Analysis:

i)

From the calibration of the potentiometer, the following data was collected which relates the display voltage V to the angular position of the arm, theta:

Angular PositionΘ (radians)

Voltage (Volts)

π/4 1.284

- π/4 -1.301

If angular position is plotted as a function of voltage, the slope of the line will give the potentiometer gain Kpot:

-1.5 -1 -0.5 0 0.5 1 1.5

-1-0.8-0.6-0.4-0.2

00.20.40.60.8

1

f(x) = 0.607631873987066 x

Angular Position vs Voltage

Voltage (V)

Angu

lar P

ositi

on (R

ad)

From the trend line with an intercept at 0, it is clear that the voltage gain across the potentiometer is 0.607.

Similarly, measured voltage across the tachometer is related to its angular velocity by a constant gain Ktach. To determine angular velocity the tachometer was programmed to make 20 revolutions and the time was recorded. From this data the average time required to make one revolution and subsequently the angular velocity is calculated as follows:

Number of Revolutions: 20Average Voltage Reading: 1.34 VoltsTime: 82.26 seconds

ω=20∗2 π82.26

=1.527 rad / s

From this we determine Ktach by dividing angular velocity by average voltage:

K tach=1.527

rads

1.34V=1.14

ii) The open loop response of the servomotor to a step input is shown below:

As one might expect the system is very sensitive to parameter changes and oscillates even as it approaches steady state, in part due to the influence of the electrical element. From here, we can approximate a steady state value of 1.58.

If final value theorem is applied to the first order model of this system, we have:

lims→ 0

Kτs+1

=K

Hence, K=1.58. To estimate the time constant tau, we take the sample point (0.027,0.9435) since this range removes much of the oscillation observed as steady state is approached. Plugging into the generic response gives:

y=K (1−e−tτ )

0.9435=1.58(1−e−0.027τ )

Solving for tau yields τ=0.029s

iii) Plugging in values obtained for K and tau into a first order model gives the transfer function:

Matlab then simulates the response to a unitary step as:

We note that the curve in this response is much smoother than what was observed from actual experimental data and removes the influence of oscillations as steady state was approached. General system parameters such as settling time are preserved thought it may be more difficult to spot on the graph from part ii.

iv) We begin by generating a transfer function for the block scheme shown below:

The nested loop will have a transfer function of form:

C ( s )=G (s )

1+H ( s)G (s )=

Kτs+1

1+K K D

τs+1

=K

τs+(1+K K D)

Combining in series with the proportional gain Kp and integrator in a closed loop configuration gives:

T ( s )=

K K p

τ s2+ (1+K K D ) s

1+K K p

τ s2+(1+K K D ) s

=K K p

τ s2+(1+K K D ) s+K K p

By comparing with a generic second order transfer function we have:

K K p

τ s2+ (1+K K D ) s+K K p

= ω2

s2+2 ζω+ω2

or:

2 ζ ω=1+K K D

τω=√ K K p

τ

From the design condition where overshoot cannot exceed 5% we have:

ln (0.05 )= −πζ

√1−ζ 2

ln (0.05 )2=ζ 2 (π2+ ln (0.05 )2 )

ζ=√ ln (0.05 )2

(π2+ ln ( 0.05 )2 )=0.6901

similarly a the design condition that requires a settling time under 1 second gives:

4ζω

= 40.6901ω

=1

Solving gives ω=5.67 rad/s. Plugging into the previously derived equations and solving for gains Kp and KD

gives:

K p=ω2 τK

=(5.67 )2 (0.029 )

1.58=0.59

K D=1−2 ζωτ

−K=

1−2 (0.6901 ) (5.67 ) (0.029 )−1.58

=−0.489

The system response to varying reference positions is shown below:

If the fourth plateau from the left is examined it can be shown that system requirements are satisfied by this response. None of the peaks between smooth plateaus corresponding to steady state position last longer than 1 second, hence the settling time requirement is met. However, it appears that the overshoot in the system surpasses the 5% limitation. In many cases the difference between the peak and the desired reference position is greater than the expected 5%, in part due to the simplification made regarding the order of the motor's transfer function.

v) For a PID controller, we add the element KI/s in parallel with Kp in the block diagram for a PD controller. Modifying the transfer function to include this term yields:

T ( s )=K K p s+K K I

τ s3+ (1+K K D ) s2+K K p s+K K I

We then compare the denominator to the standard form for a third order system whose behavior mimics a second order response:

D (s )=( s+ζω ) ( s2+2 ζω+ω2 )=s3+3 ζω s2+(2 ζ ω2+ω2 ) s+(ζ ω3)

Which yields three equations as follows:

(1+K K D )τ

=3 ζω

K K p

τ=( 2 ζ ω2+ω2 )

K K I

τ=ζ ω3

Plugging in using zeta =0.6901 for an arbitrary settling time T=0.3s :

ω= 4T sζ

= 40.6901∗0.3

=19.32rad /s

Solving for each gain gives:

K D=0.101, K P=16.31 ,K I=91.34

which gives response:

As shown, the response will settle within 0.3s at a steady state value of 1 due to a unitary input. This gives a higher overshoot than expected because the equations used to solve for zeta and omega apply to a second order system while in actuality the model reveals that this was a third order. Hence we expect to see deviation from what was shown in calculations.