What's the System Transfer Function H(z) for y[n] = −1/2y[n − 1] + x[n]?

[crazycool2]

Homework Statement

x[n] = aⁿu[n]
A discrete system
y[n] = −1/2y[n − 1] + x[n];
where x[n] and y[n] in- and output of the system, respectively.
Find the system transfer function H(z), and sketch its zeros and poles
in the z-plane

Homework Equations

u[n] is the unit step function

The Attempt at a Solution

I transformed x[n] to X(z)
but what confuses is the-1/2y[n-1],
which basically means the previous -half of the previous output plus the input is equal to out put. but I don't know how to represent the previous output.

 

[rude man]

Do you know the time-shift theorem? And don't forget the initial condition.

And BTW I suggest rewriting the equation so as to put it in standard form:
y(n+1) + by(n) = x(n+1), n = 0, 1, ... in your case.

 

[crazycool2]

I'm not so sure about the time shifting theorem, but I used the Transfer function H(z)=Y(z)/X(z). my X(z) is z/(z-a) and I tried Y(z) to be z+1/2 but I'm unsure. after that I supposed to sketch it's Zeros and poles. Am I on the right track?

 

[rude man]

I'm not so sure about the time shifting theorem, but I used the Transfer function H(z)=Y(z)/X(z). my X(z) is z/(z-a) and I tried Y(z) to be z+1/2 but I'm unsure. after that I supposed to sketch it's Zeros and poles. Am I on the right track?

You got X(z) right but I don't know where you're going with the rest. Certainly not with z+ 1/2.

Question to you: if Z{y(n)} = Y(z), what is Z{y(n+1)}?

Other than that I refer you to my previous post. I can't really say more without violating the rules of this forum - until you get past this point.

 

[crazycool2]

well in the textbook I'm using it's written

Z [x (n − n0)] = z^-n0X(z);

and if Z(Y[n])= Y(z) then Z(Y[n+1])= z¹Y(z)

that's gives me y[n]+1/2y[n-1] = z⁰Y(z) + 1/2z^-1Y(z)

that's where I go Y(z) = 1+1/2z^-1. I don't understand where I am going wrong

 

[rude man]

well in the textbook I'm using it's written

Z [x (n − n0)] = z^-n0X(z);

and if Z(Y[n])= Y(z) then Z(Y[n+1])= z¹Y(z)

that's gives me y[n]+1/2y[n-1] = z⁰Y(z) + 1/2z^-1Y(z)

that's where I go Y(z) = 1+1/2z^-1. I don't understand where I am going wrong

You book should say
If Z{y(n)} = Y(z) then
Z{y(n+1)} = z{Y(z) - y(0)}
which is actually quite close to your second equation above, except for the initial condition y(0).

Distinguish between y and Y. Y is the z transformed version of y. I notice you sometimes use them interchangeably. Not good.

Again - I urge you to rewrite your original equation in the conventional way, with n, n+1 and n+2 terms instead of n-1, n and n+1. Reason is I don't want to get confused myself & come up with the wrong answer. It's very straight-forward.

 

[crazycool2]

Z [x (n − n0)] = z^-n0X(z);
y[n]+1/2y[n-1]=x[n]
→y[n+1]0+1/2y[n]=x[n+1]
Z-transform and get
→ z¹Y(z) + 1/2z⁰Y(z)=z¹X(z)
→ (z+1/2])Y(z)=zX(z)

given that H(z) = X(z)/Y(z)

z/(z+1/2)=X(z)/Y(z)=H(z)

I hope this is correct now?
though I'm still not getting the initial point part.

 

[rude man]

Z [x (n − n0)] = z^-n0X(z);
y[n]+1/2y[n-1]=x[n]
→y[n+1]0+1/2y[n]=x[n+1]

This is now correct (I assume the 0 after the first term is a typo?)

Z-transform and get
→ z¹Y(z) + 1/2z⁰Y(z)=z¹X(z)
→ (z+1/2])Y(z)=zX(z)

Also, the right-hand side is X(z), not zX(z). In you case x[n] = aⁿ⁺¹u[n+1] so the Z transform of that term is X(z) = zZ{aⁿu(n)}. But here you're dealing with any x[n] since you're after the transfer function H = Y/X so you ignore the particular form of x[n]. See below.

given that H(z) = X(z)/Y(z)
[/quote]
H= Y/X, not X/Y.

z/(z+1/2)=X(z)/Y(z)=H(z)
I hope this is correct now?
though I'm still not getting the initial point part.

Never mind about the initial condition. In a transfer function it is assumed that all initial conditions are zero. I should have noticed at the start that you're looking for a transfer function rather than a general solution of a finite-difference equation.

Including x[n] = aⁿu[n] is a red herring in this problem. If you're looking for H = Y/X then the particular form of X is irrelevant. X is the input so Y = HX no matter what X happens to be. So your final equation is correct if you transpose X and Y and get rid of the numerator z.

If you did have to go on and solve for y[n] given the particular input x[n] then you'd have to worry about Z{x[n]}.

 

[rude man]

EDIT:

Z [x (n − n0)] = z^-n0X(z);
y[n]+1/2y[n-1]=x[n]
→y[n+1]0+1/2y[n]=x[n+1]

This is now correct (I assume the 0 after the first term is a typo?)

Z-transform and get
→ z¹Y(z) + 1/2z⁰Y(z)=z¹X(z)
→ (z+1/2])Y(z)=zX(z)

Correct.

given that H(z) = X(z)/Y(z)
[/quote]
H= Y/X, not X/Y.

z/(z+1/2)=X(z)/Y(z)=H(z)
I hope this is correct now?

Almost. If you change X(z)/Y(z) to Y(z)/X(z).

though I'm still not getting the initial point part.[/QUOTE]

Never mind about the initial condition. In a transfer function it is assumed that all initial conditions are zero. I should have noticed at the start that you're looking for a transfer function rather than the general solution of the full finite-difference equation.

Including x[n] = aⁿu[n] is a red herring in this problem. If you're looking for H = Y/X then the particular form of X is irrelevant. X is the input so Y = HX no matter what X happens to be. If you did have to go on and solve for y[n] given the particular input x[n] then you'd have to worry about Z{x[n+1]}.

Also: I think I confused you unencessarily by having you rewrite the equation so that n=0 is the lowest n.

You could have left it as y[n] + (1/2)y[n-1] = x[n]
then use Z{y[n-1]} = z^-1Z{y[n]}. I apologize for this unnecessary detour.