# E12 Assignment 13

Description:

Problems:
1. Sinusoidal Steady State of First Order System

A system has transfer function

Determine the sinusoidal steady state output if the input is:

1. x(t)=sin(0.1·t)
2. x(t)=cos(t+45°)
3. x(t)=3·sin(10·t-15°)
4. x(t)=4·sin(100·t)
1. Does the system exhibit high pass or low pass behavior (i.e., does it preferentially transmit high or low frequency inputs)?

Solution:

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.

2. Bode Plot, Distinct, Real Poles and Zeros

Consider the transfer function .

1. 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.
2. Using the Bode plot estimate the sinusoidal steady state output if the input is cos(5t).
3. Verify by calculating H(j5).  Note: Matlab's "freqresp" command does this, or you can evaluate with your calculator or by hand).
4. Use Matlab to plot the Bode diagram using the "bode" command.  Use the "grid" command to add a grid (useful for the next part).
5. Transfer your asymptotic approximation onto the Matlab plot.
Solution:

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→1012/20=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.

3. Bode, Origin Zero, Repeated Poles, Negative Sign
1. Sketch (by hand) the Bode plot for .  Suitable paper is here
2. Verify with Matlab.
Solution:

a)
Constant=-0.025     Magnitude=0.025=-32dB,  phase = ±180°
Zero at the origin (slope = +20dB/dec, phase=90°)
Double pole at s=-2, slope = -40dB/dec, phase goes from 0 to -180°.

b) >> H=tf(-0.1*[1 0],[1 4 4]); bode(H); grid

4. Hipass from Lopass

The transfer function

represents a low pass transfer function.  It's Bode plot is given below.

It seems reasonable to assume that we could make a high pass transfer function by subtracting a low pass transfer function from 1.

As ω→0 HLP(jω)=1, so  HHP(jω)=0.  As ω→∞ HLP(jω)=0, so  HHP(jω)=1.

a) Write HHP(s) in standard form (i.e., a single numerator and denominator polynomial).

b) Draw the Bode plot of HHP(s); use the axes given above showing HLP(s).

c) Sketch the impulse response of HLP(s).

d) On the same axes, sketch the impulse response of HHP(s).

Solution:

a)

b)

```HLP = tf(1,[0.1 1]);
HHP = tf([0.1 0],[0.1 1]);
bode(HLP,HHP);  legend('H_{LP}(s)','H_{HP}(s)');  grid```

c)         See part d for graph.

d)

```t=linspace(-0.1,0.7,1000);
hlp=10*exp(-10*t).*(t>=0);  plot(t,hlp,'b','Linewidth',2); hold on;
% Plotting highpass is more involved because it involves an impulse.
plot(t,-hlp,'r:','Linewidth',2);  % Plot exponential
stem(0,1,'r:^','Linewidth',2,'Markerfacecolor','r');

legend('h_{LP}(t)','h_{HP}(t)');  grid; hold off
axis([-0.1 0.7 -11 11])```
5. PWM via Fourier and Bode

It is possible to encode the value of an analog quantity within a digital signal using a technique called Pulse Width Modulation.  With this technique, the amplitude of the analog quantity is encoded in the width of the digital signal.  This problem will show how the amplitude of the original signal can be determined from the pulse width modulated signal.

The diagram below shows a pulse train, yp(t), of height A, period T, and pulse width Tp.  In this case the duty cycle is 40%  (Tp/T=0.4).  As the analog quantity varies, Tp (but not T) will vary.   To represent a quantity of 0.1, the width of the signal is Tp=0.1T...

a) Show that the Fourier Series coefficients of yp(t) are given by .   You can start with the fact that for a single pulse .  This requires no integration.

Since the function is even the cn are real we can represent yp(t) by a Fourier cosine series

We use this signal as input to a system with transfer function

b) Sketch the magnitude Bode Plot of H(s) and explain why, if ωLP << ω0, the output will be approximately equal to the average value of the input a0=ATp/T.  (What happens to signals at frequency ω0 or higher?)

c) In particular show that the constant term (a0) is passed by the system without attenuation (what is the gain of the system as frequency goes to 0?).

d) Show that the magnitude of the first harmonic, a1, is maximized when Tp=T/2 (i.e., the duty cycle is 50%).  Since the first harmonic is the larges of all the harmonics, and is attenuated the least by the system (it is the lowest frequency), this should be the worst case scenario in terms of variations, or ripples, in the output.  This requires no differentiation.

e) If Tp=T/2 what should the value of ω0be in relation to ωLP such that the output at the frequency of the first harmonic is only 1% of the output due to the DC component (i.e., a0).  The output of the system will then be almost constant with an amplitude determined by the pulse width.

Solution:

a)

b) If ω0>>ωLP then all the harmonics of the series will be greatly attenuated leaving only the constant component (a0) which is not changed since the gain is 0 dB (quantity=1) as ω→0.

c) Since H(0)=1, the constant signal (frequency=0) passes without attenuation.

d)

This is maximized when or Tp=T/2

e)  The magnitude of the first harmonic (in the output) is the magnitude of the harmonic (frequency=ω0) times the magnitude of the transfer function evaluated at jω0.

We want to find the value of such that this 1% of a0=A/2.

6. Find System From Bode - Real poles

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.

c) What are ζ and ω0?

Solution:

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)