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Sinusoid Steady-State Analysis Objectives: 1) Be able to perform a phasor transform and its inverse; 2) Be able to phasor-transform a circuit; 3) Solve arbitrarily complex circuits with sinusoidal sources using phasor method. Sinusoid Steady-State Analysis Review the terminology of sinusoids v(t) = Vmcos(t + ) V •Vm : Amplitude or magnitude • : angular frequency [rad/s] •t : time [s] • : phase angle [deg] •f = /2 : frequency [Hz] •T = 1/f : period [s] What is the frequency of the sinusoid given below? i(t) = 36 cos(4t + 45) mA A. 4t B. 4 Hz C. 4 rad/s What is the phase angle of the sinusoid given below? v(t) = 50 cos(3000t 60) V A. 60 deg B. 60 rad C. 60 deg Suppose we want to find the value of the sinusoidal current described below at t = 5 ms. When we substitute this value for t into the sinusoid, the argument for the cosine function has what units? i(t) = 0.2 cos(50t + 45) A A. degrees B. radians C. A mix of degrees and radians Sinusoid Steady-State Analysis Evaluating a sinusoid at a specified time i(0.005) = 0.2 cos(50t + 45) A 50 has the units rad/s 50t has the units radians 45 has the units degrees Need to convert 50t to degrees, or 45 to radians, and set your calculator appropriately! 2/360 = 50(0.005)/x x = 45 Thus, i(0.005) = 0.2 cos(45 + 45) A = 0.2 cos(90) = 0 A Sinusoid Steady-State Analysis Suppose v(t) = Vmcos(t + ) V in the circuit below: The equation for i(t) is L di (t ) Ri (t ) Vm cos(t ) dt The solution of this equation for i(t) is i (t ) where Vm R (L) 2 2 cos( )e ( R L ) t tan 1 (L R ) Vm R (L) 2 2 cos(t ) Consider the equation for the current in the RL circuit with the sinusoidal source: i (t ) Vm R (L) 2 2 cos( )e ( R L ) t Vm R (L) 2 2 cos(t ) Which term(s) go to 0 as t ∞? A. The first term B. The second term C. Both terms D. Neither term Sinusoid Steady-State Analysis i (t ) Vm R (L) 2 2 cos( )e ( R L ) t Vm R (L) 2 2 cos(t ) •The first term is the transient term – it decays to 0 as t goes to infinity •The second term is the steady-state term – it persists for all time greater than 0. This term • Is sinusoidal • Has the same frequency as the input voltage • Has a different magnitude and phase angle compared to the input voltage Sinusoid Steady-State Analysis The circuit analysis techniques we are studying will allow us to calculate the sinusoidal steady-state response of the circuit – that is, the circuit’s response to a sinusoidal input once the transient response has effectively decayed to 0. This is also called the AC Steady-State (ACSS) response. Sinusoid Steady-State Analysis Phasor • A complex number in polar form, with a magnitude and phase angle • Derived from a sinusoid using the phasor transform Euler' s identity : Now cos Ree j e j cos j sin v (t ) Vm cos(t ) Vm Ree j (t ) Vm Ree jt e j ReVm e jt e j Sinusoid Steady-State Analysis Phasor transform • Extracts a sinusoid’s magnitude and phase angle • Transforms a function of time into a function of frequency V PVm cos(t ) Vme j Vm Inverse phasor transform • Turns a phasor back into a sinusoid, if someone tells you the frequency v(t ) P-1 V P-1 Vm Vm cos(t ) Suppose i(t) = 36 cos(4t + 45) mA Then the phasor transform of i(t) is A. 36 mA B. 3645 mA C. 364t mA Sinusoid Steady-State Analysis Adding or subtracting sinusoids in the time domain is hard (you need trig identities!) But if you phasor-transform the sinusoids, it is easy to combine them. For example, i (t ) 5 cos(300t 36.87) 10 cos(300t 53.13) A I 536.87 10 53.13 11.18 26.57 A i (t ) P 1I 11.18 cos(300t 26.57) A Sinusoid Steady-State Analysis Consider the equations that relate voltage and current in a resistor, inductor, and capacitor in the time domain, and in the phasor domain. Sinusoid Steady-State Analysis Resistors: Time domain Phasor (frequency) domain P i (t ) I m cos(t ) v(t ) Ri (t ) RI m cos(t ) I I m V RI RI m Note that the voltage and current phasors for a resistor have the same phase angle – therefore we say that the voltage and current in a resistor are “in phase”. Sinusoid Steady-State Analysis Phasor (frequency) domain Inductors: Time domain P i (t ) I m cos(t ) v (t ) L di (t ) dt I I m V LIm e j ( 90 ) LIm e j e j 90 LIm sin(t ) LIm e j ( j ) LIm cos(t 90) jLIm e j jLI In an inductor, the voltage leads the current by 90 Sinusoid Steady-State Analysis Phasor (frequency) domain Capacitors: Time domain P v (t ) Vm cos(t ) i (t ) C dv (t ) dt V Vm I CVm e j ( 90 ) CVm e j e j 90 CVm sin(t ) CVm e j ( j ) CVm cos(t 90) jCVm e j jCV In a capacitor, the voltage lags the current by 90 The impedance of a resistor is A. Always a real number B. Always an imaginary number C. A complex number with real and imaginary parts The impedance of an inductor is A. A real number B. A positive imaginary number C. A negative imaginary number The impedance of a capacitor has the units A. Farads [F] B. Henries [H] C. Ohms [] Impedance is a phasor. A. True B. False Sinusoid Steady-State Analysis Summary: • In the time domain • Resistor: • inductor: • Capacitor: v(t) = Ri(t) v(t) = Ldi(t)/dt i(t) = Cdv(t)/dt • In the phasor domain • • • • • V = ZI Z is impedance, defined as the ratio of V to I Z has the units Ohms [] Resistor: ZR = R Inductor: ZL = jL Capacitor: ZC = 1/jC = -j/C Sinusoid Steady-State Analysis Summary: • In the time domain • • • • Ohm’s law (only for resistors): v = Ri KVL (around a loop): v1 + v2 + … + vn = 0 KCL (at a node): i1 + i2 + … + in = 0 These three laws lead to all other time-domain circuit analysis techniques • In the phasor domain • • • • Ohm’s law (for R, L, C): V = ZI KVL (around a loop): V1 + V2 + … + Vn = 0 KCL (at a node): I1 + I2 + … + I n = 0 These three laws mean we can use all time-domain circuit analysis techniques in the phasor domain! Sinusoid Steady-State Analysis Circuit in the time domain P Hard – calculus! Solution in the time domain Circuit in the phasor domain Easy – algebra! P -1 Solution in the phasor domain Sinusoid Steady-State Analysis Steps in ACSS Analysis: 1. Redraw the circuit (the phasor transform does not change the components or their connections). 2. Phasor transform all known v(t) and i(t). 3. Represent unknown voltages and currents with V and I. 4. Replace component values with impedance (Z) values. 5. Use any circuit analysis method(s) to write equations and solve them with a calculator. 6. Inverse-transform the result, which is a phasor, back to the time domain. Sinusoid Steady-State Analysis Example 9.7 is(t) = 8cos(200,000t) A. Find v(t), i1(t), and v2(t) in the steady-state. 1. Redraw the circuit. 2. Phasor transform all known voltages and currents. 3. Represent unknown voltages and currents with phasor symbols The next step is to calculate the impedance of all resistors, inductors, and capacitors. Actually, the impedance of the resistors doesn’t require calculation. A. True B. False The impedance of inductors and capacitors is calculated using A. The source frequency B. The component value C. Both A and B Sinusoid Steady-State Analysis Example 9.7 is(t) = 8cos(200,000t) A. Find v(t), i1(t), and v2(t) in the steady-state. 4. Calculate the impedances of all resistors, inductors, and capacitors. Z L jL j (200,000)( 40 ) j8 j j ZC j5 C (200,000)(1 ) Sinusoid Steady-State Analysis Example 9.7 is(t) = 8cos(200,000t) A. Find v(t), i1(t), and v2(t) in the steady-state. 5. Use DC circuit analysis techniques to find the phasor V. • In this problem, we’ll simplify the circuit by replacing all impedances by one equivalent impedance, and use Ohm’s law for phasors. Z eq 10 || (6 j8) || j5 (4 j 3) V ZI (4 j 3)(80) (32 j 25)V Sinusoid Steady-State Analysis Example 9.7 is(t) = 8cos(200,000t) A. Find v(t), i1(t), and v2(t) in the steady-state. 6. Inverse phasor-transform the result to get back to the time domain. V (32 j 24)V 40 36.87 V v(t ) P 140 36.87 40 cos(200,000t 36.87) V The most direct way to calculate the phasor I1 is A. Voltage division B. Current division C. Source transform Sinusoid Steady-State Analysis Example 9.7 is(t) = 8cos(200,000t) A. Find v(t), i1(t), and v2(t) in the steady-state. 5. Use DC circuit analysis techniques to find the phasor I1. • Use current division. I1 Z eq Z1 Is (4 j 3) (80) 10 (3.2 j 2.4) A Sinusoid Steady-State Analysis Example 9.7 is(t) = 8cos(200,000t) A. Find v(t), i1(t), and v2(t) in the steady-state. 6. Inverse phasor-transform the result to get back to the time domain. I1 (3.2 j 2.4) A 4 36.87 A i1 (t ) P 14 36.87 4 cos(200,000t 36.87) A The most direct way to calculate the phasor V2 is A. Voltage division B. Current division C. Node voltage method Sinusoid Steady-State Analysis Example 9.7 is(t) = 8cos(200,000t) A. Find v(t), i1(t), and v2(t) in the steady-state. 5. Use DC circuit analysis techniques to find the phasor V2. • Use voltage division. ZL j8 V2 V (32 j 24) ZR ZL 6 j8 (32 j 0) V Sinusoid Steady-State Analysis Example 9.7 is(t) = 8cos(200,000t) A. Find v(t), i1(t), and v2(t) in the steady-state. 6. Inverse phasor-transform the result to get back to the time domain. V2 (32 j 0) V 320 V v2 (t ) P 1320 32 cos 200,000t V Sinusoid Steady-State Analysis Example 9.7 Suppose we can vary the frequency of the current source. What frequency will cause is(t) and v(t) to be in-phase in the steady-state? If is(t) and v(t) are in-phase in the steady-state, this means A. Their phase angles are the same. B. The equivalent impedance seen by the current source has a phase angle of 0. C. The equivalent impedance seen by the current source is purely resistive. D. All of the above. Sinusoid Steady-State Analysis 1 1 1 jC Z eq 10 6 jL (6 jL)(6 jL) 10(6 jL) 10( jC )(6 jL)(6 jL) 10(6 jL)(6 jL) 10(6 jL)(6 jL) 10(6 jL)(6 jL) 36 2 L2 60 10 jL 10 jC (36 2 L2 ) 10(36 2 L2 ) 1 10L 10C (36 2 L2 ) Im 0 2 2 10(36 L ) Z eq L C 36 40 1 36 2 2 2 L ( 40 ) 10L 10C (36 2 L2 ) 0 50,000 rad/s Sinusoid Steady-State Analysis Example 9.7 Suppose we can vary the frequency of the current source. What frequency will cause is(t) and v(t) to be in-phase in the steady-state? For 50,000 rad/s Z L jL j (50,000)( 40 ) j 2 P j j ZC j 20 C (50,000)(1 ) Z eq 10 || (6 j 2) || j 20 4