Digital Systems Basic Circuit Theory Review I CMPE 650 Properties of Sinusoids Relationship: ω = 2πf f ( t ) = V m cos ( ωt + φ ) f(ωt) A ω is the angular frequency in radians/sec φ is the argument or phase angle (in radians) Period T = 2π/ω = 1/f v2 = Vmsin(ωt + φ) v1 = Vmsin(ωt) π 2π 3π φ -A YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC + - 1 (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Properties of Sinusoids Trig identities sin ( ωt + 90° ) = cos ( ωt ) sin ( ωt – 90° ) = – cos ( ωt ) cos ( ωt + 90° ) = – sin ( ωt ) cos ( ωt – 90° ) = sin ( ωt ) sin ( ωt + 180° ) = – sin ( ωt ) cos ( ωt + 180° ) = – cos ( ωt ) sin ( A + B ) = sin A cos B + cos A sin B sin ( A – B ) = sin A cos B – cos A sin B cos ( A + B ) = cos A cos B – sin A sin B cos ( A – B ) = cos A cos B + sin A sin B Don’t confuse with complex plane described below Adding two sinusoids of same ω +cos(ωt) A cos ( ωt ) + B sin ( ωt ) = C cos ( ωt – θ ) C = 2 A +B 2 θ = tan – 90° – 1 B -- A +sin(ωt) YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC 2 (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Complex Numbers Complex numbers: rectangular form Im jy z = x + jy j = –1 1 --- = – j j z 2 j = –1 r θ 3 j = –j 4 x j = 1 Polar form z = x + jy = r ∠θ = r cos ( θ ) + jr sin ( θ ) z = z ∠θ r = 2 2 x +y –1 y θ = tan -- x rect-to-polar YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 or UMBC – 1 y x = r cos ( θ ) z = x + jy θ = tan y = r sin ( θ ) z = -x + jy θ = 180° – tan polar-to-rect (-, -) 3 180 + (+, -) - x – 1 y - x 360 - (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Complex Numbers Complex conjugate * z = x – jy = r ∠– θ = e – jθ (Last form is exponential form) Sum/difference: use rectangular form. Mult/div: use polar form r1 z1 z 1 z 2 = r 1 r 2 ∠( θ 1 + θ 2 ) ----- = ----- ∠( θ 1 – θ 2 ) r2 z2 Exponential form jθ (similar to polar form) z = re from Euler’s formula e e jθ – jθ = cos ( θ ) – j sin ( θ ) Adding: Subtracting: YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 jθ cos ( θ ) = Re ( e ) = cos ( θ ) + j sin ( θ ) UMBC jθ sin ( θ ) = Im ( e ) – jθ 1 jθ cos ( θ ) = --- ( e + e ) 2 1 jθ – jθ sin ( θ ) = ----- ( e – e ) 2j 4 (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Complex Numbers Useful identities Assume z = x + jy = r ∠θ * 2 2 zz = x + y = r z = x + jy = n 2 re n jθ ⁄ 2 r ∠( θ ⁄ 2 ) = n n jnθ n z = ( x + jy ) = r ∠nθ = r e jθ ln ( re ) = ln ( r ) + ln e e e e jπ jθ = ln ( r ) + jθ = –1 j2π = e jπ ⁄ 2 j0 = e = 1 j90° = j (1 + j) j = ---------------2 – jπ ⁄ 2 j270° = e = –j e e jπ ⁄ 4 Re ( e YLAND BA L MO U M B C TI UN RE COUNT Y IVERSITY O F AR 1966 = e j45° ( α + jω )t Im ( e M = r ( cos ( nθ ) + j sin ( nθ ) ) ( α + jω )t UMBC (IMPORTANT IDENTITIES FOR US) 2 j = (1 + j) = αt jwt 1 --- = j αt ) = Re ( e e ) = e ) = Im ( e e ) = e αt jwt 5 αt (1 – j) – j = ---------------2 cos ( ωt ) sin ( ωt ) (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Phasors Sines and cosines can be expressed in terms of phasors. A phasor is a complex number that represents the amplitude and phase of a sinusoid. Phasors based on Euler’s identity e jθ = cos ( θ ) + j sin ( θ ) and e – jθ = cos ( θ ) – j sin ( θ ) So cos(θ) and sin(θ) represent the Re and Im parts of ejθ. We can write a sinusoid v ( t ) = V m cos ( ωt + φ ) = Re ( V m e jφ jωt v ( t ) = Re ( V m e e ) j ( ωt + φ ) ) Arg includes phase and time part Defining the phasor as the magnitude and the non-time part of the arg v ( t ) = Re ( V e YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC jωt ) V = V me 6 jφ = V m ∠φ (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Phasors The plot a rotating phasor: sinor Vejωt Rotation at ω (rad/s) Vm Im Re φ Vm Projection on real axis v ( t ) = Re ( V e jωt ) t1 t at t = t1 -Vm Note value of sinor at t = 0 is phasor V of the sinusoid v(t). NOTE: the term ejωt is implicitly present whenever a sinusoid is expressed as a phasor. This is where the third parameter, ω, is realized -- in the time part. It is the third constant associated with the sinusoid. YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC 7 (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Phasors To obtain the sinusoid associated with a phasor, multiply phasor V by ejωt and take the Re part. v ( t ) = Re ( V e jωt ) = Re ( V m e j ( ωt + φ ) ) = Re ( V m cos ( ωt + φ ) + jV m sin ( ωt + φ ) ) v ( t ) = V m cos ( ωt + φ ) To obtain the phasor associated with a sinusoid, first write the sinusoid in cosine form (so it can be written as the real part of a complex number), and take out the time factor. v ( t ) = V m cos ( ωt + φ ) = Re ( V m e j ( ωt + φ ) jφ jωt ) = Re ( V m e e ) = V me jφ = V m ∠φ Phasors are complex, and therefore can be represented in rectangular, polar or exponential form. They are vectors (magnitude and direction) and can be plotted as such. The phasor domain is also known as the frequency domain. YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC 8 (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Phasors Taking the derivative of a sinusoid dv ------ ( V m cos ( ωt + φ ) ) = – ωV m sin ( ωt + φ ) = ωV m cos ( ωt + φ + 90 ) dt jωt jφ j90° jωt dv ------ ( V m cos ( ωt + φ ) ) = Re ( ωV m e e e ) = Re ( jωV e ) dt The relationships of derivative and integral dv ------ ⇔ jωV dv V -----v d t ⇔ ∫ jω These relationships are useful in computing the steady-state response (where we don’t need to know the initial values). Phasors are also useful for summing sinusoids of the same frequency. YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC 9 (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Phasors Important FACTs to remember: • v(t) is time dependent, while V is NOT. • v(t) is ALWAYS real, while V is generally complex. • Phasor analysis applies only when frequency is constant and allows two or more sinusoids to be manipulated ONLY if they have the same frequency. Examples – sin A = cos ( A + 90° ) v = – 4 sin ( 30t + 50° ) V v = 4 cos ( 30t + 50° + 90° ) = 4 cos ( 30t + 140° ) V = 4 ∠140° V I = – 3 + j4 A I = 5 ∠126.87° i ( t ) = 5 cos ( ωt + 126.87° ) A (Phasors can be in any of the three forms) – j20° since j = 1 ∠90° V = j8e V V = j8 ∠– 20° = ( 1 ∠90° ) ( 8 ∠– 20° ) = 8 ∠90° – 20° = 8 ∠70° v ( t ) = 8 cos ( ωt + 70° ) V YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC 10 (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Phasors Used to solve an integrodifferential equation Solve for i(t) di 4i + 8 ∫ i dt – 3 ----- = 50 cos ( 2t + 75° ) dt 8I 4I + ------ – 3 jωI = 50 ∠75° jω Convert to phasor domain ω is 2 and- 1/j = -j. I ( 4 – j4 – j6 ) = 50 ∠75° 50 ∠75° 50 ∠75° I = ------------------- = --------------------------------------- = 4.642 ∠143.2° 10.77 ∠( – 68.2 )° 4 – j10 2 2 4 + 10 = 10.77 – 1 10 φ = tan ------ = 360° – 68.2° = 291.8° 4 m = φ = 75° – 291.8° = – 216.8 ° + 360 = 143.2° So far we’ve shown how to represent a voltage and current in phasors. What about the passive elements, R, L and C YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC 11 (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Phasors Applied to R, L and C circuits Need to transform the current/voltage relationships for R, L and C into the frequency domain. For R i = I m cos ( ωt + φ ) v = iR = RI m cos ( ωt + φ ) V = RI m ∠φ I = I m ∠φ V = RI Assume current through a resistance R And voltage across it The phasor for this voltage Phasor for the current Current and voltage are in phase Im V I Re YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC 12 (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Phasors Applied to R, L and C circuits For L i = I m cos ( ωt + φ ) di v = L ----- = – ωLI m sin ( ωt + φ ) dt di v = L ----- = ωLI m cos ( ωt + φ + 90° ) dt V = ωLI m e j ( φ + 90° ) Assume current through inductor L Voltage across it – sin ( A ) = cos ( A + 90° ) jφ j90° = ωLI m e e V = jωLI Into a phasor e j90° = j Im Voltage leads the current by 90 degrees for inductor V I Re The j term translates into a phase shift in the time domain. YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC 13 (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Impedance For C I V = ----------jωC dv i = C -----dt or I = jωCV Here, current leads voltage by 90 degrees The first form relates V and I similar to resistance. These can be written in terms of their phasor voltage to phasor current V ---- = R I V ---- = jωL I V 1 ---- = ----------jωC I V Z = ---I or V –j ---- = -------I ωC Impedance Z is the ratio of phasor voltage to current in Ωs NOTE: the impedance Z is NOT a phasor -- it does not correspond to a sinusoidally varying quanity. YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC 14 (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Impedance Example v = 12 cos ( 60t + 45° ) L = 0.1 H V = jωLI Find steady-state current with w = 60 rad/s and V = 12 ∠45° 12 ∠45° 12 ∠45° V I = ---------- = --------------------- = ------------------- = 2 ∠– 45 ° A j60 ( 0.1 ) 6 ∠90° jωL So we see the magnitude of the impedance is 6 Ωs and the inductor causes a phase shift in the current (it lags by 90 degrees). Z = R + jX Z = R – jX R = Re ( Z ) X = Im ( Z ) resistance reactance Reactance can be negative (capacitive, current lags) or positive (inductive). Other forms Z = R + jX = Z ∠θ Z = 2 R +X 2 R = Z cos ( θ ) YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC 15 θ = tan –1 X --- R X = Z sin ( θ ) (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Admittance Admittance Y is the reciprocal of impedance, measured in siemens (S) 1 I Y = --- = ---Z V Y = G + jB G = Re(Y) is called conductance (mohms) B = Im(Y) is called susceptance 1 G + jB = ---------------R + jX 1 R – jX R – jX G + jB = ---------------- ---------------- = -------------------2 2 R + jX R – jX R +X R G = -------------------2 2 R +X Relating impedance with admittance –X B = -------------------2 2 R +X So, G does NOT equal 1/R as was true for resistive circuits. YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC 16 (3/18/08) Digital Systems Basic Circuit Theory Review I CMPE 650 Examples 5Ω V s = 10 ( ∠0° )V v s = 10 cos ( 4t ) + - 0.1F The impedance is 1 1 Z = R + ----------- = 5 + ------------------ = 5 – j2.5 Ω j4 ( 0.1 ) jωC And current Vs 10 ( 5 + j2.5 ) 10 ∠0° I = ------ = ------------------- = ------------------------------ = 1.6 + j0.8 = 1.789 ∠26.57° 2 2 5 – j2.5 Z 5 + 2.5 OR Vs 10 ∠0° I = ------ = ----------------------------- = 1.789 ∠26.57° 5.6 ∠– 26.57 Z Voltage across capacitor I 1.789 ∠26.57° 1.789 ∠26.57° V = IZ C = ----------- = ---------------------------------- = ---------------------------------- = 4.47 ∠– 63.43° jωC j4 ( 0.1 ) 0.4 ∠90° Time domain i ( t ) = 1.789 cos ( 4t + 26.57° ) v ( t ) = 4.47 cos ( 4t – 63.43° ) YLAND BA L U M B C MO UN RE COUNT Y IVERSITY O F AR TI M 1966 UMBC 17 (3/18/08)
© Copyright 2024 Paperzz