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 in terms of their input-output behavior when specific test signals are applied to the input. For continuous time (CT) systems the most important test signals are the unit step u(t), the unit impulse δ(t), and sinusoids of the form cos(ωt θ). Initially, the most logical approach is to analyze the input-output behavior directly in the time domain. For the important class of linear and time-invariant (LTI) systems it is however often easier and more intuitive to work with transformations of the time domain signals. The most general approach for causal CT LTI systems (which includes all systems that are described by linear differential equations with constant coefficients) is to use the unilateral Laplace transform and the equivalence h(t) H(s) between the unit impulse response h(t) in the time domain and the system function H(s) in the Laplace domain. In this setting the steady-state response to sinusoidal input signals is then just the special case of evaluating H(s) at s = jω, i.e., along the vertical axis in the s-plane. The quantity H(jω) with magnitude H(jω) and phase H(jω) is called the frequency response of the system characterized by h(t) H(s). Correspondingly, the transformation from h(t) to H(jω) is called a transformation from the time to the frequency domain. Working in the frequency domain makes it possible to characterize LTI systems in terms of their frequency-selective properties as lowpass filters (LPF), bandpass filters (BPF), highpass filters (HPF), etc.. First Order CT LTI Systems The differential equation that relates the input x(t) and the output y(t) of a general first order CT (continuous time) LTI (linear and time-invariant) system is of the form y () (t) a y(t) = b x () (t) b x(t), where x () (t) and y () (t) denote the first derivative with respect to t of x(t) and y(t), respectively. Assuming zero initial conditions (i.e., x( )=y( )=), this transforms into s Y (s) a Y (s) = b s X(s) b X(s) = (s a ) Y (s) = (b s b ) X(s), in the Laplace domain. The system function H(s) is then H(s) = Y (s) X(s) = b s b s a. One type of block diagram implementation, called controller canonical form, of this system function is shown in the following figure.
X(s) b sq(s) s Y (s) b Q(s) a To analyze this, label the output of the integrator with Q(s) and then express the integrator input as s Q(s) = X(s) a Q(s) = Q(s) = s a X(s). The quantity q(t) Q(s) is the state variable of the system and the denominator of Q(s) determines the pole location of the system. The output Y (s) is obtained from Q(s) as Y (s) = b s Q(s) b Q(s) = (b s b ) Q(s) = b s b s a X(s), and thus H(s) = (b s b )/(s a ) as claimed. Another form to implement this system function is called observer canonical form. To derive it, start from s Y (s) a Y (s) = b s X(s) b X(s), and then rearrange terms so that those with the same power of s can be combined in the following manner ( Y (s) b X(s) ) s = b X(s) a Y (s) = Y (s) b X(s) = ( b X(s) a Y (s) ) s. To convert this into a block diagram, start from an integrator and label the output Y (s) b X(s), as shown in the block diagram below. X(s) b b s Y (s) b X(s) Y (s) a Then add b X(s) to the integrator output to obtain Y (s) and use the formula given above to compute the integrator input. 2
Poles and Zeros. The first order system function can be expressed in different ways as H(s) = b s b s a = K s z s p = G s/( z ) s/( p ), where p = a and z = b /b are the pole and the zero of H(s), and K = b and G = b /a are constant gain factors. A typical pole-zero plot of a first order CT system is shown below. Im{s} s-plane o z p Re{s} The locations of the poles and zeros of a CT LTI system determine its stability, its frequency response, and its invertibility. A causal LTI system with rational system function is bounded-input bounded-output (BIBO) stable if and only if all poles of H(s) are in the left half (to the left of the vertical axis) of the s-plane. The zeros of H(s), however, can be anywhere in the s-plane without affecting system stability. Inverse System. The inverse system of a first order system with gain K or G, pole p and zero z has system function H (s) = K s p s z = G s/( p ) s/( z ). Note that in general an inverse system may be unstable and/or non-causal..2 Second Order CT LTI Systems The differential equation of a general second order CT system with input x(t) and output y(t) is y (2) (t) a y () (t) a 2 y(t) = b x (2) (t) b x () (t) b 2 x(t), with corresponding system function H(s) = Y (s) X(s) = b s 2 b s b 2 s 2 a s a 2. 3
To obtain a blockdiagram of this system in controller canonical form, start from Y (s) = b s 2 b s b 2 X(s) = (b s 2 s 2 b sb ) X(s) = (b a s a 2 s 2 s 2 b sb ) Q(s), a s a 2 where Q(s) = which has blockdiagram: s 2 a s a 2 X(s) = s 2 Q(s) = X(s) a s Q(s) a 2 Q(s), X(s) s 2 Q(s) s sq(s) s Q(s) a a2 The output Y (s) is obtained from this as Y (s) = (b s 2 b s b 2 ) Q(s) = b s 2 Q(s) b s Q(s) b 2 Q(s). This leads to the blockdiagram in controller canonical form shown below. Y (s) b b b 2 X(s) s 2 Q(s) s sq(s) s Q(s) a a 2 Another form of drawing the blockdiagram is the observer canonical form. To obtain this form, start from Y (s) = b s 2 b s b 2 s 2 a s a 2 X(s) = (s 2 a s a 2 ) Y (s) = (b s 2 b s b 2 ) X(s), and then solve for Y (s) s 2 as Y (s) s 2 = b 2 X(s) a 2 Y (s) ( b X(s) a Y (s) ) s b X(s) s 2. Now divide both sides by s 2 so that ( (b2 Y (s) = X(s) a 2 Y (s) ) ) s b X(s) a Y (s) s b X(s). The implementation of this in the form of a blockdiagram is the observer canonical form shown below. 4
X(s) b 2 b b s Q (s) s Y (s) Q (s) a 2 a Natural Frequency and Damping Ratio. To classify second order systems and gain some physical insight into their behavior, it is advantageous to use the following alternate form for the denominator of H(s) H(s) = Y (s) X(s) = b s 2 b s b 2 s 2 a s a 2 = b s 2 b s b 2 s 2 2ζω s ω 2 where ζ is called the damping ratio and ω is called the undamped natural frequency. The poles p, p 2 of H(s) are p, p 2 = ζ ω ± ζ 2 ω. Depending on the value of ζ, three cases are distinguished:. Overdamped, ζ >. In this case p = ζω ζ 2 ω = α, p 2 = ζω ζ 2 ω = α 2, and both poles are real-valued. 2. Critically Damped, ζ =. In this case H(s) has a real-valued double pole at p = p 2 = ω = α. 3. Underdamped, ζ <. In this case there are two complex conjugate roots p = ζ ω j ζ 2 ω = α jβ, and p 2 = ζ ω j ζ 2 ω = α jβ, where α = ζ ω is called the neper frequency and β = ζ 2 ω is called the radian frequency. Examples of pole plots for the three cases are shown below. For system stability all poles must lie in the left half of the s-plane. If ζ, both poles are located on a semi-circle with radius ω., 5
Overdamped, ζ=.5, ω =.5 2 Critically Damped, ζ=, ω =.5 2 Underdamped, ζ=.5, ω =.5 2 Im{s} Im{s} Im{s} 2 4 2 Re{s} 2 4 2 Re{s} 2 4 2 Re{s}.3 Frequency Response The Laplace transform of a complex exponential function is e ±jωt u(t), ROC: Re{s}>. s jω Therefore cos(ωt) u(t) = ejωt e jωt u(t) 2 s s 2 ω 2 = s, ROC: Re{s}>. (s jω)(s jω) The response y(t) Y (s) of a first order system with system function H(s) = b s b s a, to a sinusoidal input x(t) = cos(ωt) u(t) can then be computed from where and Y (s) = X(s) H(s) = s b s b = k (s jω)(s jω) s a s jω k s jω k 2, s a k = (s jω) Y (s) s=jω = k 2 = (s a ) Y (s) s= a = (b s b ) s s jω H(s) s=jω = 2 H(jω). Transformed back into the time domain this yields for t > s s 2 ω 2 s= a = a2 b a b a 2 w 2, y(t) = k e jωt k e jωt k }{{} 2 e at = Re{H(jω) e jωt } k 2 e at. 2 Re{k e jωt } 6
If the pole (p = a ) of H(s) is in the left half of the s-plane then a > and the transient term k 2 e a t as t (i.e., the system is stable). Thus, the steady-state part of the response is y ss (t) = Re{H(jω) e jωt } = Re{ H(jω) e j H(jω) e jωt } = H(jω) cos ( ωt H(jω) ), where H(jω) is the magnitude of the frequency response and H(jω) is the phase of the frequency response. For an underdamped (ζ < ) second order system the system function is H(s) = b s 2 b s b 2 s 2 2ζω s ω 2 = b s 2 b s b 2 (s α jβ)(s α jβ), where α = ζω and β = ζ 2 ω. In this case the Laplace transform of the response to x(t) = cos ωt u(t) is where and In the time domain Y (s) = X(s) H(s) = s (s jω)(s jω) b s 2 b s b 2 (s α jβ)(s α jβ) = k s jω k s jω k 2 s α jβ k 2 s α jβ, k = (s jω) Y (s) s s=jω = H(s) = s jω s=jω 2 H(jω), k 2 = (s α jβ) Y (s) s= αjβ. y(t) = k e jωt k e jωt k }{{} 2 e αt e jβt k2 e αt e jβt = Re{H(jω) e }{{} jωt } 2 e αt Re{k 2 e jβt }. 2 Re{k e jωt } 2e αt Re{k 2, e jβt } If the complex conjugate poles of H(s) at α ± jβ are in the left half of the s-plane then α > and the second term in y(t), which is multiplied by e αt, vanishes as t. Thus, the steady-state part of the sinusoidal response is again y ss (t) = Re{H(jω) e jωt } = Re{ H(jω) e j H(jω) e jωt } = H(jω) cos ( ωt H(jω) )..4 Frequency Response Plots Since H(jω) is complex-valued in general, two 2-dimensional graphs are needed to display H(jω) versus ω. The magnitude (or amplitude) and the phase of a sinusoid are physically meaningful quantities and therefore the two graphs most frequently used are the magnitude H(jω) and the phase H(jω) of the frequency response. In Matlab the command 7
Hjw = freqs(b,a,w); %Frequency response H(jw) can be used to evaluate H(jω) at the points specified in w. The vectors b and a contain the coefficients b, b,..., and a, a,..., of the numerator and denominator polynomials of H(s). The magnitude and the phase of Hjw are then obtained using abs(hjw) and angle(hjw), respectively. Example: The magnitude and the phase of the frequency response H(jω) of the first order CT LTI system with system function H(s) = 5 s 5 s 5, are shown in the next two plots. The plots were produced with the Matlab commands b = [5 5]; %H(s) numerator coefficients a = [ 5]; %H(s) denominator coefficients w = ; %Start frequency in rad/sec w2 = 2; %End frequency in rad/sec w = linspace(w,w2,5); %Linearly spaced w vector Hjw = freqs(b,a,w); %Frequency response H(jw) subplot(2) plot(w,abs(hjw), -b ),grid %Magnitude plot subplot(22) plot(w,(8/pi)*angle(hjw), -g ),grid %Phase plot 8
5 Frequency Response H(jω), b=[5, 5], a=[, 5] 4 H(jω) 3 2 2 4 6 8 2 4 6 8 2 5 4 H(jω) [deg] 3 2 2 4 6 8 2 4 6 8 2 ω [rad/sec] One of the problems with this graph is that it is difficult to see what H(jω) is for small ω, e.g., for ω = rad/sec. Another problem is that it is not possible to tell what H(jω) is for large ω, e.g., ω =. The frequency response of filters often needs to be specified and measured over a wide range of frequencies and magnitudes and that is difficult to visualize using a plot with linearly scaled axes. Using a logarithmic scale for the frequency axis and decibels (db) for the magnitude (2 log ( H(jω) )) often yields much more informative graphs. The following Matlab script was used to compute and display the frequency response of the same system, but with a logarithmic frequency axis and magnitude in db. b = [5 5]; %H(s) numerator coefficients a = [ 5]; %H(s) denominator coefficients w = ; %Start frequency in rad/sec w2 = ; %End frequency in rad/sec w = logspace(log(w),log(w2),5); %Logarithmically spaced w vector Hjw = freqs(b,a,w); %Frequency response H(jw) subplot(2) %Magnitude in db vs. log(w) semilogx(w,2*log(abs(hjw)), -b ),grid subplot(22) %Phase in deg vs. log(w) semilogx(w,(8/pi)*angle(hjw), -g ),grid 9
Now the frequency response plots look like this: 4 Frequency Response H(jω), b=[5, 5], a=[, 5], z =, p = 5 2 2log H(jω) [db] 8 6 4 2 2 3 5 4 H(jω) [deg] 3 2 2 3 ω (log scale) [rad/sec] Note that, not only do the graphs provide more information, but the display also shows some symmetries that were not apparent from the linearly scaled graphs..5 Bode Plots Consider a CT system with system function H(s) = K (s z )(s z 2 ) (s z M ) (s p )(s p 2 ) (s p N ) = G ( s z )( s z 2 ) ( s z M ) ( s p )( s p 2 ) ( s z N ), where z i and p i are the zeros and poles of H(s), respectively, and K and G are (high and low frequency) gain factors. The frequency response of a stable CT LTI system is obtained by evaluating H(s) at s = jω. Using the second form given above, the magnitude of the frequency response is H(jω) = G jω z jω z 2 jω z M jω p jω p 2 jω p N,
and the phase of the frequency response is H(jω) = G M m= If H(jω) is expressed in decibel (db) then 2 log H(jω) = 2 log G ( jω z m ) M m= N n= 2 log jω z m ( jω z n ). N n= 2 log jω p n, and thus both, the magnitude and the phase plot, can be constructed by adding the contribution from each zero and subtracting the contribution from each pole individually. Plots of 2 log H(jω) and of H(jω) versus log ω are called Bode plots in honor of Hendrik W. Bode. If the poles and zeros are real, then the magnitude (in db) and the phase plot can be approximated by straight line segments versus a logarithmic ω axis as follows. Let r i be either a pole (r i = p i ) or a zero (r i = z i ) of H(s). Then, assuming that r i is real, jω/( r i ) can be expressed as jω/( r i ) = {, if ω < (ω/r i ) 2 ri, ω / r i, if ω > r i. In decibels, this becomes 2 log ( jω/( r i ) ) {, if ω < ri, 2 log ( ω / r i ), if ω > r i. If the root r i is real and in the left half of the s-plane, then r i < and the phase is (jω/( r i ) ) = tan ω/( r i) = tan ω r i. This can be approximated by, if ω <. r i, ( ω ) (jω/( r i ) ) log 45, if. r i ω < r i, r i 9, if ω r i. If r i > and real (which can occur for a zero in the right half of the s-plane), then the phase is (jω/( r i ) ) = tan ω/( r i) = tan ω, r i with corresponding straight line segment approximation, if ω <. r i, (jω/( r i ) ) ( ω ) log 45, if. r i ω < r i, r i 9, if ω r i.
Example: The system function H(s) = 25 s 2 (s )(s 5) = s/2 (s/ )(s/5 ), has a zero at s = 2, and poles at s = and s = 5. The Bode plots of the magnitude and phase of the frequency response H(jω) straight line approximations and actual values) are shown in the graphs below. 5 Bode Plots for H(s)=b(s)/a(s), b=[25, 5], a=[, 6, 5] 2 log H(jω) [db] 5 5 5 2 3 H(jω) [deg] 4 2 2 4 6 8 Straight Line Approx. Actual Value 2 3 ω [rad/sec] (log scale) Note that the phase starts at, goes up to 4 and then gradually approaches 9. Example: The following system function has a zero at s = 2 in the left half of the s-plane H(s) = 25 s 2 (s )(s 5) = ( ) s/ 2 (s/ )(s/5 ), and two poles at s = and s = 5. Bode plots of the magnitude and the phase of the frequency response of this system (straight line approximations and actual values) are shown in the next two graphs. 2
5 Bode Plots for H(s)=b(s)/a(s), b=[25, 5], a=[, 6, 5] 2 log H(jω) [db] 5 5 5 2 3 2 5 Straight Line Approx. Actual Value H(jω) [deg] 5 5 2 3 ω [rad/sec] (log scale) Note that, because of the zero in the right half of the s-plane, the phase starts at 8 and gradually decreases to 9. Special Case: Zero(s) at s =. If there is a zero of multiplicity L at zero then H(s) is of the form H(s) = G sl ( s z )( s z 2 ) ( s z M ) ( s p )( s p 2 ) ( s, z N ) and thus 2 log H(jω) = 2 log G 2L log ω and H(jω) = G 9 L where H(jω) is in degrees. M m= M 2 log jω N 2 log z jω, m p n m= ( jw z m ) N n= n= ( jw p n ), Special Case: Pole(s) at s =. If there is a pole of multiplicity L at zero then H(s) is of the form H(s) = G ( s z )( s z 2 ) ( s z M ) s L ( s p )( s p 2 ) ( s z N ), 3
and thus 2 log H(jω) = 2 log G and H(jω) = G where H(jω) is in degrees. M m= M m= 2 log jω z m 2L log ω ( jw z m ) 9 L N n= N n= ( jw p n ), 2 log jω p n, Special Case: Complex Pole or Zero. Let r i = α ± jβ be a complex valued pole or zero with magnitude r i = α 2 β 2. Then jω r i = The magnitude squared of this is jω α jβ = α2 β 2 βω jαω α 2 β 2 = r i 2 βω jαω r i 2. jω 2 = ( r i 2 βω) 2 α 2 ω 2 = r i 4 2 r i 2 βω (β 2 α 2 )ω 2 = r i 2 ω 2 2βω. r i r i 4 r i 4 r i 2 The term 2βω can be neglected if ω r i and if ω r i. This leads to the approximation { jω 2 r i 2 ω 2, if ω < ri, r i r i 2 ω 2 / r i 2, if ω > r i. Straight-line approximations to the phase contribution of a complex pole or zero are not very good if the system is underdamped. If α r i (close to critically damped case) then β so that ( jω ) ( = tan r i and thus the approximation ( jω ) ( ω ) tan r i r i can be used. αω r i 2 βω.6 Frequency Response of Second Order Systems ) ( ω ) tan, r i, if ω <. r i, ( ω ) log 45, if. r i ω < r i, r i 9, if ω r i, The system function and the frequency response of a second order CT LPF with passband gain of are H LP (s) = ω 2 s 2 2ζω s ω 2 at s = jω H LP (jω) = 4 ω 2 ω 2 ω 2 j 2 ζω ω.
The magnitude (squared) H LP (jω) 2 and the phase H LP (jω) of the frequency response are therefore H LP (jω) 2 ω 4 ( 2 = (ω 2 ω 2 ) 2 4 ζ 2 ω 2 ω, and H 2 LP (jω) = tan ζω ω ). ω 2 ω 2 Note that H LP (jω ) = 9. Depending on the value of ζ, the magnitude H LP (jω) of the frequency response may have an overshoot near ω. A typical example is shown in the graphs below. Second Order CT LPF Frequency Response, ω = [rad/sec], ζ=.3 2 log ( H(jω) ) [db] 2 3 4 2 3 5 H(jω) [deg] 5 2 2 3 ω [rad/sec] (log scale) For a second order BPF H BP (s) and H BP (jω) are H BP (s) = 2 ζω s s 2 2 ζω s ω 2, and H BP (jω) = j 2 ζω ω ω 2 ω 2 j 2 ζω ω. The expressions for the magnitude (squared) and the phase of the frequency response are H BP (jω) 2 = 4 ζ 2 ω 2 ω 2 (ω 2 ω 2 ) 2 4 ζ 2 ω 2 ω, and H 2 BP (jω) = π ( 2 2 ζω ω ) tan. ω 2 ω 2 If ω = ω then H BP (jω) = and H BP (jω) = max ω H BP (jω) =. To find the -3dB bandwidth of the BPF, solve H BP (jω 3 ) 2 = 4 ζ 2 ω 2 ω 2 3 (ω 2 ω 2 3) 2 4 ζ 2 ω 2 ω 2 3 5 = 2,
for ω 3. Multiplying both sides by the product of the denominators yields 8 ζ 2 ω 2 ω 2 3 = (ω 2 ω 2 3) 2 4 ζ 2 ω 2 ω 2 3 = 4 ζ 2 ω 2 ω 2 3 = (ω 2 ω 2 3) 2 = ±2 ζω ω 3 = ω 2 ω 2 3 = ω 2 3 ± 2 ζω ω 3 ω 2 =. The solutions (four because of the ± sign of the ω 3 term) of the quadratic equation in ω 3 are ω 3 = ζω ± ζ 2 ω 2 ω 2 = ( ± ζ 2 ζ ) ω. The two positive solutions are ω 3 = ( ζ2 ζ ) ω, and ω 32 = ( ζ2 ζ ) ω. The -3dB bandwidth BW 3 is the difference between these two values, i.e., BW 3 = ω 32 ω 3 = 2 ζω for 2 nd order BPF. An example of the frequency response of a BPF with ω = [rad/sec] and ζ =.5 is shown below. The -3dB bandwidth of this filter is BW 3 = [rad/sec]. Second Order CT BPF Frequency Response, ω = [rad/sec], ζ=.5 2 log ( H(jω) ) [db] 2 3 4 2 5 H(jω) [deg] 5 2 ω [rad/sec] (log scale) The system function and the frequency response of a HPF are given by H HP (s) = s 2 s 2 2 ζω s ω 2 and H HP (jω) = 6 ω 2 ω 2 ω 2 j 2 ζω ω.
The magnitude (squared) and the phase of the frequency response are therefore H HP (jω) 2 = ω 2 ( 2 (ω 2 ω 2 ) 2 4 ζ 2 ω 2 ω, and H 2 HP (jω) = π tan ζω ω ). ω 2 ω 2 At ω = ω H HP (jω ) = 9. An example with ω = 3 [rad/sec] and ζ =.5 is shown in the following graphs. Second Order CT HPF Frequency Response, ω =3 [rad/sec], ζ=.5 2 log ( H(jω) ) [db] 2 3 4 5 6 2 3 2 5 H(jω) [deg] 5 2 3 ω [rad/sec] (log scale).7 Step and Impulse Response of Second Order Systems The unit step response of a second order LPF with system function H(s) = ω 2 s 2 2ζω s ω 2 = ω 2 (s α jβ) (s α jβ), where α = ζω, β = ζ 2 ω, and α 2 β 2 = ω 2, is obtained by computing the inverse Laplace transform of G(s) = s H(s) = ω 2 s (s α jβ) (s α jβ) = A s B s α jβ B s α jβ. 7
The coefficients A and B for the partial fraction expansion are A = s G(s) s= = ω2 α 2 β 2 =, B = (s α jβ) G(s) s= αjβ = ω2 /2β β jα = ω2 /2β (α/β)) β2 α 2 e j(πtan = 2 (α/β)) ζ 2 e j(πtan. In the time domain this translates into g(t) = ( A B e αt e jβt B e αt e jβt) u(t) = ( A e αt 2Re{B e jβt } ) u(t), and, with the values for A and B substituted, g(t) = ( e ζω t ζ 2 cos ( ζ2 ω t π tan (ζ/ ζ 2 ) )) u(t). A typical plot of g(t) for an underdamped LPF is shown in the following plot. g(t) g max t t max In many applications it is important to know how much overshoot, defined as max t g(t) g(t) t the step response has. To determine the time t max at which the overshoot occurs, find the time instants for which dg(t)/dt is equal to zero. Since dg(t)/dt = h(t), the next step is to derive the unit impulse response from H(s) = ω 2 s 2 2ζω s ω 2 = ω 2 (s α jβ) (s α jβ) = C s α jβ C s α jβ, where C = (s α jβ) H(s) s= αjβ = ω2 2 jβ = ω 2 ζ 2 e jπ/2. 8
Therefore, for t >, h(t) = C e αt e jβt C e αt e jβt = e αt Re{2C e jβt } = ω e αt ζ 2 cos ( βt π/2 ) = ω e αt ζ 2 sin βt. This has zero crossings at t =, π/β, 2π/β, etc. The maximum of g(t) occurs at t max = π/β. Substituting this value for t in g(t) yields g max = g(t max ) = ( e π α/β ζ 2 cos ( tan ζ ζ 2 ) ). Now note that if tan ζ = φ = tan φ = ζ ζ 2 ζ 2 = cos φ = ζ 2, and therefore cos ( tan ζ ) = ζ 2 = g max = e π ζ/ ζ 2. ζ 2 The overshoot of g(t) in percent is then e π ζ/ ζ 2..8 Measuring Frequency Response in Simulink To measure the frequency response of a CT Simulink model an input signal x(t) = cos ω t needs to be used for each frequency ω of interest. Rather than using a separate simulation for each ω to determine H(jω ) and H(jω ), it would be nice to sweep the frequency of the sinusoid at the input of the system over a specified range of frequencies. This is not too difficult to achieve in Matlab, but requires an extension of the concept of frequency to the concept of instantaneous frequency. The general expression for a sinusoid with (fixed) frequency ω is x(t) = A cos(θ(t)), where θ(t) = ω t θ(), where θ() is the initial phase at t =. The frequency of this sinusoid is the derivative of its argument with respect to t, i.e., dθ(t) = ω. dt More generally, the instantaneous frequency ω i (t) of a sinusoid x(t) = cos(θ(t)) with time-varying frequency is defined as ω i (t) = dθ(t) dt, and thus θ(t) = t ω i (τ) dτ θ(). 9
In Matlab the integration over ω i (τ) is replaced by the cumsum command as shown in the following script file that generates a chirp signal, i.e., a sinusoid whose frequency is swept from an intial value of ω min to a final value of ω max. Fs = 44; %Sampling rate tlen = 2; %Duration in sec wmin = ; %Start frequency [rad/sec] wmax = ; %End frequency [rad/sec] tt = [:round(tlen*fs)-]/fs; %Time axis w = linspace(wmin,wmax,length(tt)); %Lin-spaced w axis theta = cumsum(w)/fs; %Phase theta(t) for chirp signal xt = cos(theta); %Sweeping cosine (chirp) signal This script file generates a sinusoid of length 2 seconds with linearly increasing instantaneous frequency from to rad/sec, using a sampling rate of F s = 44 samples per second. An example of a chirp signal with instantaneous frequency from to rad/sec is shown versus ω i in the graph below. Chirp Signal: cos(θ(t)), ω min =, ω max = [rad/sec], Lin Spaced ω.5 cos(θ(t)).5 2 3 4 5 6 7 8 9 ω i [rad/sec] To listen to a chirp signal in Matlab use wmin=, wmax= to generate xt, followed by the command sound(xt,fs). Frequency responses are very often plotted versus a logarithmic scale for ω. The following graph shows a sinusoid whose instantaneous frequency was increased logarithmically from to rad/sec and which is also plotted versus a logarithmically spaced horizontal axis. 2
Chirp Signal: cos(θ(t)), ω min =, ω max = [rad/sec], Log Spaced ω.5 cos(θ(t)).5 2 ω (log spaced) [rad/sec] i The net effect is that lower frequencies are changed less fast than higher frequencies in the chirp signal. The commands that were used to produce the logarithmic chirp are Fs = 44; %Sampling rate tlen = 2; %Duration in sec wmin = ; %Start frequency [rad/sec] wmax = ; %End frequency [rad/sec] tt = [:round(tlen*fs)-]/fs; %Time axis w = logspace(log(wmin),log(wmax),length(tt)); %Log-spaced w axis theta = cumsum(w)/fs; %Phase theta(t) for chirp signal xt = cos(theta); %Sweeping cosine (chirp) signal One possibility to make a plot of the magnitude of the frequency response of a CT system in Simulink is to use a chirp signal (with logarithmically increasing frequency) as input and to display the absolute value of the output signal y(t) in decibel. If 2 log ( y(t) ) is plotted versus logarithmically spaced instantaneous frequency ω i, the envelope of the resulting waveform looks similar to a Bode plot of 2 log ( H(jω) ). The commands to get this started in MATLAB for H(s) = (5s 5)/(s 5) are: 2
Fs = 8; %Sampling rate tlen = 2; %Time duration for chirp wmin = ; %Start frequency wmax = ; %End frequency b = [5 5]; %Numerator of H(s) a = [ 5]; %Denominator of H(s) tt = [:round(tlen*fs)-]/fs; %Time axis for chirp w = logspace(log(wmin),log(wmax),length(tt)); %Log-spaced w axis theta = cumsum(w)/fs; %Phase theta(t) for chirp signal xt = cos(theta-pi/2); %Sweeping cosine (chirp) signal t = tt ; %t vector for Simulink u = xt ; %u vector for Simulink sim( CTmodel,[t() t(end)]); %Run Simulink model %with ode3 fixed-step solver %and step size /Fs semilogx(w,2*log(abs(yout)), -m ),grid ylim([- 2]) %Display - db... 2 db The resulting graph is shown below, together with a pole-zero plot in the s-plane and a plot of the unit step response g(t). The dashed blue line in the frequency response is the straight line Bode plot approximation to the magnitude of the frequency response. 2 Frequency Response, b=[5, 5], a=[, 5], z =, p = 5, Fs=8 Hz 5 2 log y(t) 5 5 2 3 ω i (log scale) [rad/sec] 5 s Plane: Pole Zero Plot 5 Unit Step Response 4 Im{s} g(t) 3 2 5 5 5 Re{s} 2 4 6 8 Time [ms] 22
.9 Pole-Zero Plots in Matlab The figure below shows a pole-zero plot for the system function H(s) = 25 s 5 s 2 7 s, which has zeros z = 2 and z 2 = ±, and poles p = 2 and p 2 = 5. 5 s Plane: Pole Zero Plot Im{s} 5 5 5 Re{s} To make such a plot in MATLAB, the following commands can be used b = [ 25 5]; %Numerator of H(s) a = [ 7 ]; %Denominator of H(s) zz = roots(b); %Zeros of H(s) pp = roots(a); %Poles of H(s) mmx = max(abs([zz;pp])); %Maximum magnitude j = sqrt(-); %Square root of - cir = exp(j*(pi/8)*[:36]); %Unit circle subplot(22) plot(mmx*real(cir),mmx*imag(cir), :k,... real(zz),imag(zz), ob,real(pp),imag(pp), xr ) grid axis square xlabel( Re\{s\} ),ylabel( Im\{s\} ) title( s-plane: Pole-Zero Plot ) figure(gcf) Note that plotting a circle with radius equal to the largest magnitude of all finite poles and zeros (mmx*real(cir),mmx*imag(cir), :k ), followed by the axis square command, ensures the same (auto-)scaling along the horizontal and vertical axes. 23
. Application: Envelope and Phase Detection In the frequency domain linear systems are characterized by their (steady-state) magnitude and phase response to sinusoids. In practice, a function generator is used to feed a signal x(t) = cos ωt to the input of a system under test (SUT), and the frequency ω is swept over a range from ω to ω 2. The output from the SUT is y(t) = A(ω) cos(ωt θ(ω)), where A(ω) is the amplitude response of the SUT at ω and θ(ω) is the phase response of the SUT at ω. Plotting those two quantities versus ω then yields the magnitude and the phase response of the SUT. An interesting question is how to process the sinuoidal waveform y(t) to extract A(ω) and θ(ω). The blockdiagram below shows how this can be done using a combination of linear and nonlinear building blocks. 2 v i (t) LPF at ω L v Li (t) cos ωt H(s) under test A cos(ωt θ) sin ωt -2 v q (t) LPF at ω L v Lq (t) The quantities v i (t) and v q (t) are computed as v i (t) = 2 cos ωt A cos(ωt θ) = A [cos θ cos(2ωt θ)], v q (t) = 2 sin ωt A cos(ωt θ) = A [sin θ sin(2ωt θ)]. If v i (t) and v q (t) are passed through a LPF with cutoff frequency ω L ω, then the cos(2ωt θ) and the sin(2ωt θ) terms are removed and v Li (t) and v Lq (t) become v Li (t) = A cos θ, and v Lq (t) = A sin θ. From this the magnitude A and the phase θ can be computed as ( A = vli 2 (t) v2 Lq (t), and θ = vlq (t) ) tan v Li (t) Note that there is some processing delay through the LPFs which imposes some limit on how fast ω can be swept over a desired frequency range.. 24
2 Prelab Questions P. Estimate System Function from Step Response. Determine the system function H(s) of a second order CT system with the following unit step response g(t). 3 Unit Step Response of 2 nd Order CT LPF 2.5 2.5 g(t).5.5...2.3.4.5.6.7.8 t [sec] P2. Unit Impulse Response of Bandpass Filter. Determine the unit impulse response h(t) of the bandpass filter with system function H(s) = 2 ζω s s 2 2 ζω s ω 2. P3. Second Order System with Noise. A general second order CT system has system function H(s) = Y (s) X(s) = b s 2 b, s b 2, s 2 a s a 2 and therefore or s 2 Y (s) a s Y (s) a 2 Y (s) = b s 2 X(s) b s X(s) b 2 X(s), y (2) (t) a y () (t) a 2 y(t) = b x (2) (t) b x () (t) b 2 x(t). This differential equation can be implemented using either integrators or differentiators and an interesting question is which of the two versions performs better in practice. (a) The block diagram below shows an implementation using integrators. X(s) D (s) D 2 (s) ω 2 s 2ζω s Y (s) ω 2 25
Show that, if D (s) = and D 2 (s) =, then H(s) = Y (s)/x(s) for this implementation is equal to the system function given above. (b) Real integrators are noisy and this can be modeled by adding disturbance signals D (s) and D 2 (s) to the inputs of the integrators. Determine Y (s) as a function of X(s), D (s), and D 2 (s). (c) The next block diagram shows an implementation of a second order CT system using differentiators. Y (s) D (s) D 2 (s) X(s) ω 2 ω 2 s s 2ζω Show that, if D (s) = and D 2 (s) =, then H(s) = Y (s)/x(s) for this implementation is equal to the same system function as in (a). (d) Real differentiators are noisy and this can be modeled by adding disturbance signals D (s) and D 2 (s) to the inputs of the differentiators. Determine Y (s) as a function of X(s), D (s), and D 2 (s). (e) Noise tends to have a lot of high frequency components. Which of the transfer functions in (b) and (d) acts more like a highpass filter for D (s) and D 2 (s)? What conclusions can you draw? 3 Lab Experiments E. Step, Impulse, Frequency Response of LPFs. The main goal of this experiment is to look at the step response and the frequency response of different lowpass filters (LPF) with the goal of selecting the best one to be used in experiment 2 for measuring the magnitude and the phase of the frequency response of a system under test. Use the model shown below to simulate a first order system. 26
For a second order system the model shown next can be used in Simulink. (a) Consider the first order LPF with system function H(s) = ω L s ω L. Use ω L = 5 rad/sec and a sampling rate of F s = 8 Hz to generate the unit step response of this system and the magnitude of the frequency response in the range ω =... rad/sec. Use the technique explained in the section Measuring Frequency Response in Simulink to generate a sweeping chirp signal for the input of the Simulink model and then display the absolute value of the sinusoid at the output of the Simulink model in decibels. To run the simulation use fixed-step ode3 solver with a fixed step size of /F s. Make a labeled display that shows the magnitude of the frequency response, a pole-zero plot, and the step response, similar to the example shown in part (b) or in the introduction. 27
(b) Measure and plot the step response and the magnitude of the frequency response in the range ω =... rad/sec for a second order LPF with ω = 5 rad/sec and for ζ =, / 2, /2,.. Use a sampling rate of F s = 8 Hz and generate a chirp signal of length 2 sec. Use the fixed-step ode3 solver and a fixed step size of /F s to run the simulations. For which of the four values of ζ is the system rise time t r (from the first time the step response is above % to the time when it stays above 9smallest? What is the value of this t r? Compare the results with the first order LPF in (a). An example of how to display your simulation results is shown below for ω = 5 rad/sec and ζ =.2. Frequency Response, b=[,, 25], a=[, 2, 25], pp=[ 48.9898i, 48.9898i], Fs=8 Hz 2 log y(t) 2 3 4 5 2 3 ω i (log scale) [rad/sec] 5 s Plane: Pole Zero Plot 2 Unit Step Response.5 Im{s} g(t).5 5 5 5 Re{s} 2 4 6 Time [ms] (c) Consider now the cascade (or series connection) of the first order LPF and a second order LPF to form a third order LPF. Assuming that ω L = ω = 5 rad/sec, how should ζ of the second order component LPF be chosen to minimize the rise time t r? How should zeta be chosen to have a maximally flat magnitude of the frequency response up to ω? Compare the performance of the third order LPF to the performance of the second and first order LPFs. E2. Extracting Magnitude and Phase of Frequency Response. Build the following model for measuring the magnitude and the phase of a system under test (SUT) in Simulink. 28
Note that the H(s) under test and the LPF blocks are Transfer Function blocks from the Continuous library in Simulink. (a) Use the best LPF with ω = 5 rad/sec that you found in E for the two lowpass filters (LPF cos and LPF sin). The best LPF is the one that has a fast rise time t r and an (almost) flat magnitude of the frequency response in the passband. Use a BPF with ω = rad/sec and ζ =.5 for the SUT to test the magnitude/phase frequency response measurement system. Sweep the frequency over the range ω =... [rad/sec] and note that now both a cosine and a sine chirp signal are needed. Use a sampling frequency of F s = 6 Hz or more for the chirp signals and a duration of about 2 sec. In Simulink use the fixed-step ode3 solver with a step size of /F s. Here is what the magnitude and phase plots for the BPF should look like. 29
Magnitude and Phase of Frequency Response of BPF Filter, ω = [rad/sec], ζ=.5 2 log ( H(jω) ) [db] 2 3 4 5 2 3 4 5 H(jω) [deg] 5 2 3 4 ω [rad/sec] (log scale) (b) Design second order CT filters with the following specifications: (i) LPF with ω = 3 rad/sec and ζ such that the overshoot in H(jω) near ω is 3 db. (ii) BPF with ω = rad/sec and ζ such that the -4 db bandwidth is 3 rad/sec. (iii) HPF with ω = 2 rad/sec and ζ such that the overshoot in H(jω) near ω is db. In all three cases plot the magnitude (in db) and the phase (in degrees) of the frequency response for the range ω =... using the Simulink model given in (a). Once the design constraints are met, make also labeled plots of the step response g(t) of each system and make a pole-zero plot for each system. Which of the three systems has the poles closest (as a fraction of ω ) to the vertical axis? Why? (c) Measure and plot the frequency response of the following two system functions H (s) = s ω s ω, and H 2 (s) = s2 2 ζω s ω 2 s 2 2 ζω s ω 2 Choose ω = [rad/sec], use F s = 6 Hz, and plot the magnitude (in db) and the phase (in degrees) of the frequency response over the range ω =... [rad/sec]. For H 2 (s) use ζ =, /2,.. What is the distinguishing feature of these two systems?. 3
(d) Just to convince yourself that the filters in (c) are actually quite useful building blocks, combine H (s) and H 2 (s) as shown in the Simulink model below which tests the system H (s) = H 2 (s) H (s). Instead of addition one could also use subtraction and test the system H (s) = H 2 (s) H (s) whose Simulink model is shown next. Use again ω = [rad/sec] and F s = 6 Hz in both cases and measure and plot the magnitude (in db) and the phase (in degrees) of the frequency response of both systems 3
over the range ω =... [rad/sec]. How should ζ be chosen (try ζ =, /2,.) if the goal is to obtain a filter whose frequency response magnitude goes as quickly as possible from a passband gain of db to a stopband attenuation of -4 db? What type of filters are obtained from H (s) and H (s)? What are the orders of these filters? E3. Using Differentiators to Implement CT Systems. The goal of this experiment is to compare the performance of second order CT LPFs, built either from integrators or differentiators, in the presence of noise. For simplicity, it is assumed that H(s) = ω 2 s 2 2ζω s ω 2, with ω =, and ζ = 2. The Simulink model for the integrator implementation is shown below. The main input is In and the main (LPF) output is Out. Output Out 2 is the BPF output of the circuit and Out 3 is the HPF output. The inputs In 2 and In 3 can be used to model real integrators that have a noisy input stage by injecting a random noise signal. The next figure shows how to implement the same circuit using differentiators instead of integrators. 32
The main input is again In and the main output is Out. The BPF output is Out 2 and the HPF output is Out 3. The intent of inputs In 2 and In 3 is to be able to model real differentiators with noisy input stages by injecting random noise signals. (a) Set the noise inputs In 2 and In 3 to zero and verify that both, the integrator and the differentiator models, behave like identical LPFs with ω = and ζ = /2. Use a sweeping cosine input with ω =.... as input and a sampling frequency of Hz. The time duration for the sweeping cosine should be about 2 sec. Note: The Derivative block in Simulink is only an aproximation to the ideal s-domain derivative s. It computes the derivative of an input x(t) as x () (t) = dx(t) dt x(t) x(t ) where = /F s if a fixed step size of /F s is chosen. Thus, the output of the Derivative block at time t depends on the input at time t. But in the differentiator blockdiagram implementation of a second order CT system there are two feedback loops from the outputs of the differentiators back to the input of the system. This creates an algebraic loop which Simulink tries to solve iteratively by adjusting the feedthrough variable (the one whose output at time t depends directly on the input at time t) until convergence occurs. Since this can slow down the simulation considerably, Simulink issues a warning similar to the one shown below., 33
For the purposes of this experiment you can safely ignore this warning and proceed with the simulation. (b) To generate Gaussian noise for the simulation of the (real-world) noisy integrators and differentiators, use the Matlab code tlen2 = 2; %Time duration for chirp An =... %Set the noise amplitude here tt2 = [:round(tlen2*fs)-]/fs; %Time axis for chirp nt2 = An*randn(size(tt2)); %Noise signal 2 where the noise amplitude An is adjusted to obtain a desired signal-to-noise ratio (SNR) at the integrator/differentiator inputs. For example, if An=.3536 is specified for the integrator implementation of the LPF, then graphs of the frequency response and the step response should look similar to the following: 34
Frequency Response of Integrator (SNR=6 db) Version of Order 2 CT LPF 2 log y(t) 2 3 4 5 ω i (log scale) [rad/sec].5 Unit Step Response g(t).5.5 2 4 6 8 2 4 6 8 2 Time [sec] The effect of the noise is clearly visible in both graphs, especially towards higher frequencies. Compare this to the next figure (which is essentially noise-free) for which the same Simulink model, but with An=.7, was used: Frequency Response of Integrator (SNR=4 db) Version of Order 2 CT LPF 2 log y(t) 2 3 4 5.5 ω i (log scale) [rad/sec] Unit Step Response g(t).5.5 2 4 6 8 2 4 6 8 2 Time [sec] 35
Your task: Use a value of approximately. as a starting point for An and run a simulation for both the integrator and the differentiator implementation of the second order CT LPF specified earlier. Make plots of the sinusoidal response over the range ω =.... and of the step response, similar to the ones shown above. Then use x(t) = sin t as input to measure the SNR in db (i.e., log (P x /P n ), where P x is the average signal power and P n is the average noise power) at the inputs of the integrators/differentiators. Adjust this SNR to 2 db by adjusting the value of An. Then measure the SNR at the output of the LPF, again for x(t) = sin t. What observations do you make? From your simulations it should now be obvious why integrators are used in blockdiagrams and not differentiators. Compare also to the results of prelab problem P3. Notes: The average power P x of a CT waveform x(t) for t t < t 2 is defined as P x = P x (t, t 2 ) = t2 x(t) 2 dt. t 2 t t If x(t) is sampled with sampling rate F s and n = t F s, n 2 = t 2 F s, then t 2 t t2 t x(t) 2 dt F s n 2 n n 2 n=n x(n/f s ) 2 F s = n 2 x n 2, n 2 n n=n where x n = x(n/f s ). To measure SNRs for LTI systems, a useful approach is to measure the average signal power without noise being present and the average noise power without the signal being present. Then, assuming that the signal and the noise are statistically independent, the SNR can be computed as the ratio between the two powers measured. c 22 2, P. Mathys. Last revised: 2-7-, PM. 36