EEET2197 TUTORIAL THREE

EEET2197 CONTROL SYSTEMS

TUTORIAL 3 - SOLUTION.
Question 1.
A linear dynamic system is characterised by the following transfer function:
α
G (s ) =
2
+
s +1 s + 2
For what range of real values for α will the unit step response of the system exhibit undershoot?
[Answer]
If α > 0 , then the unit step response of the system will be uniformly positive with no undershoot.
Undershoot will result if and only if the transfer function includes a RHP (non-minimum phase)
zero. To test for this, combine the partial fractions to form the pole-zero transfer function form:
2(s + 2 ) + α (s + 1) (2 + α )s + (4 + α )
G (s ) = =
(s + 1)(s + 2) (s + 1)(s + 2)
The system zero is therefore:
z = − (4 + α ) (2 + α )
This zero will be positive (i.e. a RHP or non-minimum phase zero) if and only if:
− 4 < α < −2
Question 2.
A linear 2nd order system is governed by the following differential equation :
d 2 y (t ) dy (t )
2
+5 + 6 y (t ) = 6u (t )
dt dt
Find the response of this system for a cosine input signal of the form:
u (t ) = 10 cos(4t )
[Answer]
The Laplace transform of the input signal is given by:

U (s ) = L{u (t )} =
10 s
s + 16
2

Therefore the system response to this input is given by:

   10 s 
Y (s ) = G (s )U (s ) =  2
6
 s + 5s + 6   s + 16 
2

 A Ds 
Y (s ) = 
B C
+ + 2 + 2 
 s + 2 s + 3 s + 16 s + 16 
The A and B coefficients are found using residue theory as:
 60 s   60 s 
A = lim   = −6 B = lim   = 7. 2
s → −2
(
 (s + 3) s 2 + 16 ) s → −3
(
 (s + 2 ) s 2 + 16 )


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EEET2197 TUTORIAL THREE

The C and D coefficients are found as:

( ) ( )
A(s + 3) s 2 + 16 + B(s + 2 ) s 2 + 16 + (C + Ds )(s + 2 )(s + 3) = 60 s
− A × 48 − 32 × B 6 × 48 − 32 × 7.2
For s = 0: C= = = 9.6 = 2.4 × 4
6 6
60 + (6 )(4 )(17 ) − (7.2 )(3)(17 ) − (9.6 )(12 )
For s = 1: D= = −1.2
12
Giving:
 6 7.2 2.4 × 4 1.2 s 
Y (s ) = − + + 2 − 2 
 s + 2 s + 3 s + 16 s + 16 
Taking the inverse Laplace transform:
y (t ) = L−1{Y (s )} = 7.2e −3t − 6e −2t + 2.4 sin (4t ) − 1.2 cos(4t )

Question 3.
Determine the magnitude and phase response of the 2nd order system defined in question 2 at the
frequency ω = 4 rad/s . How does this result relate to the solution obtained in question 2?
[Answer]
The transfer function can be obtained as:
Y (s )
G (s ) =
6
= 2
U (s ) s + 5s + 6
Hence:

G ( jω ) =
6 6
=
( jω ) + j5ω + 6 j5ω + 6 − ω 2
2
( )
At ω = 4 rad/s :

G( j 4) =
6
= 0.2683∠ − 116.6°
j 20 − 10
The input phasor is:

U ( j 4) =
10 j 0°
e
2
Hence:

Y ( j 4) = G ( j 4 )U ( j 4 ) =
2.683
∠ − 116.6°
2
From the solution to question 2, the sinusoidal steady state solution is:
y SS (t ) = 2.4 sin (4t ) − 1.2 cos(4t ) = A cos(4t + δ )

tan (− δ ) =
2.4
A = 2.4 2 + 1.2 2 = 2.683 ⇒ δ = −116.6°
− 1.2

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EEET2197 TUTORIAL THREE

Hence the phasor solution:

2.683
YSS = ∠ − 116.6°
2
The frequency response of a systems at a given frequency is therefore the sinusoidal steady state
solution when excited by that frequency, or the phasor solution for that frequency.
Question 4.
Consider the feedback control system below:

Determine the damping ratio and the natural frequency of this system for an integral gain constant
of K I = 2.5 , and hence calculate the % overshoot and rise time that this control system will exhibit.
[Answer]
The closed loop transfer function can be obtained as:
Y (s ) GH (s ) 10 K I 10 K I
= = = 2
R(s ) 1 + GH (s ) s (s + 5) + 10 K I s + 5s + 10 K I
Comparing to the canonical second order system gives:
5
ωo = 10 K I , ζ =
2 10 K I

Substituting in K I = 2.5 gives:

5
ωo = 10 × 2.5 = 5 rad/s , ζ = = 0.5
2 10 × 2.5
To calculate overshoot and rise-time, we require β and ωd as:

(
β = sin −1 1 − 0.52 =) π
3
ωd = ωo 1 − ζ 2 = 5 × 1 − 0.52 = 4.33rad/s

Hence:
πζ π ×0.5
− −
1−ζ 2
Mp =e =e 1− 0.5 2
= 0.163 or 16.3%

π −π 3
tr = = 0.484 s
4.33

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EEET2197 TUTORIAL THREE

Question 5.
Consider the feedback control system below:

Determine whether this control system is capable of tracking the following reference signals
without steady state error. Apply the internal model principle to explain your results.
(a) Unit Step Function
[Answer]
Closed loop transfer function for the error:
GH (s )
Err (s ) = Y (s ) − R(s ) = R(s ) − R(s ) = R(s )
1
1 + GH (s ) 1 + GH (s )
s(s + 20 )
Err (s ) = R(s )
s(s + 20 ) + K p
Applying the final value theorem for a unit step function:
 s (s + 20 ) 1 
err (∞ ) = lim{sErr (s )} = lims =0
s →0 s →0  s (s + 20 ) + K s 
 p

Hence this system will follow a unit step function without steady state error. This is because the
plant contains a pole at the origin (i.e.1 s ) , which is the signal corresponding to the unit step
function as required according to the internal model principle.

(b) Ramp Function

[Answer]
Applying the final value theorem for a ramp function:
 s(s + 20 ) 1   s(s + 20 ) 
err (∞ ) = lim{sErr (s )} = lims  = lim 2 
 s(s + 20 ) + K p s  s →0  s (s + 20 ) + K p s 
s →0 s →0 2

Applying L’Hôpital’s rule :

 s(s + 20)   2 s + 20  20
lim 2
s →0 s (s + 20 ) + K s 
 = lim 2
s →0  3s + 40 s + K 
=
 p   p Kp

Hence this system will have a steady state error when following a ramp function. This is
(
because neither the plant nor the controller contains a double-pole at the origin i.e. 1 s 2 , which )
is the signal corresponding to the ramp function.

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EEET2197 TUTORIAL THREE

(c) Sinusoidal Excitation

[Answer]
The error transfer function response at a frequency ω is given by:
Err ( jω )
=
jω ( jω + 20 )
=
j 20ω + 20 − ω 2 ( )
R ( jω ) jω ( jω + 20 ) + K p j 20ω + K p − ω 2 ( )
Err ( jω ) ω 4 + 360ω 2 + 400
=
R ( jω ) ω 4 + (400 − 2 K p )ω 2 + K p2

This is non-zero for all ω , and hence the system exhibits steady-state error under sinusoidal
excitation. Again this is because of the absence of a sinusoidal system in the controller and plant
transfer functions.
Question 6.
Consider the feedback control system below:

Determine whether this control system is capable of tracking the following a sinusoidal reference at
the frequency ω = 10 rad/s without steady state error.
[Answer]
The error transfer function response at a frequency ω is defined according to:
GH (s )
Err (s ) = Y (s ) − R(s ) = R(s ) − R(s ) = R(s )
1
1 + GH (s ) 1 + GH (s )

Err (s )
=
(
(s + 20) s 2 + 10 )
(
R(s ) (s + 20 ) s 2 + 10 + sK p )
Err ( jω )
=
(
( jω + 20) − ω 2 + 10 )
(
R( jω ) ( jω + 20 ) − ω 2 + 10 + jωK p )
At the frequency ω = 10 rad/s is given by:

(
Err jω = j 10 ) (j
10 + 20  − 10 + 10 

2


)
= =0
(
R jω = j 10 ) ( )
j 10 + 20  − 10 + 10  + jωK p
2

 
Hence this system will follow a sinusoidal reference with the frequency ω = 10 rad/s without
any steady state error.

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EEET2197 TUTORIAL THREE

Question 7.
Obtain the transfer function for the circuit below, relating the output voltage v2 (t ) to the input
voltage v(t ) .

[Answer]
Applying the unit current method:
I L2 = 1 Vc 2 = R2 + sL2

I c 2 = Vc 2 sC 2 = s 2 L2 C 2 + sR2 C 2

I R1 = I L 2 + I C 2 = s 2 L2 C 2 + sR2 C 2 + 1

V = Vc 2 + I R1 R1 = s 2 L2 C 2 R1 + s(L2 + R2 R1C 2 ) + R1 + R2

Therefore the output voltage transfer function is:

V2 (s ) R2
=
V (s ) s 2 L2 C 2 R1 + s(L2 + R2 R1C 2 ) + R1 + R2

Question 8.
Using mesh or nodal analysis techniques, obtain the transfer function for the circuit below, relating
the voltage across the resistor R2 to the input voltage v(t ) .

[Answer]
The nodal equations are most easily obtained as:
1 1 
V   + + sC1 − sC1   VL 
R  =  1
R sL1  
 1 
s (C1 + C 2 ) +
1  V 
0  − sC1  R2 
 R2 

Inverting:

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EEET2197 TUTORIAL THREE

 
 s(C1 + C 2 ) + R
1
 VL  1
sC1  V 
 =  2   R1 
VR2   1   1  2 2   
+ sC1   0 
1 1
+ sC1  s(C1 + C 2 ) +
1 +
 +  − s C1 
sC1

 R1 sL1  R2  R1 sL1

 
 s(C1 + C 2 ) + R
1
 VL  sL1 R2 R1
sC1  V 
 =  2   R1 
VR2  s L1C1C 2 R1 R2 + s L1 (R2 R1 (C1 + C 2 ) + R1C1 ) + sL1 + R1
3 2
 sC1
1
+
1  
+ sC1   0 
 R1 sL1 


 V 
 s(C1 + C 2 ) +
1
 VL   
sL1 R2 R1  R2  R1 
 =
VR2  s L1C1C 2 R1 R2 + s L1 (R2 R1 (C1 + C 2 ) + R1C1 ) + sL1 + R1
3 2  V 
 sC1 
 R1 

Extracting the transfer function for VR2 V gives:

VR2 (s ) s 2 C1 L1 R2
=
V (s ) s L1C1C 2 R1 R2 + s L1 (R2 R1 (C1 + C 2 ) + R1C1 ) + sL1 + R1
3 2

B. P. McGrath August 2016