StepByStep_PulseSp03(1pp)

# StepByStep_PulseSp03 - ANALYSIS OF CIRCUITS WITH ONE ENERGY STORING ELEMENT CONSTANT INDEPENDENT SOURCES A STEP-BY-STEP APPROACH THIS APPROACH

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Unformatted text preview: ANALYSIS OF CIRCUITS WITH ONE ENERGY STORING ELEMENT CONSTANT INDEPENDENT SOURCES A STEP-BY-STEP APPROACH THIS APPROACH RELIES ON THE KNOWN FORM OF THE SOLUTION BUT FINDS THE CONSTANTS K1 , K 2 ,t USING BASIC CIRCUIT ANALYSIS TOOLS AND FORGOES THE DETERMINATION OF THE DIFFERENTIAL EQUATION MODEL x (t ) = K1 + K 2e - t t ,t > 0 K1 is the steady state value of the variable and can be determined analyzing the circuit in steady state x (0+ ) is the initial condition and provides the second equation to compute the constants K1 , K 2 t is the time constant and can be determined using Thevenin across the energy storing element CIRCUITS WITH ONE ENERGY STORING ELEMENT Obtaining the time constant: A General Approach RTH a a Circuit with resistances and sources Inductor or Capacitor b VTH + iR ic + - Thevenin b Representation of an arbitrary circuit with one storage element RTH a Inductor or Capacitor + KCL@ node a ic + iR = 0 Capacitive Circuit t = RTH C Inductive Circuit RTH a + vR - t= Use KVL + vL vR + vL = vTH vR = RTH iL diL vL = L dt diL L + RTH iL = vTH dt dvC L VTH + ic = C dt iL b v -v b iR = C TH Case 1.1 Case 1.2 RTH Voltage across capacitor Current through inductor dv v -v C C + C TH = 0 dt RTH æ L ö diL v dvC ÷ ç + iL = TH = iSC RTH C + vC = vTH ç R ÷ dt RTH dt è TH ø VTH C vc _ L RTH THE STEPS STEP 1. THE FORM OF THE SOLUTION x ( t ) = K1 + K 2 e - t t ,t > 0 K1 = x (¥); K1 + K 2 = x (0+ ) DETERMINE x (0+ ) STEP 2: DRAW THE CIRCUIT IN STEADY STATE PRIOR TO THE SWITCHING AND DETERMINE CAPACITOR VOLTAGE OR INDUCTOR CURRENT STEP 3: DRAW THE CIRCUIT AT 0+. THE CAPACITOR ACTS AS A VOLTAGE SOURCE. THE INDUCTOR ACTS AS A CURRENT SOURCE. DETERMINE THE VARIABLE AT t=0+ DETERMINE x (¥) STEP 4: DRAW THE CIRCUIT IN STEADY STATE AFTER THE SWITCHING AND DETERMINE THE VARIABLE IN STEADY STATE. STEP 5: DETERMINE THE TIME CONSTANT t = RTH C circuit with one capacitor t= L RTH circuit with one inductor STEP 6: DETERMINE THE CONSTANTS K1, K2 K1 = x (¥), K1 + K 2 = x (0+ ) LEARNING EXAMPLE FIND i (t ), t > 0 LEARNING EXAMPLE FIND i (t ), t > 0 STEP 3 : Determine i (0+ ) USE A CIRCUIT VALID FOR t=0+. THE CAPACITOR ACTS AS SOURCE STEP 1 : i (t ) = K1 + K 2e - t t ,t > 0 STEP 2 : Initial voltage across capacitor USE CIRCUIT IN STEADY STATE PRIOR TO THE SWITCHING KVL i= 24V = 2mA = i (0-) vC ( 0 + ) = v c ( 0 - ) 12kW vc (0-) = 36V - (2mA)(2k ) = 32[V ] i (0 + ) = 32V 16 = mA 6k 3 NOTES FOR INDUCTIVE CIRCUIT (1)DETERMINE INITIAL INDUCTOR CURRENT IN STEP 2 (2)FOR THE t=0+ CIRCUIT REPLACE INDUCTOR BY A CURRENT SOURCE STEP 4 : Determine i (¥) STEP 5 : Determine time constant STEP 6 : Determine K1 , K 2 STEP 6 : Determine K1 , K 2 STEP 4 : Determine i (¥) USE CIRCUIT IN STEADY STATE AFTER SWITCHING (STEP 1) i (t ) = K1 + K 2e - t t (STEP 3) i (0+ ) = (STEP 4) i (¥) = \ K2 = 36 i (¥) = mA 8 STEP 5 : Determine time constant Capacitive circuit : t = RTH C 36 mA = K1 8 16 36 5 -= 386 = K1 + K 2 FINAL ANSWER C = 100m F t = (1.5 ´ 103 W)(100 ´ 10-6 F ) = 0.15 s t 36 5 - 0.15 i (t ) = + e ,t > 0 86 NOTE: FOR INDUCTIVE CIRCUIT L t= RTH ORIGINAL CIRCUIT RTH RTH = 2k || 6k = 1.5kW 16 mA 3 ,t > 0 USING MATLAB TO DISPLAY FINAL ANSWER ì 2mA t £ 0 ï i (t ) = í 36 5 - t + e 0.15 , t > 0 ï8 6 î Command used to define linearly spaced arrays » help linspace LINSPACE Linearly spaced vector. LINSPACE(x1, x2) generates a row vector of 100 linearly equally spaced points between x1 and x2. LINSPACE(x1, x2, N) generates N points between x1 and x2. See also LOGSPACE, :. Script (m-file) with commands used. Prepared with the MATLAB Editor %example6p3.m %commands used to display funtion i(t) %this is an example of MATLAB script or M-file %must be stored in a text file with extension ".m” %the commands are executed when the name of the M-file is typed at the %MATLAB prompt (without the extension) tau=0.15; %define time constant tini=-4*tau; %select left starting point tend=10*tau; %define right end point tminus=linspace(tini,0,100); %use 100 points for t<=0 tplus=linspace(0,tend, 250); % and 250 for t>=0 iminus=2*ones(size(tminus)); %define i for t<=0 iplus=36/8+5/6*exp(-tplus/tau); %define i for t>=0 plot(tminus,iminus,'ro',tplus,iplus,'bd'), grid; %basic plot command specifying %color and marker title('VARIATION OF CURRENT i(t)'), xlabel('time(s)'), ylabel('i(t)(mA)') legend('prior to switching', 'after switching') axis([-0.5,1.5,1.5,6]);%define scales for axis [xmin,xmax,ymin,ymax] FIND v (t ), t > 0 LEARNING EXAMPLE STEP 3 : Determine v (0+ ) STEP 1 : v ( t ) = K1 + K 2e - t t ,t > 0 STEP 2 : Initial inductor current FIND v (t ), t > 0 LEARNING EXAMPLE STEP 3 : Determine v (0+ ) Use circuit at t=0+. Inductor is replaced by current source STEP 1 : v ( t ) = K1 + K 2e - t t ,t > 0 V1 STEP 2 : Initial inductor current Use circuit in steady state prior to switching I1 = 24 = 4mA 6 i L (0 - ) = 6 I1 6+3 20 V1 - 24 V1 V1 8 + + + = 0 V1 = [V ] 3 4 6 12 3 v (0+ ) = 24[V ] - V1 = 6k || 3k 8 i L (0-) = mA 3 52 [V ] 3 STEP 4 : DETERMINE v (¥) STEP 5 : DETERMINE TIME CONSTANT STEP 6 : DETERMINE K1 , K 2 STEP 4 : DETERMINE v (¥) USE CIRCUIT IN STEADY STATE AFTER SWITCHING STEP 6 : DETERMINE K1 , K 2 K1 = v (¥) = 24[V ] (step 4) 52 = K1 + K 2 (step 3) 3 52 20 K1 = - 24 = - [V ] 3 3 v (0 + ) = t 20 ANS : v (t ) = - + 24e 2 , t > 0 3 v (¥) = 24[V ] STEP 5 : DETERMINE TIME CONSTANT t= 4H = 2s 2W L Inductive Circuit : t = RTH RTH = 4 || 6 || 12 RTH = 2W ORIGINAL CIRCUIT LEARNING EXTENSION FIND vO ( t ), t > 0 STEP 3 : DETERMINE vO (0+ ) - t t STEP 1 : vo ( t ) = K1 + K 2e , t > 0 STEP 2 : INITIAL CAPACITOR VOLTAGE LEARNING EXTENSION FIND vO ( t ), t > 0 v1 - 12 v1 - 8 v1 + + = 0 */6 3 2 2 8v1 - 48 = 0 Þ v1 = 6[V ] v 2 = 2[V ] Þ vC (0-) = vC (0+ ) = 10[V ] KCL @ v1 : STEP 3 : DETERMINE vO (0+ ) + t =0+ + 10V - - t t STEP 1 : vo ( t ) = K1 + K 2e , t > 0 STEP 2 : INITIAL CAPACITOR VOLTAGE 12V + v2 + vC ( 0 - ) - v1 1 v2 = (12 - v1 ) 3 vC (0-) = 12 - v2 vO ( 0 + ) - vO ( 0 + ) = t <0 + 2 (10) = 5[V ] 2+2 STEP 4 : DETERMINE vO (¥) STEP 5 : DETERMINE TIME CONSTANT STEP 6 : DETERMINE K1 , K 2 ORIGINAL CIRCUIT STEP 4 : DETERMINE vO (¥) STEP 6 : DETERMINE K1 , K 2 + vO ( ¥ ) 24 2 vO (¥) = (12) = [V ] 5 5 STEP 5 : DETERMINE TIME CONSTANT Capacitive Circuit : t = RTH C 8 C = 2F Þ t = s 5 RTH 4 RTH = 1 || 4 = W 5 K 1 = vO ( ¥ ) = 24 [V ] 5 1 vO (0+ ) = 5[V ] = K1 + K 2 Þ K 2 = [V ] 5 ANS : vO ( t ) = 24 1 +e 55 - t 8 5 [V ]; ORIGINAL CIRCUIT t >0 STEP 3 : DETERMINE vO (0+ ) FIND vO (t ), t > 0 - t t STEP 1 : vO (t ) = K1 + K 2e , t > 0 STEP 2 : INITIAL INDUCTOR CURRENT iL (0+) STEP 3 : DETERMINE vO (0+ ) FIND vO (t ), t > 0 iL (0+) vO ( 0 + ) - t t STEP 1 : vO (t ) = K1 + K 2e , t > 0 STEP 2 : INITIAL INDUCTOR CURRENT vO i L (0-) v - 12 v - ( -4) v O +O + O =0 2 2 2 8 vO = [V ] 3 4 i L (0-) = i L (0+) = [ A] 3 8 vO (0+ ) = 2i L (0+ ) = [V ] 3 STEP 4 : DETERMINE vO (¥) STEP 5 : DETERMINE TIME CONSTANT STEP 6 : DETERMINE K1 , K 2 ORIGINAL CIRCUIT STEP 4 : DETERMINE vO (¥) STEP 6 : DETERMINE K1 , K 2 vO ( ¥ ) K1 = vO (¥) = 6[V ] (step 4) 8 vO ( + ) = = K1 + K 2 (step 3) 3 8 10 K 2 = - 6 = - [V ] 3 3 t 10 ANS : vO (t ) = 6 - e 0.5 , t > 0 3 2 (12) = 6[V ] 2+2 STEP 5 : DETERMINE TIME CONSTANT vO ( ¥ ) = RTH Inductive Circuit L t= RTH RTH = 4W t= 2 = 0. 5 s 4 ORIGINAL CIRCUIT LEARNING EXAMPLE FIND vO (t ), t > 0 STEP 1 : vO (t ) = K1 + K 2e - t t ,t >0 STEP 3 : DETERMINE vO (0+ ) STEP 2 : DETERMINE i L (0+ ) LEARNING EXAMPLE FIND vO (t ), t > 0 STEP 1 : vO (t ) = K1 + K 2e - t t ,t >0 STEP 3 : DETERMINE vO (0+ ) STEP 2 : DETERMINE i L (0+ ) i L (0-) = i L (0+ ) = 3[ A] 18[V ] iA = 3A 6[V ] + vO (0+ ) = 18[V ] STEP 4 : DETERMINE vO (¥) STEP 5 : DETERMINE TIME CONSTANT OPEN CIRCUIT VOLTAGE STEP 4 : DETERMINE vO (¥) vO (¥ ) = 27[V ] vB OPEN CIRCUIT VOLTAGE + vOC - vO (¥) KVL ' iA = vB 4 ' v B - ( -2i A ) ' iA KV L ' 2i A 9[V ] v B - 36 v B + + = 0 * / 12 2 4 6 ' ' 11v B + 4i A = 36 ´ 6 v B = 18[V ], i A = 4.5[ A] STEP 5 : DETERMINE TIME CONSTANT inductive circuit L t= RTH Circuit with dependent sources v RTH = OC i SC i" = 6 A A 12V + vOC = 24 + 12 = 36[V ] ORIGINAL CIRCUIT SHORT CIRCUIT CURRENT ORIGINAL CIRCUIT SHORT CIRCUIT CURRENT NOTE: FOR THE INDUCTIVE CASE THE CIRCUIT USED TO COMPUTE THE SHORT CIRCUIT CURRENT IS THE SAME USED TO DETERMINE vO (¥ ) i SC i1 36 = 2(i1 + i2 ) + 4i1 36 = '' 2(i1 + i2 ) + 6i2 - 2i A' '' i A' '' i A' = i1 i2 i SC = ''' 2i A 36 [ A] 8 vOC = 36[V ] ü 3 ý Þ RTH = 8W L = 3 H Þ t = s i SC = 36 / 8[ A]þ 8 STEP 6 : DETERMINE K1 , K 2 vO (¥ ) = 27 = K1 (step 4) vO (0+ ) = 18 = K1 + K 2 Þ K 2 = -9[V ] (step 3) - ANS : vO (t ) = 27 - 9e t 3 8 , t >0 ORIGINAL CIRCUIT LEARNING EXTENSION FIND vo (t ), t > 0 STEP 2 : DETERMINE CAPACITOR VOLTAGE AT t = 0 + STEP 3 : DETERMINE vO (0+ ) STEP 4 : DETERMINE vO (¥) STEP 1 : vO (t ) = K1 + K 2e - t t , t >0 FIND vo (t ), t > 0 LEARNING EXTENSION STEP 1 : vO (t ) = K1 + K 2e - t t , t >0 STEP 2 : DETERMINE CAPACITOR VOLTAGE AT t = 0 + -24[V ] + KVL v A = 12[V ] + vC ( 0 - ) vC (0-) = 24 + 24 + 12 = 60[V ] vC ( 0 + ) = vC ( 0 -) - STEP 3 : DETERMINE vO (0+ ) vO = vC Þ vO (0+ ) = 60[V ] STEP 4 : DETERMINE vO (¥) i =0 vA = 0 vO ( ¥ ) vO (¥ ) = 24[V ] STEP 5 : DETERMINE TIME CONSTANT OPEN CIRCUIT VOLTAGE vO ( ¥ ) SHORT CIRCUIT CURRENT i SC KVL STEP 6 : DETERMINE K1 , K 2 ORIGINAL CIRCUIT STEP 5 : DETERMINE TIME CONSTANT capacitive circuit Þ t = RTH C OPEN CIRCUIT VOLTAGE RTH = vOC i SC + vOC = vO (¥) = 24[V ] vOC vO (¥ ) SHORT CIRCUIT CURRENT 2W i SC KVL STEP 6 : DETERMINE K1 , K 2 K1 = vO (¥) = 24 (step 4) vO (0+ ) = 60 = K1 + K 2 (step 3) K 2 = 36[V ] ANS : vO (t ) = 24 + 36e - t 12 ,t >0 2i SC - 24 - 2v A = 0 v A = -2i SC i SC = 4[ A] RTH = 24 = 6W 4 t = 6W ´ 2 F = 12 s ORIGINAL CIRCUIT STEP 2: Initial inductor current Inductor example FIND vO (t ), t > 0 2W 4W t <0 2H + 6V - 2W + vO - STEP 1: Form of the solution STEP 3: Determine output at 0+ (inductor current is constant) STEP 4: Find output in steady state after the switching STEP 5: Find time constant after switch STEP 6: Find the solution STEP 2: Initial inductor current Inductor example FIND vO (t ), t > 0 2W 4W t <0 2H + 6V - 2W 2W 4W + iL (0-) vO - vO (t ) = K1 + K 2 e 2W + 6V STEP 3: Determine output at 0+ (inductor current is constant) STEP 1: Form of the solution - iL ( 0 - ) = 3 A t <0 2W t t 4W t =0+ 3A + 6V - 2W + vO (0+) = -6V vO ( 0 + ) _ STEP 4: Find output in steady state after the switching 2W 4W 4W t >0 + 2W 6V - t =0 + 2H vO ( ¥ ) _ 2W 4W t >0 + 6V - + 2W 6V - + v (t ) O _ vO (¥) = 0 STEP 5: Find time constant after switch RTH 2W 2W L t= RTH RTH = 8W t = 0.25 s STEP 6: Find the solution K1 + K 2 = vO (0+) = -6V K1 = vO (¥) = 0 vO (t ) = -6e - t 0.25 ;t >0 v O ( t ) = -6e - 4 t ; t > 0 Pulse Response Step by Step PULSE RESPONSE WE STUDY THE RESPONSE OF CIRCUITS TO A SPECIAL CLASS OF SINGULARITY FUNCTIONS ì0 t < 0 u( t ) = í î1 t > 0 CURRENT STEP VOLTAGE STEP TIME SHIFTED STEP LEARN BY DOING PULSE SIGNAL i (t ) = 10[u( t ) - u(t - 0.01)](mA ) v (t ) = -10[u(t - 1) - u(t - 2)](V ) PULSE AS SUM OF STEPS NONZERO INITIAL TIME AND REPEATED SWITCHING t dx + x = fTH ; x ( t0 + ) = x0 dt x (t ) = e - t -t0 t 1t x (t0 ) + ò e tt t-x t fTH ( x )dx 0 x(t ) = K1 + K 2e - t -t0 t ; t ³ t0 RESPONSE FOR CONSTANT SOURCES This expression will hold on ANY interval where the sources are constant. The values of the constants may be different and must be evaluated for each interval The values at the end of one interval will serve as initial conditions for the next interval LEARNING EXAMPLE FIND THE OUTPUT VOLTAGE vO ( t ); t > 0 t > 0.3 Þ v (t ) = 0 to = 0.3 v (0.3+ ) = 4(1 - e o " " v O ( t ) = K1 + K 2 e m - ( t - 0.3) t' 0.3 0.4 ) t ' = 0.4 " " vo (¥) = 0 Þ K1 = 0 K 2 = vo (0.3+ ) = 2.11(V ) vo (t ) = 2.11e t < 0 Þ v ( t ) = 0 Þ vO ( t ) = 0 vO ( 0 + ) = 0 t t > 0 Þ v (t ) = 9V ' ' v o ( t ) = K1 + K 2 e t t = RTH C = (6k || 12k ) ´ 100 m F = 0.4 s 8 ` ' vo (¥ ) = (9) = K1' vo (0+ ) = K1 + K 2 = 0 10 + 8 t æ ö ç 1 - e 0 .4 ÷ vo ( t ) = 4 ç ÷ è ø - - t - 0.3 0.4 ; t > 0.3 EXAMPLE FOR 0 < t < 0.5 (switch at b) to = 0 THE SWITCH IS INITIALLY AT a. AT TIME t=0 IT MOVES TO b AND AT t=0.5 IT MOVES BACK TO a. FIND vO (t ) , t > 0 12V b+ vS (t ) 20m F - + t ' K 2e t ' v (0+ ) = 12[V ] = K1' + K 2 vO (¥ ) = 0 = K1' t = (10kW)(20 m F ) = 0.2 s 10kW a + - vO ( t ) = K1' vO ( t ) = 12e + vO - t 0.2 , 0 < t < 0.5 FOR t > 0.5 (switch at a ) to = 0.5 - v s (t ) vO (0.5+ ) = vO (0.5-) = 12e 12 " " v O ( t ) = K1 + K 2 e - - 0.5 0.2 = 0.985 ( t - 0.5) t' " " vO (0.5+ ) = 0.985 = K1 + K 2 " vO (¥ ) = 12 = K1 0.5 t ( s) " K 2 = 0.985 - 12 = -11.015 Piecewise constant source ON EACH INTERVAL WHERE THE SOURCE IS CONSTANT THE OUTPUT IS OF THE FORM vO (t ) = K1 + K 2e - t -tO t vO ( t ) = 12 - 11.015e - t - 0.5 0.2 , t > 0.5 The constants are determined in the usual manner USING MATLAB TO DISPLAY OUTPUT VOLTAGE %pulse1.m % displays the response to a pulse response tmin=linspace(-0.5,0,50); %negative time segment t1=linspace(0,0.5,50); %first segment t2=linspace(0.5, 1.5,100); %second segment vomin=12*ones(size(tmin)); vo1=12*exp(-t1/0.2); %after first switching vo2=12-11.015*exp(-(t2-0.5)/0.2); %after second switching plot(tmin,vomin,'bo',t1,vo1,'rx',t2,vo2,'md'),grid title('OUTPUT VOLTAGE'), xlabel('t(s)'),ylabel('Vo(V)') First Order ...
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## This note was uploaded on 12/01/2011 for the course EE 2120 taught by Professor Aravena during the Fall '08 term at LSU.

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