3 Electrical Circuits 3. Basic Concepts Electric charge coulomb of negative change contains 624 0 8 electrons. Current ampere is a steady flow of coulomb of change pass a given point in a conductor in second. (amperes) (coulombs) (3.) (seconds) Time varying current ( + ) () () lim 0 (3.2) Voltage Voltage dierence between two points is the work in joules required to move coulomb of charge from one point to the other. (volts) (joules) (3.3) (coulombs) Power The rate at which something either absorbs or produces energy. The power absorbed by an electric element is the product of the voltage and the current. (Watts) (joules) (volts) (amperes) (3.4) (seconds) 3.2 Circuit Elements Ideal voltage source 7
Sec 3.2. Circuit Elements Resistor Capacitor Inductor i R vr i C v i C v Current Source Voltage Sourse DC Voltage Sourse GND i v 5V Figure 3.: Basic circuit elements. v v i (pure AC) v I V I (pure DC) v I V I + v i t Figure 3.2: DC signals versus AC signals. Voltage source provides a specified voltage across the two terminals and does not depend on the current flowing through the source. The output impedance of ideal voltage source is zero. The current flowing through an ideal voltage source is completely determined by the circuit connected to the source. Ideal current source Current source provides a specified current and does not depend on the voltage across the source. The output impedance of ideal current source is infinite. The voltage across an ideal current source is completely determined by the circuit connected to the source. DC signals v.s. AC signals Figure 3.2 shows the definitions of DC signals and AC signals. Resistor Resistance is the property of materials that resists the movement of electrons. 8
ecture 3. Electrical Circuits Ohm s aw (volts) (ohms) (amperes) Conductance is the inverse of resistance. For parallel resistors, (3.5) + 2 + 3 + (3.6) When two resistors are connected in parallel, the equivalent resistance is smaller than any of the two resistors. 2 + 2 2 + 2 2 (3.7) For series resistors, + 2 + 3 + (3.8) Capacitor Capacitance is the ability of a capacitor to store charges on its two conductors. (farad) (coulombs) (volts) (3.9) For time-varying voltage () across the capacitor The capacitor is an open circuit, i.e., () 0,when () isaconstant. The voltage across the capacitor can not jump. However, the current flowing through the capacitor does not have such constraint. () lim 0 () (3.0) For parallel capacitors, () Z () (3.) + 2 + 3 + (3.2) For series capacitors, Inductor + 2 + 3 + (3.3) 9
Sec 3.3. Circuit aws For time-varying current () flowing through the inductor Theinductorisashortcircuit, i.e., () 0,when () isaconstant. The current flowing through the inductor can not jump. However, the voltage across the inductor does not have such constraint. () (flux linkages) lim 0 lim 0 lim ( + ) () 0 () (3.4) For parallel inductors, () Z () (3.5) For series capacitors, + 2 + 3 + (3.6) + 2 + 3 + (3.7) Dual circuit Capacitor and Inductor formulas are the same except that the symbols dier. Capacitor () () + () (3.8) Inductor () () + () (3.9) 3.3 Circuit aws Kirchho s voltage law ( KV ) The algebraic sum of the voltages around any closed loop of a circuit is zero. Kirchho s current law ( KC ) The algebraic sum of the currents entering every node must be zero. 20
ecture 3. Electrical Circuits i S vo v i O S C R R Current Source Parallel v C R Voltage Source Serial i R Figure 3.3: Similarity between capacitor and inductor. 3.4 Network Theorems Definition 3. inear circuit is formed by interconnecting the terminals of independent sources, controlled sources, and linear passive elements to form one or more closed paths. inear passive elements include Resistor, Capacitor, andinductor. The characteristics of these elements satisfy the conditions of linearity. Resistor () () 2 () 2 () (3.20) ()+ 2 () ( ()+ 2 ()) Capacitor () ( ()) 2 () ( 2()) (3.2) ()+ 2 () ( ()+ 2 ()) Inductor () 2 () ()+ 2 () Z 0 Z 0 Z 0 () 2 () (3.22) ( ()+ 2 ()) Theorem 3.2 In a linear network containing multiple sources, the voltage across or current 2
Sec 3.4. Network Theorems R R2 V A I R3 A V R R2 (a) V A R 3 V + R R2 (b) I R A 3 V 2 (c) Figure 3.4: Example of linear circuit with multiple independent sources. through any passive element may be found as the algebraic sum of the individual voltages or currents due to each of the independent sources action along, with all other independent sources deactivated. Voltage source is deactivated by replacing it with a short circuit. Current source is deactivated by replacing it with an open circuit. Controlled sources remain active when the superposition theorem is applied. Example 3.3 Given the circuit in Figure 3.4, find the voltage across the resistor 3 using the superposition theorem of linear network.. The voltage across the resistor 3 is the superposition of the voltage when each independent source actions alone, as shown in Figure 3.4 (b) and (c). + 2 (3.23) 2. The contribution of the voltage source The current source is replaced with an open circuit. 3. The contribution of the current source 3 + 2 + 3 (3.24) 22
ecture 3. Electrical Circuits The voltage source is replaced with an short circuit. 2 3 + 2 + 3 (3.25) 3.4. Equivalent Circuits of One-Port Networks inear Network A 2 (a) inear Network B Z TH V TH 2 (b) inear Network B I N YN 2 inear Network B (c) Figure 3.5: Equivalent circuits. (a) The original circuit. (b) Thevenin s equivalent. (c) Norton s equivalent. Equivalent circuit Areductionof a complex linear circuit into a simpler form. Amodelof a complex linear circuit contained in a black box. Theorem 3.4 Thevenin s theorem states that an arbitrary linear, one port network such as network A in Figure 3.5 (a) can be replaced at terminals 2 with an equivalent seriesconnected voltage source and impedance as in Figure 3.5 (b). is the open-circuit voltage of network A at terminals 2. is the ratio of the open-circuit voltage over short circuit current determined at terminals 2. The equivalent impedance looking into network A through terminals 2with all independent sources deactivated. Voltage sources are replaced by short circuits. Current sources are replaced by open circuits. 23
Sec 3.4. Network Theorems Theorem 3.5 Norton s theorem states that an arbitrary linear, one port network such as network A in Figure 3.5 (a) can be replaced at terminals 2 with an equivalent parallelconnected current source and admittance as in Figure 3.5 (c). is the short-circuit current flowing through terminals 2duetonetworkA. is the ratio of short-circuit current over open-circuit voltage at terminals 2. Conversion of equivalent circuits. Any method for determining is equally valid for finding (3.26) (3.27) (3.28) Example 3.6 In Figure 3.6, 4, 2, 2, 2 3, find the Thevenin s equivalent circuit and Norton s equivalent circuit for the network to the left of terminals 2 R R2 V A I A 2 Figure 3.6: Examples of Thevenin s and Norton s equivalent circuits.. Thevenin s equivalent is the open-circuit voltage at terminals 2 + 4+48 (3.29) is the ratio of the open-circuit voltage over short circuit current determined at terminals 2 with network B disconnected. By the superposition of the short-circuit current caused by and the 24
ecture 3. Electrical Circuits short circuit current can be found. 2 ( )+( ) +2 + +2 (3.30) 8 5 By definition, can be derived as 2 5. Alternatively, can be found as the equivalent impedance for the circuit to the left of terminals 2 is replaced with short circuit. is replaced with open circuit. + 2 5 (3.3) 2. Norton s equivalent is the short-circuit current at terminals 2 whichcanbederivedasineq. (3.30). is the ratio of the short-circuit current over the open-circuit voltage with network B disconnected. From Eq. (3.29), the open circuit voltage is 8. Thus, 8 8 5 5 Alternatively, 5 02 3.4.2 Equivalent Circuits of Two-Port Networks A two-port network is an electrical circuit or device with two pairs of terminals. Figure 3.7 depicts a two-port linear network. Only two of the four variables, 2,, 2 can be independent. I I 2 + V - inear + V2 Network - I I 2 Figure 3.7: Two-port linear network. Characterization of Two-Port Networks parameters (, 2 depend on, 2 ) 25
Sec 3.4. Network Theorems The four parameters represent impedance. 2 and 2 are transfer impedances. Each of the parameters can be evaluated by open-circuiting an appropriate port of the network. + 2 2 2 2 + 22 2 (3.32) parameters in matrix form. 2 0 2 2 0 2 2 2 0 (3.33) 2 2 0 2 " # " #" # 2 (3.34) 2 2 22 2 parameters (, 2 depend on, 2 ) The four parameters represent admittance. 2 and 2 are transfer admittances. Each of the parameters can be evaluated by short-circuiting an appropriate port of the network. + 2 2 2 2 + 22 2 (3.35) 2 0 2 2 0 2 2 2 0 (3.36) 2 2 0 2 26
ecture 3. Electrical Circuits parameters in matrix form. " # " 2 #" # 2 2 22 2 (3.37) Relation between parameters and parameters. " # " # " 2 2 22 2 2 22 2 22 22 2 2 2 parameters (, 2 depend on, 2 ) The represents impedance. The 22 represents admittance. Parameters and 2 are obtained by short-circuiting port 2. Parameters 2 and 22 are obtained by open-circuiting port. # (3.38) + 2 2 2 2 + 22 2 (3.39) 2 0 2 2 0 2 2 2 0 (3.40) 22 2 0 2 parameters in matrix form. " 2 # " 2 2 22 #" 2 # (3.4) Example 3.7 In Figure 3.8, 0, 2 6, find the parameters and the parameters for the network 27
Sec 3.4. Network Theorems I R I 2 + + V - 0.3I a I a R 2 V 2 - I I 2 Figure 3.8: Example of two-port linear network. From Eq. (3.33), the parameters can be obtained as follows. 2 0 (0 + 6) 6 23 +03 3 2 0 6 03 0 3 2 +03 3 2 2 2 0 22 2 2 0 6 6 +03 3 462 6 6 462 +03 3 23 From Eq. (3.40), the parameters can be calculated as follows. 2 0 0 2 0 2 03 2 2 2 2 2 0 22 2 2 0 +03 2 Equivalent Circuits of Special Two-Port Networks 3 6 05 3 6 027 T-Model network. (Figure 3.9 (a)) 2and 3 can be derived from the parameters of a two-port networks. " # " # 2 +3 3 (3.42) 2 22 3 2+3 -Model network. (Figure 3.9 (b)) 28
ecture 3. Electrical Circuits and can be derived from the parameters of a two-port networks. " # " # 2 + (3.43) 2 22 Transformation from -Model to T-Model. + 2 3 + + + + + + (3.44) Transformation from T-Model to -Model. 3+2+23 2 3+2+23 3+2+23 3 (3.45) I I 2 Z Z2 V Z3 V 2 (a) T-Model Network I Zc I 2 V Za Zb V2 (b) Model Network Figure 3.9: Equivalent circuits of two port networks. (a) T-Model Network. (b) Pi- Model Network. 29
Sec 3.5. aplace Transforms 3.5 aplace Transforms aplace Transform A mathematical tool for representing system transfer functions of causal TI systems. A generalization of Fourier transform. The result of Fourier transform can be obtained by evaluating the aplace transform along the imaginary axis. Definition 3.8 Given () with () 0for 0, its aplace Transform, which is defined as follows, is a complex function over a complex-number domain. () [()], Z 0 () Z 0 () (3.46) Definition 3.9 Correspondingly, the Inverse aplace Transform is as follows: () [ ()] I 2 () (3.47) Example 3.0 Given () &() (() () 0for 0) and their aplace Transforms () &(),the following shows the properties of aplace Transform. inearity () ()+() () ()+() (3.48) Time Derivatives () () () () (0 ) (3.49) () () Time Integrals () () (0 ) 2 (0 ) (0 ) () Z 0 () + (0 ) () () + (0 ) (3.50) Time Scaling () () ( ) where 0 (3.5) Time Delay () ( 0 ) () 0 () where 0 0 (3.52) 30
ecture 3. Electrical Circuits Multiplication () () () () () (3.53) Shift Convolution Product () () () () () () ()() () ( ) (3.54) () ()() (3.55) () I 2 ()( ) (3.56) Example 3. aplace Transform Pairs () 2 cos ( 0 + ) () () (! ) + cos 0 2 + 2 0 0 sin 0 2 + 2 0 ( + 0 ) + ( 0 ) 2 cos ( 0 tan ) ( ) 2 + 2 0 3.6 Equivalent Circuits in aplace Domain Capacitor characteristic in aplace domain. Admittance of capacitor in aplace domain is Footnote () () () ( () (0 )) (3.57) Figure 3.0 (a) depicts the KC equivalent circuit in aplace domain. characteristic in aplace domain. 3
Sec 3.6. Equivalent Circuits in aplace Domain I C (s) / sc I C (s) V (s) C sc Cv( 0 ) C V C (s) v C (0 s ) (a) (b) I C (s) s I ( s) (s) V / s i (0 ) s (s) V i(0 ) (c) (d) Figure 3.0: (a) KC equivalent circuit of a capacitor in aplace domain. (b) KV equivalent circuit of a capacitor in aplace domain. (c) KC equivalent circuit of an inductor in aplace domain. (b) KV equivalent circuit of an inductor in aplace domain. Impedance of capacitor in aplace domain is () Z Z 0 () (3.58) () + (0 ) () ()+ (0 ) Figure 3.0 (b) depicts the KV equivalent circuit in aplace domain. Inductor characteristic in aplace domain. Admittance of inductor in aplace domain is () Z Z 0 () (3.59) () + (0 ) () ()+ (0 ) Figure 3.0 (c) depicts the KC equivalent circuit in aplace domain. characteristic in aplace domain. Impedance of inductor in aplace domain is () () () ( () (0 )) (3.60) Figure 3.0 (d) depicts the KV equivalent circuit of an inductor in aplace 32
ecture 3. Electrical Circuits domain. 3.7 System Transfer Function in Time and aplace Domains Two ways to deduce the system transfer function in time and aplace domains. Start in time domain and then take the aplace transform. Start in aplace domain and use Inverse aplace transform to return to time domain. Example 3.2 Consider the RC circuit in Figure 3., the input is the current source () and the output () is the voltage across the RC components. Find out the system transfer function and the outputs w.r.t. dierent inputs. is R C vo Figure 3.: The RC circuit to be solved in time domain and aplace domain. By writing the node equation with KC in time domain, we obtain the system equation. () () + () (3.6) By taking aplace transform to both sides, the system equation in aplace domain can be derived. () () + ( () (0 )) (3.62) The system transfer function in aplace domain. () () () (0 )0 +() + (3.63) 33
Sec 3.7. System Transfer Function in Time and aplace Domains Initial Value (Driving Free) Response Given () 0 () 0, the output can be obtained from Eq. (3.62). 0 () + ( () (0 )) () (0 )) +() (3.64) The time domain response can be calculated by taking Inverse aplace transform. () (0 ) (3.65) Impulse Response Given () (); (), the output can be obtained from Eq. (3.63). For simplicity, we do not consider initial condition, i.e., (0 )lim 0 () 0. () (3.66) + The time domain response can be calculated by taking Inverse aplace transform. () (3.67) Step Response Given () (); (), the output can be obtained from Eq. (3.63). For simplicity, we do not consider initial condition, i.e., (0 )lim 0 () 0. () ( +()) (3.68) +() The time domain response can be calculated by taking Inverse aplace transform. The capacitor is like an open circuit once it is fully charged. () ( ) (3.69) Sinusoidal Response Given () cos( 0 ) (); () ( 2 + 2 0), the output can be obtained from Eq. (3.63). For simplicity, we do not consider initial condition, i.e., (0 )lim 0 () 0. () + + + (3.70) 2 + 2 0 + 2 + 2 0 The time domain response can be calculated by taking Inverse aplace transform. 34
ecture 3. Electrical Circuits The term ( + ) stands for transient response, which can be transformed into exponentially decaying time function. 2 +( 0 ) 2 The term (+ ) ( 2 + 2 0) represent the steady state response, which shall yield the time-domain signal. + 2 + 2 0 2 + 2 0 + 2 + 2 0 r ³ 2 0 2 + 0 tan ( μ ) 0 ³ + () cos ( 0 )+ 0 sin ( 0 ) 0 2 ³ ³ μ 2 2 0 + 0 0 ', ' 0 () cos ( 0 ) 0 cos ( 0 ) (3.7) At low frequency, the capacitor is similar to an open circuit. 0 À ' 0, ' () sin ( 0 0) cos 0 0 2 At high frequency, the capacitor is similar to a short circuit. The circuit acts as a low-pass filter. Example 3.3 Consider the RC circuit in Figure 3.2, the input is the current source () and the output () is the voltage across the RC components. Find out the system transfer function and the outputs w.r.t. dierent inputs. is R C vo Figure 3.2: The RC circuit to be solved in time domain and aplace domain. Start in Time Domain 35
Sec 3.7. System Transfer Function in Time and aplace Domains i S (t) i R (t) i C (t) i (t) R C v o (t) Figure 3.3: Time domain analysis. By writing the node equation with KC in time domain, we obtain where () ()+ ()+ () (3.72) () () () () () Z () Z 0 () + (0 ) (3.73) The system equation in time domain. () () + () + Z Apply the aplace transform to both sides of Eq.(3.74). 0 () + (0 ) (3.74) () () + ( () (0 )) + ()+ (0 ) (3.75) where (), [ ()] and () [ ()]. The system transfer function in aplace domain. Assume no initial condition, i.e., (0 ) lim 0 () 0and (0 ) lim 0 () 0. () () ( + + ) (3.76) Start in aplace Domain 36
ecture 3. Electrical Circuits R s I S (s) sc (0 ) V o (s) Cv( 0 ) o i s Figure 3.4: aplace domain analysis. Replace all the circuit elements by their KC equivalent circuits in aplace domain. () (0 ) (3.77) ()+ (0 ) () ()+ ()+ () () + () (0 ) μ + ()+ (0 ) (3.78) The system transfer function. Impulse Response () ()+ ()+ () (0 )0 (0 )0 () + ()+ () () ( + + ) () (3.79) Impulse response is the system output w.r.t an input signal of impulse function. It is also the transfer function of the system which can be used to find system poles and zeros. () (); () Assume no initial condition. () ( + + ) () ( + + ) (3.80) 2 + + 37
Sec 3.7. System Transfer Function in Time and aplace Domains System poles and zeros are the roots of denominator and numerator of the system transfer function. Zeros, the roots of (). Poles, the roots of (). () () (), () () (3.8) The simplest way to obtain () is by analyzing the system impulse response in aplace domain. () () () 2 + + () () (3.82) The time-domain expression of (), [ ()] depends on the nature of the system poles, which are the roots of denominator. et the system characteristic equation be equal to 0. Rewrite it as The roots are 2 + + 0 (3.83) 2 + + 0with,, (3.84) and 0 ± 2 4 2 r 2 ± ( 2 )2 4 (3.85) 4 2 r 2 ± ( 2 )2 Thenatureoftheroots(, 0 ) shall depend on the value of the critical expression. 2, ( 2 )2 ( 2 )2 ( ) (3.86) ( 2 )2 38
ecture 3. Electrical Circuits Impulse Response of RC Circuit 0.8 0.6 0.4 vo(t) 0.2 0 2 4 6 8 0 2 4 0.2 0.4 Underdamped Criticall Damped Overdamped t Figure 3.5: Comparison of system impulse responses in dierent cases. Traditionally, we define two more parameters. Neper frequency, 2 2 Resonant frequency 0 (3.87) 0, r (3.88) We can now rewrite the roots of the characteristics equation. and 0 ± q 2 2 0 (3.89) Three distinct cases are possible with respect to and 0 of and 0. Overdamped, p 2 2 0 is real, i.e., 2 2 0 and 0 are two real, distinctand negative values. () can be factorized as follows. ( 0 )and ( 0 ) depending on the values 39
Sec 3.7. System Transfer Function in Time and aplace Domains Both and are real numbers. () ( )( 0 ) + 0 + + + 0 (3.90) By taking the inverse aplace transform, the response of () (withoutinitial condition) can be formulized as follows: () + 0 (3.9) Figure 3.5 depicts the () in the overdamped case. Underdamped, p 2 2 0 is imaginary, i.e., 2 2 0 and 0 are two complex conjugate poles. The damped resonant frequency, p 2 0 2 is real. and 0 ± (3.92) () can be factorized as follows. and are complex conjugates. () ( ( + )) ( ( )) ( + ) + ( ) (3.93) (3.94) Taking inverse aplace transform, 0 () canbeobtained. Note that ( + ) and( ) are real number since and are complex conjugates. () (+) + ( ) + [ (cos + sin )+ (cos sin )](3.95) [( + )cos + ( )sin ] Figure 3.5 depicts the () in the underdamped case. Critically damped, p 2 2 00, i.e., 2 2 0 40
ecture 3. Electrical Circuits j Critically damped - s P s P ' 0 Overdamped Underdamped +j -j Figure 3.6: ocations of poles and zeros. and 0. () ( ()) 2 + + Taking inverse aplace Transform, 0 () canbeobtained. ( + ) 2 (3.96) 0 () + (3.97) Figure 3.5 depicts the 0 () in the case of critically damped. Summary The positions of poles determine the nature of System Response. and 0 ± 2 2 with 2 and 0! Response is overdamped if 2 2 0 0and, 0 are two distinct real numbers. Response is underdamped if 2 2 0 0and, 0 are complex conjugate numbers. Response is critically damped if 2 2 0 and, 0 are real and the same. Thepositionsofsystempolesmovealongthepath(knownasRootocus) shown in the following diagram. Sinusoidal Response Sinusoidal response is the system output w.r.t an input signal of sinusoidal function. () cos() () or () [ ()] ( 2 + 2 ) 4
Sec3.8.BodePlots For simplicity, we do not consider initial condition. () 2 ( 2 + 2 )( 2 + + ) 2 ( )( 0 )(2 + 2 ) ( ) + ( 0 (3.98) ) + + ( 2 + 2 ) Instead of solving for the coecients analytically by hand (which is a practically impossible task), we shall deduce the time and frequency domain responses by reasoning. Time-Domain Response The terms ( )and( 0 ) stand for transient response. Both can be transformed into damped response in time with exponentially decaying envelop. where (3.99) 2 The term ( + ) ( 2 + 2 ) represent the steady state response, which shall yield the time-domain signal 0 cos( ). 0 2 + 2 tan ( ) + ( 2 + 2 ) ( 2 + 2 ) + ( 2 + 2 ) cos()+ sin() 0 cos( ) (3.00) 3.8 Bode Plots A graph used to show the frequency response of an TI system. To plot the magnitude in decibels (db) and use a log scale for The log scale helps to compress a wide range of data. Both the magnitude and the phase of the transfer function versus the angular frequency System transfer function () can be written as a product of factors of the following items.. Constant factor 2. Poles or zeros at the origin, ± 3. Real poles or zeros, (+) ± 4. Complex-conjugate poles or zeros, ( 2 2 +2+) ±,where is the damping 42
ecture 3. Electrical Circuits ratio and 0 () () (3.0) () The magnitude of () in db allows us to plot the factors individually and sum the results to obtain the complete plot. Given () asfollows, () 20log 0 () (3.02) () ( + )( + 2 ) ( + )( + 2 ) (3.03) () can be obtained by plotting each factor individually () 20log 0 () 20log 0 ( + )( + 2 ) ( + )( + 2 ) (3.04) 20log 0 +20log 0 + +20log 0 + 2 20 log 0 + 20 log 0 + 2 Bode plot for each of the factors.. Constant factor,. Magnitude The magnitude 20 log 0 is a constant. Phase The phase is a constant and equal to 0 ( 0 )or±80 ( ± ) depending on whether is positive or negative, respectively. 2. Poles or zeros at the origin, ± Magnitude The magnitude is a straight line that intersects the axis (0dB) at and has a slope of ±20dB/decade. Phase The phase is a constant and equal to ±90. () () ± ( 2 ) ± (3.05) () 20log 0 ± ±20 log 0 (3.06) 43
Sec3.8.BodePlots 3. Real poles or zeros, (+) ± Poles or zeros locate at ]() ±(2) ±90 (3.07) () ( +) ± ± (3.08) () 20log 0 μq +() 2 ± ±0 log 0 +() 2 (3.09) Magnitude ow frequency response ( ; ): A horizontal line of 0dB () ±0 log 0 () 0 db (3.0) High frequency response ( À ; À ): A straight line of slope ±20dB/decade that intersects 0dB when () ±0 log 0 () 2 μ ±20 log 0 μ ±20 log 0 log 0 (3.) Corner frequency response () ±0 log 0 2±3dB (3.2) Phase () 0 ±320 07079 (negative) or 425 ( positive ) (3.3) ]() ± ± tan (3.4) ow frequency: ]() 0 Corner frequency: ]() ±45 High frequency: ]() ±90 4. Complex conjugate poles or zeros, ( 2 2 +2 +) ± For simplicity, we consider only the case of a single pair of complex conjugate poles. 44
ecture 3. Electrical Circuits If the poles are repeated by times, all coordinates on the curves will be multiplied by. If we have zeros instead of poles, curves are mirror images through the axis. () (3.5) 2 2 +2 + () 20 log 0 ³ 2 2 2 +4 2 2 2 2 0 log 0 ³ 2 2 2 +4 2 2 2 (3.6) Magnitude ow frequency response ( ; ): A horizontal line of 0dB High frequency response ( À ; À ): A straight line of slope ±40dB/decade that intersects 0dB when ³ 2 () 0 log 0 2 2 2 +4 2 2 ³ 2 ' 0 log 2 0 2 2 +4 2 2 ' 40 log 0 () μ 40 log 0 log 0 (3.7) Phase 0 ]() tan ]() 80 + tan 2 2 2 (3.8) 2 2 2 (3.9) Example 3.4 Consider the transfer function () ( + ) where k and F, express its frequency responses including both the magnitude and the phase responses with Bode Plot. () can be firstly written as follows. () + (3.20) The factor Magnitude is a constant and equal to 20 log 0 0 3 60dB. Phase is a constant and equal to 0 The factor ( + ) 45
Sec3.8.BodePlots Magnitude is a constant and equal to 0dB. Phase is around 0 Magnitude is a straight line of slope 20dB/decade. Phase is around 45 when Phase is around 90 when À 60 Bode Diagram Magnitude (db) 50 40 30 20 0 Phase (deg) -45-90 0 0 2 0 3 0 4 0 5 Frequency (rad/sec) Figure 3.7: Bode plot for the transfer function () ( + ) Example 3.5 Consider the transfer function () ( + ) where k and F, express its frequency response including both the magnitude and the phase responses with Bode Plot. () can be firstly written as follows. () + (3.2) The factor Magnitude is a constant and equal to 20 log 0 0 3 60dB Phase is a constant and equal to 0 The factor Magnitude is a straight line of slope 20dB/decade and intersects the -axis at Phase is 90. 46
ecture 3. Electrical Circuits The factor ( + ) Magnitude is a constant and equal to 0dB. Phase is around 0 Magnitude is a straight line of slope 20dB/decade. Phase is around 45 when Phase is around 90 when À 0 Bode Diagram Magnitude (db) -0-20 -30-40 90 Phase (deg) 60 30 0 0 0 2 0 3 0 4 0 5 Frequency (rad/sec) Figure 3.8: Bode plot for the system transfer function () ( + ) Example 3.6 Consider the transfer function () ( 2 + + ) where k, Fand, express its frequency response including both the magnitude and the phase responses with Bode Plot. () can be firstly written as follows. () ( 2 + + ) (3.22) 2 + + The factor Magnitude is a constant and equal to 20 log 0 0dB Phase is a constant and equal to 0 The factor Magnitude is a straight line that intersects the axis (0dB) at andhas a slope of 20dB/decade. 47
Sec3.8.BodePlots Phase is a constant and equal to 90. The factor ( 2 + +) Magnitude is a horizontal line of 0dB Phase is tan 2 2 2. À Magnitude is a straight line of slope 40dB/decade that intersects 0dB when Phase is 80 + tan 2 2 2 60 Bode Diagram 50 Magnitude (db) 40 30 20 0 0 90 Phase (deg) 45 0-45 -90 0 0 2 0 3 0 4 0 5 Frequency (rad/sec) Figure 3.9: Bode plot for the system transfer function () ( 2 + + ) 48