HW5, MAE 171. Spring 2005. UCI

Nasser Abbasi

Problem 6-1


Figure

part a

Calculate and plot the unit-step response at the sampling instances for the case when MATH

First find the open loop $G\left( z\right) $ MATH

MATH

Hence

MATH

Hence the closed loop transfer function is, noting that MATH

MATH

For $T=2\sec $ we get from (1)

MATH

Now since the input is a unit step, then MATH

Hence the output MATH

MATH

Hence

MATH

From tables, the inverse Z transform of MATH $\ $ for $n=1,2,3,\cdots $

Hence MATH

And MATH MATH

$\ \allowbreak \ $

Hence The inverse Z transform of $C\left( z\right) $ is MATH

This is a plot of $c[n]$


Figure

Part(b)

Calculate the system unit step response for the analog system. Plot the response on the same graph as the result of (a).

The closed loop transfer function is

MATH

Hence MATH

MATH

MATH

Hence MATH

Here is a graph which shows the analog and discrete responses





Figure




Part(c)

Let MATH and $T=0.4\sec $. Calculate the unit step response and plot these on the same plot

From equation (1) in part(a), we have

MATH

for $T=0.4$

MATH

Now since the input is a unit step, then MATH

Hence the output MATH

MATH

Hence

MATH

From tables, the inverse Z transform of MATH $n=1,2,3,\cdots $

Hence MATH

And MATH MATH

Hence The inverse Z transform of $C\left( z\right) $ is MATH

The following is the plot comparing discrete system response for $T=2$ and $0.4\sec $ with the pure analog system.

Notice the effect of changing $T$ on the discrete response. For larger $T$ the system had a larger overshoot and was less damped. As sampling rate increased the response exhibited over damping.





Figure

Problem 6-12

Consider the sampeld-data systems with the following char. equation

(1) $z-0.999=0$

(2) $z-0.99=0$

(3) $z-0.9=0$

(4) $z+0.9=0$

(5) MATH

(6) $z^{2}-1=0$

(7) $z^{2}-2z+0.99=0$

(8) $z^{2}-1.2z+0.7=0$

Solution Table

This table summarizes the results.

system $T=0.1$ $T=1$
$T_{s}\sec $ $T_{r}\sec $ $M_{p}$ $T_{s}\sec $ $T_{r}\sec $ $M_{p}$
$1$ $z-.999$ $400$ N/A N/A $4000$ N/A N/A
$2$ $z-.99$ $40$ N/A N/A $400$ N/A N/A
$3$ $z-.9$ $3.8$ N/A N/A $38$ N/A N/A
$4$ $z+.9$ $3.79$ $0.05$ $89.9\,\%$ MATH $0.5$ $89.9\,\%$
$5$ $z^{2}-1.85z+.854$ MATH N/A N/A $8.\,\allowbreak 16$ N/A N/A
$6$ $z^{2}-1$ N/A N/A N/A N/A N/A N/A
$7$ $z^{2}-2z+.99$ N/A N/A N/A N/A N/A N/A
$8$ $z^{2}-1.2z+0.7$ MATH $0.233$ $48.3\,\,\%$ $22.\,\allowbreak 4$ $2.\,\allowbreak 34$ $48.3\,\,\%$

From the table above we observe that the $T_{r}$ (rise time) and $T_{s}$ (settling time) varies with $T$ only. While $M_{p}$ is not affected.

As $T$ becomes larger (slower sampling), the response will take longer to rise and to settle. The ratio of the change in $T$ is the same ratio by which the response will take longer to settle for these systems. To classify the responses, this table and the plots show the classification. This Next plot is the transient response characteristics of the systems for $T=0.1\sec $


Figure

system $T=0.1$
$1$ $z-.999$ stable
$2$ $z-.99$ stable
$3$ $z-.9$ stable
$4$ $z+.9$ stable,exponential damped sinusoidal
$5$ $z^{2}-1.85z+.854$ stable,damped
$6$ $z^{2}-1$ marginal unstable.
$7$ $z^{2}-2z+.99$ unstable, exponential
$8$ $z^{2}-1.2z+0.7$ stable, exponential damped sinusoidal

Solution

part(a)

What can you determine about the system natural response characteristic for each system for $T=0.1\sec $?

For second order system we can determine the transient response characteristic as follows.

From the location of the discrete transfer function poles, we determine $\zeta ,\omega _{n}$ using the three equations below. And in turn from $\zeta ,\omega _{n}$ we can determine, using the simplified equations, the rise time $T_{r}$, the maximum overshoot $M_{p}$ and the settling time $T_{s}.$ In addition we can find the time constant $\tau $ from knowing $\zeta ,\omega _{n}$

Let a Z system pole be expressed as MATH, then these are the equations to use: (see book page 216 for the derivation)

MATH

We can now determine $T_{r},M_{p},T_{s}$ for the system response to a unit step.

For $T_{s}$ use MATH

For $M_{p}$ (since this is a unit step response, hence the final value is 1), we get

MATH

For $T_{r}\ $use

MATH

where MATH


Figure

For a first order discrete system, the response will not exhibit oscillations if the root is positive. So for a first order system with positive root we can only say something about $\tau $, the time constant (which is the time required for response to reach $63\%$ of its final value.)

Assume root of Z system is at MATH but since this is a real positive root, then $\theta =0$, and $z_{1}=r$ and since MATH, we get that MATH where $r$ here is the magnitude of $z_{1}$

Hence from

MATH

We see that $a=-\frac{\ln r}{T}$ hence since $\tau =\frac{1}{a}$ then we get MATH

Now, For a first order discrete system, the response will exhibit oscillations if the root is negative.

Assume root of Z system is at MATH but since this is a real negative root, then $\theta =\pi $, and $z_{1}=-r$ and since MATH, we get that MATH where $r$ here is the magnitude of $z_{1}$ and hence since we are taking the natural log of a negative value, then $s$ is a complex value say MATH, from which we find $\omega _{n}$ and $\zeta $ and then we can find the rest of the transient response properties since MATHand we find $\zeta $ from MATH

Now we can start to solve the problems

(1) $z-0.999=0$

Since is a first order system. root $r=0.999$ hence from MATH we get

MATH

(2)$z-0.99=0$

Since is a first order system. root $r=0.99$ hence from MATH we get

MATH

Notice that for a small change in the location of the root ($10\%$ change), the system now responds $10$ times as fast.

(3)$z-0.9=0$

Since is a first order system. root $r=0.9$ hence from MATH we get

MATH

Notice that for a small change in the location of the root ($10\%$ change), the system now responds $10$ times as fast.

(4)$z+0.9=0$

Since is a first order system. since root $r=-0.9$ hence transient response will oscillate.

MATH

To evaluate $\ln (-.9),$sinceMATH

Let MATH

Hence MATH

From (1) we get$\ $MATH

Hence MATH

Hence

$\omega _{d}=10\pi $ and $\sigma =-1.0536$
so MATH

and MATH

Hence now we can obtain the transient response characteristics.

MATH

MATH

MATH

MATH

MATH

(5)MATH, Solution is: MATH

Since there are no complex poles, and both poles positive, then system will not oscillate even though it is a second order system. Hence we know that $M_{p}=0$ and $\omega _{n}=0$ and $T_{r}=\infty $ in this case since we will use $T_{r}$ here as the time to go from $0$ to $100\%$ of the final value.

using MATHthere are 2 time constants since there are 2 roots, we get MATH

and MATH

We pick the larger time constant since that is associated with the dominant pole.

hence

MATH

(6) $z^{2}-1=0$

poles are at $\pm 1$

This is a marginally unstable system. For an input of $1/s$, we see that the pole of the input equals the pole of the system at $s=0$, hence resonance will occur. So not applicable to determine the transient characteristic.

(7) $z^{2}-2z+0.99=0$, Solution is: MATH

We have a pole outside the unit circle. Hence this is unstable system. So not applicable to determine the transient characteristic.

(8) $z^{2}-1.2z+0.7=0$, Solution is: MATH

Here the pole is MATH where $r=0.8366$ and $\theta =0.7711 $ rad$=44^{0}$

MATHMATH

MATH

MATH

MATH

part(b)

Redo part(a) for $T=1$

(1) $z-0.999=0$

Since is a first order system. root $r=0.999$ hence from MATH we get

MATH

(2)$z-0.99=0$

Since is a first order system. root $r=0.99$ hence from MATH we get

MATH

(3)$z-0.9=0$

Since is a first order system. root $r=0.9$ hence from MATH we get

MATH

(4)$z+0.9=0$

As from part (a), we get

MATH

Hence

$\omega _{d}=\ \pi $ and $\sigma =-0.10536$
so MATH

and MATH

Hence now we can obtain the transient response characteristics.

MATH

MATH

MATH

MATH

MATH

(5)MATH, Solution is: MATH

using MATHthere are 2 time constants since there are 2 roots, we get MATH

and MATH

We pick the larger time constant since that is associated with the dominant pole.

hence

MATH

(6) $z^{2}-1=0$

poles are at $\pm 1$

This is a marginally unstable system. For an input of $1/s$, we see that the pole of the input equals the pole of the system at $s=0$, hence resonance will occur. So not applicable to determine the transient characteristic.

(7) $z^{2}-2z+0.99=0$, Solution is: MATH

We have a pole outside the unit circle. Hence this is unstable system. So not applicable to determine the transient characteristic.

(8) $z^{2}-1.2z+0.7=0$, Solution is: MATH

Here the pole is MATH where $r=0.8366$ and $\theta =0.7711 $ rad$=44^{0}$

MATH MATH

MATH

MATH

MATH




Part(c)

From the table above we observe that the $T_{r}$ (rise time) and $T_{s}$ (settling time) varies with $T$ only. While $M_{p}$ is not affected.

Problem 6-10

The following system has parameters

MATH

part(a)

Find $\zeta ,\omega _{n}$ and $\tau $ for the sample period $T=0.5\sec $

Find open loop $G\left( z\right) $ then the closed loop MATH then find the zeros of the characteristic equation, and from these find $\zeta ,\omega _{n}$ and $\tau $ using the relations MATH

MATH

MATH

MATH

So

MATH

Characteristic equation MATH, Solution is: MATH

Hence $r=0.8346$ and $\theta =0.4543$ rad

Hence

MATH

Part(b)

redo part (a) with $T=0.1\sec $

MATH

MATH

So

MATH

Characteristic equation MATH, Solution is: MATH, Solution is: $\ $ MATH

Hence $r=0.95366$ and $\theta =0.0877$ rad

Hence

MATH

part(c)

repeat part(a) for analog system.

MATH

Hence MATH

$\omega _{n}=1$
and MATH
$\zeta =0.5$

Hence MATH

Part(d)

Give table listing the three parameters as function of sampling frequency $f_{s}=\frac{1}{T}$. State the result of decreasing the sampling frequency on the parameters.




Solution summary table

T sec $f_{s}$ $\zeta $ $\omega _{n}$ rad/sec $\tau \sec $
$1$ $1$ $0.25$ $0.9191$ $4.36$
$0.5$ $2$ $0.369\,77$ $0.977\,9$ $2.766$
$0.1$ $10$ $0.475\,85\,$ $0.997\,\,$ $2.108$
analog $0.5$ $1$ $2$

We see as the sampling frequency $f_{s}$ becomes smaller, then the damping ratio decreases, this means the response will become more oscillatory in the transient stage with larger overshoot.

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