Overview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros)


 Virginia Owen
 2 years ago
 Views:
Transcription
1 Overview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros) J. McNames Portland State University ECE 222 Bode Plots Ver
2 Transfer Function Review x(t) H(s) y(t) Recall that if H(s) is known and x(t) =A cos(ωt + φ), then we can find the steadystate solution for y(t): y ss (t) =A H(jω) cos (ωt + φ + H(jω)) J. McNames Portland State University ECE 222 Bode Plots Ver
3 x(t) Bode Plots H(s) y(t) Bode plots are standard method of plotting the magnitude and phase of H(s) Both plots use a logarithmic scale for the xaxis Frequency is in units of radians/second (rad/s) The phase is plotted on a linear scale in degrees Magnitude is plotted on a linear scale in decibels H db (jω) 2 log 1 H(jω) J. McNames Portland State University ECE 222 Bode Plots Ver
4 Decibel Scales It is important to become adept at translating between amplitude, H(jω), and decibels, H db (jω). Amplitude ( H(jω) ) Decibels (2 log 1 H(jω) ) 1 2 log 1 1 = 1 2 log 1 1 = 1 2 log 1 1 = 1 2 log 1 1 =.1 2 log 1.1 =.1 2 log 1.1 =.1 2 log 1.1 = log 1 2 = log 1 2 = log 1 2 = J. McNames Portland State University ECE 222 Bode Plots Ver
5 Example 1: Bode Plots 2 nf 1kΩ 1 kω v s (t) R L + v o (t)  1. Find the transfer function of the circuit shown above. 2. Generate the bode plot. J. McNames Portland State University ECE 222 Bode Plots Ver
6 Example 1: Workspace J. McNames Portland State University ECE 222 Bode Plots Ver
7 Example 1: Bode Plot Active Lowpass RC Filter H(jω) (db) H(jω) (degrees) Frequency (rad/s) J. McNames Portland State University ECE 222 Bode Plots Ver
8 w = logspace(1,5,5); H = 5e3./(j*w + 5e3); subplot(2,1,1); h = semilogx(w,2*log1(abs(h))); set(h, LineWidth,1.4); ylabel( H(j\omega) (db) ); title( Active Lowpass RC Filter ); set(gca, Box, Off ); grid on; set(gca, YLim,[5 25]); Example 1: MATLAB Code subplot(2,1,2); h = semilogx(w,angle(h)*18/pi); set(h, LineWidth,1.4); ylabel( \angle H(j\omega) (degrees) ); set(gca, Box, Off ); grid on; set(gca, YLim,[85 185]); xlabel( Frequency (rad/s) ); J. McNames Portland State University ECE 222 Bode Plots Ver
9 Example 2: Bode Plots 2 μf 1kΩ 1 μf 1kΩ v s (t) + R L  1. Find the transfer function of the circuit shown above. 2. Generate the bode plot. J. McNames Portland State University ECE 222 Bode Plots Ver
10 Example 2: Workspace J. McNames Portland State University ECE 222 Bode Plots Ver
11 H(jω) (degrees) H(jω) (db) Example 2: Bode Plot Active Lead/Lag RC Filter (rad/s) Frequency J. McNames Portland State University ECE 222 Bode Plots Ver
12 w = logspace(1,5,5); H = 2*(j*w+5)./(j*w + 1); Example 2: MATLAB Code subplot(2,1,1); h = semilogx(w,2*log1(abs(h))); set(h, LineWidth,1.4); ylabel( H(j\omega) (db) ); title( Active Lead/Lag RC Filter ); set(gca, Box, Off ); grid on; set(gca, YLim,[2 8]); subplot(2,1,2); h = semilogx(w,angle(h)*18/pi); set(h, LineWidth,1.4); ylabel( \angle H(j\omega) (degrees) ); set(gca, Box, Off ); grid on; set(gca, YLim,[1816]); xlabel( Frequency (rad/s) ); J. McNames Portland State University ECE 222 Bode Plots Ver
13 Bode Plot Approximations Until recently (late 198 s) bode plots were drawn by hand There were many rulesofthumb, tables, and template plots to help Today engineers primarily use MATLAB, or the equivalent Why discuss the old method of plotting by hand? It is still important to understand how the poles, zeros, and gain influence the Bode plot These ideas are used for transfer function synthesis, analog circuit design, and control systems We will discuss simplified methods of generating Bode plots Based on asymptotic approximations J. McNames Portland State University ECE 222 Bode Plots Ver
14 Alternate Transfer Function Expressions There are many equivalent expressions for transfer functions. H(s) = N(s) D(s) = b ms m + b m 1 s m b 1 s + b a n s n + a n 1 s n a 1 s + a = b m s ±l (s z 1)(s z 2 )...(s z m ) a n (s p 1 )(s p 2 )...(s p n ) ( )( ) ( ) 1 s = ks ±l z 1 1 s z s z m ( )( ) ( ) 1 s p 1 1 s p s p n This last expression is called standard form The first step in making bode plots is to convert H(s) to standard form J. McNames Portland State University ECE 222 Bode Plots Ver
15 Magnitude Components Consider the expression for the transfer function magnitude: H db (jω) = 2log 1 H(jω) = 2log 1 k s±l (1 s z 1 )...(1 s (1 s p 1 )...(1 s p n ) z m ) s=jω jω 1 ±l z = 2log 1 k jω jω z m 1 jω p jω p n = 2log 1 k ±l2 log 1 ω +2 log 1 1 jω z log 1 1 jω z m 2 log 1 1 jω p 1 2 log 1 1 jω p n J. McNames Portland State University ECE 222 Bode Plots Ver
16 Magnitude Components Comments H db (ω) = 2log 1 k ±l2 log 1 ω +2 log 1 1 jω z log 1 1 jω z m 2 log 1 1 jω p 1 2 log 1 1 jω p n Thus, H db (ω) can be written as a sum of simple functions This is similar like using basis functions {δ(t),u(t),& r(t)} to write an expression for a piecewise linear signal We will use this approach to generate our piecewise linear approximations of the bode plot Note that there are four types of components in this expression Constant Linear term Zeros Poles J. McNames Portland State University ECE 222 Bode Plots Ver
17 Magnitude Components: Constant H(jω) (db) ω (rad/sec) 4 The constant term, 2 log 1 k, is a straight line on the Bode plot. J. McNames Portland State University ECE 222 Bode Plots Ver
18 Magnitude Components: Linear Term H(jω) (db) ω (rad/sec) 4 The linear term, ±l 2 log 1 ω, is a line on the magnitude plot with a slope equal to ±l 2 db per decade. The xaxis intercept occurs at ω =1rad/s. Plot the bode magnitude plots for H(s) =s, 1 s,s2, 1 s 2. J. McNames Portland State University ECE 222 Bode Plots Ver
19 Magnitude Components: Real Zeros Consider two limiting conditions for a term containing a zero, 2 log 1 1 jω z First condition: ω z lim 2 log ω z 1 1 jω z = Thus, if ω z 1, then2 log 1 Second condition: ω z 1 jω z ω lim z 2 log 1 1 jω z. =2log1 jω z =2log 1 ω 2 log 1 z ω Thus, if z 1, then this term is linear (on a log scale) with a slope of 2 db per decade and an xaxis intercept at ω = z. J. McNames Portland State University ECE 222 Bode Plots Ver
20 Magnitude Components: Real Zeros Continued H(jω) (db) ω (rad/sec) 4 Our piecewise approximation joins these two linear asymptotic approximations at ω = z. Plot the piecewise approximation of the term 2 log 1 1 jω z. J. McNames Portland State University ECE 222 Bode Plots Ver
21 Magnitude Components: Real Zeros Continued 2 5 Bode Magnitude Real Zero: z = ±1 rad/s 4 3 Mag (db) Frequency (rad/sec) The approximation is least accurate at ω = z. The true magnitude is 3 db higher than the approximation at this corner frequency. J. McNames Portland State University ECE 222 Bode Plots Ver
22 Magnitude Components: Real Poles H(jω) (db) ω (rad/sec) 4 Consider two limiting conditions for a term containing a pole, 2 log 1 1 jω p This is just the negative of the expression for a zero The piecewise approximation is the mirror image of that for a zero J. McNames Portland State University ECE 222 Bode Plots Ver
23 Magnitude Components: Real Poles Continued 1 Bode Magnitude Real Pole: p = 1 rad/s 1 Mag (db) Frequency (rad/sec) The approximation is least accurate at ω = p. The true magnitude is 3 db less than the approximation at this corner frequency. J. McNames Portland State University ECE 222 Bode Plots Ver
24 Complex Poles & Zeros Complex poles and zeros require special attention Will discuss later You will not be expected to plot approximations with complex poles or zeros on exams There are essentially 3 steps to generating piecewise linear approximations of bode plots 1. Convert to standard form 2. Plot the components 3. Graphically add the components together J. McNames Portland State University ECE 222 Bode Plots Ver
25 Example 3: Magnitude Components Draw the piecewise approximation of the bode magnitude plot for H(s) = (s + 1)(s + 1)2 1s 2 (s + 1) J. McNames Portland State University ECE 222 Bode Plots Ver
26 Example 3: Workspace H(jω) 6 (db) ω (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
27 Example 3: Solution Bode Magnitude Example Mag (db) Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
28 Example 4: Magnitude Components Draw the piecewise approximation of the bode magnitude plot for H(s) = 1 11 s(s + 1) (s + 1)(s + 1)(s +1, ) 2 J. McNames Portland State University ECE 222 Bode Plots Ver
29 Example 4: Workspace H(jω) 6 (db) ω (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
30 Example 4: Solution Mag (db) Bode Magnitude Example Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
31 Phase Components H(jω) = k (jω)l (1 jω z 1 )...(1 jω z m ) (1 jω p 1 )...(1 jω p n ) Each of these terms can be expressed in polar form: a + jb = Ae jθ. Note that (jω) l = ω l j l = ω l (e j π 2 ) l = ω l e j π 2 l H(jω) = k e jη kπ (ωl e jl π 2 )N 1 e jθ 1...N m e jθ m D 1 e jφ 1...Dn e jφ n = k ω l N 1...N m D 1...D n exp ( j(η k π + l π 2 + θ θ n φ 1 φ n ) ) where η k = { k 1 k< J. McNames Portland State University ECE 222 Bode Plots Ver
32 Phase Components: Comments H(jω) = η k π + l π 2 + θ θ n φ 1 φ n ( = η k π + l π jω ) ( jω ) z 1 z m ( 1 jω ) ( 1 jω ) p 1 p n Thus the phase of H(jω) is also a linear sum of the phases due to each component We will consider each of the four components in turn Constant Linear term Zeros Poles J. McNames Portland State University ECE 222 Bode Plots Ver
33 H(jω) 18 9 Phase Components: Constant (deg) 9 ω (rad/sec) 18 The complex angle of the constant term, k, is either if k> or 18 if k<. J. McNames Portland State University ECE 222 Bode Plots Ver
34 H(jω) 18 9 Phase Components: Linear Term (deg) 9 ω (rad/sec) 18 The linear term, (jω) l = j l = l 9, is a constant multiple of 9 Plot the bode phase plots for H(s) =s, 1 s,s2, 1 s 2. J. McNames Portland State University ECE 222 Bode Plots Ver
35 Phase Components: Real Zeros Consider three limiting conditions for a term containing a zero, 1 jω z First condition: ω z lim ( 1 jω ) ω z z = ). ω Thus, if z 1, then ( 1 jω z Second condition: ω = z ( ) 1 jω ω= z z = (1 jη z )= η z 45 where η z = sign(z) Third condition: ω z ω lim z ( 1 jω z Thus, if ω z 1, then ( 1 jω z ) = ( jω z ) ηz 9. ) = ηz 9 J. McNames Portland State University ECE 222 Bode Plots Ver
36 Phase Components: Real Zeros Continued H(jω) 18 9 (deg) 9 ω (rad/sec) 18 Our piecewise approximation joins these three linear asymptotic approximations at ω =1 1 z and ω =1 z. Plot the piecewise approximation of the term ( 1 jω ) z. Assume that z is in the lefthand plane (i.e. Re{z} < ) J. McNames Portland State University ECE 222 Bode Plots Ver
37 Phase Components: Real Zeros in Left Plane Phase (deg) Frequency (rad/sec) The approximation is least accurate at ω =.1 z and ω =1 z. J. McNames Portland State University ECE 222 Bode Plots Ver
38 Phase Components: Real Zeros Continued 2 H(jω) 18 9 (deg) 9 ω (rad/sec) 18 If the zero is in the right half plane (i.e. Re{z} > ), then the phase approaches 9 asymptotically. Plot the piecewise approximation of the term ( 1 jω ) z. Assume that z is in the right half plane. J. McNames Portland State University ECE 222 Bode Plots Ver
39 Phase Components: Real Zeros in Right Plane Phase (deg) Frequency (rad/sec) The approximation is least accurate at ω =.1 z and ω =1 z. J. McNames Portland State University ECE 222 Bode Plots Ver
40 Phase Components: Real Poles Real poles in the left half plane have the same phase as real zeros in the right half plane. We will only discuss poles in the left half plane because only these systems are stable. First condition: ω p lim ω p ( 1 jω p ) = Second condition: ω = p ( 1 jω p = (1 sign(p) j) = (1 + j) = 45 ) ω= p Third condition: ω p ( lim 1 jω ω p p ) = ( jω p ) = j = 9 J. McNames Portland State University ECE 222 Bode Plots Ver
41 Phase Components: Real Poles H(jω) 18 9 (deg) 9 ω (rad/sec) 18 Plot the piecewise approximation of the term that p is in the lefthand plane (i.e. Re{p} < ) ( 1 jω p ). Assume J. McNames Portland State University ECE 222 Bode Plots Ver
42 Phase Components: Real Poles in Left Plane Phase (deg) Frequency (rad/sec) The approximation is least accurate at ω =.1 p and ω =1 p. J. McNames Portland State University ECE 222 Bode Plots Ver
43 Example 5: Phase Components H(jω) (deg) (rad/sec) Draw the piecewise approximation of the bode phase plot for H(s) = (s + 1)(s + 1)2 1s 2 (s + 1) J. McNames Portland State University ECE 222 Bode Plots Ver
44 Example 5: Solution 5 Phase (deg) Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
45 Example 6: Phase Components H(jω) (deg) (rad/sec) Draw the piecewise approximation of the bode phase plot for H(s) = 1 11 s(s + 1) (s + 1)(s + 1)(s +1, ) 2 J. McNames Portland State University ECE 222 Bode Plots Ver
46 Example 6: Solution 1 5 Phase (deg) Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
47 Example 7: Magnitude H(jω) 6 (db) ω (rad/sec) Plot the magnitude of H(s) =1 s +1 s + 1 J. McNames Portland State University ECE 222 Bode Plots Ver
48 Example 7: Phase H(jω) (deg) (rad/sec) Plot the phase of H(s) =1 s +1 s + 1 J. McNames Portland State University ECE 222 Bode Plots Ver
49 Example 7: Solution 4 Bode Example 1 Mag (db) Phase (deg) Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
50 Example 8: Magnitude H(jω) 6 (db) ω (rad/sec) Plot the magnitude of H(s) = 1 s 1 s + 1 J. McNames Portland State University ECE 222 Bode Plots Ver
51 Example 8: Phase H(jω) (deg) (rad/sec) Plot the phase of H(s) = 1 s 1 s + 1 J. McNames Portland State University ECE 222 Bode Plots Ver
52 Example 8: Solution 4 Bode Example 2 Mag (db) Phase (deg) Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
53 1 mh 1 mf Example 9: Circuit Example + v s (t) 11 Ω v o (t)  Draw the straightline approximations of the transfer function for the circuit shown above. Hint: Recall from one of the Transfer Functions Examples R L H(s) = s s 2 + R L s + 1 LC J. McNames Portland State University ECE 222 Bode Plots Ver
54 Example 9: Magnitude H(jω) 6 (db) ω (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
55 Example 9: Phase H(jω) (deg) (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
56 Example 9: Solution 2 Bode Example 3 Mag (db) Phase (deg) Frequency 1 2 (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
57 Complex Poles Complex poles can be expressed in the following form: C(s) = ω 2 n s 2 +2ζω n s + ω 2 n = 1 1+2ζ s ω n + ( ) 2 s ω n ω n is called the undamped natural frequency ζ (zeta) is called the damping ratio The poles are p 1,2 =( ζ ± ζ 2 1) ω n If ζ 1, the poles are real If <ζ<1, the poles are complex If ζ =, the poles are imaginary: p 1,2 = ±jω n If ζ<, the poles are in the right half plane (Re{p} > ) and the system is unstable J. McNames Portland State University ECE 222 Bode Plots Ver
58 Complex Poles Continued The transfer function C(s) can also be expressed in the following form C(s) = 1 1+2ζ s ω n + ( ) 2 = s ω n 1 1+ s Qω n + ( ) 2 s ω n where Q 1 2ζ The meaning of Q, thequality factor, will become clear in the following slides. 1 C(jω)= 1 ( ) 2 ω ω n + jω Qω n J. McNames Portland State University ECE 222 Bode Plots Ver
59 Complex Poles Magnitude 2 log 1 C(jω) = 2log 1 1 ( ) 2 ω + jω ω n Qω n 1 ( ) 2 = 2 log 1 1 ω2 ωn 2 + ( ω Qω n ) 2 For ω ω n, 2 log 1 C(jω) 2 log 1 1 =db For ω ω n, 2 log 1 C(jω) 2 log 1 ω 2 ω 2 n = 4 log 1 ω ω n db At these extremes, the behavior is identical to two real poles. J. McNames Portland State University ECE 222 Bode Plots Ver
60 For ω = ω n, Complex Poles Magnitude Continued ( ) 2 2 log 1 C(jω) = 2 log 1 1 ω2 ωn 2 + ( ω Qω n 2 log 1 C(jω n ) = 2 log 1 1 Q =2log 1 Q = Q db ) 2 J. McNames Portland State University ECE 222 Bode Plots Ver
61 Mag (db) Complex Poles Magnitude Complex Poles Q =.1 Q =.5 Q =.77 Q = 1 Q = 2 Q = 1 Q = Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
62 For ω ω n, For ω ω n, For ω = ω n, Complex Poles Phase 1 C(jω) = ( ) 2 ω 1 ω n + jω Qω n C(jω) 1 ω Qω n C(jω) 1 = = Qω n ω = 1= 18 C(jω) 1 Qj = 1 j = j = 9 At the extremes, the behavior is identical to two real poles. At other values of ω near ω n, the behavior is more complicated. J. McNames Portland State University ECE 222 Bode Plots Ver
63 Phase (deg) Complex Poles Phase Complex Poles Q =.1 Q =.5 Q =.77 Q = 1 Q = 2 Q = 1 Q = Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
64 Complex Zeros Left half plane: Inverted magnitude of complex poles Inverted phase of complex poles Right half plane: Inverted magnitude of complex poles Same phase of complex poles This is the same relationship real zeros had to real poles J. McNames Portland State University ECE 222 Bode Plots Ver
65 Complex Zeros Magnitude 8 Complex Zeros 6 4 Mag (db) Q =.1 Q =.5 Q =.77 Q = 1 Q = 2 Q = 1 Q = Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
66 Complex Zeros Phase 18 Complex Zeros Phase (deg) Q =.1 Q =.5 Q =.77 Q = 1 Q = 2 Q = 1 Q = Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
67 Complex Poles Maximum What is the frequency at which C(jω) is maximized? C(jω) = C(jω) = 1 ( ) ( ) 2 1+ jω Qω n + jω ω n 1 ( ) 2 ( 1 ω2 + ω 2 n ) 2 ω Qω n For high values of Q, the maximum of C(jω) > 1 This is called peaking The largest Q before the onset of peaking is Q = This curve is said to be maximally flat This is also called a Butterworth response In this case, C(jω n ) = 3 db and ω n is the cutoff frequency J. McNames Portland State University ECE 222 Bode Plots Ver
68 Complex Poles Maximum Continued If Q>.77, the maximum magnitude and frequency are as follows: ω r = ω n 1 1 Q 2Q 2 C(jω r ) = 1 1 4Q 2 ω r is called the resonant frequency or the damped natural frequency As Q, ω r ω n For sufficiently large Q (say Q>5) ω r ω n C(jω r ) Q Peaked responses are useful in the synthesis of highorder filters Complex zeros (in the left half plane) have the inverted magnitude and phase of complex poles J. McNames Portland State University ECE 222 Bode Plots Ver
69 5 mh R Complex Poles Example + v s (t) 2 nf v o (t)  Generate the bode plot for the circuit shown above. H(s) = 1 LC s 2 + R L s + 1 LC = ω 2 n s 2 +2ζω n s + ω 2 n = ω 2 n s 2 + ω n Q s + ω 2 n J. McNames Portland State University ECE 222 Bode Plots Ver
70 Complex Poles Example Continued 1 1 ω n = =1k rad/s LC ζ = R R LC = CL = R.1 2L 2 1 L L Q = LC R = 1 C R = 5 R R =5Ω ζ =.5 Q = 1 Very Light Damping R =5Ω ζ =.5 Q =1 Light Damping R = 77 Ω ζ =1.41 Q =.77 Strong Damping R =1kΩ ζ =1 Q =.5 Critical Damping R =5kΩ ζ =5 Q =.1 Over Damping J. McNames Portland State University ECE 222 Bode Plots Ver
71 Complex Poles Example Continued 2 4 Resonance Example Mag (db) R = 5 Ω R = 5 Ω 8 R = 77 Ω R = 1 kω 1 R = 5 kω R = 5 kω Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
72 Complex Poles Example Continued 3 Resonance Example 2 4 Phase (deg) R = 5 Ω R = 5 Ω R = 77 Ω R = 1 kω R = 5 kω R = 5 kω Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
73 Complex Poles Example Continued 4 Output (V) Resonance Example R = 5 Ω R = 5 Ω R = 77 Ω R = 1 kω R = 5 kω R = 5 kω Time (sec) J. McNames Portland State University ECE 222 Bode Plots Ver
Dynamic circuits: Frequency domain analysis
Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution
More informationCore Concepts Review. Orthogonality of Complex Sinusoids Consider two (possibly nonharmonic) complex sinusoids
Overview of ContinuousTime Fourier Transform Topics Definition Compare & contrast with Laplace transform Conditions for existence Relationship to LTI systems Examples Ideal lowpass filters Relationship
More informationLecture 16 FREQUENCY RESPONSE OF SIMPLE CIRCUITS
Lecture 6 FREQUENCY RESPONSE OF SIMPLE CIRCUITS Ray DeCarlo School of ECE Purdue University West Lafayette, IN 47907285 decarlo@ecn.purdue.edu EE202, Frequency Response p 2 R. A. DeCarlo I. WHAT IS FREQUENCY
More informationRadar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.
Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter
More informationEE221 Circuits II. Chapter 14 Frequency Response
EE22 Circuits II Chapter 4 Frequency Response Frequency Response Chapter 4 4. Introduction 4.2 Transfer Function 4.3 Bode Plots 4.4 Series Resonance 4.5 Parallel Resonance 4.6 Passive Filters 4.7 Active
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect
More informationMAS107 Control Theory Exam Solutions 2008
MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION 2008 MAS07 Control Theory Exam Solutions 2008 Geir Hovland, Mechatronics Group, Grimstad, Norway June 30, 2008 C. Repeat question B, but plot the phase curve
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (HS 25) Solution Exercise Set Loop Shaping Noele Norris, 9th December 26 Exercise (A Nonminimum Phase System) To increase the rise time of the system, we
More informationEE221 Circuits II. Chapter 14 Frequency Response
EE22 Circuits II Chapter 4 Frequency Response Frequency Response Chapter 4 4. Introduction 4.2 Transfer Function 4.3 Bode Plots 4.4 Series Resonance 4.5 Parallel Resonance 4.6 Passive Filters 4.7 Active
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (Autumn 27) Solution Exercise Set 2 Loop Shaping clruch@ethz.ch, 8th December 27 Exercise (A Nonminimum Phase System) To decrease the rise time of the system,
More informationSolution: K m = R 1 = 10. From the original circuit, Z L1 = jωl 1 = j10 Ω. For the scaled circuit, L 1 = jk m ωl 1 = j10 10 = j100 Ω, Z L
Problem 9.9 Circuit (b) in Fig. P9.9 is a scaled version of circuit (a). The scaling process may have involved magnitude or frequency scaling, or both simultaneously. If R = kω gets scaled to R = kω, supply
More informationH(s) = 2(s+10)(s+100) (s+1)(s+1000)
Problem 1 Consider the following transfer function H(s) = 2(s10)(s100) (s1)(s1000) (a) Draw the asymptotic magnitude Bode plot for H(s). Solution: The transfer function is not in standard form to sketch
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are
More informationSignals and Systems. Lecture 11 Wednesday 22 nd November 2017 DR TANIA STATHAKI
Signals and Systems Lecture 11 Wednesday 22 nd November 2017 DR TANIA STATHAKI READER (ASSOCIATE PROFFESOR) IN SIGNAL PROCESSING IMPERIAL COLLEGE LONDON Effect on poles and zeros on frequency response
More informationFrequency Response part 2 (I&N Chap 12)
Frequency Response part 2 (I&N Chap 12) Introduction & TFs Decibel Scale & Bode Plots Resonance Scaling Filter Networks Applications/Design Frequency response; based on slides by J. Yan Slide 3.1 Example
More informationLab 3: Poles, Zeros, and Time/Frequency Domain Response
ECEN 33 Linear Systems Spring 2 2 P. Mathys Lab 3: Poles, Zeros, and Time/Frequency Domain Response of CT Systems Introduction Systems that are used for signal processing are very often characterized
More informationDesigning Information Devices and Systems II Fall 2018 Elad Alon and Miki Lustig Discussion 5A
EECS 6B Designing Information Devices and Systems II Fall 208 Elad Alon and Miki Lustig Discussion 5A Transfer Function When we write the transfer function of an arbitrary circuit, it always takes the
More informationECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]
ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative
More informationOverview of Sampling Topics
Overview of Sampling Topics (Shannon) sampling theorem Impulsetrain sampling Interpolation (continuoustime signal reconstruction) Aliasing Relationship of CTFT to DTFT DT processing of CT signals DT
More informationResponse to a pure sinusoid
Harvard University Division of Engineering and Applied Sciences ES 145/215  INTRODUCTION TO SYSTEMS ANALYSIS WITH PHYSIOLOGICAL APPLICATIONS Fall Lecture 14: The Bode Plot Response to a pure sinusoid
More informationChapter 8: Converter Transfer Functions
Chapter 8. Converter Transfer Functions 8.1. Review of Bode plots 8.1.1. Single pole response 8.1.2. Single zero response 8.1.3. Right halfplane zero 8.1.4. Frequency inversion 8.1.5. Combinations 8.1.6.
More informationIf you need more room, use the backs of the pages and indicate that you have done so.
EE 343 Exam II Ahmad F. Taha Spring 206 Your Name: Your Signature: Exam duration: hour and 30 minutes. This exam is closed book, closed notes, closed laptops, closed phones, closed tablets, closed pretty
More informationME 375 EXAM #1 Friday, March 13, 2015 SOLUTION
ME 375 EXAM #1 Friday, March 13, 2015 SOLUTION PROBLEM 1 A system is made up of a homogeneous disk (of mass m and outer radius R), particle A (of mass m) and particle B (of mass m). The disk is pinned
More informationPlan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.
Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequencyresponse design method Goal: wrap up lead and
More informationFrequency Response Analysis
Frequency Response Analysis Consider let the input be in the form Assume that the system is stable and the steady state response of the system to a sinusoidal inputdoes not depend on the initial conditions
More information8.1.6 Quadratic pole response: resonance
8.1.6 Quadratic pole response: resonance Example G(s)= v (s) v 1 (s) = 1 1+s L R + s LC L + Secondorder denominator, of the form 1+a 1 s + a s v 1 (s) + C R Twopole lowpass filter example v (s) with
More informationControl Systems I Lecture 10: System Specifications
Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
More informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More informationFrequency domain analysis
Automatic Control 2 Frequency domain analysis Prof. Alberto Bemporad University of Trento Academic year 20102011 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 20102011
More informationFirst and Second Order Circuits. Claudio Talarico, Gonzaga University Spring 2015
First and Second Order Circuits Claudio Talarico, Gonzaga University Spring 2015 Capacitors and Inductors intuition: bucket of charge q = Cv i = C dv dt Resist change of voltage DC open circuit Store voltage
More informationRLC Circuits and Resonant Circuits
P517/617 Lec4, P1 RLC Circuits and Resonant Circuits Consider the following RLC series circuit What's R? Simplest way to solve for is to use voltage divider equation in complex notation. X L X C in 0
More informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOTLOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOTLOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the rootlocus
More informationJ. McNames Portland State University ECE 223 Sampling Ver
Overview of Sampling Topics (Shannon) sampling theorem Impulsetrain sampling Interpolation (continuoustime signal reconstruction) Aliasing Relationship of CTFT to DTFT DT processing of CT signals DT
More information'XNH8QLYHUVLW\ (GPXQG73UDWWU6FKRRORI(QJLQHHULQJ. EGR 224 Spring Test II. Michael R. Gustafson II
'XNH8QLYHUVLW\ (GPXQG73UDWWU6FKRRORI(QJLQHHULQJ EGR 224 Spring 2017 Test II Michael R. Gustafson II Name (please print) In keeping with the Community Standard, I have neither provided nor received any
More information'XNH8QLYHUVLW\ (GPXQG73UDWWU6FKRRORI(QJLQHHULQJ. ECE 110 Fall Test II. Michael R. Gustafson II
'XNH8QLYHUVLW\ (GPXQG73UDWWU6FKRRORI(QJLQHHULQJ ECE 110 Fall 2016 Test II Michael R. Gustafson II Name (please print) In keeping with the Community Standard, I have neither provided nor received any assistance
More informationCHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION
CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION Objectives Students should be able to: Draw the bode plots for first order and second order system. Determine the stability through the bode plots.
More informationSinusoidal steadystate analysis
Sinusoidal steadystate analysis From our previous efforts with AC circuits, some patterns in the analysis started to appear. 1. In each case, the steadystate voltages or currents created in response
More informationRefinements to Incremental Transistor Model
Refinements to Incremental Transistor Model This section presents modifications to the incremental models that account for nonideal transistor behavior Incremental output port resistance Incremental changes
More informationHomework 7  Solutions
Homework 7  Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More informationROOT LOCUS. Consider the system. Root locus presents the poles of the closedloop system when the gain K changes from 0 to. H(s) H ( s) = ( s)
C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s)  H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closedloop system when the gain K changes from 0 to 1+ K G ( s)
More information'XNH8QLYHUVLW\ (GPXQG73UDWWU6FKRRORI(QJLQHHULQJ. EGR 224 Spring Test II. Michael R. Gustafson II
'XNH8QLYHUVLW\ (GPXQG73UDWWU6FKRRORI(QJLQHHULQJ EGR 224 Spring 2018 Test II Michael R. Gustafson II Name (please print) In keeping with the Community Standard, I have neither provided nor received any
More informationFrequency response. Pavel Máša  XE31EO2. XE31EO2 Lecture11. Pavel Máša  XE31EO2  Frequency response
Frequency response XE3EO2 Lecture Pavel Máša  Frequency response INTRODUCTION Frequency response describe frequency dependence of output to input voltage magnitude ratio and its phase shift as a function
More informationLecture 4: RLC Circuits and Resonant Circuits
Lecture 4: RLC Circuits and Resonant Circuits RLC series circuit: What's V R? Simplest way to solve for V is to use voltage divider equation in complex notation: V X L X C V R = in R R + X C + X L L
More informationR. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder
R. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder 8.1. Review of Bode plots Decibels Table 8.1. Expressing magnitudes in decibels G db = 0 log 10
More informationClass 13 Frequency domain analysis
Class 13 Frequency domain analysis The frequency response is the output of the system in steady state when the input of the system is sinusoidal Methods of system analysis by the frequency response, as
More informationSTABILITY. Have looked at modeling dynamic systems using differential equations. and used the Laplace transform to help find step and impulse
SIGNALS AND SYSTEMS: PAPER 3C1 HANDOUT 4. Dr David Corrigan 1. Electronic and Electrical Engineering Dept. corrigad@tcd.ie www.sigmedia.tv STABILITY Have looked at modeling dynamic systems using differential
More informationELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques [] For the following system, Design a compensator such
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 2: Drawing Bode Plots, Part 2 Overview In this Lecture, you will learn: Simple Plots Real Zeros Real Poles Complex
More informationFREQUENCYRESPONSE DESIGN
ECE45/55: Feedback Control Systems. 9 FREQUENCYRESPONSE DESIGN 9.: PD and lead compensation networks The frequencyresponse methods we have seen so far largely tell us about stability and stability margins
More informationOverview of DiscreteTime Fourier Transform Topics Handy Equations Handy Limits Orthogonality Defined orthogonal
Overview of DiscreteTime Fourier Transform Topics Handy equations and its Definition Low and high discretetime frequencies Convergence issues DTFT of complex and real sinusoids Relationship to LTI
More informationStep Response Analysis. Frequency Response, Relation Between Model Descriptions
Step Response Analysis. Frequency Response, Relation Between Model Descriptions Automatic Control, Basic Course, Lecture 3 November 9, 27 Lund University, Department of Automatic Control Content. Step
More informationMAE 143B  Homework 9
MAE 143B  Homework 9 7.1 a) We have stable firstorder poles at p 1 = 1 and p 2 = 1. For small values of ω, we recover the DC gain K = lim ω G(jω) = 1 1 = 2dB. Having this finite limit, our straightline
More informationFast Fourier Transform Discretetime windowing Discrete Fourier Transform Relationship to DTFT Relationship to DTFS Zero padding
Fast Fourier Transform Discretetime windowing Discrete Fourier Transform Relationship to DTFT Relationship to DTFS Zero padding Fourier Series & Transform Summary x[n] = X[k] = 1 N k= n= X[k]e jkω
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall K(s +1)(s +2) G(s) =.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering. Dynamics and Control II Fall 7 Problem Set #7 Solution Posted: Friday, Nov., 7. Nise problem 5 from chapter 8, page 76. Answer:
More informationAPPLICATIONS FOR ROBOTICS
Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table
More informationLecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types
Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This
More informationLecture 9 Infinite Impulse Response Filters
Lecture 9 Infinite Impulse Response Filters Outline 9 Infinite Impulse Response Filters 9 FirstOrder LowPass Filter 93 IIR Filter Design 5 93 CT Butterworth filter design 5 93 Bilinear transform 7 9
More information100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) =
1 AME 3315; Spring 215; Midterm 2 Review (not graded) Problems: 9.3 9.8 9.9 9.12 except parts 5 and 6. 9.13 except parts 4 and 5 9.28 9.34 You are given the transfer function: G(s) = 1) Plot the bode plot
More informationFast Fourier Transform Discretetime windowing Discrete Fourier Transform Relationship to DTFT Relationship to DTFS Zero padding
Fast Fourier Transform Discretetime windowing Discrete Fourier Transform Relationship to DTFT Relationship to DTFS Zero padding J. McNames Portland State University ECE 223 FFT Ver. 1.03 1 Fourier Series
More informationMAE 143A: Homework 7 Solutions
MAE 143A: Homework 7 Solutions Jacob Huffman and Elena Menyaylenko Date Due: February 6 th, 13 Problem 1: (i) Lets utilize the formula cos(a + B) = cos A cos B sin A sin B. (ii) u(t) = Acos(ωt + φ) u(t)
More informationLaplace Transform Analysis of Signals and Systems
Laplace Transform Analysis of Signals and Systems Transfer Functions Transfer functions of CT systems can be found from analysis of Differential Equations Block Diagrams Circuit Diagrams 5/10/04 M. J.
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency ResponseDesign Method
.. AERO 422: Active Controls for Aerospace Vehicles Frequency Response Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to
More informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuoustime, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering
More informationAutomatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year
Automatic Control 2 Loop shaping Prof. Alberto Bemporad University of Trento Academic year 21211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21211 1 / 39 Feedback
More informationME 375 Final Examination Thursday, May 7, 2015 SOLUTION
ME 375 Final Examination Thursday, May 7, 2015 SOLUTION POBLEM 1 (25%) negligible mass wheels negligible mass wheels v motor no slip ω r r F D O no slip e in Motor% Cart%with%motor%a,ached% The coupled
More informationECSE 4962 Control Systems Design. A Brief Tutorial on Control Design
ECSE 4962 Control Systems Design A Brief Tutorial on Control Design Instructor: Professor John T. Wen TA: Ben Potsaid http://www.cat.rpi.edu/~wen/ecse4962s04/ Don t Wait Until The Last Minute! You got
More informationSinusoidal SteadyState Analysis
Chapter 4 Sinusoidal SteadyState Analysis In this unit, we consider circuits in which the sources are sinusoidal in nature. The review section of this unit covers most of section 9.1 9.9 of the text.
More informationINTRODUCTION TO DIGITAL CONTROL
ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a lineartimeinvariant
More informationProblem Weight Score Total 100
EE 350 EXAM IV 15 December 2010 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score 1 25 2 25 3 25 4 25 Total
More informationEE 4343/ Control System Design Project LECTURE 10
Copyright S. Ikenaga 998 All rights reserved EE 4343/5329  Control System Design Project LECTURE EE 4343/5329 Homepage EE 4343/5329 Course Outline Design of Phaselead and Phaselag compensators using
More informationSteady State Frequency Response Using Bode Plots
School of Engineering Department of Electrical and Computer Engineering 332:224 Principles of Electrical Engineering II Laboratory Experiment 3 Steady State Frequency Response Using Bode Plots 1 Introduction
More informationControl for. Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e
Control for Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e Motion Systems m F Introduction Timedomain tuning Frequency domain & stability Filters Feedforward Servooriented
More information(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:
1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.
More informationVALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203. DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING SUBJECT QUESTION BANK : EC6405 CONTROL SYSTEM ENGINEERING SEM / YEAR: IV / II year
More informationFrequency Response. Re ve jφ e jωt ( ) where v is the amplitude and φ is the phase of the sinusoidal signal v(t). ve jφ
27 Frequency Response Before starting, review phasor analysis, Bode plots... Key concept: smallsignal models for amplifiers are linear and therefore, cosines and sines are solutions of the linear differential
More information2.010 Fall 2000 Solution of Homework Assignment 1
2. Fall 2 Solution of Homework Assignment. Compact Disk Player. This is essentially a reprise of Problems and 2 from the Fall 999 2.3 Homework Assignment 7. t is included here to encourage you to review
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 22: The Nyquist Criterion Overview In this Lecture, you will learn: Complex Analysis The Argument Principle The Contour
More informationTransfer func+ons, block diagram algebra, and Bode plots. by Ania Ariadna Bae+ca CDS Caltech 11/05/15
Transfer func+ons, block diagram algebra, and Bode plots by Ania Ariadna Bae+ca CDS Caltech 11/05/15 Going back and forth between the +me and the frequency domain (1) Transfer func+ons exist only for
More informationThe Frequencyresponse Design Method
Chapter 6 The Frequencyresponse Design Method Problems and Solutions for Section 6.. (a) Show that α 0 in Eq. (6.2) is given by α 0 = G(s) U 0ω = U 0 G( jω) s jω s= jω 2j and α 0 = G(s) U 0ω = U 0 G(jω)
More informationFrequency Response Techniques
4th Edition T E N Frequency Response Techniques SOLUTION TO CASE STUDY CHALLENGE Antenna Control: Stability Design and Transient Performance First find the forward transfer function, G(s). Pot: K 1 = 10
More information(a) Find the transfer function of the amplifier. Ans.: G(s) =
126 INTRDUCTIN T CNTR ENGINEERING 10( s 1) (a) Find the transfer function of the amplifier. Ans.: (. 02s 1)(. 001s 1) (b) Find the expected percent overshoot for a step input for the closedloop system
More informationRaktim Bhattacharya. . AERO 632: Design of Advance Flight Control System. Preliminaries
. AERO 632: of Advance Flight Control System. Preliminaries Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. Preliminaries Signals & Systems Laplace
More informationTest II Michael R. Gustafson II
'XNH8QLYHUVLW\ (GPXQG73UDWWU6FKRRORI(QJLQHHULQJ EGR 224 Spring 2016 Test II Michael R. Gustafson II Name (please print) In keeping with the Community Standard, I have neither provided nor received any
More informationEE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions
EE C28 / ME C34 Fall 24 HW 6.2 Solutions. PI Controller For the system G = K (s+)(s+3)(s+8) HW 6.2 Solutions in negative feedback operating at a damping ratio of., we are going to design a PI controller
More informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : 0. LS_D_ECIN_Control Systems_30078 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: Email: info@madeeasy.in Ph: 04546 CLASS TEST 089 ELECTRONICS ENGINEERING
More informationHomework Assignment 3
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 1 Homework Assignment 3 Assigned September 30, 2008. Due in lecture October 7, 2008. Note that you must include all of your work to obtain full
More informationStudio Exercise Time Response & Frequency Response 1 st Order Dynamic System RC LowPass Filter
Studio Exercise Time Response & Frequency Response 1 st Order Dynamic System RC LowPass Filter i i in R out Assignment: Perform a Complete e in C e Dynamic System Investigation out of the RC LowPass
More informationThe requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot  in time domain
Compensators To improve the performance of a given plant or system G f(s) it may be necessary to use a compensator or controller G c(s). Compensator Plant G c (s) G f (s) The requirements of a plant may
More informationMAE 143B  Homework 7
MAE 143B  Homework 7 6.7 Multiplying the first ODE by m u and subtracting the product of the second ODE with m s, we get m s m u (ẍ s ẍ i ) + m u b s (ẋ s ẋ u ) + m u k s (x s x u ) + m s b s (ẋ s ẋ u
More informationBIOEN 302, Section 3: AC electronics
BIOEN 3, Section 3: AC electronics For this section, you will need to have very present the basics of complex number calculus (see Section for a brief overview) and EE5 s section on phasors.. Representation
More informationFrequency Response DR. GYURCSEK ISTVÁN
DR. GYURCSEK ISTVÁN Frequency Response Sources and additional materials (recommended) Dr. Gyurcsek Dr. Elmer: Theories in Electric Circuits, GlobeEdit, 2016, ISBN:9783330713413 Ch. Alexander, M. Sadiku:
More informationTransfer function and linearization
Transfer function and linearization Daniele Carnevale Dipartimento di Ing. Civile ed Ing. Informatica (DICII), University of Rome Tor Vergata Corso di Controlli Automatici, A.A. 2425 Testo del corso:
More informationOutline. Classical Control. Lecture 1
Outline Outline Outline 1 Introduction 2 Prerequisites Block diagram for system modeling Modeling Mechanical Electrical Outline Introduction Background Basic Systems Models/Transfers functions 1 Introduction
More information2.161 Signal Processing: Continuous and Discrete Fall 2008
MT OpenCourseWare http://ocw.mit.edu 2.6 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Massachusetts
More informationCompensator Design to Improve Transient Performance Using Root Locus
1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning
More informationChapter 2. Classical Control System Design. Dutch Institute of Systems and Control
Chapter 2 Classical Control System Design Overview Ch. 2. 2. Classical control system design Introduction Introduction Steadystate Steadystate errors errors Type Type k k systems systems Integral Integral
More informationPoles and Zeros and Transfer Functions
Poles and Zeros and Transfer Functions Transfer Function: Considerations: Factorization: A transfer function is defined as the ratio of the Laplace transform of the output to the input with all initial
More informationAssignment 4 Solutions ContinuousTime Fourier Transform
Assignment 4 Solutions ContinuousTime Fourier Transform ECE 3 Signals and Systems II Version 1.01 Spring 006 1. Properties of complex numbers. Let c 1 α 1 + jβ 1 and c α + jβ be two complex numbers. a.
More informationPROBLEMS OF CHAPTER 4: INTRODUCTION TO PASSIVE FILTERS.
PROBEMS OF CHAPTER 4: INTRODUCTION TO PASSIVE FITERS. April 4, 27 Problem 4. For the circuit shown in (a) we want to design a filter with the zeropole diagram shown in (b). C X j jω e g (t) v(t)  σ (a)
More information