ECE 209-02 Class#1Lecture Notes 1Example 1.3ECE 209-02 Class#1 Lecture Notes 1 1. Sinusoidal Steady-State Analysis 1.1. Sinusoidal Signal We will consider circuit with time-varying voltage or current sources. In particularly, we will focus on sinusoidal sources and their effect on the circuit. We can express a sinusoidal varying function with either the sine function or the cosine function, but we will use the cosine function tvVmφ/ωT Let’s write a sinusoidal varying voltage as )cos()(φω+= tVtvm where • Vm >0 is the maximum amplitude of the sinusoidal voltage. • ω>0 is the frequency in radian/sec, which is Tfππω22 == (radian/second) • T is the period which the sinusoidal function in seconds. • f is the frequency of the sinusoidal function in Hertz(Hz). πω21==Tf >0 • φ is phase angle or phase shift in radians-common conventionπφπ≤<−or πφ20 ≤≤− . It determines the value of sinusoidal function at t=0. If φ is positive the sinusoidal function shifts to the left. If φ is negative the function shifts to the right. If φ is given in degree, we may need to convert it into radians. Because of ωt is in radians. So that Number of radians =180πx( number of degree) The root mean square (rms) value of the function is ∫+=TttrmsdtvTV0021 1-1which can be can be written as ∫++=TttmrmsdttVTV00)(cos122φω ++=+=∫∫++ TttmTttmrmsdttTVdttVTV0000)22cos(2121)(cos1222φωφω {}∫∫++++=TttTttmrmsdttdtTV0000)22cos(22φωV []++=+TttmrmstTTV0)22sin(2122φωωV []+−+++= )22sin()242sin(212002φωφπωωttTTVmrmsV += )0(2122ωTTVmrmsV 2mrmsV=V Example 1.1: A sinusoidal voltage V )30120cos(300otv +=π • The period of the voltage:πω120= rd/s. Tπω2=, 6012==ωπT s. • The frequency 601==Tf Hz • The rms value of v Vrms13.2122300==V • The magnitude value of v at t=2.778 ms: Vvooo0)3060cos(300)30)002778.0(120cos(300 =+=+=π 1-21.2 The Phasor representation of sinusoidal signals Euler’s formula is known that for θ real θθθsincos jej+= , where 1−=j Thus, we see that is a complex number {}θθcos=ℜje and {}θθsin=ℑje We can write a sinusoidal voltage function as {}{}tjjmtjmmeeVeVtVtvωφφωφωℜ=ℜ=+=+ )()cos()( where Vm represents the amplitude and ejφ represents the angle of the voltage function. The both together is called as phasor representation or phasor transform of sinusoidal function. V= P{=φjmeV )cos(φω+tmV } We can also express the phasor in rectangular form V=φφsincosmmjV+V or v= jyx+ Therefore, =mV |V|22yx += is the magnitude and )/arctan( xy=φis the phase Also, we can give the phasor representation in angle form Dφφ∠=mjmVeV Im ReVm φ The inverse phasor transfer form is: V(t) =P-1{}{})cos(φωωφφ+=ℜ= tVeeVeVmtjjmjm 1-3Example 1.2: RLVs i(t) Let say that {}tjjmseeVtvωφℜ=)( Kirchhoff’s voltage law to the circuit will be {}tjjmeeVRidtdiLωφℜ=+ We know that steady-state solution for I(t) will be in the following form {}tjjmsseeItiωβℜ=)( , and {}tjjmsseeIJdttdiωβωℜ=)( and apply that to differential equation {}{}{}tjjmtjjmtjjmeeVeeIReeLIJωφωβωβωℜ=ℜ+ℜ tjjmtjjmeeVeeIRLjωφωβω=+ )( φφφβωωωωjmjmjmjmeLRLVjeLRRVLjReVeI2222)()( +−+=+= {}βjmeIti1P)(−= in time domain • The phasor transform can apply to calculate sum of sinusoidal function easily if the sinusoidal functions have the same frequency For )(...)()()(21tvtvtvtvn+++= v=v1+v2+…+vn so that {}tjetω)...vvv()(n21++ℜ=v • if a is real number av(t) can be written as {}tjeatavωv)( ℜ= 1-4Example 1.3 )30cos(20)(1−= ttvω )60cos(40)(2+= ttvω a. Solve by using trigonometric identities b. Solve by using the phasor concepts Remember the function angle relation yxyxyxyxyxyxyxyxyxyxyxyxsincoscossin)sin(sincoscossin)sin(sinsincoscos)cos(sinsincoscos)cos(−=−+=++=−−=+ a. Expand v1(t) and v2(t): DD30sinsin2030coscos20)(1tttvωω+= DD60sinsin4060coscos40)(2tttvωω−= tttvtvtvωωsin)60sin4030sin20(cos)60cos4030cos20()()()(21DDDD−++=+= tttvωωsin64.24cos32.37)( −= 2237.32 24.64 44.72+= 37.32 24.64( ) 44.72 cos sin44.72 44.72vt t tωω=− []( ) 44.72 cos33.43cos sin 33.43sinvt t tωω=− ( ) 44.72cos( 33.43 )vt tω=+D 33.43° 44.7224.64 b. DDD43.3372.4464.2432.37)64.3420()1032.17(60403020vvv21∠=+=++−=∠+−∠=+=jjj 37.32 in phasor representation )44.33cos(72.44)(D+= ttvω 1-5v2(t) v(t) v1(t) 1-6 >> f=1000; >> w=2*pi*f; >> phase=30; >> theta=phase*pi/180; >> t=0:1e-5:2e-3; >> v1=20*cos(w*t-theta); >> plot(t,v1) >> grid on; >> hold on; >> phase2=60; >> theta2=phase2*pi/180; >> v2=40*cos(w*t+theta2); >> plot(t,v2,'r') >> >> v=v1+v2; >> plot(t,v,'g') >> >> xlabel ('t[s]') >> ylabel ('v_1(t), v_2(t), v(t)') >>Note: The time derivatives of sinusoidal function