Mixing
difference equation, LTI system, Z-transform, impulse response
exercises 2 and 3 instructions are in picture
Here is a z-table that I can use to make inverse z-transforms
I did questions 2a and 2b, but I don't know much about how to do the final 2c (I got stuck there)
I did exercise 3 and drew the block diagram as best as I understood it.
I could use maybe some help for 2c
2.a) answer
Find H(z) transfer function, and check stability from poles (inside unit circle or not)
firstly try to get difference equation
$y[n]=1/4*(x[n]+x[n-1]+x[n-2]+x[n-3])$
Z-transform both sides
$Y(Z)= 1/4 * (X(Z)+z^{-1}*X(z)+z^{-2}*X(z)+z^{-3}*X(z))$
After factorizing and shuffling some terms I get that
$ frac{Y(z)}{X(z)}=H(z)= 1/4 * (1+z^{-1}+z^{-2}+z^{-3})=frac{z^3+z^2+z+1}{4z^3}$
we see from $4z^3=0 $ that poles are zero so the system is stable (is that right interpretation???)
2.b) answer
impulse response of the system, I think that this means that when we want to get the impulse response of the system, what we could do is maybe inverse Z transform the transferfunction => so we get h[n] = impulse response
I was looking at my earlier notes and there was mention that when input sequence $x[n] =delta[n]$, implies that $y[n]=h[n] $= impulse response
Can we just plug in the $x[n]=delta[n]$ and see what comes out for $y[n]$???
with the plug-in technique we just get $h[k]=1/4*(delta[k]+delta[k-1]+delta[k-2]+delta[k-3])$
with the inverse Z-transforming the H(z) technique, I get as follows
$(H(z)= 1/4*(1+z^{-1}+z^{-2}+z^{-3})) $
I think, we are supposed to identify
$z^{-k_0}*1 = z^{-k_0}*X(z)$
from the z-table, and also use linearity of inverse-Z-transform to each sumterm
applying the identities from the z-table, we should get
$h[k] = 1/4* (delta[k]+u[k-1]delta[k-1]+u[k-2]delta[k-2]+u[k-3]delta[k-3])$
looks like basically the same answer, I don't think those unit steps multiplied change the result that much, except ensure causality, but regular deltafunction should also be zero, at negative indexes
2.c) answer
I think one tactic might be to Z-transform unit step $Z(u[n])=frac{z}{z-1}=X(z)$
multiply that with transfer function, and then you get Y(z)
inverse Z transform should give output sequence...
$Y(z)=H(z)*X(z)=frac{z^3+z^2+z+1}{4z^2(z-1)}$
I got stuck at that inverse-Z transform... Can you at least verify that the Y(z) is correct at this stage? I think Ive used this same tactic of inverting from the Z-domain into the k-domain to get the output sequences but sometimes I get stuck at the inverse Z-transform...
with the plugging in technique you just get
$y[k]=1/4*(u[k]+u[k-1]+u[k-2]+u[k-3])$
3. answer
Fibonacci
This was goddam difficult also to do.
first start by Z-transform both sides
$Z(y[n+2])-Z(y[n+1])=Z(y[n])
$
$Y(z)=z^2Y(z) -z^2-z-zY(z)+z $
$Y(z)=z^2Y(z) -z^2-zY(z) $
$Y(z)=frac{z^2}{z^2-z-1}$
Then our teacher gave us tip that we should divide both sides by z, so that it maybe helps to find partial fractions expansion such that we find something in the z-table.
So, first divide both sides by z, and also solve roots of the denominator polynomial $z^2-z-1$ , in order to factorize
$Y(z)/z= frac{A}{z-b}+frac{B}{z-a}= frac{z}{ (z-frac{-sqrt5+1}{2}) (z-frac{sqrt5+1}{2}) }$
so that
$a= frac{sqrt5 +1}{2}$
$b= frac{-sqrt5+1}{2}
$
for the partial Fraction coefficients in capital letters, I obtained with the "equating coefficients technique" something very ugly as follows
$A=frac{-sqrt5+5}{10}$
$B=frac{sqrt5+5}{10}$
Then, finally we can have
$Y(z)/z = A/(z-b)+B/(z-a)$
$Y(z) = Az/(z-b)+Bz/(z-a)$
from the z-table we can identify exponential function j^k == z/(z-j)
$
y[k] = A*b^k+B*a^k
$
I'm too lazy to even write that out but I checked that fibonacci y[k] with my calculator and it seems to work, where k>=0.
I was pretty mindblown by the idea that Fibonacci recursive sequence can be made with explicit formula like this,though...
I had to google about that fibonacci block diagram because I didn't understand what is the input sequence x[n] with fibonacci.
But, apparently you can model the fibonacci system if you put x[n]= unit impulse = deltafunction
so then the entire difference equation changes its form slightly from the recursive definition as such...
$y[n]= u[n-1]*y[n-1]+ u[n-2]*y[n-2]+delta[n]$
But, I tabulated some values from n = [-1,0,1,2,3,4,5...] and it seems to be same as the regular fibonacci values
recurrence-relations fibonacci-numbers z-transform
asked Nov 21 '18 at 10:23
Late347
366
add a comment |
exercises 2 and 3 instructions are in picture
Here is a z-table that I can use to make inverse z-transforms
I did questions 2a and 2b, but I don't know much about how to do the final 2c (I got stuck there)
I did exercise 3 and drew the block diagram as best as I understood it.
I could use maybe some help for 2c
2.a) answer
Find H(z) transfer function, and check stability from poles (inside unit circle or not)
firstly try to get difference equation
$y[n]=1/4*(x[n]+x[n-1]+x[n-2]+x[n-3])$
Z-transform both sides
$Y(Z)= 1/4 * (X(Z)+z^{-1}*X(z)+z^{-2}*X(z)+z^{-3}*X(z))$
After factorizing and shuffling some terms I get that
$ frac{Y(z)}{X(z)}=H(z)= 1/4 * (1+z^{-1}+z^{-2}+z^{-3})=frac{z^3+z^2+z+1}{4z^3}$
we see from $4z^3=0 $ that poles are zero so the system is stable (is that right interpretation???)
2.b) answer
impulse response of the system, I think that this means that when we want to get the impulse response of the system, what we could do is maybe inverse Z transform the transferfunction => so we get h[n] = impulse response
I was looking at my earlier notes and there was mention that when input sequence $x[n] =delta[n]$, implies that $y[n]=h[n] $= impulse response
Can we just plug in the $x[n]=delta[n]$ and see what comes out for $y[n]$???
with the plug-in technique we just get $h[k]=1/4*(delta[k]+delta[k-1]+delta[k-2]+delta[k-3])$
with the inverse Z-transforming the H(z) technique, I get as follows
$(H(z)= 1/4*(1+z^{-1}+z^{-2}+z^{-3})) $
I think, we are supposed to identify
$z^{-k_0}*1 = z^{-k_0}*X(z)$
from the z-table, and also use linearity of inverse-Z-transform to each sumterm
applying the identities from the z-table, we should get
$h[k] = 1/4* (delta[k]+u[k-1]delta[k-1]+u[k-2]delta[k-2]+u[k-3]delta[k-3])$
looks like basically the same answer, I don't think those unit steps multiplied change the result that much, except ensure causality, but regular deltafunction should also be zero, at negative indexes
2.c) answer
I think one tactic might be to Z-transform unit step $Z(u[n])=frac{z}{z-1}=X(z)$
multiply that with transfer function, and then you get Y(z)
inverse Z transform should give output sequence...
$Y(z)=H(z)*X(z)=frac{z^3+z^2+z+1}{4z^2(z-1)}$
I got stuck at that inverse-Z transform... Can you at least verify that the Y(z) is correct at this stage? I think Ive used this same tactic of inverting from the Z-domain into the k-domain to get the output sequences but sometimes I get stuck at the inverse Z-transform...
with the plugging in technique you just get
$y[k]=1/4*(u[k]+u[k-1]+u[k-2]+u[k-3])$
3. answer
Fibonacci
This was goddam difficult also to do.
first start by Z-transform both sides
$Z(y[n+2])-Z(y[n+1])=Z(y[n])
$
$Y(z)=z^2Y(z) -z^2-z-zY(z)+z $
$Y(z)=z^2Y(z) -z^2-zY(z) $
$Y(z)=frac{z^2}{z^2-z-1}$
Then our teacher gave us tip that we should divide both sides by z, so that it maybe helps to find partial fractions expansion such that we find something in the z-table.
So, first divide both sides by z, and also solve roots of the denominator polynomial $z^2-z-1$ , in order to factorize
$Y(z)/z= frac{A}{z-b}+frac{B}{z-a}= frac{z}{ (z-frac{-sqrt5+1}{2}) (z-frac{sqrt5+1}{2}) }$
so that
$a= frac{sqrt5 +1}{2}$
$b= frac{-sqrt5+1}{2}
$
for the partial Fraction coefficients in capital letters, I obtained with the "equating coefficients technique" something very ugly as follows
$A=frac{-sqrt5+5}{10}$
$B=frac{sqrt5+5}{10}$
Then, finally we can have
$Y(z)/z = A/(z-b)+B/(z-a)$
$Y(z) = Az/(z-b)+Bz/(z-a)$
from the z-table we can identify exponential function j^k == z/(z-j)
$
y[k] = A*b^k+B*a^k
$
I'm too lazy to even write that out but I checked that fibonacci y[k] with my calculator and it seems to work, where k>=0.
I was pretty mindblown by the idea that Fibonacci recursive sequence can be made with explicit formula like this,though...
I had to google about that fibonacci block diagram because I didn't understand what is the input sequence x[n] with fibonacci.
But, apparently you can model the fibonacci system if you put x[n]= unit impulse = deltafunction
so then the entire difference equation changes its form slightly from the recursive definition as such...
$y[n]= u[n-1]*y[n-1]+ u[n-2]*y[n-2]+delta[n]$
But, I tabulated some values from n = [-1,0,1,2,3,4,5...] and it seems to be same as the regular fibonacci values
recurrence-relations fibonacci-numbers z-transform
asked Nov 21 '18 at 10:23
Late347
366
add a comment |
exercises 2 and 3 instructions are in picture
Here is a z-table that I can use to make inverse z-transforms
I did questions 2a and 2b, but I don't know much about how to do the final 2c (I got stuck there)
I did exercise 3 and drew the block diagram as best as I understood it.
I could use maybe some help for 2c
2.a) answer
Find H(z) transfer function, and check stability from poles (inside unit circle or not)
firstly try to get difference equation
$y[n]=1/4*(x[n]+x[n-1]+x[n-2]+x[n-3])$
Z-transform both sides
$Y(Z)= 1/4 * (X(Z)+z^{-1}*X(z)+z^{-2}*X(z)+z^{-3}*X(z))$
After factorizing and shuffling some terms I get that
$ frac{Y(z)}{X(z)}=H(z)= 1/4 * (1+z^{-1}+z^{-2}+z^{-3})=frac{z^3+z^2+z+1}{4z^3}$
we see from $4z^3=0 $ that poles are zero so the system is stable (is that right interpretation???)
2.b) answer
impulse response of the system, I think that this means that when we want to get the impulse response of the system, what we could do is maybe inverse Z transform the transferfunction => so we get h[n] = impulse response
I was looking at my earlier notes and there was mention that when input sequence $x[n] =delta[n]$, implies that $y[n]=h[n] $= impulse response
Can we just plug in the $x[n]=delta[n]$ and see what comes out for $y[n]$???
with the plug-in technique we just get $h[k]=1/4*(delta[k]+delta[k-1]+delta[k-2]+delta[k-3])$
with the inverse Z-transforming the H(z) technique, I get as follows
$(H(z)= 1/4*(1+z^{-1}+z^{-2}+z^{-3})) $
I think, we are supposed to identify
$z^{-k_0}*1 = z^{-k_0}*X(z)$
from the z-table, and also use linearity of inverse-Z-transform to each sumterm
applying the identities from the z-table, we should get
$h[k] = 1/4* (delta[k]+u[k-1]delta[k-1]+u[k-2]delta[k-2]+u[k-3]delta[k-3])$
looks like basically the same answer, I don't think those unit steps multiplied change the result that much, except ensure causality, but regular deltafunction should also be zero, at negative indexes
2.c) answer
I think one tactic might be to Z-transform unit step $Z(u[n])=frac{z}{z-1}=X(z)$
multiply that with transfer function, and then you get Y(z)
inverse Z transform should give output sequence...
$Y(z)=H(z)*X(z)=frac{z^3+z^2+z+1}{4z^2(z-1)}$
I got stuck at that inverse-Z transform... Can you at least verify that the Y(z) is correct at this stage? I think Ive used this same tactic of inverting from the Z-domain into the k-domain to get the output sequences but sometimes I get stuck at the inverse Z-transform...
with the plugging in technique you just get
$y[k]=1/4*(u[k]+u[k-1]+u[k-2]+u[k-3])$
3. answer
Fibonacci
This was goddam difficult also to do.
first start by Z-transform both sides
$Z(y[n+2])-Z(y[n+1])=Z(y[n])
$
$Y(z)=z^2Y(z) -z^2-z-zY(z)+z $
$Y(z)=z^2Y(z) -z^2-zY(z) $
$Y(z)=frac{z^2}{z^2-z-1}$
Then our teacher gave us tip that we should divide both sides by z, so that it maybe helps to find partial fractions expansion such that we find something in the z-table.
So, first divide both sides by z, and also solve roots of the denominator polynomial $z^2-z-1$ , in order to factorize
$Y(z)/z= frac{A}{z-b}+frac{B}{z-a}= frac{z}{ (z-frac{-sqrt5+1}{2}) (z-frac{sqrt5+1}{2}) }$
so that
$a= frac{sqrt5 +1}{2}$
$b= frac{-sqrt5+1}{2}
$
for the partial Fraction coefficients in capital letters, I obtained with the "equating coefficients technique" something very ugly as follows
$A=frac{-sqrt5+5}{10}$
$B=frac{sqrt5+5}{10}$
Then, finally we can have
$Y(z)/z = A/(z-b)+B/(z-a)$
$Y(z) = Az/(z-b)+Bz/(z-a)$
from the z-table we can identify exponential function j^k == z/(z-j)
$
y[k] = A*b^k+B*a^k
$
I'm too lazy to even write that out but I checked that fibonacci y[k] with my calculator and it seems to work, where k>=0.
I was pretty mindblown by the idea that Fibonacci recursive sequence can be made with explicit formula like this,though...
I had to google about that fibonacci block diagram because I didn't understand what is the input sequence x[n] with fibonacci.
But, apparently you can model the fibonacci system if you put x[n]= unit impulse = deltafunction
so then the entire difference equation changes its form slightly from the recursive definition as such...
$y[n]= u[n-1]*y[n-1]+ u[n-2]*y[n-2]+delta[n]$
But, I tabulated some values from n = [-1,0,1,2,3,4,5...] and it seems to be same as the regular fibonacci values
recurrence-relations fibonacci-numbers z-transform
asked Nov 21 '18 at 10:23
Late347
366
exercises 2 and 3 instructions are in picture
Here is a z-table that I can use to make inverse z-transforms
I did questions 2a and 2b, but I don't know much about how to do the final 2c (I got stuck there)
I did exercise 3 and drew the block diagram as best as I understood it.
I could use maybe some help for 2c
2.a) answer
Find H(z) transfer function, and check stability from poles (inside unit circle or not)
firstly try to get difference equation
$y[n]=1/4*(x[n]+x[n-1]+x[n-2]+x[n-3])$
Z-transform both sides
$Y(Z)= 1/4 * (X(Z)+z^{-1}*X(z)+z^{-2}*X(z)+z^{-3}*X(z))$
After factorizing and shuffling some terms I get that
$ frac{Y(z)}{X(z)}=H(z)= 1/4 * (1+z^{-1}+z^{-2}+z^{-3})=frac{z^3+z^2+z+1}{4z^3}$
we see from $4z^3=0 $ that poles are zero so the system is stable (is that right interpretation???)
2.b) answer
impulse response of the system, I think that this means that when we want to get the impulse response of the system, what we could do is maybe inverse Z transform the transferfunction => so we get h[n] = impulse response
I was looking at my earlier notes and there was mention that when input sequence $x[n] =delta[n]$, implies that $y[n]=h[n] $= impulse response
Can we just plug in the $x[n]=delta[n]$ and see what comes out for $y[n]$???
with the plug-in technique we just get $h[k]=1/4*(delta[k]+delta[k-1]+delta[k-2]+delta[k-3])$
with the inverse Z-transforming the H(z) technique, I get as follows
$(H(z)= 1/4*(1+z^{-1}+z^{-2}+z^{-3})) $
I think, we are supposed to identify
$z^{-k_0}*1 = z^{-k_0}*X(z)$
from the z-table, and also use linearity of inverse-Z-transform to each sumterm
applying the identities from the z-table, we should get
$h[k] = 1/4* (delta[k]+u[k-1]delta[k-1]+u[k-2]delta[k-2]+u[k-3]delta[k-3])$
looks like basically the same answer, I don't think those unit steps multiplied change the result that much, except ensure causality, but regular deltafunction should also be zero, at negative indexes
2.c) answer
I think one tactic might be to Z-transform unit step $Z(u[n])=frac{z}{z-1}=X(z)$
multiply that with transfer function, and then you get Y(z)
inverse Z transform should give output sequence...
$Y(z)=H(z)*X(z)=frac{z^3+z^2+z+1}{4z^2(z-1)}$
I got stuck at that inverse-Z transform... Can you at least verify that the Y(z) is correct at this stage? I think Ive used this same tactic of inverting from the Z-domain into the k-domain to get the output sequences but sometimes I get stuck at the inverse Z-transform...
with the plugging in technique you just get
$y[k]=1/4*(u[k]+u[k-1]+u[k-2]+u[k-3])$
3. answer
Fibonacci
This was goddam difficult also to do.
first start by Z-transform both sides
$Z(y[n+2])-Z(y[n+1])=Z(y[n])
$
$Y(z)=z^2Y(z) -z^2-z-zY(z)+z $
$Y(z)=z^2Y(z) -z^2-zY(z) $
$Y(z)=frac{z^2}{z^2-z-1}$
Then our teacher gave us tip that we should divide both sides by z, so that it maybe helps to find partial fractions expansion such that we find something in the z-table.
So, first divide both sides by z, and also solve roots of the denominator polynomial $z^2-z-1$ , in order to factorize
$Y(z)/z= frac{A}{z-b}+frac{B}{z-a}= frac{z}{ (z-frac{-sqrt5+1}{2}) (z-frac{sqrt5+1}{2}) }$
so that
$a= frac{sqrt5 +1}{2}$
$b= frac{-sqrt5+1}{2}
$
for the partial Fraction coefficients in capital letters, I obtained with the "equating coefficients technique" something very ugly as follows
$A=frac{-sqrt5+5}{10}$
$B=frac{sqrt5+5}{10}$
Then, finally we can have
$Y(z)/z = A/(z-b)+B/(z-a)$
$Y(z) = Az/(z-b)+Bz/(z-a)$
from the z-table we can identify exponential function j^k == z/(z-j)
$
y[k] = A*b^k+B*a^k
$
I'm too lazy to even write that out but I checked that fibonacci y[k] with my calculator and it seems to work, where k>=0.
I was pretty mindblown by the idea that Fibonacci recursive sequence can be made with explicit formula like this,though...
I had to google about that fibonacci block diagram because I didn't understand what is the input sequence x[n] with fibonacci.
But, apparently you can model the fibonacci system if you put x[n]= unit impulse = deltafunction
so then the entire difference equation changes its form slightly from the recursive definition as such...
$y[n]= u[n-1]*y[n-1]+ u[n-2]*y[n-2]+delta[n]$
But, I tabulated some values from n = [-1,0,1,2,3,4,5...] and it seems to be same as the regular fibonacci values
recurrence-relations fibonacci-numbers z-transform
recurrence-relations fibonacci-numbers z-transform
asked Nov 21 '18 at 10:23
Late347
366
asked Nov 21 '18 at 10:23
Late347
366
asked Nov 21 '18 at 10:23
Late347
366
asked Nov 21 '18 at 10:23
Late347
366
asked Nov 21 '18 at 10:23
Late347
366
366
add a comment |
add a comment |
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elite level block diagram drawn with paint3D...
I wish I had a program that could easily draw these kind of schematics/block diagrams in nicer style...
answered Nov 21 '18 at 10:26
Late347
366
add a comment |
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elite level block diagram drawn with paint3D...
I wish I had a program that could easily draw these kind of schematics/block diagrams in nicer style...
answered Nov 21 '18 at 10:26
Late347
366
add a comment |
elite level block diagram drawn with paint3D...
I wish I had a program that could easily draw these kind of schematics/block diagrams in nicer style...
answered Nov 21 '18 at 10:26
Late347
366
add a comment |
elite level block diagram drawn with paint3D...
I wish I had a program that could easily draw these kind of schematics/block diagrams in nicer style...
answered Nov 21 '18 at 10:26
Late347
366
elite level block diagram drawn with paint3D...
I wish I had a program that could easily draw these kind of schematics/block diagrams in nicer style...
answered Nov 21 '18 at 10:26
Late347
366
answered Nov 21 '18 at 10:26
Late347
366
answered Nov 21 '18 at 10:26
Late347
366
answered Nov 21 '18 at 10:26
Late347
366
366
add a comment |
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