A system has transfer function .

Determine the sinusoidal steady state output if the input is:

- x(t)=sin(0.1·t)
- x(t)=cos(t+45°)
- x(t)=3·sin(10·t-15°)
- x(t)=4·sin(100·t)

- Does the system exhibit high pass or low pass behavior (i.e., does it preferentially transmit high or low frequency inputs)?

Note: it isn't necessary to draw the Bode plot for this problem, you can simply calculate the magnitude and phase of H(jω) at each frequency.

a) H=0.01∠89° (see below). Output is 0.01·sin(0.1·t+89°).

>> H=0.1j/(10+0.1j) H = 0.0001 + 0.0100i >> abs(H) ans = 0.0100 >> angle(H)*180/pi ans = 89.4271

b) H=0.1∠84° (see below). Output is 0.1·cos(t+129°).

>> H=1j/(10+1j) H = 0.0099 + 0.0990i >> abs(H) ans = 0.0995 >> angle(H)*180/pi ans = 84.2894

c) H=0.71∠45° (see below). Output is 2.1·sin(10·t+30°).

>> H=10j/(10+10j) H = 0.5000 + 0.5000i >> abs(H) ans = 0.7071 >> angle(H)*180/pi ans = 45

d) H=1∠6° (see below). Output is 4·sin(100·t+6°).

>> H=100j/(10+100j) H = 0.9901 + 0.0990i >> abs(H) ans = 0.9950 >> angle(H)*180/pi ans = 5.7106

e) The high frequency inputs are transmitted without attenuation. This is a high pass system.

Consider the transfer function .

- Plot the asymptotic (straight line approximations) magnitude and phase plots for a Bode diagram. Assume 3.2 is halfway between 1 and 10 on a log scale. Suitable paper is here.
- Using the Bode plot estimate the sinusoidal steady state output if the input is cos(5t).
- Verify by calculating H(j5). Note: Matlab's "freqresp" command does this, or you can evaluate with your calculator or by hand).
- Use Matlab to plot the Bode diagram using the "bode" command. Use the "grid" command to add a grid (useful for the next part).
- Transfer your asymptotic approximation onto the Matlab plot.

a)

Constant=31 Magnitude=31≈30dB, phase =
0°

Zero at the s=-10 (i.e., break frequency is 10 rad/sec). Slope is
+20 dB/dec, phase goes from 0 to 90°

Pole at s=-1, and at s=-3.2 (i.e., break frequencies at 1 and 3.2 rad/sec).
Slope from each is -20 db/dec, and phase goes from 0 to -90°

b) At ω=5, H(jω)≈12 dB→10^{12/20}=4.0,
∠H(jω)≈-100° Output=4.0cos(5t-100°).

c) Output=3.7cos(5t-109°);

>> H=tf(10*[1 10],[1 4.2 3.2]); >> H_5rad=freqresp(H,5) H_5rad = -1.2333 - 3.4816i >> abs(H_5rad) ans = 3.6936 >> angle(H_5rad)*180/pi ans = -109.5058

d) >> bode(H); grid

e) See part a.

- 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.
- Verify with Matlab.

a)

s^{2}+2s+25 yields ω_{0}=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 ω_{0}=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

a) Find the differential equation for the system with Bode plot shown.

b) Show a system of your own design that has the given Bode plot (e.g., mass-spring-dashpot, resistor-capacitor-inductor. For this problem the system can even be resistor-capacitor (two capacitors) or resistor-inductor).

c) What are ζ and ω_{0}?

a,c) DC gain=1 (0 dB) and there is a double pole at w0=10 rad/sec (mag drops 40 dB/dec, and phase goes from 0 to -180°.

b)

Consider the transfer function .

- Show that the peak of |G(jω)| occurs at

Hint: if X is positive, then the extrema of X occurs at the same place as the extrema of X^{2}(or, the extrema of √X occurs at the same place as that of X), and the peak of X occurs at the same place as the minimum of 1/X. - Show that the peak has amplitude .

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

The peak due to a second order underdamped system, , occurs at frequency and has amplitude . We often use approximations when sketching by hand.

a) For the approximation, for what value of ζ is the approximation accurate to within 10%?

b) For the approximation , for what value of ζ is the approximation accurate to within 10%?

c) If we say that, for hand plots, we only need to plot a resonant peak if it is bigger than 6 dB (recall that 6 dB is about a factor of two), what is the corresponding value of ζ?

The results of this problem indicate that if we want to make a hand plot, we can use the given approximations for resonant frequency and peak amplitude.

a)

b)

c)

Since we only have a peak for ζ<1/√2=0.71, we use the lower value of ζ=0.26 (so we don't need to plot for ζ>0.26).

In other words, if the peak is large enough to plot in a hand plot, the value of ζ is small enough that we can use approximations for the height of the peak and the frequency at which the peak occurs.

Consider a system transfer function with Transfer Function

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

- In response to a step input, the steady state value of the system is zero if a
_{0}=0, otherwise it is a_{0}/b_{0}.- That the value of the transfer function goes to zero at low frequencies (s→0) if a
_{0}=0, otherwise it is a_{0}/b_{0}.- In response to a step input, the value at t=0
^{+}is zero if m<n, otherwise it is a_{n}/b_{n}.- That the value of the transfer function goes to zero at high frequencies (s→∞) if m<n, otherwise it is a
_{n}/b_{n}.- (optional) If m<n then the first n-m-1 derivatives of the step response are zero at t=0
^{+}.

a)

b)

c)

d)

e)

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

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, ω_{n}=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,....)).

% 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) G_{LP}. 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) G_{BP}. 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) G_{HP}. 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) G_{N}. 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.