"a) Explain the strengths of instrumentation amplifiers compared to differential amplifiers in signal conditioning.\r\n\r\nb) A 50 V range voltmeter is connected across the terminals A and B of the circuit shown in Figure 1 The voltmeter has a resistance of 1000 kΩ.\r\n\r\ni) Find the open circuit voltage and output impedance of the circuit.\r\n\r\nii) Find the reading of the voltmeter, the percentage loading error and the accuracy.\r\n\r\n1376_Figure1.jpg\r\n\r\nFigure: Loading effect due to shunt connected voltmeter\r\n\r\nc) The signal conditioning circuit shown in Figure 2 is used to measure the strain resulting from deflection of a steel bar The voltage Vo, is measured to be 100 mV and the gauge factor is 2.0. Modulus of elasticity for steel is 30 x 106psi (pounds per square inch).\r\n\r\nThe resistances are R = 25kΩ and aR = 50Ω.\r\n\r\nCalculate (i) gain of the instrumentation amplifier (ii) change in resistance, ΔR (iii) ΔR/R (iv) strain ΔL/L and (v) stress.a) A continuous time control system is described by Figure 1. C(s) is a linear controller transfer function Consider a plant with the following first order transfer function: G(s) = b/s+a\r\n\r\n1187_Figure.jpg\r\n\r\nFigure 1: Block diagram of a continuous time control system\r\n\r\ni) Prove that a proportional plus integral controller is sufficient to achieve zero steady state error for this system with a step input.\r\n\r\nii) If a=2 and b=1, calculate the proportional and integral gains of C(s) to make the closed loop damping ratio equal to 0.8 and natural frequency equal to 2 radis.\r\n\r\nb) A continuous time control system is described by Figure 1. Consider a plant with the following second order transfer function:\r\n\r\nG(s) = c/(s2+as+b)\r\n\r\nC(s) represents a PID controller. Prove that the PID controller achieves zero steady state error for this system with a step input.\r\n\r\nc) A continuous time control system is described by Figure 1. Design a PID controller by applying the Reaction Curve method. The process unit step response is shown in Figure 2.\r\n\r\na) i) Draw the block diagram of a typical discrete time control system.\r\n\r\nii) Explain the abasing phenomenon in discrete time control systems. How is the abasing problem solved in practice?\r\n\r\nb) For the control system block diagram of Figure:\r\n\r\ni) Determine the open loop discrete time transfer function.\r\n\r\nii) Determine the closed loop discrete time transfer function of the system.\r\n\r\niii) Find the range of the sampling time period (T) for stability.\r\n\r\niv) Calculate the critical frequency.\r\n a) What is the Nyquist sampling theorem in discrete time systems?\r\n\r\nIn practice, how should the sampling frequency be chosen in discrete time control systems?\r\n\r\nb) Given the discrete-time system:\r\n\r\ny(k + 1) - y(k) = u(k + 1)\r\n\r\nfind the system transfer function and its response to a sampled unit step.\r\n\r\nc) A linear system is controlled using a discrete time controller shown in Figure 1.\r\n\r\n959_Figure.jpg\r\n\r\nFigure 1: Block diagrarn of a discrete time control system\r\n\r\ni) Write the equation of the discrete time open loop transfer function.\r\n\r\nii) Write the characteristic equation for this system.\r\n\r\niii) For a sampling time period T = 1sec: calculate the critical gain of this system The root locus leaves the unit circle at z = -1.\r\nANSWER TO QUESTION 1.1a) NYQUIST SAMPLING THEOREM:”A CONTINUOUS TIME SIGNAL MAYBE COMPLETELYREPRESENTATED IN ITS SAMPLES & RECOVERED BACKIF THE SAMPLING FREQUENCY Fs> OR EQUAL TO 2 Fm WHERE Fs ISTHE SAMPLING FREQUENCY & Fm IS THE MAXIMUMFREQUENCYPRESENT IN THE SIGNAL.THE STATEMENT CAN BE BROKEN INTO TWOPARTS.THE FIRST PART REPRESENTS THE REPRESENTATION OF SIGNALS IN ITSSAMPLES & MAXIMUM SAMPLING RATE REQUIRED TO REPRESENT ACONTINUOUS TIME SIGNAL INTO ITS SAMPLES.THE SECOND PART OF THE THEOREM REPRESENTS RECONSTRUCTIONOF THE ORIGNAL SIGNAL FROM ITS SAMPLES. IT GIVES SAMPLING RATEREQUIRED FOR SATISFACTORY RECONSTRUCTION OF SIGNAL FROM ITSSAMLPES.DISCRETE TIME SAMPLING: Sampling frequency needs to be large enough in comparison with themaximum rate of change of f(t). • Otherwise, high frequencycomponents will be mistakenly interpreted as low frequencies in thesampled sequence. Example: f(t) = 3 cos 2pt + cos 20pt + p 3 ! for ? = 0.1 s we obtain f*k+ = 3 cos(0.2pk) + cos 2pk + p 3 ! f*k+= 3 cos(0.2pk) + 0.5 The high frequency component appears as asignal of low frequency (here zero). This phenomenon is known asaliasing. ANSWER TO Ib)Y(k+1)-y(k)= u(k+1) EQUATION1IN PARTICULAR INSTEAD OF DIFFERENTIAL OPERATOR WE CAN FORDISCRETE TIME SYSTEMS DEFINE THE FORWARD SHIFT OPERATORqSUCH THAT:qy(k)= y(k+1) EQUATION 2PUTTING EQUATION 2 IN EQUATION 1 WE GETY(k+1) -y(k+1)/q=u(k+1)Y(k+1) x (1-1/q)=u(k+1)Y(k+1)/u(k+1) =1/(1-1/q) =q/(q-1)SO TANSFER FUNCTION = y(k+1)/u(k-!) = q/q-1 RESPONSE TO A SAMPLED UNIT STEP INPUTIn analogy with continuous-time systems, the response of a discrete- time systems is obtained by solving the difference equation. As a simpleexample, we can determine the output of the first-order system. Y(k+1) + ay(k) = bu(k) By applying the system equation recursively to compute y1, y2, . . . , ykwe obtainKk-1 k-n-1Y(k)=(-a) y(0)+? (-a)bu(n) n=0In particular, the step response for the input u(k)=u(step) k-1 k-n-1 Y(k) =k ? (-a)b u(n) n=023 k-1= ( 1 - a + (-a)+ (-a)+ · · · + (-a ) b u(step)k= 1 + (-a) b u(step)1 + a Our Equation isY(k+1)- y(k)=bu(k+1) a=-1, b =1y(k)= infinity,as denominator is zero.Response is unstable. SECOND METHOD:Our Equation is:Y(k+1)-ay(k) = bu(k)Y(k+1)=ay(k)+bu(k)We can z Transform the difference equation and get a transfer functionThat is described by the difference equation,Zy[z} =ay[z]+bu[z]Y[z]/U[Z] =b/z-aIn this case a =1,b=1.Y[z]/U[z]=b/z-a= 1/z-1This gives us following observations:a)The transfer function is a function of z and not s.b)The transfer function has one pole and it is at z=1.The analysis breaks down.Both methods give us unstable response to a unit step input.ANSWER TO QUESTION 1c)i)OPEN LOOP TRANSFER FUNCTION=k/s x1- e to power-Ts/sx1/s+2 =kx (1-e to power –Ts) x1sx s x s+ 13ii)CHARACTERISTIC EQUATION: THE CHARACTERISTIC EQUATION IS OF THE FORM1 +C(z)H(z) G(z)= 01+ L(z)= 0WHERE L(z) IS POPULARLY KNOWN AS LOOP TRANSFER FUNCTION.FROM ABOVE EQUATION WE CAN WRITE L(z) =-1CHARACTERISTIC EQUATION IN S DOMAIN WOULD BE1+ K(1-e to pwer Ts)/s x s x (s+1)NOW LAPLACE TRANSFER OF 1/S=z/z-1 (1- e to power Ts)/s x(s+1)=(1-e to power Ts)[ 1/s-1/s+1]=(1-e to power Ts)[z/z-1 –z/z-e to power minusT]=(1-z to power minus 1[z/z-1 –z/z-e to power –T]SO CHACTERISTIC EQUATION IN Z minus T IS1+kz[1- e to power T]/[z—1][z-e to power minus T]WHEN T =1 SECONDG[z]=kz[1-e to powerminus 1)/[z-1][z-e to power minus ] e to power minus 1= 0.3679Therefore 1- e to power minus 1 = 1- 0.3679 =0.6321G[z] =kz x 0.6321/ [z-1] x [z-0.3679]BREAK AWAY/ BREAK IN POINTSZ square =0.3679,Z1=0.6065,Z2=-0.6065Putting Z =-1 (Given)in 1+G(z)=0Absolute value of 0.6321 x -1/(-1-1)x(-1 -0.3679=1/KcCritical Gain Kc= Absolute Value of 2 x1.3679/0.6321Critical Gain Kc =4.328 ANSWER TO QUESTION2 i)BLOCK DIAGRAM OF A TYPICAL DISCRETE TIME CONTROLSYSTEMDISCRETE TIME SYSTEM MAYBE DEFINED AS A SYSTEM IN WHICHASSOCIATED SIGNALS ARE ARE ALSO DISCRETE TIME SIGNALS. THIS MEANS THAT IN A DISCRETE TIME SYSTEM, THE INPUT &OUTPUT ARE BOTH DISCRETE TIME SIGNALS.---------------------------------------- BLOCK DIAGRAM OF X(n)--------- ?DISCRETETIME -------------- ? Y(n) SYSTEM---------------------------------------FOR EXAMPLE MICROPROCESSORS, SEMICONDUCTORS, MEMORIES AND SHIFT REGISTERS ARE DISCRETE TIME SIGNALS.QUANTISED DISCRETE TIME IS THE OPERATING WAND SENSING OFDEPARTMENT STORE CASH REGISTAR.ii)ALIASING :Sampling frequency needs to be large enough in comparison with themaximum rate of change of f(t). • Otherwise, high frequencycomponents will be mistakenly interpreted as low frequencies in thesampled sequence. Example: f(t) = 3 cos 2pt + cos 20pt + p 3 !for ? = 0.1 s we obtain f*k+ = 3 cos(0.2pk) + cos 2pk + p 3 ! f*k+= 3 cos(0.2pk) + 0.5 The high frequency component appears as asignal of low frequency (here zero). This phenomenon is known asaliasing. bi)OPEN LOOP DISCRETE TIME TRANSFER FUNCTION:Y(s)/R(s) = Z(OH)s xG(s)Y(s)/R(s) = (1 -e to power –Ts) x3 / s x(s+1)bii)CLOSED LOOP TRANSFER FUNCTION:Y(s)/R(s)=Z(OH)X G(s)/ 1+Z(OH) X G(s)Y(s)/R(s)= (1 -e to power –Ts) x 3 Sx( s+1)1 +(1 –e to power –Ts)x 3 s x(s + 1) ANSWER TO QUESTION 2 iii) 2iv) CALCULATION OF CRITICAL FREQUENCY:A CRITICAL FREQUENCY IS DEFINED TO BE A VALUE OF OMEGA FOR WHICHPI (OPEN LOOP} OMEGA=180 DEGREESTHIS FREQUENCY IS ALSO REFFERED TO AS PHASE CROSS OVER FREQUENCY.IN OUR CASE OPEN LOOP TRANSFER FUNCTIONL(s) = Z(OH)XG(s)L(s) = (1- e to power – Ts)x3/s x( s+1)e to power – Ts=1/ (1 + St/n)to power ne to power – Ts= 1-Ts+T sq. xs sq./ 2 factorial - T cube x s cube/3 factorial THIS SERIES CAN BE TRUNCATED FOR DESIGN USE & IS VALID FOR LOW VALUE Te to power- Ts=1-Ts1- e to power -Ts = 1-(1- Ts)=TsSoL(s)= Tsx3( s+ 1)=3T/(s+1)CHANGING stojwL(jw) = 3T/ (jw + 1)Pi (open loop) ( angle omega) = -tan inverse omega= - 180 degrees180 degrees= pie radiansTan inverse omega=180 degrees= pie radiansTan omega= tan ( pie)Omega =Pie radians/ second =3.14 radians per second = critical frequencyCRITICAL FREQUENCY= 3.14 RADIANS/ SECOND ANSWER TO QUESTION 3CONTROL SYSTEMS ARE JUDGED ACCORDING TO THEIR PERFORMANCE IN THREE AREAS:1)TRANSIENT RESPONSE.2) STABILITY.3)STEADY STATE ERRORWE WILL DETERMINE STEADY STATE ERROR OF TOTAL RESPONSE BY LETTINGT APPROACHINFINITY & DETERMINE e INFINITY.WE WILL USE FINAL VALUE THEOREMf infinity=Lim s F(s) as s approaches 0Integral (Lim 0to infinity) [d/dt f(t)]e to power –st dt =Sf(s) -f(0)Lim s approaches 0 of Integral (Lim 0to infinity) [d/dt f(t)]e to power –st dt =Sf(s) -f(0)= Integral Lim (0 to onfinity) d/dt f(t) dt =f(t) o to infinityF(infinity) - f(0) = Lim sF(s) as s approaches 0 – f(0)f(infinity) =Lim s F(s) as s approaches 0Proportional Integral (PI) ControllerThe proportional integral controller produces an output, which is thecombination of outputs of the proportional and integral controllers.u(t)=KPe(t)+KI ?e(t)dt u (t) =KP e(t)+KI?e(t) d t Apply Laplace transform on both sides -U(s)=(KP+KIs)E(s)U(s)=(KP+KIs)E(s)U(s)E(s)=KP+KIsU(s)E(s)=KP+KIsTherefore, the transfer function of proportional integral controlleris KP+KIsKP+KIs.The block diagram of the unity negative feedback closed loop controlsystem along with the proportional integral controller is shown in thefollowing figure.The proportional integral controller is used to decrease the steady stateerror without affecting the stability of system.E(s)=R(s)/ 1+ C(s)x G(s)C(s)= Kp+ Ki/s for aproportional integrator controllerG(s)= b/s + a GIVENR(s) =1/sfor a unit step inputWe wil establish E(s) in terms of system & its input.E(s) =R(s)-Y(s)Y(s) =E(s)C(s)G(s)E(s) =R(s)– E(s)C(s)G(s) E(s)+E(s)C(s)G(s)= R(s)E(s) ( 1+C(s) G(s)= R(s)E(s)= R(s)/1+C(s)G(s)Putting the values in above equationE(s)=i/s/1 + (Kp + Ki/s) x 1/(s+2)Applying FinalValue Theoremf(infinity) =Lim s F(s) as s approaches 0f(infinity)=Lim s E(s) as s aporaches 0Ess=Lim s xi/s/1 + (Kp + Ki/s) x (s+2) as s approaches 0Ess=Lim 1/1 + (Kp + Ki/s) x (s+2) as s approaches 0Ess=Lim 1/1+Kp x s + 2 Kp+ Ki + 2 Ki/s as s approaches 0Ess=1/ 0 + 2Kp + 2Ki+ infinityEss=1/infinityEss=0So Steady State Error with Proportional + Integral is sufficient to giveZero output with a steady state input.In the above equation Kp is proportional gain & Ki is integral gain.If a =2,b=1, G(s) becomes1/s + 2C(s)= (Kp+Ki/s) for a proportional+ integral controllerY(s)/R(s) = Cs) x G(s)/1 + C(s) G(s)Y(s)/R(s) = (Kp + Ki/s) x 1/s+2// 1 + (Kp + Ki/s x 1/s+2Y(s)/R(s) = Kpxs +Ki//sx(s+2)+Kpxs + Ki Y(s)/R(s) = Kpxs + Ki// s square + (2+Kp)xs+KiSo characteristic equation is s square+ (2 + Kp)xs+ Kind This is compared with standard 2 order equation which iss square+2x epsillonx omega nxs+ omega n squareWe get (2+Kp)xs=2xepsillonx omega nxs2+Kp=2xepsillon nxomega nNow epsilon = 0.8, omega n= 2 rad/secondgivenTherefore 2 + Kp= 2x 0.8x 2=3.2ThereforeKp = 3.2-2= 1.2AndKi= omega n square = 2 square=4So proportional gain=1.2& integral gain= 4ANSWER TO QUESTION 3 b)Proportional Integral Derivative (PID) ControllerThe proportional integral derivative controller produces an output, which isthe combination of the outputs of proportional, integral and derivativecontrollers.u(t)=KPe(t)+KI ?e(t)dt+KDde(t)dt u(t )=K P e ( t )+ K I ? e (t )dt +K D de (t )dtApply Laplace transform on both sides -U(s)=(KP+KIs+KDs)E(s)U(s)=(KP+KIs+KDs)E(s)U(s)E(s)=KP+KIs+KDsU(s)E(s)=KP+KIs+KDsTherefore, the transfer function of the proportional integral derivativecontroller is KP+KIs+KDsKP+KIs+KDs. 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