E12 Assignment 14

Description: 

Problems: 
1. Bode Plot of Time Delay

Recall that multiplication by in the Laplace domain e-sT corresponds to a time delay of T seconds in the time domain. Determine a set of rules for drawing the magnitude and phase of e-sT on a Bode plot. Note: there are no approximations for both high and low frequency asymptotes as with other type of terms (i.e., simple poles and zeros) we have used.

a) Fill in the table for a time delay of T=0.02 Sec.

ω
(rad/sec)

(dB) 

(degrees) 
0.1    
1    
10    
100    
942    

b) Draw the Bode plot for .  Suitable paper is here (there is an expanded scale for phase).  Note: the phase is dropping linearly with ω, but it appears exponential on the graph because of the logarithmic axis for ω.

c) You can create this system with a delay in Matlab  thusly: "H=tf(100,[1 10],'InputDelay',0.02)".  Do so and verify your hand-drawn diagram.

d) If the input is sin(300t) what is the output?  (Use the Bode plot to find magnitude and phase).

e)  If the input is sin(300t) what is the output?  (Evaluate the transfer function as s=300j to find magnitude and phase (Matlab's "freqresp" command does this)).  If there is a difference between this answer and your previous answer, explain it.

Solution: 

a) Magnitude is 1=0 dB, and phase = -ωT rad = -ωT·180/π degrees.

ω
(rad/sec)
 
(dB) 

(radians) 

(degrees) 
0.1 1 0 -0.002 -0.1146
1 1 0 -0.02 -1.146
10 1 0 -0.2 -11.46
100 1 0 -2.0 -114.6
942 1 0 -18.84 -1075

b)        Individual terms (constant, pole, time delay) are shown in different colors below.

 

c)

>> H=tf(100,[1 10],'InputDelay',0.02)
                                     100
Transfer function:   exp(-0.02*s) * ------
                                    s + 10
>> bode(H); grid

 

d) From Bode plot: -9.41 dB = 10^(-9.41/20) = 0.34          Output = 0.34 sin(300t -426°).

e) See below.  Magnitude is the same, but phase is different by 360° (which is unimportant because angle is only unique to within multiples of one rotation (i.e, 360° or 2π radians)).  The answers are slightly different from previous because part d used values at ω=295 rad/sec.

>> H300=freqresp(H,300)
H300 =   0.1037 - 0.3166i

>> abs(H300)
ans =    0.3331

>> angle(H300)*180/pi
ans =  -71.8655

>> ans-360
ans = -431.8655
2. Height of Underdamped Peak

Consider the transfer function .

  • Show that the peak of |G(jω)| occurs at
    Hint: if X is positive, then the peak of X occurs at the same place as the peak of X2, and the peak of X occurs at the same place as the minimum of 1/X.
  • Show that the peak has amplitude .
Solution: 

a)  .  Peak of |G(jω)| is at same frequency as minimum of 1/|G(jω)|^2.  Find this by differentiation:

b)

We can pull some terms out of the square root

3. Tabulating zeta for which Bode approximations are valid

The peak due to a second order underdamped system, , occurs at frequency  and has amplitude

We often use the approximations and

Compare these exact expressions to the approximations given below by filling in the table.

ζ ωr
(exact)
ωr
(approx)
|Gjωr|
(exact, dB)
|Gjωr|
(approx, dB)
0.1      
0.2      
0.3      
0.4      
0.5      
0.6      
0.7      
1/√2      

What can you say about the application of the approximations when sketching a Bode plot by hand?

Solution: 
ζ ωr
(exact)
ωr
(approx)
|Gjωr|
(exact, dB)
|Gjωr|
(approx, dB)
0.1 0.99ω0 14.0 14.0
0.2 0.96ω0 8.13 7.96
0.3 0.91ω0 4.85 4.44
0.4 0.83ω0 2.70 1.94
0.5 0.71ω0 1.24 0
0.6 0.53ω0 0.35 -1.58
0.7 0.14ω0 0.00 -2.92
1/√2 0 0.00 -3.01

For cases in which there is any significant peak (i.e., ζ≤0.4), the peak is close to ω0, and the height is close to 1/(2ζ)

4. Complex Bode Plot
  1. Sketch (by hand) the Bode plot for .  Suitable paper is here.  You may assume that the phase drops from 0° to -180° instantaneously at the break frequency of the underdamped poles.
  2. Verify with Matlab.
Solution: 

a)  

s2+2s+25 yields ω0=5, ζ=0.2

Constant = 3.2 ≈ 10 dB;
Double zero at the origin yields +40 dB/dec with phase of +180° (or -180°)
Pole at s=-100 yields slope of -20 dB/dec and phase from 0 to -90°
Underdamped poles with  ω0=5, ζ=0.2 gives a peak of about 1/2ζ=2.5=8 dB.

b)  >> H=tf(8000*[1 0 0],conv([1 100],[1 2 25])); bode(H); grid

5. Transfer Function Properties (1)

Consider a system transfer function with Transfer Function

Assume none of denominator coefficients are zero, and that m≤n.  Show that

  1. In response to a step input, the steady state value of the system is zero if a0=0, otherwise it is a0/b0.
  2. That the value of the transfer function goes to zero at low frequencies (s→0) if a0=0, otherwise it is a0/b0.
  3. In response to a step input, the value at t=0+ is zero if m<n, otherwise it is an/bn.
  4. That the value of the transfer function goes to zero at high frequencies (s→∞) if m<n, otherwise it is an/bn.
  5. (optional) If m<n then the first n-m-1 derivatives of the step response are zero at t=0+.
Solution: 

a)

b)

c)

d)

e)

lower order derivatives will also be zero, but any derivative of higher order will not be equal to zero.

6. Transfer Function Properties (2)

Verify that each of the five results of the previous problem is correct by plotting the step response and Bode plots of the four Transfer Functions given below.  Use ζ=0.25, ωn=10.   You needn't do any calculations.  Use Matlabs "tf," "step" and "bode" command.  Note that each command can plot multiple system responses simultaneously (i.e., step(sys1, sys2,....)).

Solution: 
% Set values.
w=10; z=0.25;   
% Now define transfer functions
GLP=tf([0 0 w*w],[1 2*z*w w*w]);
GBP=tf([0 2*z*w 0],[1 2*z*w w*w]);
GHP=tf([1 0 0],[1 2*z*w w*w]);
GN=tf([1 0 w*w],[1 2*z*w w*w]);

%% Step response
step(GLP,GBP,GHP,GN);
legend('G_{LP}','G_{BP}','G_{HP}','G_N')

%% Bode
bode(GLP,GBP,GHP,GN);
legend('G_{LP}','G_{BP}','G_{HP}','G_N')

a) GLP. Step response has final value of 1, initial value of 0, and initial slope of zero.   Bode=1 at low frequencies and 0 at high frequencies.

b) GBP. Step response has final value of 0, initial value of 0, and initial slope not equal to zero.   Bode=0 at low frequencies and 0 at high frequencies.

c) GHP. Step response has final value of 0, initial value of 1, and initial slope not equal to zero.   Bode=0 at low frequencies and 1 at high frequencies.

d) GN. Step response has final value of 1, initial value of 1, and initial slope not equal to zero.   Bode=1 at low frequencies and 1 at high frequencies.

7. Matching Pole Zero, Xfer Function, Step Response and Bode Plot

For this problem, consult the pdf file given here.  It shows pole-zero plots, step responses, and bode plots.  Each is associated with one of the transfer functions shown below.  Fill in the table below with the label of the corresponding pole-zero (a-f), step (a-f) or bode plot (A-F).

Transfer Function Pole-Zero Plot Step Response Bode Plot
   
     
     
     
     
     
Solution: 

Note you can tell ε is second order and χ is first order, because the derivative of the function in ε is zero as t→0 in addition to the value of the function itself.

Transfer Function Pole-Zero Plot Step Response Bode Plot
d χ E
f δ A
b ε F
e α B
c β D
a φ C
8. Heating Food in a Microwave

The next problems investigate why microwave ovens are faster than conventional ovens using a simplistic, but not altogether unreasonable, model.  The task is to heat a bolus of meat about the size and shape of an orange; we will model the bolus as two masses as shown.  The outer region (Region 1) has a thermal capacitance, C1=3000 J/°K.  The inner region (Region 2) has a thermal capacitance, C2=1000 J/°K.  The thermal resistance between the external environment and Region 1 is R1a=2°K/W.  The resistance between the inner and outer regions is R12=0.1°K/W.   A diagram is shown.

Note: the mechanism of heat transfer in a microwave oven is radiation - but we can assume this is just a constant power input

  1. Draw an equivalent electrical circuit that represents:
    1. the bolus in a conventional oven in which the temperature outside the bolus is controlled.  We will assume the temperature outside the bolus starts at ambient temperature (we consider this 0°).  When we put the food in the oven, we will model this as a step change in external temperature.
    2. the bolus in a microwave oven in which power is deposited directly in region 1 (but not in region 2 because the depth of penetration of microwaves in tissue is limited to 1-2 cm).  Assume the temperature in the oven is constant and equal to ambient temperature.  When we turn the microwave on it immediately begins depositing power in region 1 (but the air inside the microwave does not heat up appreciably).
  2. Develop a state space model (with two outputs - the temperature of each region) for
    1. the bolus in a conventional oven.  The input is the external temperature.
    2. the bolus in a microwave oven.  The input is the power from the microwave.
  3. If the bolus is initially at ambient temperature (call this 0°) and is suddenly placed in a conventional oven at 200 °K above ambient:
    1. plot the temperature of the two regions.
    2. approximately how long does it take region 2 to reach 50° above ambient?
  4. If the bolus is initially at ambient temperature (call this 0°) and is suddenly placed in a microwave oven that generate 1500 Watts:
    1. plot the temperature of the two regions.
    2. approximately how long does it take region 2 to reach 50° above ambient?
Solution: 

a)  Diagram on left shows conventional oven (oven temperature is input), diagram at the right shows a microwave oven (input power is input).

b) Note that the state space systems are the same except for the B matrix.  This is because the systems are identical with zero input (the voltage source becomes a short circuit, and the current source becomes an open circuit).

Convential Oven   Microwave Oven
Energy Balance
a=0)
 
  State Space

c)
>> C1=3000; C2=1000; R12=0.1;  R1a=2;
>> A=[-(1/R1a+1/R12)/C1 1/R12/C1; 1/R12/C2 -1/R12/C2];
>> B=[1/R1a/C1 0]';  C=[1 0; 0 1;];  D=0;
>> sys=ss(A,B,C,D);
>> sys.OutputName={'Temp 1','Temp 2'}; sys.InputName='Input TempPower';
>> sys.StateName={'T1','T2'};
>> step(200*sys)    % Note step size is 200.

 

It takes about 2380 seconds (≈40 minutes) to reach 50 degrees.

d) We need only change the "B" matrix and the amplitude of the input

>> sys.b=[1/C1 0]'; 
>> step(1500*sys)

Zoom in for detail.  It takes about 205 seconds (≈3.5 minutes) to reach 50 degrees.

9. Hair Dryer in a box

Just for fun  (nothing to solve)

I used numbers for heat transfer between a steel box and air and got roughly the same results.

Solution: 

Answer