Hi,


This is a review quesiton that I am unsure of the correct answer

If I have a discrete time unit step input x(kT) and the output y(kT) was

T=0.5, 2T= 3T= 4T= 5T= 6T= 7T=1

Would the z transform of the output signal y(kT) look something like

\(y(kT)=\frac{1}{0.5-z^-1}+\frac{1}{1-z^-2}+\frac{1}{1-z^-3}\frac{1}{1-z^-4}+\frac{1}{1-z^-5}+\frac{1}{1-z^-6}+\frac{1}{1-z^-7}\)

Thanks for any help..

 

It is difficult to understand what you mean because your functions are defined improperly. Do you mean y(T)=0.5, y(2T)=1, y(3T)=1,...etc?

If so the z-transform would be:

y(Z) = 0.5*Z^-1 + Z^-2 + Z^-3 + Z^-4.... etc

Do you need the z-transform in closed form?

 

Let's say at k=1 you have y(kT)=0.5 and for all future values of k (i.e., 2 to infinity) y(kT)=1. I have attached an example of how you can get that in closed form.

 

Sorry for the misinterpretation of the question what you have wrote in the pdf is correct.


So you have found is y(z) how to do you then bring this back to a discrete time system transfer function G(z)=Y(z)/X(z). What is the X(z) part?

Also if this was to be re re-written in to differential equation form so you can check the result use the first four terms(k=0,1,2,3) of the signals for a unit step input.

Thanks for your help

 

Your input, X(KT), was the unit step function. So you need to take the z-transform of a unit step function to get X(z). I can't remember off of the top of my head but I think the z-transform of the unit step is -

X(z)=\(\frac{z}{z-1}\)

 

Tskaggs,

so by using the valve of X(z) and then substituting it into the equation i have come up with the follow for G(z) does this look correct?



\( X(z) =\frac{z}{z-1}\)

\( G(z)= \frac{\frac{0.5(z+1)}{z(z-1)}}{\frac{z}{z-1}}\)

\( G(z)= \frac{0.5(z+1)}{z(z-1)} \times \frac{z-1}{z}\)

\( G(z)= \frac{0.5(z+1)}{z^2}\)

I appreciate your help

 

That answer looks legitimate to me.

 

thanks for your help

 

Just to add to this and change the system to a difference equation, is this done correclty

\( \frac {0.5(z+1)}{z2} \)
\( \frac {0.5(1-z^-1)}{z^-2} \)

\( 0.5(1-z^-^1)X(z)= z^2 Y(z)\)

\( (y(k)-y(k-1)= 0.5x(k-1)\)

\( y(k)= 0.5x(k-1)+y(k-1)\)

 

Just to add to this and change the system to a difference equation, is this done correclty

\( \frac {0.5(z+1)}{z2} \)
\( \frac {0.5(1-z^-1)}{z^-2} \)

\( 0.5(1-z^-^1)X(z)= z^2 Y(z)\)

\( (y(k)-y(k-1)= 0.5x(k-1)\)

\( y(k)= 0.5x(k-1)+y(k-1)\)

You lost me when you went from the first step to the second. Something seems odd about this transformation.

 

The first line is the transfer function G(z) and then the next I changed it to back to an equation where z was raised to the minus 1. Maybe I didnt not complete this step correctly. I have done this as this is what the difference equation has done in the text book

\( \frac {0.5(z+1)}{z2} \)


\( G(z) = \frac {Y(z)}{R(z)} =\frac {0.5(1-z^-^1)}{z^-^2} \)


\( 0.5(1-z^-^1)X(z)= z^-^2 Y(z)\)

\( (y(k)-y(k-2)= 0.5x(k-1)\)

\( y(k)= 0.5x(k-1)+y(k-2)\)


Have I explained myself better this time.

 

I think you are transforming incorrectly-

\(\frac{0.5(z+1)}{z^{2}}=\frac{0.5(z+1)*z^{-1}}{z^{2}*z^{-1}}=\frac{0.5(1+z^{-1})}{z}\)