ECE-205 : Dynamical SystemsHomework #10Due : Thursday February 17 at the beginning of class1) For the following systems, determinei) the settling time based on the pole locationsii) the magnitude and phase of0)(H jw from the pole-zero diagram (graphically)iii) the steady state output of the systema) 2, ( ) 4cos(3 ) ( )( 5 )( 5 )( )sx t t u ts j s jH s+=+ + +=- b) 5( ) , ( ) 5sin(4 30 ) ( )( 2 6 )( 2 6 )osH s x t t u ts j s j+= = ++ + + -c) 30( ) , ( ) 10cos(2 45 ) ( )( 1)( 4 4 )( 4 4 )oH s x t t u ts s j s j= = -+ + + + -Answers: 0.8, ( ) 0.42cos(3 4.15 )2, ( ) 1.11sin(4 34.97 )4, ( ) 4.16cos(2 138.17 )os ssos ssos ssT y t tT y t tT y t t= = -= = += = -2) For the following transfer functions, using the provided Bode plotsi) determine the correct form for each building block, i.e, 221 21, ) 1( ,nnn nsK s sztw w+ +� �� �� �+ii) sketch and label each building block (magnitude and phase)iii) sketch the straight line approximation to the Bode plot (magnitude and phase) by adding the building block componentsa) 10010( )H ss += b) 1())( 00sH ss+= c) ( 10)( )( 100)sH ss s+=+ d) 21000( 10)( )( 1000)sH ss+=+13) The following three figures display the magnitude of six transfer functions. All of the poles and zeros of thesetransfer functions are in the left half plane (these are minimum phase transfer functions). All of the magnitudes, poles, and zeros are either zero or simple powers of 10 (1 1 210 ,1,10 ,10-etc). Estimate the transfer functions. 100101102103-100-50050100Magnitude (dB)Frequency (rad/sec)100101102103-120-100-80-60-40-200Magnitude (dB)Frequency (rad/sec)Figure 1: Problem 3, Systems a and b2101102103-1001020304050Magnitude (dB)Frequency (rad/sec)10-1100101102103104105-40-20020406080Magnitude (dB)Frequency (rad/sec)Figure 2: Problem 3, Systems c and d310-1100101102103104-80-60-40-2002040Magnitude (dB)Frequency (rad/sec)10-1100101102103104-120-100-80-60-40-200Magnitude (dB)Frequency (rad/sec)Figure 3: Problem 3, Systems e and fAnswers: 22 22 22 2 510.1 1100 1100, ( ) , ( ) 10 11001 10.1 1 0.1 1110 10, ( )1 11 1 11 11 1 110 1001000 100 1(0)( , )00) (H sH s HsH s H s ss ss sH ss ss s ss� �+� �� �� �= = +� �� �� � � �+ +� � � �� � � �=� �� �� � � �� �+ ++ + +� �� �� � � �� �� �� �� � �=��=�=44) The cutoff frequency of a filter is the frequency at which the amplitude of the transfer function is equal to12of its maximum value. This is the half power or 3 dB point. For each of the following systems, determine the cutoff frequency1 1( ) , ( ) , ( )a b cbH s H s H ss p as p as p= = =+ + +Answers: ,c cppaw w= =5) For the following four systems, determine the bandwidth of the system. For lowpass systems, the bandwidth is equal to the cutoff frequency. For bandpass systems, the bandwidth is the difference between the low and the high frequency cutoffs. For the systems below, determine the bandwidth of the systems. Answers: 65 Hz, 110 Hz, 470 Hz, 25 Hz100101102-10-8-6-4-202|Ha(j)|(dB)Frequency (Hz)100101102-10-8-6-4-202|Hb(j)| (dB)Frequency (Hz)101102103-8-6-4-202468|Hc(j)| (dB)Frequency (Hz)101102103-8-6-4-202468|Hd(j)| (dB)Frequency (Hz)56) Consider the following simple feedback control block diagram. The plant is ( )42pG ss=+. a) What is the bandwidth of the plant alone (assuming there is no feedback) b) Assuming a proportional controller, ( )c pG s k=, determine the closed loop transfer function,0( )G s c) Assuming a proportional controller, ( )c pG s k=, determine the value of pkso the bandwidth of the closed loopsystem is 24 rad/sec.d) Assuming the proportional controller from problem c, determine the settling time and the steady state error for a unit step.Partial Answers: 4, 10, 1/6 7) In this problem we will relate the settling time of a system with the bandwidth of a system if there are not complex poles. Consider a transfer function for a first order system,( )pG sBs=+. Show that for this system, theproduct of the (2%) settling time and the bandwidth is equal to 4. Thus to decrease the settling time you need to increase the bandwidth, and visa versa.8) (Matlab Problem #1) Use Matlab to construct bode plots of the transfer functions you derived in problem 3. Print out each plot and turn it in.Matlab Hints:(1) If we want to plot a Bode plot of the transfer function 10.1 1101( )1100sssG� �+� �� ��=�+� �� �from 210w-=to 510w =, then we would type the following in MatlabG = tf(0.1*[1/10 1],[1/100 1]);Bode(G,logspace(-2,5,1000); grid;(2) If we have a transfer function with repeated poles, such as63411( )1001110ssG s� �+� �� �� �+� �� �=we want Matlab to do the work of multiplying these out. To multiply polynomials in Matlab, we use the command conv, which performs a discrete-time convolution. To construct the above transfer function then, we would type something like the following:np = [1/100 1];n = conv(np,np); % raised to the second powern = conv(n,np); % raised to the third powerdp = [1/10 1];d = conv(dp,dp); % raised to the second powerd = conv(d,dp); % raised to the third powerd = conv(d,dp); % raised to the fourth powerG = tf(n,d);9) (Matlab Problem #2) The Matlab routine homework10.m plots the magnitude of the frequency response of the transfer function based on the location of the poles and zeros entered in lines 8 and 9 of the code. This magnitude plot is not in Db and the amplitude has been scaled to have a maximum value of 1 for all of the plots.All you are asked to do in this problem is change the location of the poles and the zeros, run the code, and turn in your plots. You can save your plots to a word file and then turn that in , but do not put more than three plots in a page, and the grader must be able to read them. However, try to think about the pole-zero plots you made inproblem 1 as you do the problem.a) Run the code with poles at 10 , 5 10 , 1 110 0j j j� - � - �- and no zeros (leave the variable zeros as an empty array, do not put in a zero). Note that as the magnitude of the imaginary part of the pole gets significantly larger than the magnitude of the real part, we see well defined peaks in the frequency response. Turn in all three plots.b) Leave the poles at 1 10 j- � and add zeros first at 0, then at 0 and 5 j�, then at 0 and 15 j�. Note that at the frequency where a zero of the transfer function occurs, the output of the transfer