EE 230 Lecture 24. Waveform Generators. - Sinusoidal Oscillators

EE 230 Lecture 24 Waveform Generators - Sinusoidal Oscillators

Quiz 18 Determine the characteristic equation for the following network without adding an excitation. C R L

And the number is? 1 3 8 2? 6 5 4 9 7 7

Quiz 18 Determine the characteristic equation for the following network without adding an excitation. V X C R L Solution: V X 1 G+sC+ sl 0 1 1 CR LC 2 VX s +s + = 0 2 1 1 D s = s +s + CR LC

Review from Last Lecture X IN Linear Network X OUT Ns T s = Ds Theorem: The poles of any transfer function of a linear system are independent of where the excitation is applied and where the response is taken provided the dead network for the systems are the same. Or, equivalently, Theorem: The characteristic equation D(s) of a linear system are independent of where the excitation is applied and where the response is taken provided the dead network for the systems are the same.

Review from Last Lecture Poles of a Network X IN Linear Network X OUT Ns T s = Ds X IN =0 Linear Network X OUT Ds Dead Network

Review from Last Lecture Theorem: The characteristic polynomial D(s) of a system can be obtained by assigning an output variable to the dead network of the system and using circuit analysis techniques to obtain an expression that involves only the output variable expressed as X 0 F(s)=0 where F(s) is a polynomial. When expressed in this form, the characteristic polynomial of the system is D(s)=F(s)

Review from Last Time: Pole Locations of Waveform Generators C R Inverting Integrator V OUT2 R 1 R 2 Noninverting Comparator with Hysteresis θ = V OUT1 R1 R R 1 2 p = 1-θ 1 θ RC V OUT1 C R V OUT2 p = R2 1 R RC 1 R1 R 2 Im Re Both have a single pole on the positive real axis

Sinusoidal Oscillators The previous two circuits provided square wave and triangular ( triangularish ) outputs - previous two circuits had a RHP pole on positive real axis What properties of a circuit are needed to provide a sinusoidal output What circuits have these properties

What properties of a circuit are needed to provide a sinusoidal output? Insight into how a sinusoidal oscillator works Characteristic Equation Requirements for Sinusoidal Oscillation (Sec 13.1) Barkhausen Criterion (Sec 13.1)

Insight into how a sinusoidal oscillator works X IN Ns T s = Ds X OUT Linear Network {p 1, p 2,... p n} If e(t) is any excitation, no matter how small, can be expressed as N E E E s = D s E s = N s s-x s-x... s-x 1 2 m Then the output can be expressed, in the s-domain, as N s N s E R s =E s T s = D E s D s

X IN Insight into how a sinusoidal oscillator works Linear Network X OUT {p, p,... p } Ns T s = 1 2 n Ds N s N s E R s = D E s D s Using a partial fraction expansion for R(s) obtain a a a b b b R s =...... s-p s-p s-p s-x s-x s-x 1 2 n 1 2 m 1 2 n 1 2 m r t = L -1 R s Then the output can be expressed, in the time domain, as r t a e + a e +...+ a e + b e + b e +...+ b e p1t p2t pnt x1t x2t xmt 1 2 n 1 2 m If the excitation is very small and vanishes, or is zero, the term due to the excitation will vanish

X IN Insight into how a sinusoidal oscillator works Linear Network X OUT Ns T s = 1 2 n Ds {p, p,... p } E N s N s R s = D s D s E Under the assumption that the excitation vanishes, r t a e + a e +...+ a e p1t p2t pnt 1 2 n every pole can be expressed as the sum of a real part and imaginary part p k= α k+ jβk (where the real part or the imaginary part may be 0) r t a e e + a e e +...+ a e e 1t j 1t 2t j 2t nt j nt 1 2 n

X IN Insight into how a sinusoidal oscillator works Linear Network X OUT Ns T s = 1 2 n Ds {p, p,... p } E N s N s R s = D s D s E Theorem: If a linear network has a pole p k= α k+ jβk with a non-zero imaginary part, then its complex conjugate the network. p k= αk - jβk is also a pole of That is, all poles that are not on the real axis appear as complex-conjugate pairs Im Re

X IN Insight into how a sinusoidal oscillator works Linear Network X OUT Ns T s = 1 2 n Ds {p, p,... p } E N s N s R s = D s D s E r t a e e + a e e +...+ a e e 1t j 1t 2t j 2t nt j nt 1 2 n The terms on the right can be simply regrouped and renamed as ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1t j 1t j j ˆ rt j 1t j j ˆ 3t 3t ˆ 3t 3t... ˆ kt kt ˆ kt kt 1 2 3 4 k k+1 r t a e e a e e + a e e a e e a e e a e e +...+ aˆ e... aˆ e ˆ t k+2 n k+2 n ˆ t The first group of terms all correspond to complex conjugate pair poles and the second group to those that lie on the real axis

X IN Insight into how a sinusoidal oscillator works Linear Network X OUT Ns T s = 1 2 n Ds {p, p,... p } E N s N s R s = D s D s E r t aˆ e e aˆ e e + aˆ e e aˆ e e aˆ e e aˆ e e ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1t j 1t 1t j 1t 3t j 3t 3t j 3t kt j kt kt j kt 1 2 3 4... k k+1 +...+ aˆ e... aˆ e ˆ t k+2 n k+2 n ˆ t Theorem: The coefficient in the partial fraction expansion corresponding to a pole with a non-zero imaginary part is the complex conjugate of the term corresponding to the complex conjugate pole. r t aˆ e e aˆ e e + aˆ e e aˆ e e aˆ e e aˆ e e ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1t j 1t ' 1t j 1t 3t j 3t ' 3t j 3t kt j kt ' kt j kt 1 1 3 3... k k +...+ aˆ e... aˆ e ˆ t k+2 n k+2 n ˆ t

X IN Insight into how a sinusoidal oscillator works Linear Network X OUT Ns T s = 1 2 n Ds {p, p,... p } E N s N s R s = D s D s E r t aˆ e e aˆ e e + aˆ e e aˆ e e aˆ e e aˆ e e ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1t j 1t ' 1t j 1t 3t j 3t ' 3t j 3t kt j kt ' kt j kt 1 1 3 3... k k +...+ aˆ e... aˆ e ˆ t k+2 n k+2 n ˆ t Consider now one of the complex conjugate pair terms aˆ e e aˆ e e k ˆ ˆ kt j kt ' kt j kt k aˆ e e aˆ e e =2 aˆ e cos t ˆ ˆ ˆ ˆ ˆ ˆ kt j kt ' kt j kt kt k k k k ˆ1t ˆ ˆ3t ˆ ˆkt... ˆ 1 1 3 3 k k r t 2 aˆ e cos t + 2 aˆ e cos t 2 aˆ e cos t +...+ aˆ e... aˆ e ˆ t k+2 n k+2 n ˆ t

X IN Insight into how a sinusoidal oscillator works Linear Network X OUT Ns T s = 1 2 n Ds {p, p,... p } E N s N s R s = D s D s E ˆ1t ˆ ˆ3t ˆ ˆkt... ˆ 1 1 3 3 k k r t 2 aˆ e cos t + 2 aˆ e cos t 2 aˆ e cos t +...+ aˆ e... aˆ e ˆ t k+2 n k+2 n ˆ t Consider now three cases, the real part of the pole is negative, the real part of the pole is positive, and the real part of the pole is 0 It the real part of the pole is negative, the corresponding term in r(t) will vanish It the real part of the pole is positive, the corresponding term in r(t) will diverge to +/- It the real part of the pole is zero, the corresponding term in r(t) will be a sinusoidal waveform that will persist forever The condition for sinusoidal oscillation should be apparent!

Sinusoidal Oscillation Im Re A circuit with a single complex conjugate pair of poles on the imaginary axis at +/- jβ will have a sinusoidal output given by X t =2 aˆ sin( t+ ) OUT k The frequency of oscillation will be β rad/sec but the amplitude and phase are indeterminate

Sinusoidal Oscillation Criteria A network that has a single complex conjugate pair on the imaginary axis at jω and no RHP poles will have a sinusoidal output of the form X 0 (t)=asin(ωt+θ) A and θ can not be determined by properties of the linear network

Characteristic Equation Requirements for Sinusoidal Oscillation Linear Network X OUT Characteristic Equation Oscillation Criteria: If the characteristic equation D(s) has exactly one pair of roots on the imaginary axis and no roots in the RHP, the network will have a sinusoidal signal on every nongrounded node.

Barkhausen Oscillation Criteria Consider a basic feedback amplifier X IN + - A X OUT β A FB 1 A Aβ Aβ is termed the loop gain

Barkhausen Oscillation Criteria X IN + - A X OUT A β A FB 1 Aβ Barkhausen Oscillation Criteria: A feedback amplifier will have sustained oscillation if Aβ=-1 There are many subtly different ways various authors present Barkhausen criteria but all invariably state, in some way, that must have Aβ=-1 Some state it must occur at only one frequency, others make comments about waveshape, Sedra and Smith convey right idea but are neither rigorous or completely correct, and most other authors have a similar problem

If sinusoidal oscillation is required, be very careful about using Barkhausen Criteria

Characteristic Equation Oscillation Criteria (CEOC) If the characteristic equation D(s) has exactly one pair of roots on the imaginary axis and no roots in the RHP, the network will have a sinusoidal signal on every nongrounded node. Sinusoidal Oscillator Design Strategy Build networks with exactly one pair of complex conjugate poles slightly in the RHP and use nonlinearities in the amplifier part of the network to limit the amplitude of the output. i.e. p=α ±jβ where α is very small but positive Nonlinearity in amplifier will result in a small amount of distortion Frequency of oscillation will deviate slightly from β Poles must be far enough in the RHP so process and temperature variations do not cause movement back into LHP because if that happened, oscillation would cease!

Sinusoidal Oscillator Design Strategy Build networks with exactly one pair of complex conjugate poles slightly in the RHP and use nonlinearities in the amplifier part of the network to limit the amplitude of the output. i.e. p=α ±jβ where α is very small but positive Im Re Know Barkhausen Crieteria to answer interviewers questions but use CEOC criteria to design sinusoidal oscillators!

Sinusoidal Oscillator Design Consider the following circuit: V 0 R 2 C 2 R 1 C 1 KV 0 By KCL sc G V G +sc + = KV sc G 2 2 2 2 0 1 1 0 G 2+sC2 G 2+sC2 1 1 1-K 1 2 V0 s +s + + + = 0 R2C2 R1C 1 R2C1 R1R 2C1C 2 1 1 1-K 1 R C R C R C R R C C 2 D s = s +s + + + 2 2 1 1 2 1 1 2 1 2

Sinusoidal Oscillator Design Consider the following circuit: V 0 R 2 C 2 R 1 C 1 KV 0 1 1 1-K 1 R C R C R C R R C C 2 D s = s +s + + + 2 2 1 1 2 1 1 2 1 2 If the coefficient of the s term is set to 0, will have cc poles at s = j 1 R R C C 1 2 1 2 Thus, it the coefficient of the s term vanishes, it will be a sinusoidal oscillator of frequency ω = OSC 1 R R C C 1 2 1 2

Sinusoidal Oscillator Design Consider the following circuit: V 0 R 2 C 2 R 1 C 1 KV 0 1 1 1-K 1 R C R C R C R R C C 2 D s = s +s + + + Oscillation Condition: 2 2 1 1 2 1 1 2 1 2 1 1 1-K + + R C R C R C 2 2 1 1 2 1 0 ω = OSC 1 R R C C 1 2 1 2 This is achieved by having K satisfy the equation C R K + C R 1 1 2 2 1

Sinusoidal Oscillator Design Consider the special practical case where R 1 =R 2 =R and C 1 =C 2 =C: V 0 R C C1 R2 K 1 + 3 C R 2 1 R C KV 0 1 ω OSC = RC This is termed the Wein Bridge Oscillator One of the most popular sinusoidal oscillator structures Practically make K slightly larger than 3 and judiciously manage the nonlinearities to obtain low distortion

The Wein-Bridge Oscillator Another Perspective: V 0 R 2 C 2 R 1 C 1 KV 0 Z 1 V 0 Z 2 R 2 C 2 C1 R2 K 1 + 3 C R 2 1 R 1 C 1 KV 0

The Wein-Bridge Oscillator Another Perspective: V 0 R 2 C 2 R 1 Z 1 R 2 C 2 V 0 Z 2 C 1 KV 0 R 1 C 1 KV 0 C1 R2 K 1 + 3 C R 2 1 V 0 + K Z 2 V X Note this is a feedback amplifier with gain K and Z1 β = Z +Z 1 2 Z 1 Lets check Barkhausen Criteria for this circuit

End of Lecture 24

The Wein-Bridge Oscillator Lets check Barkhausen Criteria for this circuit V 0 R 2 C 2 + K V X R 1 C 1 KV 0 V 0 Z 2 Z β = Z +Z Z 1 1 1 2 C1 R2 K 1 + 3 C R 2 1 Z KR C s Kβ = Z +Z s R R C C +s R C +R C +R C +1 1 1 1 2 1 2 1 2 1 2 1 2 1 1 2 2 Setting Kβ= -1, obtain KR C s 1 1 2 s R1R 2C1C 2+s R1C 2+R1C 1+R2C 2 +1 1 2 s R1R 2C1C 2+s R1C 2+R1C 1+R2C2 KR1C 1 +1=0

The Wein-Bridge Oscillator Lets check Barkhausen Criteria for this circuit V 0 R 2 C 2 + K V X R 1 C 1 KV 0 V 0 Z 1 Z 2 Z 1 β = Z 1 +Z 2 C1 R2 K 1 + 3 C R 2 1 Putting in s= jω, obtain the Barkhausen criteria 2 1-ω R1R 2C1C 2 +j ω R1C 2+R1C 1+R2C2 KR1C 1 =0 Solving, must have (from the imaginary part) C R K + C R 1 1 2 2 1 And this will occur at the oscillation frequency of (from real part) ω = OSC 1 R R C C 1 2 1 2

The Wein-Bridge Oscillator R 1 R 2 C2 Basic implementation for the equal R, equal C case V 0 C 1 KV 0 + K V X (2+ε 1 )R 1 V 0 Z 1 Z 2 R 1 V 0UT R C R C 1 ω OSC = RC

The Wein-Bridge Oscillator R 1 (2+ε 1 )R 1 Amplifier Transfer Characteristics R C V 0UT V OUT R C 3 V SATH 1 1 ω OSC = RC V IN V SATL Slope slightly larger than 3 Amplitude of oscillation will be approximately V SATH (assuming V SATH =-V STATL ) Distortion introduced by the abrupt nonlinearities when clipping occurs

The Wein-Bridge Oscillator R 1 (2+ε 1 )R 1 V 0UT Amplifier Transfer Characteristics R C V OUT R C V SATH 1 3 1 ω OSC = RC V IN V OUT V SATH 1 3 V SATL V IN V SATL Abrupt nonlinearities cause distortion Better performance (reduced nonlinearity) can be obtained by introducing less abrupt nonlinearities to limit amplitude