Chapter 9.Sinusoidal Analysis PDF

Chapter 9: Sinusoidal Steady-State Analysis Course Instructor Prof. Adel Abdennour Electrical Engineering Department Islamic University in Madinah https://sites  Main points to be covered in this Chapter: Review of Sinusoidal Functions &...

Chapter 9: Sinusoidal Steady-State Analysis Course Instructor Prof. Adel Abdennour Electrical Engineering Department Islamic University in Madinah https://sites  Main points to be covered in this Chapter: Review of Sinusoidal Functions & Complex Numbers Sinusoidal Response, Phasors, Passive elements in frequency domain V-I relationship for a resistor, inductor, and capacitor Concept of Impedance and Reactance KVL & KCL laws in the frequency domain Series and Parallel Impedance and Admittance Wye and Delta transformations Mesh analysis and nodal equation in phasor representation Thevenin and Norton Equivalents Prepared by prof. Draou/ prof. Abdennour 2  9.1 The Sinusoidal Source A Sinusoidal voltage source produces a voltage that varies sinusoidally with time A Sinusoidal current source produces a current that varies sinusoidally with time The sine and the cosine functions are both called sinusoidal functions. Prepared by prof. Draou/ prof. Abdennour 3  9.1 The Sinusoidal Source v  Vm coswt    1 f  T T: is referred as the Period of the function f: is the number of cycles/s, or the frequency Φ: is known as the phase angle of the sinusoidal voltage Prepared by prof. Draou/ prof. Abdennour 4  9.1 The Sinusoidal Source   2f  2 / T (radians/ sec) Prepared by prof. Draou/ prof. Abdennour 5  9.1 The Sinusoidal Source Another important characteristic is the rms value (root mean squared value) v  Vm coswt    t 0 T 1 Vrms   V 2 m cos 2 wt   dt T t0 Vm Hence the rms value of v is: Vrms  2 Prepared by prof. Draou/ prof. Abdennour 6  Example Prepared by prof. Draou/ prof. Abdennour 7 Prepared by prof. Draou/ prof. Abdennour 8  Example Prepared by prof. Draou/ prof. Abdennour 9 Prepared by prof. Draou/ prof. Abdennour 10  9.2 The Sinusoidal Response vs  Vm cost    If the initial current is zero, we can find i(t) for t>0. From KVL, we get: di L  Ri  Vm cost    dt Prepared by prof. Draou/ prof. Abdennour 11  9.2 The Sinusoidal Response The solution of i would be:  Vm cos   e cost      ( R / L )t Vm i  R  L 2 2 2 R  L 2 2 2 Where θ is the angle whose tangent is ωL/R The first term of the above equation is referred to as: transient component of the current. The second term of the above equation is referred to as: steady-state component of the current. In this chapter we focus on the steady state component Prepared by prof. Draou/ prof. Abdennour 12  9.3 The Phasor The phasor is a complex number representing the amplitude and phase angle of a sinusoidal function.  j It is rooted in Euler’s identity: e  cos  j sin  Where cos   e j   is known as the real part And sin  e  is the imaginary part j As an example:    v  Vm cost     Vm e j t    Vm e jt e j  Moving terms around, we get:  j v   Vme e jt  Prepared by prof. Draou/ prof. Abdennour 13  9.3 The Phasor The quantity Vm e j is a complex number representing the amplitude and phase angle of the sinusoidal function. It is known as the phasor representation, or phasor transform of the sinusoidal function. Thus, P: is the phasor transform from of the sinusoidal function from the time domain to the complex domain (frequency domain) V  Vme  PVm cost    j Prepared by prof. Draou/ prof. Abdennour 14  9.3 The Phasor V  Vme j  PVm cost    is also known as the polar form of a phasor, it can be transformed to a rectangular form: V  Vm cos   jVm sin  Both forms are very useful in circuit applications The abbreviation of the angle notation extensively j used is: 1  1e o We can also use the inverse phasor transform: As an example, if : V  100  26 (phasor), then we o can write: v  100cos t  26 o   Going from phasor form to time domain is referred to: inverse phasor transform Prepared by prof. Draou/ prof. Abdennour 15  9.3 The Phasor AP 9.1 Find the phasor transform of each trigonometric function: a) v  170 cos(377 t  40o ) V b) i  10 sin(1000 t  20o ) A c)   i  5 cos(t  36.87 o )  10 cos(t  53.13o ) A d)  v  300 cos(20000 t  45o )  100 sin( 20000 t  30o ) mV a) 170  40o V b) 10  70o A c) 11.18  26.57o A d) 339.90  61.51o mV Prepared by prof. Draou/ prof. Abdennour 16  9.3 The Phasor AP 9.2 Find the time domain expression corresponding to each phasor: a) V  18.6  54o V b) V  (2045o  50  30o ) mA c) V  20  j80  3015o  V a ) 18.6 cos(t  54o )V b) 48.81cos(t  126.68o ) mA c ) 72.79 cos(t  97.08o )V Prepared by prof. Draou/ prof. Abdennour 17  9.4 The Passive Circuit Elements in the Frequency Domain 9.4.1 The V-I relationship for a resistor If i  I m cost  i   I m  i Where I m is the maximum amplitude of the current. And  i is the phase angle of the current. Then the voltage would be: v  RI m cost  i   RI m cost   i  In phasor form: V  RI me ji  RI mi  RI  V  RI Prepared by prof. Draou/ prof. Abdennour 18  9.4 The Passive Circuit Elements in the Frequency Domain 9.4.1 The V-I relationship for a resistor The figure shows both voltage and current waveforms. NOTE: The signals are said to be in phase because they both have zeros, maxima, and minima at the same instant of time. Prepared by prof. Draou/ prof. Abdennour 19  9.4 The Passive Circuit Elements in the Frequency Domain 9.4.2 The V-I relationship for an inductor Similarly for a current that flows through an inductor with an expression as: i  I m cos t  i   The expression for the voltage would be:  LI m sint   i  di vL dt  LI m cost   i  90o  The phasor representation of the voltage would be: V  LI m e j ( i 90o ) j i  LI m e e  j 90o Note:  j 90o j i e  cos 90o  j sin 90o  jLI m e  jLI j Prepared by prof. Draou/ prof. Abdennour 20  9.4 The Passive Circuit Elements in the Frequency Domain 9.4.2 The V-I relationship for an Inductor The above equation can be rewritten as: V  (L90 o ) I m  i  LI m ( i  90 o ) The frequency domain equivalent circuit of the inductor: The figure illustrates the concept of voltage leading current or current lagging voltage. Prepared by prof. Draou/ prof. Abdennour 21  9.4 The Passive Circuit Elements in the Frequency Domain 9.4.3 The V-I relationship for a Capacitor For a current flowing through a capacitor: dv iC with v  Vm cos(t   v ) dt i  CVm sin(t   v )  CVm cos(t   v  90)  I  CVme j v e 90 j  I  jCVm e j v  jCV 1 V  I jC The equivalent circuit in the phasor domain Prepared by prof. Draou/ prof. Abdennour 22  9.4 The Passive Circuit Elements in the Frequency Domain 9.4.3 The V-I relationship for a Capacitor The voltage across the capacitor lags behind the current by exactly 90º. It can be shown that: 1 V   90o I m  i o C Im V ( i o  90o ) C The figure illustrates the concept of current leading voltage or voltage lagging current. Prepared by prof. Draou/ prof. Abdennour 23  9.4 The Passive Circuit Elements in the Frequency Domain 9.4.4 Impedance and Reactance All the equations above are of the form: V=ZI, where Z represents the impedance of the circuit in Ω. The impedance is the ratio of a circuit voltage phasor to its current phasor. Thus,  The impedance of a resistor is R  The impedance of an inductor is jωL  The impedance of a capacitor is 1/jωC Prepared by prof. Draou/ prof. Abdennour 24  9.4 The Passive Circuit Elements in the Frequency Domain 9.4.4 Impedance and Reactance Impedance in frequency domain is analogous to resistance, inductance, and capacitance in time domain. The imaginary part of the impedance is called Reactance. Circuit element Impedance Reactance Resistor R - Inductor jωL ωL Capacitor -j/ωC -1/ωC Prepared by prof. Draou/ prof. Abdennour 25  9.4 The Passive Circuit Elements in the Frequency Domain AP 9.3 The current in the 20 mH inductor is 10cos(10,000t+30º) mA. Calculate (a) the inductive reactance; (b) the impedance of the inductor; (c) the phasor voltage V, and (d) the steady state expression for v(t) a ) 200 b) j 200 c ) 2120o V d ) 2 cos(10,000t  120o )V Prepared by prof. Draou/ prof. Abdennour 26  AP 9.4 The voltage across the terminals of the 5μF capacitor is 30cos(4,000t+25º) V. Calculate (a) the capacitive reactance; (b) the impedance of the capacitor; (c) the phasor current I, and (d) the steady state expression for i(t) a )  50 b)  j 50 c ) 0.6115o A d ) 0.6 cos(4000t  115o ) A Prepared by prof. Draou/ prof. Abdennour 27  9.5 Kirchhoff’s laws in the Frequency Domain 9.5.1 Kirchhoff’s Voltage Law Kirchhoff’s voltage law: v1  v2  v3 ....... vn  0, In the sinusoidal steady state, it becomes complex: vm1 cos(t  1 )  vm2 cos(t  2 ) ...  vmn cos(t  n )  0, Using Euler’s identity we can write:       vm1 e j1 e jt   vm2 e j 2 e jt ...   vmn e j n e jt  0,     vm1 e e j1 jt  vm2 e ej 2 jt j n ....  vmn e e jt  0 Factoring the term e jt from each term yields   (vm1 e j1  vm2 e j 2  vm3 e j 3 ....  vmn e j n )e j t  0 Prepared by prof. Draou/ prof. Abdennour 28  The last equation can be rewritten as,   (V1  V2  V3 ....  Vn )e j t  0 j t Since e 0 We conclude that the KVL in the frequency domain is given by: V1  V2  V3 ....  Vn  0 Prepared by prof. Draou/ prof. Abdennour 29  9.5 Kirchhoff’s laws in the Frequency Domain 9.5.2 Kirchhoff’s Current Law A similar derivation applies to a set of sinusoidal currents. Thus if: i1  i2  i3 ....... in  0, Then, the KCL in the frequency domain is given by: I1  I 2  I 3 ....  I n  0 Prepared by prof. Draou/ prof. Abdennour 30  9.5 Kirchhoff’s laws in the Frequency Domain AP 9.5 Four branches terminate at a common node. The reference direction of each branch current (i1, i2, i3, and i4) is toward the node. i2 If i1  100 cos(t  25o ) i1 i3 Find i4 i2  100 cos(t  145 ) o i3  100 cos(t  95o ) i4 Solution i4  0 Prepared by prof. Draou/ prof. Abdennour 31  9.6 Series, Parallel, and Delta-Wye Simplifications 9.6.1 Combining Impedances in Series and Parallel Vab  Z1 I  Z 2 I ....  Z n I Vab  ( Z1  Z 2 ....  Z n ) I Vab Z ab   Z1  Z 2 ....  Z n I Prepared by prof. Draou/ prof. Abdennour 32  Example 9.6 vs  750cos(5000t  30o )V Solution: Prepared by prof. Draou/ prof. Abdennour 33  Combining Impedances in Parallel V V V V I  I1  I 2 ....  I n    ....  Z ab Z1 Z 2 Zn This can be expressed in terms of admittance (Y), defined as the reciprocal of impedance: 1 1 1 1   ....  Z ab Z1 Z 2 Zn Prepared by prof. Draou/ prof. Abdennour 34  Combining Impedances in Parallel Admittance is a complex number: Note that: 1 Y   G  jB ( Siemens) Yab  Y1  Y2 ....  Yn Z Where G is called the conductance and B is called the susceptance Circuit element Admittance Susceptance Resistor G - Inductor -j/ωL -1/ωL Capacitor jωC ωC Solve example 9.7 Prepared by prof. Draou/ prof. Abdennour 35  AP 9.6 Vs  125  60o C Find   5000rad / s a) The value of capacitance that yields a steady state output current i with a phase angle of -105º b) The magnitude of the steady state output current i Solution: 3 j 1 Z  90  j (5000)(32 10 )   90  j (160  ) 5000C 5000C V I   i   v   Z  105o  60o   Z   Z  45o Z 1  Re(  Re(Z) )  Im(  Z = Im(Z)Z )  90  160   C  2.86F 5000C 125 Now : Z  90  j 90  127.2845o  I   0.982A 127.28 Prepared by prof. Draou/ prof. Abdennour 36  9.6.2 Delta-to-Wye Transformations The Δ-to-Y transformation of impedances: Y impedances as fct of Δ Δ impedances as fct of Y Z1  Zb Zc Z1Z 2  Z 2 Z 3  Z 3 Z1 Za  Z a  Zb  Zc Z1 Z2  Zc Za Z Z  Z 2 Z 3  Z 3 Z1 Zb  1 2 Z a  Zb  Zc Z2 Z3  Z a Zb Z Z  Z 2 Z 3  Z 3 Z1 Zc  1 2 Z a  Zb  Zc Z3 (Solve example 9.8) Prepared by prof. Draou/ prof. Abdennour 37  AP 9.9 Use a Δ-to-Y transformation to find the current I. Prepared by prof. Draou/ prof. Abdennour 38  Solution: (50)(10) 5  50  40  10 (50)(40) 20  50  40  10 (40)(10) 4  50  40  10 Prepared by prof. Draou/ prof. Abdennour 39  9.7 Source Transformations and Thevenin-Norton Equivalent Circuits The Thevenin-Norton equivalent circuits are also valid for the frequency domain circuits. Note: Circuit may contain dependeant or independent sources Frequency-domain source Frequency-domain thevenin and transformation Norton equivalent 40 Prepared by prof. Draou/ prof. Abdennour Solve examples 9.9 and 9.10  AP 9.10 Use source transformations to find v0(t) v1  240 cos(4000t  53.13o ) v2  96 sin( 4000t ) Solution: Source transformation: 8/22/2023 Prepared by prof. Draou/ prof. Abdennour 41 Prepared by prof. Draou/ prof. Abdennour 42  AP 9.11 Find the Thevenin equivalent. Solution: v1 vth Voltage Division of the 10Ix V1=20Ix Prepared by prof. Draou/ prof. Abdennour 43  To find Zth, remove the independent current source and apply a test volyage Vt vt  Ix  0  Prepared by prof. Draou/ prof. Abdennour 44  9.8 The Node Voltage Method The basic concepts of the node voltage method can also be applied for frequency domain circuits. V1 V2 Example 9.11 Node 1: Node 2: Solve for v1 and v2, then for all the branch currents Prepared by prof. Draou/ prof. Abdennour 45  AP 9.12 Use node voltage to find the expression for v(t) is  10 cos(t ) ω=50krad/s vs  100 sin(t )  Prepared by prof. Draou/ prof. Abdennour 46  9.9 The Mesh Current Method The basic concepts of the mesh current method can also be applied for frequency domain circuits. Example 9.12 I1 I2 Mesh 1: Mesh 2: Solve for I1 and I2, then for all the node voltages Prepared by prof. Draou/ prof. Abdennour 47  AP 9.13 Use Mesh current to find the phasor current I. Mesh 1: Mesh 2: Ia Ib Ic Mesh 3: Solving the three equations yields: Prepared by prof. Draou/ prof. Abdennour 48  9.12 Phasor Diagrams A phasor diagram shows the magnitude and phase of each quantity in the complex plane. The complex number Examples of phasors Prepared by prof. Draou/ prof. Abdennour 49  Example 9.15 Use the phasor diagram to find the value of R that will make iR lag is by 45o when w=5krad/s 4Vm-Vm A phasor with 45o has equal real and imaginary values: Vm 1   3Vm  R   R 3 Prepared by prof. Draou/ prof. Abdennour 50  After this review, you should be able to: Represent a complex number on the complex plane Perform basic complex number operations Convert complex numbers between the Rectangular and the Exponential form Use the rectangular form for addition and subtraction of complex numbers Use exponential form for multiplication and division of complex numbers Prepared by prof. Draou/ prof. Abdennour 51  Switch between the instantaneous representation and the phasor representation Determine the impedance and admittance of the resistor, the inductor and the capacitor Combine Impedances and Admittances in series and in parallel Calculate the resistance, reactance, conductance and susceptance Switch between wye and delta configuration Analyze simple sinusoidal circuits Prepared by prof. Draou/ prof. Abdennour 52  Find the Thevenin and Norton equivalents for sinusoidal circuits Apply source transformation to sinusoidal circuits Apply the nodal and mesh analyses to sinusoidal circuits Apply and manipulate phasor diagrams This is the end of the review of Chapter 9 Prepared by prof. Draou/ prof. Abdennour 53