HW4, MAE 171. Spring 2005. UCI
Nasser Abbasi
Find the Z transform of the following functions, using the z-transform tables. Compare the pole-zero locations of and in the s-plane. Let
Note: The problem did not mention this, but since we must have same number of zeros and poles for a transfer function (this is when we consider both finite and non-finite zeros), hence, in this problem, I will find all such zeros.
From problem 3-4 (which we did in HW3),
For ,
Now to find poles/zeros of .
We can solve this directly from the expression for , but an easier method is as follows. We know the poles of to be located at distances of From the poles of for , So by knowing the poles of we find the poles of .
However, for the zeros of , there is no such correlation, but the zeros of are still periodic of period
Since we know the poles of to be located at , then we can find the poles of that are periodic with pole to be at and the poles of that are periodic with pole at to be at
Since rad/sec, hence the poles of are at
Now to find the zeros of , since , then
Hence a zero of is when or when . There is also a zero when , since that will make the denominator blow up. Since number of zeros must match the number of poles, we can generate the rest of the zeros from periodic repetition (but zeros are already at so the rest of the zeros are all at .
This tables summarizes the results
|
|||||
|
|||||
|
Roots of denominator are
Hence poles of are , and a non-finite zeros at
From problem 3-4 (which we did in HW3)
Hence for ,
Hence a finite zero is at and non-finite zero at . (complex infinity). For the poles, the roots of the denominator are found to be at
Since we know the poles of to be located at , then we can easily find the poles of that are periodic with pole to be at and the poles of that are periodic with pole to be at
Since , hence the poles of are at
Now to find the zeros of . Since then
For
Then we see that a zero of is at , another zero comes when we take the denominator to infinity, which gives a zero at
This tables summarizes the results
|
|||||
|
|||||
|
Find the system response at the sampling instances to a unit step input for the system shown. Plot versus time.
Let plant transfer function be called
But
Notice the cancellation of the zero in the plant with the pole from the ZOH.
Now since (because , hence )
Then (2) can be written as
Hence (1) can now be written as
Looking at for few values, for we get the following sequence generated
This is a plot of
Verify the answer in part(a) by determining the input to the plant and then calculating by continuous-time method.
but
Hence we see that the plant is driven by a unit step, the same as the reference signal itself.
Now that is found, we use (1) to find
Hence
Here is a plot
Find system response at the sampling instances to a unit-step input for the system shown
Where
Hence
Hence, since all are in star format, we can switch to Z domain
But and
Hence
Hence (1) can be written as
Now need to find . But
Where
But and
Hence
Hence
Substitute above into (2) we get
Now Let we get
Note that the poles of are inside the unit circuit, hence this is a a stable discrete system.
Hence
Hence
Hence
The following is a plot of the response for few values of
Now I will find since then
But since , which we found earlier
Since a delay in Z maps to in S, then the above coverts to S domain as
Now (3) can be written as
,
Hence
Since where
Now and
Hence
In particular, when we get
To verify, I plot the above result for up to 1000 and for up to 15 seconds.
This is a plot of and on the same plot to compare.
for each system express in terms of the input and the transfer functions shown
To simplify things, I'll write the symbols without the argument
Solve for noting that
Hence since then
Since now all are starred, then we can convert to domain easily
Note that since the error signal Laplace transform is the difference of 2 starred transform, then it is already starred.
Solve for
Hence now since
Since all starred, we convert to domain
solve for
Hence from we get
Hence
Solve for
Hence
Solve for
Hence since we get
Hence