855
$endgroup$
If $frac{1}{1+az}=sum b_n z^{-n}$, then plugging in $w=1/z$, we get $frac{1}{1+a/w}=sum b_n w^n$. So let us rewrite $frac{1}{1+a/w}=frac{1}{frac{w+a}{w}}=frac{w}{a+w}=(w/a)frac{1}{1-(-w/a)}=(w/a)sum (-w/a)^n$. Now just expand and plug back in for $z$.
answered Jan 15 at 0:20
AaronAaron
16.1k22855
edited Jan 15 at 1:54
answered Jan 15 at 0:20
AaronAaron
16.1k22855
answered Jan 15 at 0:20
AaronAaron
16.1k22855
answered Jan 15 at 0:20
AaronAaron
16.1k22855
16.1k22855
add a comment |
add a comment |
$begingroup$
HINT
You have
$$
frac{1}{1+az}
= frac{1}{az} left[ frac{1}{1 + 1/(az)} right]
$$
answered Jan 15 at 0:16
gt6989bgt6989b
36k22557
$endgroup$
$begingroup$
So do partial fraction decomposition on this? And it would chain into a series of these terms?
$endgroup$
– Kevin Silken
Jan 15 at 0:19
$begingroup$
@KevinSilken no, the bracket is a geometric series, like the one you did, except in $1/z$...
$endgroup$
– gt6989b
Jan 15 at 0:20
add a comment |
$begingroup$
HINT
You have
$$
frac{1}{1+az}
= frac{1}{az} left[ frac{1}{1 + 1/(az)} right]
$$
answered Jan 15 at 0:16
gt6989bgt6989b
36k22557
$endgroup$
$begingroup$
So do partial fraction decomposition on this? And it would chain into a series of these terms?
$endgroup$
– Kevin Silken
Jan 15 at 0:19
$begingroup$
@KevinSilken no, the bracket is a geometric series, like the one you did, except in $1/z$...
$endgroup$
– gt6989b
Jan 15 at 0:20
add a comment |
$begingroup$
HINT
You have
$$
frac{1}{1+az}
= frac{1}{az} left[ frac{1}{1 + 1/(az)} right]
$$
answered Jan 15 at 0:16
gt6989bgt6989b
36k22557
$endgroup$
HINT
You have
$$
frac{1}{1+az}
= frac{1}{az} left[ frac{1}{1 + 1/(az)} right]
$$
answered Jan 15 at 0:16
gt6989bgt6989b
36k22557
answered Jan 15 at 0:16
gt6989bgt6989b
36k22557
answered Jan 15 at 0:16
gt6989bgt6989b
36k22557
answered Jan 15 at 0:16
gt6989bgt6989b
36k22557
36k22557
$begingroup$
So do partial fraction decomposition on this? And it would chain into a series of these terms?
$endgroup$
– Kevin Silken
Jan 15 at 0:19
$begingroup$
@KevinSilken no, the bracket is a geometric series, like the one you did, except in $1/z$...
$endgroup$
– gt6989b
Jan 15 at 0:20
add a comment |
$begingroup$
So do partial fraction decomposition on this? And it would chain into a series of these terms?
$endgroup$
– Kevin Silken
Jan 15 at 0:19
$begingroup$
@KevinSilken no, the bracket is a geometric series, like the one you did, except in $1/z$...
$endgroup$
– gt6989b
Jan 15 at 0:20
$begingroup$
So do partial fraction decomposition on this? And it would chain into a series of these terms?
$endgroup$
– Kevin Silken
Jan 15 at 0:19
$begingroup$
So do partial fraction decomposition on this? And it would chain into a series of these terms?
$endgroup$
– Kevin Silken
Jan 15 at 0:19
$begingroup$
@KevinSilken no, the bracket is a geometric series, like the one you did, except in $1/z$...
$endgroup$
– gt6989b
Jan 15 at 0:20
$begingroup$
@KevinSilken no, the bracket is a geometric series, like the one you did, except in $1/z$...
$endgroup$
– gt6989b
Jan 15 at 0:20
add a comment |
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