Chapter 4 and 5 - Signal Theory PDF
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This document provides information on Bode diagrams, transfer functions and other topics related to signal theory and circuit analysis. The document references concepts like Laplace transforms and frequency responses, as well as calculations for different circuit configurations.
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Chapter 4 : Bode Diagram and Chapter 5 (Part I) : ELEMENTS OF SIGNAL THEORY 1 Chapter 4: Bode Diagram I. Introduction Transfer functions and frequency response are essential tools for analyzing linear systems. They allow...
Chapter 4 : Bode Diagram and Chapter 5 (Part I) : ELEMENTS OF SIGNAL THEORY 1 Chapter 4: Bode Diagram I. Introduction Transfer functions and frequency response are essential tools for analyzing linear systems. They allow us to understand how a system behaves in different frequency ranges, simplifying complex mathematical relationships into more manageable forms. By converting time-domain signals to the frequency domain, we can easily predict system outputs, assess stability. This knowledge forms the foundation for understanding Bode plots and their applications in circuit analysis. System (Linear Invariant System) Input Output II. Transfer Functions Transfer function represents mathematical relationship between input and output of linear time-invariant system Laplace transform converts time-domain signals to complex frequency domain Input-output relationship expressed as ratio of output to input in Laplace domain Transfer function denoted as 𝐻(𝑠), where 𝑠 represents complex frequency variable Transfer function is the ratio of the amplitudes between input and output in the complex frequency domain 2 Considering all initial conditions to zero. Then the transfer function H(s)) of the plant is given as Y (s ) H (s ) U(s) H(s) Y(s) U (s ) 3 The transfer function 𝐻(𝑗 𝜔) of any system is a complex number. We have three main modes of representation: the BODE plane, the BLACK plane and the NIQUYST plane. Example 1 : Consider the following electrical circuit: 𝟏 𝑽𝒐𝒖𝒕 𝒋𝝎𝑪 𝟏 𝑯 𝒋𝝎 = = = 𝑽𝒊𝒏 𝟏 𝟏 + 𝒋𝝎𝑹𝑪 𝑹 + 𝒋𝝎𝑪 𝟏 = 𝟏 + 𝒋𝟐𝝅𝒇𝑹𝑪 𝟏 So the transfer function 𝐻(𝑠) with 𝒔 = 𝐣𝝎 is given by : 𝑯(𝒔) = 𝟏 + 𝒔𝑹𝑪 Example 2 : Consider the following electrical circuits: Transfert Function : 𝐿 𝑉𝑠 𝑗𝑅𝜔 𝑉𝑠 𝑗𝑅𝐶𝜔 𝐻 𝑗𝜔 = = 𝐻 𝑗𝜔 = = 𝑉𝑒 𝐿 𝑉𝑒 1 + 𝑗𝑅𝐶𝜔 1+𝑗𝑅𝜔 4 III. Frequency Response Frequency response 𝑯 𝒋𝝎 𝒐𝒓 𝑯(𝒇)describes system's steady-state output to sinusoidal input Represents system's behavior across different input frequencies Obtained by evaluating transfer function along imaginary axis (𝒔 = 𝒋𝝎) Gain indicates amplitude ratio of output to input at specific frequency Phase shift represents time delay between input and output signals Complex frequency combines real and imaginary parts (s = σ + jω) III.1. Gain and Phase Characteristics Gain measured in decibels (dB) or as magnitude ratio Gain calculation: 20 log₁₀|H(jω)| for voltage signals, 10 log₁₀|H(jω)| for power signals Phase shift measured in degrees or radians Phase calculation: ∠𝑯(𝒋𝝎) = tan⁻¹(𝑰𝒎*𝑯(𝒋𝝎)+ / 𝑹𝒆*𝑯(𝒋𝝎)+) Gain and phase plots provide visual representation of system's frequency response Bode plots display gain and phase separately on logarithmic frequency scale 5 The cut-off pulsation ωc or The cut-off frequency 𝒇𝒄 Magnitude Response measures how a system amplifies or attenuates input signals at different frequencies Represents the ratio of output amplitude to input amplitude across the frequency spectrum Example : Consider the RC circuit shown in the following figure. Where the output voltage 𝑉𝑠 is deduced from the voltage divider rule 6 Let : To calculate the transfer function we have : and Or we can simply use the voltage divider and get The Magnitude of Transfer funtion (transmittance) or voltage gain is The Gain in dB and phase 7 III.2. Decibels Your ears respond to sound logarithmically, both in frequency and in intensity. Musical octaves are in ratios of two. "A" in the middle octave is 220 Hz, in the next, 440 Hz, then 880 Hz, etc... It takes about ten times as much power for you to sense one sound as twice as loud as another. The bel is named for Alexander Power ratio expressed in bels = Graham Bell, who did original research in hearing. It is a logarithmic expression of a unitless ratio (like the magnitude of H(ω) or gain of an amplifier). The bel unit is never actually used, instead we use the decibel (dB, 1/10th of a bel). dB are also used to express voltage and current ratios, which is related to power when squared. 𝑉2 𝑃= = 𝐼2𝑅 𝑅 8 By replacing the Laplace variable p by jω. The gain in dB is defined by : Where 𝑌(𝑗𝜔) 𝒀(𝒋𝝎) output of the 𝑮 𝒅𝑩 = 𝟐𝟎. 𝑳𝒐𝒈|𝑯(𝒋𝝎)| = 𝟐𝟎. 𝑳𝒐𝒈 system and 𝑿(𝒋𝝎) 𝑋(𝑗𝜔) the input Three solutions are used in practice to represent this complex transfer function graphically. If we plot the modulus in decibels as a function of frequency and phase as a function of frequency on a logarithmic frequency scale: this is the Bode diagram. If you plot the imaginary part as a function of the real part with frequency parameterization: this is the Nyquist diagram. If the modulus is represented as a function of phase with frequency parametrization, this is the Black diagram. 9 Example : Consider the simple circuit looked at earlier Example Consider the following transfer function (A>0) : 𝜔 1+𝑗𝜔 0 𝐻 𝑗𝜔 = 𝐴. 𝜔 So the magnitude of the transfer function (see following figure) 1+𝑗𝜔 1 is given by : 2 𝜔 1+ 𝜔0 𝐻 𝑗𝜔 = 𝐴. ⟹ 2 𝜔 1+ 𝜔1 2 2 𝜔 𝜔 𝐺𝑑𝐵 = 20𝐿𝑜𝑔10 𝐴 + 10𝐿𝑜𝑔10 1+ − 10𝐿𝑜𝑔10 1− 𝑑𝐵 𝜔0 𝜔1 10 And the argument (see following figure) is given by : 𝜔 𝜔 𝜑 = 𝑎𝑟𝑐𝑡𝑎𝑛𝑔 − 𝑎𝑟𝑐𝑡𝑎𝑛𝑔 𝜔0 𝜔1 The frequency scale is logarithmic. Frequency 𝑓 corresponds to 𝐿𝑜𝑔(𝑓). III.3. Cut-off pulsation 𝝎𝒄 or Cut-off Freuency 𝒇𝒄 The cut-off pulsation can be calculated as follows 𝐻𝑚𝑎𝑥 𝐻(𝜔𝑐 ) = 𝑤𝑒𝑟𝑒 𝐻𝑚𝑎𝑥 = 𝐻(𝜔 → 0) 2 𝑂𝑟 𝐺(𝜔𝑐 ) = 20𝑙𝑜𝑔 𝐻(𝜔𝑐 ) = −3𝑑𝐵 IV. Bode Diagram Bode plots are essential tools for analyzing frequency responses in electrical systems. They use separate graphs to show how a system's gain and phase shift change with frequency, helping engineers understand system behavior across different frequencies. Magnitude plot indicates amplification or attenuation of input signals Positive dB values on magnitude plot represent signal amplification Negative dB values on magnitude plot indicate signal attenuation Phase plot shows lead or lag between input and output signals Positive phase values indicate output signal leads input signal Negative phase values show output signal lags behind input signal 11 Example: The cut-off pulsation Study at the limits (Boundary study) When the pulsation ω tends to zero, the gain G tends to zero and the argument φ tends to zero.W hen ω tends to infinity, G tends to -∞and φ tends to -π/2. And for 𝜔 = 𝜔𝑐 ; G =-3dB and φ = -π/4. Détermination des asymptotes aux courbes 𝑮(𝝎) et 𝝋 (𝝎) For ω ≪ 𝜔0 ; G(ω) ≅ 0 dB and φ(ω) ≅ 0 𝜔 For ω ≫ 𝜔0 ; G(ω) ≅ 20𝐿𝑜𝑔 𝜔. This asymptotic straight line decreases as a function of 0 the pulsation with a slope of -20dB/decade. It passes through the point (ω0,0).and φ(ω) ≅ - π⁄2. 12 Bode Diagram Part II: Chapter 5: ELEMENTS OF SIGNAL THEORY I. Definition of Laplace Transform The Laplace Transform is an integral transformation of a function f(t) from the time domain into the complex frequency domain, giving F(s) s: complex frequency Called “The One-sided or unilateral Laplace Transform”. 13 Example 1: Determine the Laplace transform of each of the following functions shown below: Solution: 1 a) The Laplace Transform of unit step, u(t) is given by L u (t ) U (s ) 1e st dt 0 s b) The Laplace Transform of exponential function, 𝑒 −𝑎𝑡 𝑢(𝑡),a>0 is given by: t st 1 L f (t ) F (s ) e e dt 0 s c) The Laplace Transform of impulse function (Dirac Impulse), δ(t) is given by L f (t ) F (s ) 0 (t )e st dt 1 (t ) Impulse Function (or Dirac Delta Function) 14 II. Functional Transform TYPE f(t) F(s) Impulse δ(t) 1 Step u(t) 1 s Ramp t 1 s2 Exponential e at 1 s a Sine sin t s 2 2 Cosine cost s s 2 2 15 TYPE f(t) F(s) Damped ramp te at 1 s a 2 Damped sine e at sin t s a 2 2 Damped cosine sa e at cos t s a 2 2 V. Properties of Laplace Transform 1. Linearity: If 𝐹1(𝑠) and 𝐹2(𝑠) are, respectively, the Laplace Transforms of 𝑓1(𝑡) and 𝑓2(𝑡) TL a1f1 (t ) a2f 2 (t ) a1F1 (s ) a2F2 (s ) Example: 1 TL cos(t )u (t ) TL 2 e j t e j t u (t ) s s 2 2 16 2. Scaling ; If 𝐹 (𝑠) is the Laplace Transforms of 𝑓 (𝑡), then 1 s TL f (at ) F( ) a a Example: 2 TL sin(2t )u (t ) s 2 4 2 3. Time Shift : If 𝐹 (𝑠) is the Laplace Transforms of 𝑓 (𝑡), then TL f (t a ) e as F (s ) Example: s TL cos( (t a )) e as s 2 2 4. Frequency Shift : If F (s) is the Laplace Transforms of f (t), then TL e at f (t ) F (s a ) Example: s a TL e at cos(t ) (s a )2 2 17 5. Time Differentiation (Derivatives): If 𝐹 (𝑠) is the Laplace Transforms of 𝑓(𝑡), then the Laplace Transform of its derivative is df TL sF s f 0 dt initial condition at t = 0 This is a very important transform because derivatives appear in the ODEs( Ordinary Differential Equations) we wish to solve. Similarly, for higher order derivatives: d n f TL n s n F s s n 1f 0 s n 2f 0 ... sf 0 f 0 1 n 2 n 1 dt initial condition at t = 0 Special Case: All Initial Conditions are Zero 1 0 ... f n 1 0 0. Supposef 0 f Then : 𝑑 𝑛 𝑓(𝑡) 𝑇𝐿 𝑛 = 𝑠 𝑛 𝐹(𝑠) 𝑑𝑡 In process control problems, we usually assume zero initial conditions.18 6. Time Integration : If 𝐹 (𝑠) is the Laplace Transforms of 𝑓 (𝑡), then the Laplace Transform of its integral is t 1 TL f (t )dt F (s ) 0 s 7. Initial and Final Value Theorem: The initial-value and final-value properties allow us to find the initial value 𝑓(0) and 𝑓(∞) of 𝑓(𝑡) directly from its Laplace transform F(s). Initial Value: f (0) lim f (t ) lim sF (s ) t 0 s Final Value It can be used to find the steady-state value of a closed loop system (providing that a steady-state value exists. f () lim f t lim sF s t s 0 Example: Suppose, 5s 2 Y s Then, Initial Value s 5s 4 Final Value 5s 2 y 0 lim y t lim 5s 2 1 y lim y t lim 0.5 t 0 s 5s 4 t s 0 5s 4 19 IV- First-Order Differential Equations Consider a linear time-invariant system defined by: dy(t ) ay(t ) bx(t ) dt Apply the one-sided Laplace transform: sY ( s ) y (0 ) aY ( s ) bX ( s ) We can now use simple algebraic manipulations to find the solution: ( s a )Y ( s ) y (0 ) bX ( s ) y (0 ) bX ( s ) Y (s) sa sa If the initial condition is zero, we can find the transfer function: Y ( s) b H (s) X ( s) s a Why is this transfer function, which ignores the initial condition, of interest? (Hints: stability, steady-state response) Note we can also find the frequency response of the system: b H ( e j ) H ( s ) s j j a How does this relate to the frequency response found using the Fourier transform? Under what assumptions is this expression valid? 20 RC Circuit Applying the Laplace transform, we obtain : 𝑑𝑢2 𝑡 𝑇𝐿 𝑅𝐶. + 𝑢2 𝑡 = 𝑇𝐿 𝑢1 𝑡 𝑑𝑡 Assuming zero CI initial conditions, we obtain : 𝑅𝐶. 𝑠. 𝑈2 𝑠 + 𝑈2 𝑠 = 𝑈1 𝑠 So the transfer function of this system, whose differential equation is of first order, is given by the following equation : 𝑈2 𝑠 1 𝐻 𝑠 = = 𝑈1 𝑠 1 + 𝑅𝐶𝑠 The frequency response is given by: 1 𝑼𝟐 𝒋𝝎 1 1 𝑤𝑖𝑡 𝜔𝑐 = 𝐻 𝑗𝜔 = 𝐻(𝑠) = = = 𝑅𝐶 𝑠=𝑗𝜔 𝑼𝟏 𝒋𝝎 1 + 𝑅𝐶𝑗𝜔 1 + 𝑗 𝜔 𝜔𝑐 21 𝒙(𝒕) = 𝒖(𝒕) : constant supply source Example: Calculate the output voltage between the capacitor for RC circuit if the input voltage is alternating voltage givenby: 𝑣𝑖𝑛 𝑡 = 𝑥 𝑡 = 𝑉𝑚𝑎𝑥 sin(𝜔𝑡) 22 V. Second-Order Differential Equation Consider a linear time-invariant system defined by: d 2 y (t ) dy(t ) dx(t ) 2 a 1 a 0 y (t ) b1 b0 x (t ) assume x ( 0 )0 dt dt dt Apply the Laplace transform: dy(t ) s 2 Y ( s ) y (0 ) s a1 [ sY ( s ) y (0 )] a 0Y ( s) b1 sX ( s) b0 X ( s ) dt t 0 y (0 ) s y (0 ) a1 y (0 ) b1 s b0 Y (s) X (s) s a1 s a 0 2 s a1 s a 0 2 If the initial conditions are zero: What is the nature of the impulse response of this Y ( s) b s b0 system? H ( s) 2 1 X ( s) s a1 s a0 How do the coefficients a0 and a1 influence the Example:ample: impulse response? d 2 y (t ) dy(t ) 2 6 8 y (t ) 2 x (t ) H ( s ) dt 2 dt s 2 6s 8 x(t ) u (t ) X ( s ) 1 / s 2 1 0.25 0.5 0.25 Y (s) s 2 6s 8 s s s2 s4 y (t ) 0.25 0.5e 2t 0.25e 4t , t 0 23 Nth-Order Case Consider a linear time-invariant system defined by: d N y (t ) N 1 d i y (t ) M d i x(t ) ai bi (M N ) dt N i 0 dt i i 0 dt i b0 b1 s b2 s 2... bM s M H (s) a 0 a1 s a 2 s 2... s N Example: s 2 2 s 16 1 H ( s) 3 X ( s ) s 4 s 2 8s s2 s 2 2 s 16 s 1 1 2 Y ( s) H (s) X (s) 3 s 4s 8s s 2 (s 2) 4 s s 2 2 2 y (t ) e 2t cos2t sin 2t 1 2e 2t , t 0 1 2 24 SOLVING LINEAR ODE USING LAPLACE TRANSFORMS Example 1: Solve the differential equation dy 3y 0 dt given the initial condition y (0) 2. Solution: L y 3 y L 0 Laplace both sides of the DE sY ( s) y (0) 3Y ( s) 0 T L Y ( s ) s 3 2 0 Y (s) 2 T L 1 Y (s) s 3 y (t ) 2 y (t ) L 1 2e 3t s 3 25 Circuit Analysis Voltage/Current Relationships: Diff. Eq. : Laplace Transform : v(t ) Ri(t ) V ( s ) RI ( s ) dv(t ) 1 1 1 i (t ) V ( s ) I ( s ) v(0) dt C Cs s di(t ) V ( s ) LsI ( s ) Li(0) v(t ) L dt Series Connections (Voltage Divider): Z 1 (s ) V1 (s ) V (s ) Z 1 (s ) Z 2 (s ) Z 2 (s ) V2 (s ) V (s ) Z 1 (s ) Z 2 (s ) 26 Circuit Analysis Voltage/Current Relationships: Diff.Eq.: Laplace Transform: v (t ) Ri (t ) V (s ) RI (s ) dv (t ) 1 1 1 i (t ) V (s ) I (s ) v (0) dt C Cs s di (t ) V (s ) LsI (s ) Li (0) v (t ) L dt Series Connections (Voltage Divider): Z1 ( s) V1 ( s) V ( s) Z1 ( s) Z 2 ( s) Z 2 (s) V2 ( s ) V (s) Z1 ( s) Z 2 ( s) 27 Circuit Analysis Parallel Connections (Current Divider): Z 2 (s) I1 ( s) I (s) Z1 (s) Z 2 ( s) Z1 (s) I 2 ( s) I ( s) Z1 (s) Z 2 (s) Example: 1 / Cs Vc ( s) X ( s) Ls R (1 / Cs ) V ( s) 1 / LC H ( s) c 2 X ( s ) s ( R / L) s (1 / LC ) R VR ( s) X ( s) Ls R (1 / Cs ) V ( s) ( R / L) s H ( s) c 2 X ( s ) s ( R / L) s (1 / LC ) 28 Analysis of RLC Circuit (Second Order) Using KVL, we have: 𝒖𝟏 𝒕 − 𝒖𝑹 𝒕 − 𝒖𝑳 𝒕 − 𝒖𝟐 𝒕 = 𝟎 𝒖𝑹 𝒕 = 𝑹. 𝒊 𝒕 𝒅𝒖𝟐 𝒕 𝒊 𝒕 = 𝑪. 𝒅𝒕 𝒅𝒊(𝒕) 𝒖𝑳 𝒕 = 𝑳. 𝒅𝒕 We obtain : 𝒅𝟐 𝒖𝟐 𝒕 𝒅𝒖𝟐 𝒕 𝑳𝑪. 𝒅𝒕𝟐 + 𝑹𝑪. + 𝒖𝟐 𝒕 = 𝒖𝟏 𝒕 𝒅𝒕 Which is a second-order differential equation with a constant coefficient. Using the Laplace transform, we obtain d2 u2 t du2 t TL LC. + RC. + u2 t dt 2 dt 𝑼𝟐 𝒑 1 = TL u1 t 𝐻 𝑝 = = 𝑼𝟏 𝒑 1 + 𝑅𝐶𝑝 + 𝐿𝐶𝑝2 ⇒ 𝑼𝟐 𝒋𝝎 1 U2 p. LC. p2 + RC. p + 1 = U1 p 𝐻 𝑗𝜔 = = 2 𝑼𝟏 𝒋𝝎 1 + 𝑅𝐶𝑗𝜔 + 𝐿𝐶 𝑗𝜔 29 VI. The Inverse Laplace Transform Suppose F(s) has the general form of P (s ) Numerator Polynomial b0 b1s b 2s 2... bM s M F (s ) Q (s ) Deno minator Polynomial a0 a1s a2s 2... aN s N The finding the inverse Laplace transform of F(s) involves two steps: 1. Decompose F(s) into simple terms using partial fraction expansion. 2. Find the inverse of each term by matching entries in Laplace Transform Table. 30 3. Using Residus theorem to calculate the integral 1. Decompose F(s) into simple terms using partial fraction expansion. Basic idea: Expand a complex expression for F(s) into simpler terms, each of which appears in the Laplace Transform table. Then you can take the TL-1 of both sides of the equation to obtain f(t). poles Pole 𝒑𝒊 → 𝑸 𝒑 = 𝟎| 𝒑=𝒑𝒊 Multiple pole 31 Example 1 : Find the inverse Laplace transform of 3 5 6 F (s) 2 Solution: s s 1 s 4 3 1 5 1 6 1 f (t ) L L L 2 s s 1 s 4 (3 5e t 3 sin( 2t )u (t ), t 0 s 5 Example 2: Y s s 1 s 4 Perform a partial fraction expansion (PFE) To find 𝛼1 : Multiply both sides by s + 1 and s 5 let s = -1 1 2 s5 4 s 1 s 4 s 1 s 4 1 s4 s 1 3 To find 𝛼2 : Multiply both sides by s + 4 and let s = -4 s5 1 2 s 1 s 4 3 32 Example 3 : Recall that the ODE, y 6 y 11y 6 y 1 , with zero initial conditions resulted in the expression 1 Y s s s 3 6s 2 11s 6 The denominator can be factored as s s 3 6s 2 11s 6 s s 1 s 2 s 3 Note: Normally, numerical techniques are required in order to calculate the roots. 1 1 2 3 4 Y s (*) s s 1 s 2 s 3 s s 1 s 2 s 3 Solve for coefficients to get 1 1 1 1 1 , 2 , 3 , 4 6 2 2 6 Substitute numerical values into (eq *): 1/ 6 1/ 2 1/ 2 1/ 6 Y (s) -1 s s 1 s 2 s 3 Take TL of both sides: 1 1/ 6 1 1/ 2 1 1/ 2 1 1/ 6 TL Y s TL TL TL 1 TL s s 1 s 2 s 3 1 1 t 1 2t 1 3t y t e e From Table, e 33 6 2 2 6 f(t) F(s) f(t) F(s) (t ) 1 w e at sin(wt ) u( t ) 1 (s a)2 w 2 s s a e st 1 e at cos(wt ) sa (s a)2 w 2 1 s sin w cos t s2 sin(wt ) s 2 w 2 n! tn s cos w sin s n 1 cos(wt ) te at 1 s 2 w 2 s a 2 n! t n e at (s a )n 1 w sin(wt ) s 2 w 2 s cos(wt ) s 2 w 2 34 Common Transform Properties: f(t) F(s) to s f ( t t0 )u( t t0 ), t0 0 e F ( s) to s f ( t )u( t t0 ), t 0 e L[ f ( t t0 ) e at f ( t ) F (s a) d n f (t ) s n F ( s ) s n 1 f ( 0) s n 2 f ' ( 0) ... s 0 f n 1 f ( 0) dt n dF ( s ) tf ( t ) ds t 1 f ( )d F ( s) 0 s 35