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EE C128 / ME C134 Problem Set 1 Solution (Fall 2010) Wenjie Chen and Jansen Sheng, UC Berkeley

1. (10 pts) BIBO stability

The system ℎ(𝑡) = 𝑐𝑜𝑠(𝑡)𝑢(𝑡) is not BIBO stable. What is the region of convergence for 𝐻(𝑠)?

A bounded input 𝑥(𝑡) = 𝑠𝑖𝑛(𝑡)𝑢(𝑡) gives an unbounded output for this system. Determine

𝑦(𝑡) = 𝑥(𝑡) ∗ ℎ(𝑡) using Laplace transform properties.

Solution:

The Laplace transform of cos 𝑡 is 𝐻 𝑠 = cos 𝑡 𝑢(𝑡)𝑒−𝑠𝑡𝑑𝑡∞

−∞. Since cos 𝑡 𝑢 𝑡 is bounded

by a constant, the region of convergence for 𝐻(𝑠) is for all 𝑠 positive, or 𝑅𝑒 𝑠 > 0.

To calculate the convolution 𝑦(𝑡) = 𝑥(𝑡) ∗ ℎ(𝑡), we can calculate the transforms of each

function and use the properties of Laplace transforms (FPE6e P758).

𝐻 𝑠 =𝑠

𝑠2+1 , 𝑋 𝑠 =

1

𝑠2+1

Now, convolution in time becomes a product in the Laplace domain so 𝑌 𝑠 =𝑠

𝑠2+1

1

𝑠2+1=

𝑠

(𝑠2+1)2. To convert 𝑌(𝑠) back into the time domain, notice we can use the property that

multiplication by time becomes the negative derivative in the s-domain (𝑡𝑓(𝑡) becomes−𝑑

𝑑𝑠𝐹(𝑠)).

𝑌 𝑠 = −1

2

𝑑

𝑑𝑠

1

𝑠2+1= −

1

2

𝑑

𝑑𝑠𝑋(𝑠) so 𝑦 𝑡 =

1

2𝑡 sin 𝑡 𝑢(𝑡), which is unbounded as 𝑡 ∞.

2. (25 pts) Laplace transform review

For each transfer function below (all are causal and at least marginally stable), determine ℎ(𝑡)

and sketch the impulse response. (Pay attention to scales, dynamics, etc.)

Solution:

i) 𝐻1 𝑠 =1

𝑠+10 (𝑠+1)= −

1

9

1

𝑠+10+

1

9

1

𝑠+1

ℎ1 𝑡 = −1

9𝑒−10𝑡𝑢(𝑡) +

1

9𝑒−𝑡𝑢(𝑡)

ii) 𝐻2 𝑠 =𝑠+10

(𝑠+1)=

𝑠+1

𝑠+1+

9

𝑠+1= 1 +

9

𝑠+1

ℎ2 𝑡 = 𝛿 𝑡 + 9𝑒−𝑡𝑢(𝑡)

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iii) 𝐻3 𝑠 =𝑠−10

𝑠+10 (𝑠+1)=

20

9

1

𝑠+10−

11

9

1

𝑠+1

ℎ3 𝑡 =20

9𝑒−10𝑡𝑢(𝑡) −

11

9𝑒−𝑡𝑢(𝑡)

iv) 𝐻4 𝑠 =1

𝑠2+2𝑠+101=

1

(𝑠+1)2+102

ℎ4 𝑡 =1

10𝑒−𝑡sin(10𝑡)𝑢(𝑡)

v) 𝐻5 𝑠 =𝑠

𝑠2+2𝑠+101=

𝑠

(𝑠+1)2+102 =𝑠+1

(𝑠+1)2+102 −1

(𝑠+1)2+102

ℎ5 𝑡 = 𝑒−𝑡 cos 10𝑡 𝑢(𝑡) −1

10𝑒−𝑡sin(10𝑡)𝑢(𝑡)

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3. (20 pts) Equivalent models

The figure below shows a mechanical model for a simple resonant wing drive system, where a

bending actuator drives a wing through a pulley.

i) Write the transfer function relating input force 𝐹𝑆𝐿 to output velocity.

ii) Draw the equivalent electrical circuit and determine transfer function from voltage input to

current output for the circuit.

Solution:

i) Define going towards the right of the page as the positive 𝑥-direction. By observation, we can

see that a positive shift from equilibrium would create two negative forces from the two springs, a

positive velocity on the mass would also create two negative forces from the two dampers, and a

positive force from the actuator would create a positive force on the mass. Now we can write the

differential equation defining the dynamics of the system as:

𝐹𝑆𝐿 −𝐵𝑆𝐿𝑥 − 𝐵𝐿𝑥 − 𝐾𝑆𝐿𝑥 − 𝐾𝐿𝑥 = 𝑀𝐿𝑥 .

To calculate the transfer function, use the Laplace transform and solve for Y(s)

R(s)=

sX (s)

FSL (s) ,

𝐹𝑆𝐿(𝑠) − 𝑠𝐵𝑆𝐿𝑋(𝑠) − 𝑠𝐵𝐿𝑋(𝑠) − 𝐾𝑆𝐿𝑋(𝑠) − 𝐾𝐿𝑋(𝑠) = 𝑠2𝑀𝐿𝑋(𝑠)

sX(s)

FSL (s)=

𝑠

𝑠2𝑀𝐿 + 𝑠(𝐵𝑆𝐿 + 𝐵𝐿) + 𝐾𝑆𝐿 + 𝐾𝐿

ii) Using voltage as our input and current as our output, we can convert the mechanical model

into an electrical model. Force becomes voltage, velocity becomes current, masses become

inductors, springs become capacitors, and dampers become resistors. Components in parallel

become things in series and vice versa. Following these rules, we obtain the equivalent electrical

circuit:

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To solve for the transfer function, we can use Ohm’s law with impedances and simplify

𝑉 𝑠 = 𝐼 𝑠 𝐵𝑆𝐿 + 𝐼 𝑠 𝐵𝐿 +1

𝑠𝐾𝑆𝐿𝐼 𝑠 +

1

𝑠𝐾𝑆𝐼 𝑠 + 𝑠𝑀𝐿𝐼 𝑠

𝐼(𝑠)

𝑉 𝑠 =

1

𝐵𝑆𝐿 + 𝐵𝐿 +1𝑠𝐾𝑆𝐿 +

1𝑠𝐾𝑆 + 𝑠𝑀𝐿

𝐼 𝑠

𝑉 𝑠 =

𝑠

𝑠2𝑀𝐿 + 𝑠𝐵𝑆𝐿 + 𝑠𝐵𝐿 + 𝐾𝑆𝐿 + 𝐾𝑆

Which we can see is the same result as above, verifying that our models are equivalent.

4. (20 pts) Electromechanical system example

A DC motor has electrical constant 𝐾𝑒 , torque constant 𝑘𝑡 , resistance 𝑅𝑚 , inertia 𝐽𝑚 and viscous

damping 𝑏𝑚 . The motor is connected to an inertial load 𝐽𝐿 through a shaft with spring constant 𝑘𝐿

and the inertial load has viscous damping 𝐵𝐿. Write the equations of motion for the system.

Solution:

(The DC motor dynamics can be found in FPE6e P47-48.) Here, the free-body diagram for this

electromechanical system with the load inertia is shown in Fig.4.1

Figure 4.1: Electromechanical System (Standard DC Motor & Inertia Load)

where 𝜃𝑚 and 𝜃ℓ are the rotation angles of the motor and load inertia, respectively. 𝑉 is //the

input voltage of DC motor, 𝐿 is the motor inductance. All other variables are defined as in the

problem statement.

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The dynamics of the system can be derived individually for each part of the system. First we

assume the following:

𝐿 ≪ 𝑅𝑚 , so we disregard the motor inductance (treat as wire, 𝐿 = 0) Perfect efficiency of the motor and shaft transmission

The motor torque output to the rotor is 𝑇 = 𝑘𝑡𝐼𝑚 , and the back emf voltage is 𝑉𝑒 = 𝐾𝑒𝜃 𝑚 .

Analysis of the electric circuit and the application of Kirchhoff’s voltage law yield

𝑉 = 𝑅𝑚 𝐼𝑚 + 𝐿𝑑𝐼𝑚

𝑑𝑡+ 𝑉𝑒 = 𝑅𝑚 𝐼𝑚 + 𝐾𝑒𝜃 𝑚 (4.1)

The motor output torque 𝑇 is used to drive the rotor directly, accounting for the rotor rotation,

viscous damping and the shaft torsion. The torque resulted from the shaft torsion is 𝑘𝐿 𝜃ℓ − 𝜃𝑚 ,

which is to drive the load inertia. Application of Newton’s laws yields

𝐽𝑚𝜃 𝑚 + 𝑏𝑚𝜃 𝑚 = 𝑇 − 𝑘𝐿 𝜃ℓ − 𝜃𝑚 = 𝑘𝑡𝐼𝑚 − 𝑘𝐿 𝜃ℓ − 𝜃𝑚 (4.2)

𝐽𝐿𝜃 ℓ + 𝐵𝐿𝜃 ℓ = 𝑘𝐿(𝜃ℓ − 𝜃𝑚 ) (4.3)

Solve 𝐼𝑚 in (4.1) and then substitute into (4.2). It follows

𝐽𝑚𝜃 𝑚 + 𝑏𝑚 +𝑘𝑡𝐾𝑒

𝑅𝑚 𝜃 𝑚 =

𝑘𝑡𝑉

𝑅𝑚− 𝑘𝐿 𝜃ℓ − 𝜃𝑚 (4.4)

𝐽𝐿𝜃 ℓ + 𝐵𝐿𝜃 ℓ = 𝑘𝐿(𝜃ℓ − 𝜃𝑚 ) (4.5)

5. (15 pts) Electrical circuit example

For the circuit below, using ideal op-amp assumptions, determine 𝑍𝑖𝑛 𝑠 =𝑉𝑖𝑛 (𝑠)

𝐼𝑖𝑛 (𝑠), where 𝑉𝑖𝑛 and

𝐼𝑖𝑛 are voltage across and current into 𝑍𝑖𝑛 node.

Solution:

Applying the ideal op-amp assumptions, we have (FPE6e P43)

𝑖+ = 𝑖− = 0, 𝑉+ = 𝑉− (5.1)

The current through the resistance 𝑅𝐿 is 𝐼𝐿 =𝑉𝑖𝑛−𝑉−

𝑅𝐿 , the current through the capacitor 𝐶 is

𝐼𝐶 = 𝐶𝑑 𝑉𝑖𝑛−𝑉+

𝑑𝑡, and the current through the resistance 𝑅 is 𝐼𝑅 =

𝑉+

𝑅. Since 𝑖+ = 0, application of

Kirchhoff’s current law yields

𝐼𝐶 = 𝐼𝑅 ⟹ 𝐶𝑑 𝑉𝑖𝑛−𝑉+

𝑑𝑡=

𝑉+

𝑅 (5.2)

Take Laplace transform of both sides of (5.2), resulting

𝐶𝑠 𝑉𝑖𝑛 𝑠 − 𝑉+(𝑠) = 𝑉+(𝑠)/𝑅 ⟹ 𝑉+ 𝑠 =𝑅𝐶𝑠

𝑅𝐶𝑠+1𝑉𝑖𝑛 (𝑠) (5.3)

Now, the current through Zin node is

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Iin = IL + IC =𝑉𝑖𝑛−𝑉−

𝑅𝐿+

𝑉+

𝑅 (5.4)

Substituting 𝑉−(𝑠) = 𝑉+(𝑠) =𝑅𝐶𝑠

𝑅𝐶𝑠+1𝑉𝑖𝑛 (𝑠) into (5.4), we obtain

𝑅𝐿𝐶𝑠 + 1 𝑉𝑖𝑛 𝑠 = 𝑅𝑅𝐿𝐶𝑠 + 𝑅𝐿 𝐼𝑖𝑛 (𝑠) (5.5)

⟹ 𝑍𝑖𝑛 𝑠 =𝑉𝑖𝑛 (𝑠)

𝐼𝑖𝑛 (𝑠)=

𝑅𝑅𝐿𝐶𝑠+𝑅𝐿

𝑅𝐿𝐶𝑠+1 (5.6)

6. (10 pts) Block Diagram Manipulation

Using block diagram manipulation, determine 𝑌 (𝑠)/𝑅(𝑠). Hint- start reduction at pt. A.

Solution:

Using block diagram manipulation:

Starting from Point A, replace the first inner negative feedback loop with its equivalent transfer

function.

Again, at Point A, replace inner negative feedback loop with its equivalent transfer function.

Now replace the serial connection of the shaded transfer function and 𝐺2 with its equivalent

transfer function.

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Then replace the negative feedback loop with its equivalent transfer function, by moving the

feedback point to the inner loop and adding another feedforward path.

Now replace the negative feedback loop with its equivalent transfer function.

Now reduce the parallel combination part (𝐻3𝐺3) with its equivalent transfer function.

Then replace the serial connection of the shaded transfer functions with its equivalent transfer

function.

Now replace the parallel combination part with its equivalent transfer function.

Finally, the shaded transfer function corresponds to 𝑌(𝑠)

𝑅(𝑠).

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We can also determine 𝒀(𝒔)

𝑹(𝒔) by solving the equations from the block diagram.

Assign the variables 𝑇1 ,𝑇2 ,𝑇3 as in the following diagram:

We have the relations from the diagram:

𝑇1 = 𝑅 − 𝐻3𝑌 − 𝐻2𝑇3

𝑇2 = 𝑇1 −𝐻1𝑇3

𝑇3 = 𝐺1𝑇2

𝑌 = 𝐺2𝑇3 + 𝐺3𝑅

Solving the above equations, we obtain

𝑌(𝑠)

𝑅(𝑠)=

𝐺1𝐺2 + 𝐺3 + 𝐺1𝐻1𝐺3 + 𝐺1𝐻2𝐺3

1 + 𝐺1𝐻1 + 𝐺1𝐻2 + 𝐺1𝐺2𝐻3