MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.CV ( t )+-sRRC( a )( b )1122voH ( s ) = V ( s )V ( s )os( c )ZZ12+-s( a )( b )V ( s )( c )� Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 71 Reading: Nise: Sec. 2.4 • 1 Transfer Function Generation by Simplification Example 1 Find the transfer function for a “lead-lag” compensator Draw as where 1 R1Z1 = C1s � R1 = R1C1s + 1 1 R2Z2 = C2s � R2 = R2C2s + 1 1copyright cD.Rowell 2008 7–1CLI ( s )1R2RV ( s )Lii12( a )ZZ12( a )ii12IIUsing the voltage divider formed by Z1 and Z2 Z2Vo = VsZ1 + Z2 (R2CR22 s+1 ) = Vs (R1CR11 s+1 ) + ( R2CR22 s+1 ) R2R1C1s + R2 = VsR1R2(C1 + C2)s + (R1 + R2) Vo(s) R2R1C1s + R2H(s) = = Vs(s) R1R2(C1 + C2)s + (R1 + R2) Example 2 Find the transfer function VL(s)H(s) = I(s) for the circuit: We note that VL(s) = ZLi2(s) = Lsi2(s) Combine elements to let 1 1 Z1 = = R1 + Y1 Cs 1 Z2 = = R2 + Ls Y2 and use the current divider relationship at node (a). 7–2C1R2RLL12+-VY2i2(s) = Y1 + Y2 where 1 sC Y1 = = Z1 sCR1 + 1 1 1 Y2 = = . Z2 sL + R2 Then R1Cs + 1 i2(s) = I(s)Cs(R2 + Ls) + (R1Cs + 1) and LCR1s2 + Ls VL(s) = Lsi2(s) = I(s)LCs2 + C(R1 + R2)s + 1 so that the transfer function is VL(s) LCR1s2 + Ls H(s) = = I(s) LCs2 + C(R1 + R2)s + 1 2 Transfer Function Generation through Mesh (Loop) Currents This method expresses the system dynamics as a set of simultaneous algebraic equations in a set of internal mesh (or loop) currents. It is useful for complex circuits containing a voltage source. The following example sets out the method in a series of steps. Example 3 Find the transfer function of a “bridged-T” filter: Note: This is a difficult example to solve using impedance reduction methods. It is, however, well suited to the mesh current method. Draw the system as an impedance graph: 7–3ZZZZZ21534V132Define a set of (clockwise) loops as shown, ensuring that every graph branch is covered by at least one loop. The loops 1, 2 and 3 are somewhat arbitrary. We assume hypothetical continuous mesh currents i1, i2, and i3 that flow around each loop. Step 1: Write loop (compatibility) equations for each loop (using the arrows on the graphs branches to define the direction of the voltage drop): VZ1 + VZ3 − V = 0 VZ2 + VZ4 − VZ3 = 0 VZ5 + VZ2 − VZ1 = 0 Step 2: Define the mesh currents i1, i2, and i3 and write the current in each branch in terms of the mesh currents (use the arrows on the loops to define the signs): iZ1 = i1 − i3 iZ2 = i2 − i3 iZ3 = i1 − i2 iZ4 = i2 iZ5 = i3 Step 3: Write the mesh equations in terms of the mesh currents Z1(i1 − i3) + Z3(i1 − i2) = V Z2(i2 − i3) + Z4i2 − Z3(i1 − i2) = 0 Z5i3 − Z2(i2 − i3) − Z1(i1 − i3) = 0 which is a set of 3 simultaneous algebraic equations in the loop currents i1, i2, and i3. Step 4: We note that the output is VZ4 = i2Z4, therefore solve for i2 and create the transfer function in terms of the impedances. 7–4V+-RRC12voVZ1Z2Z312Example 4 Use the mesh current method to find the transfer function Vo(s)H(s) = V (s) in the “lead” network: Let Z1 = 1/Cs, Z2 = R1, and Z3 = R2. Step1: The loop equations are: VZ2 + VZ3 − V = 0 VZ1 − VZ2 = 0 Step 2: The mesh currents are i1 and i2, and iZ1 = i2 iZ2 = i1 − i2 iZ3 = i1 Step 3: Rewrite the loop equations Z2(i1 − i2) + Z3i1 = V Z1i2 − Z2(i1 − i2) = 0 or (Z2 + Z3)i1 − Z2i2 = V −Z2i1 + (Z1 + Z2)i2 = 0 In matrix form � Z2 + Z3 −Z2 �� i1 � � V � = −Z2 Z1 + Z2 i2 7–5 0LR2C1CL1 1 0 v a cp o w e r s u p p l y / r e c t i f i e rf i l t e rl o a d231tttv ( t )32v ( t )1v ( t )1A t:2A t:3A t:r i p p l eStep 4: Vo = i1Z3, so we solve for i1 (using any method). Using Cramer’s Rule: ����V −Z2 ����0 Z1 + Z2 Z1 + Z2i1 = = V���Z2 + Z3 −Z2 ���Z1Z2 + Z1Z3 + Z2Z3 � −Z2 Z1 + Z2 � so that (Z1 + Z2)Z3Vo = V Z1Z2 + Z1Z3 + Z2Z3 and Vo(s) (R1 + 1/Cs)R2H(s) = = V (s) R1/Cs + R2/Cs + R1R2 R1R2Cs + R2H(s) = R1R2Cs + (R1 + R2) Note that this problem could have be done using a voltage-divider approach Example 5 A common full-wave wave rectified dc power supply for electronic equipment is: Typical waveforms in the circuit are 7–6LR2CCLT h e v e n i n s o u r c e m o d e lp- s e c t i o n f i l t e rl o a d+-VR1'( a )( b )( c )( d )vo1ZZZZ2143V( a )( b ) ( c )( d )ZZZZ2143V12The filter acts to “smooth” the full-wave rectified waveform at (2) to produce a dc output at (3) with a very much reduced “ripple”. We use a linearized model of the transformer/rectifier circuit, with a voltage source (with a waveform as at (2) above) and a series resistor R (a Th´evenin source) - see Lecture 8 - as below: The task is to find the transfer function Vo(s)H(s) = . V (s) Solution: Combine series and parallel impedances to simplify the structure. Draw as an impedance graph where Z1 = R1 Z2 = sL 1 RL/sC2Z3 = sC2 � RL = RL + 1/sC2 1 Z4 = sC1 The system output is VZ3 = iZ3 Z3. Choose mesh loops to contact all branches as below. 7–7The loop equations are: vZ1 + vZ4 − V = 0 vZ2 + vZ2 − vZ4 = 0 and the branch currents (in terms of the mesh currents) are: iZ1 = i1 iZ2 = i2 iZ3 = i2 iZ4 = i1 − i2 Rewrite the loop equations in terms of the mesh currents: Z1i1 + Z4(i1 − i2) = V Z2i2 + Z3i2 − Z4(i1 − i2) = 0 or (Z1 + Z4)i1 − Z4i2 = V −Z4i1 + (Z2 + Z3 + Z4)i2 = 0 giving a pair of simultaneous algebraic equations in the mesh currents. Solve for iZ3 = i2 � Z1 + Z4 −Z4 � � i1 � = � V � −Z4 Z2 + …
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