HW4, MAE 171. Spring 2005. UCI

Nasser Abbasi

Problem 4-3

Find the Z transform of the following functions, using the z-transform tables. Compare the pole-zero locations of $E\left( s\right) $ and MATH in the s-plane. Let $T=0.1$

part c

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.

MATH

finite zeros of $E\left( s\right) $ at $s=-2$, non-finite zeros at $\infty $, and , poles are at $s=0,s=-1.$

From problem 3-4 (which we did in HW3), MATH

For $T=0.1$,

MATH
$\Rightarrow $
finite zero at $z=0$, non-finite zero at $\infty $, poles at MATH

Now to find poles/zeros of MATH.

We can solve this directly from the expression for MATH, but an easier method is as follows. We know the poles of MATH to be located at distances of $j\omega _{s}n\ $From the poles of $E\left( s\right) $ for MATH , So by knowing the poles of $E\left( s\right) $ we find the poles of MATH.

However, for the zeros of MATH, there is no such correlation, but the zeros of MATH are still periodic of period $j\omega _{s}$

Since we know the poles of $E\left( s\right) $ to be located at $0,-1$, then we can find the poles of MATH that are periodic with pole $0$ to be at MATH and the poles of MATH that are periodic with pole at $-1$ to be at MATH

Since MATH rad/sec, hence the poles of MATH are at

MATH
and at
MATH

Now to find the zeros of MATH, since MATH, then $\ $ MATH

Hence a zero of MATH is when $\epsilon ^{0.1s}=0$ or when $s=-\infty $. There is also a zero when $s=+\infty $, 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 $\infty $ $,$ so the rest of the zeros are all at $\infty $.

This tables summarizes the results

$E\left( s\right) $
zeros $-2,\infty $
poles $0,-1$
MATH
zeros $\pm \infty $
poles MATH
$E\left( z\right) $
zeros $0,\infty $
poles MATH

part f$\ $

MATH

Roots of denominator are MATH

Hence poles of $E\left( s\right) $ are $s=-1\pm 2j$, and a non-finite zeros at $s=\pm \infty $

From problem 3-4 (which we did in HW3)

MATH

Hence for $T=0.1$, MATH

Hence a finite zero is at $z=0,$ and non-finite zero at $z=\infty $. (complex infinity). For the poles, the roots of the denominator are found to be at MATH

Since we know the poles of $E\left( s\right) $ to be located at $-1\pm 2j$, then we can easily find the poles of MATH that are periodic with pole $-1+2j$ to be at MATH and the poles of MATH that are periodic with pole $-1-2j$ to be at MATH

Since MATH , hence the poles of MATH are at MATH

MATH
and are at

MATH

Now to find the zeros of MATH. Since MATH then MATH

For $T=0.1,$ MATH

Then we see that a zero of MATH is at $s=-\infty $, another zero comes when we take the denominator to infinity, which gives a zero at $+\infty $

This tables summarizes the results

$E\left( s\right) $
zeros $\pm \infty $
poles $-1\pm 2j$
MATH
zeros $\pm \infty $
poles MATH
$E\left( z\right) $
zeros $0,\infty $
poles MATH

Problem 4-5

part(a)

Find the system response at the sampling instances to a unit step input for the system shown. Plot $c\left( nT\right) $ versus time.

MATH




Let plant transfer function be called MATH

MATH

But MATH

Notice the cancellation of the zero in the plant with the pole from the ZOH.

Now since MATH (because MATH, hence $e[nT]=u[nT]$)

Then (2) can be written as

MATH

Hence (1) can now be written as

MATH

Looking at $c[nT]$ for few values, for $T=1$ we get the following sequence generated

MATH

This is a plot of $c[nT]$


Figure

part(b)

Verify the answer in part(a) by determining the input to the plant $m\left( t\right) $ and then calculating $c\left( t\right) $ by continuous-time method.

MATH

but MATH

Hence we see that the plant is driven by a unit step, the same as the reference signal $r\left( t\right) $ itself.

Now that $M\left( s\right) $ is found, we use (1) to find $c\left( t\right) $

MATH

Hence

MATH MATH
and from (a) we obtained that
MATH $u[k]$
. This verifies the result.

Here is a plot


Figure




Problem 4-8

Find system response at the sampling instances to a unit-step input for the system shown


Figure




MATH

Where MATH

Hence

MATH

Hence, since all are in star format, we can switch to Z domain

MATH

But MATH and MATH

Hence MATH

Hence (1) can be written as

MATH

Now need to find $N\left( z\right) $. But MATH

Where MATH

But MATH and MATH

Hence MATH

Hence MATH

Substitute above into (2) we get

MATH

Now Let $T=1$ we get

MATH

Note that the poles of $C\left( z\right) $ are inside the unit circuit, hence this is a a stable discrete system.

MATH

Hence MATH

MATH

MATH

Hence

MATH

Hence MATH

The following is a plot of the response for few values of $n$


Figure

Now I will find $c\left( t\right) ,$ since MATH then

MATH

But since MATH, which we found earlier MATH

Since a delay $z^{-1}$ in Z maps to $\epsilon ^{-Ts}$ in S, then the above coverts to S domain as MATH

Now (3) can be written as

MATH

MATH, MATH

Hence

MATH

Since MATH where MATH

Now MATH and MATH

Hence

MATH

In particular, when $T=1$ we get

MATH

To verify, I plot the above result for $n$ up to 1000 and for $t$ up to 15 seconds.

This is a plot of $c(t)$ and $c[n]$ on the same plot to compare.


Figure

Problem 5-1

for each system express $C\left( z\right) $ in terms of the input and the transfer functions shown

part(a)


Figure

To simplify things, I'll write the symbols without the argument $\left( s\right) $

MATH

Solve for $E$ noting that MATH

MATH

Hence since $C=GE^{\ast }$ then MATH

Since now all are starred, then we can convert to $z$ domain easily MATH

Part (b)


Figure

Note that since the error signal Laplace transform is the difference of 2 starred transform, then it is already starred.

MATH

Solve for $B$

MATH

Hence now sinceMATH

Since all starred, we convert to $z$ domain

MATH

part(c)


Figure

MATH

solve for $E$

MATH

Hence from $C=E^{\ast }G$ we get

MATH

Hence MATH

part(d)


Figure

MATH

Solve for $C$

MATH

Hence MATH

part(e)


Figure

MATH

Solve for $E$

MATH

Hence since $C=E^{\ast }G$ we get

MATH

Hence

MATH