The stability of linear time-invariant feedbac systems A. Theory The system is atrictly stable if The degree of the numerator of H(s) (H(z)) the degree of the denominator of H(s) (H(z)) and/or The poles of H(s) left half plane (interior of the unit circle for the discrete-time case). Wide sense stable systems the poles of H(s), s p imaginary axis jω / the poles of H(z) z p unit circle r (discrete-time). Strictly stable system if and only if the denominator of the transfer function H(s) is a Hurwitz polynomial. Q(s) strictly/wide sense Hurwitz if Has real coefficients, (strictly or not) positive + The roots of Q(s) (poles) are in the left half plane (which means Re{z} < 0) A strictly Hurwitz polynomial has all the minors of n th order, Δ n > 0 (its zeros are in the closed left half plane). A wide sense Hurwitz polynomial can have some minors zero Δ 0, and the rest positive (some of its zeros are in the closed left half plane and if there are zeros on the imaginary axis). Δ n a a a a 3 5 7 a a a a 0 4 6 3 5 4 6 3... 0... 0 0 a a a... 0 0 a a a... 0 0 0 a a... 0........ 0...... a n Y s X s H s (continuous time) + H s G s Y z H z (discrete time) X z + H z G z
H( s) G( s) respectively W( z) H( z) G( z) W s the transfer function in open loop. The open loop system is strictly stable if The roots +W(s)0 closed left half plane (the real part <0) or The roots +W(z)0 interior of the unit circle r The Nyquist s Stability Criterion Y s X s H s respectively + H s G s Y s X s H s + H s G s + H( s) G( s) R( z) + H( z) G( z) R s + doesn t have zeros in the right half plane or the imaginary axis jω R ( s) H( s) G( s) > H( jω ) G( jω ) ω (, ) Nyquist hodograph of the open loop system. The number of complete rotations around the point (-/, 0) in counter-clocwise sense is n times where n n + n, i c n i number of poles from the right half plane and n c number of poles from the imaginary axis (continuous time) + doesn t have zeros outside the unit circle r (ze jω ) H( z) G( z) R z i n number of poles outside the unit circle r, c n number of poles on the unit circle r (discrete time)
B. Problems Problem. Consider the amplifier with the transfer function Ga H( s) s + a a) Find the DC gain of the amplifier. b) What is the time constant of the system? c) The bandwidth of the amplifier is the frequency where the magnitude of the frequency response is times smaller than the DC gain. What is the bandwidth of this amplifier? d) The amplifier is placed into a feedbac loop: x(t) Ga s+ a y(t) K What is the DC gain, the time constant and the bandwidth of the feedbac system? e) Find the value for which the bandwidth of the closed loop system is double the bandwidth in open loop. Solution. a) H( jω ) L at Ga jω + a b) e σ () t. s+ a If ht () e τ σ () t t. The DC gain is H ( 0 ) G then τ is the time constant of the system at ht () Gae σ () t τ a G B H jωm H jω Ga G ω T a ω + a c) The bandwidth is. But M Ga ω + a d) + Y s X s Y s H s Y s X s H s Y s H s Y s H s X s H s 3
Ga Y( s) H( s) s+ a Hbî s X ( s) + H( s) Ga + s + a G The DC gain of the feedbac system is Hclosed ( 0) + G Ga Ga Hclosed ( s) s+ a+ Ga s+ a( + G) The time constant is τ a + G H closed ( ω) Ga jω + a + G ( ) Ga G jωcclosed + a( + G) ( + G) Ga G ( ω ) + a ( + G) ( + G) cclosed ( ωcclosed) + a ( + G) ( + G) a ( + G) ( ωcclosed) + a ( + G) ω a( + G) cclosed a e) The cutoff frequency (bandwidth) of the closed loop system: ωcclosed ω T a( + G) a. G Problem. Determine the transfer function for each system represented below. a) x(t) H 0 (s) u(t) H (s) y(t) G(s) 4
b) H 0 (z) x[n] u[n] H (z) v[n] y[n] w[n] G(z) c) x(t) u(t) v(t) w(t) H (s) H (s) y(t) G (s) G (s) Solution. a) b) U( s) X ( s) H0 ( s) + Y( s) H ( s) U( s) + G( s) H ( s) Y( s) H ( s) X ( s) H0( s) + G( s) H( s) Y( s) H( s) H0( s) X ( s) + G( s) H ( s) Y s U s Y s G s H s Y s G s H s U s H s un [ ] xn [ ] wn [ ] şi V ( z) U( z) H ( z) U( z) + G( z) H ( z) X ( z) X ( z) U( z) + G( z) H ( z), but V ( z) U z H ( z) U z X z W z X z G z V z X z G z H z U z 5
H ( z) X ( z) V ( z) + G ( z ) H ( z ) 0 + H ( z) X ( z) X ( z) H0 ( z) + + G( z) H ( z) Y( z) H ( z) H0 ( z) + X ( z) + G( z) H ( z) Y z X z H z V z c) U( s) X ( s) G ( s) Y( s) V ( s) U( s) H ( s) W ( s) V ( s) Y( s) G ( s) Y( s) W ( s) H ( s) Y( s) H( s) V( s) Y( s) G( s) Y( s) H( s) H( s) U( s) Y( s) G( s) H ( s) H ( s) X ( s) G ( s) Y( s) Y( s) G ( s) { } + + Y s H s H s G s H s G s H s H s X s Y s H s H s. X s + H s H s G s + H s G s Problem 3. Consider the system described by the linear constant-coefficients differential equation y + 6y + 9y + y x + 6x + 8x with null initial conditions. Study the stability of the system based on the Hurwitz stability criterion. Solution. + 6 + 9 + + 6 + 8 3 Y( s) s + 6s + 8s 3 X ( s) s + 6s + 9s+ 3 3 s Y s s Y s sy s Y s s X s s X s sx s The numerator has positive coefficients, which means we can apply Hurwitz criterion. a0, a 6, a 9, a3 We compute the determinant: a a3 a5 6 0 Δ 3 a0 a a4 9 0 0 a a 0 6 3 6
and the minors: Δ 6> 0 Δ 54 53 > 0 Δ 69 + 60 + 00 090 066 53Δ All the minors are positive, meaning the system is strictly stable. 3 Problem 4. Consider the following closed loop system x(t) H(s) y(t) G(s) Find the values for for which the system is stable, if s a. H( s) G( s) f. H ( s ) G ( s ) + s b. H( s) G( s) g. H ( s ) G ( s ) c. H( s) G( s) d. H( s) G( s) e. H( s) G( s) s ( s + ) + 0 s s + H s G s h. ( s + ) 3 ( s + ) 4 s + s 4 i. H( s) G( s) s + s+ (homewor). Solution. a. H( s) G( s) H jω G jω s + jω +, s p - In order to setch the hodograph we can use the Bode plots. H( jω) G( jω) ω + ω 0 log H( jω) G( jω) 0 log ( ω + ) 0 log + ω arg{ H( jω ) G( jω) } arg ( jω) arctg + arctgω 7
Counter-clocwise ω - ω0 ω± ω 8
The open loop system is stable the hodograph doesn t have to mae rotations around the point of coordinates,0 0 0 or (, 0) < > > The closed loop system is stable for >-! s The open loop system is unstable with one pole in the left half plane s p H( jω) G( jω) ω + ω arg{ H( jω ) G( jω) } arg ( + jω) π + arctg π + arctgω jω+ jω+ H( jω) G( jω) jω ω ω + ω + j ω + ω + b. H( s) G( s) ω > Re{ H( jω) G( jω )} < 0, { H( j ) G( j )} For 0 Re Im Im ω ω < 0 ω Re{ H( jω ) G( jω )} Im{ H( jω ) G( jω )} 0-0 -/ -/ 0 0 The Nyquist hodograph maes rotations in counter-clocwise sense. The system is stable if the point of coordinates,0 is encircled one time in counterclocwise sense: > > >. The system is stable for >. < 0 > 0 9
counterclocwise ω0 ω± ω 0
. The open loop system is stable. s ( s + ) + 0 H( jω) G( jω) H jω G jω jω ( jω + ) + 0 c. H( s) G( s) ω + ω + 0 ω ω 0 log H( jω) G( jω) 0 log + 0 log + 0 ω arg{ H( jω) G( jω) } arctg ( ω) arctg 0 < 0 >. > The sense of rotation of the hodograph isn t counterclocwise the point shouldn t be encircled.,0
counterclocwise ω ω0 ω0 ω H s G s s p, ±. The open loop system is unstable. s H( jω ) G( jω) Re { H( jω) G( jω) }; Im{ H( jω) G( jω )} 0 ω ω + d. ω 0 Re H jω G jω - -/ 0 { } Sensul e antiorar ω0 ω ω
The open loop system has a pole in the right half plane; the closed loop system is stable if the critical point is encircled one time in counter-clocwise: < < 0 > H s G s H( jω) G( jω) + + e. ( s ) ( jω ) ( ω) ( ω) + H jω G jω H j G j ω 0log 0log + ω arg H jω G jω arctg ω { } The open loop system is stable, therefore the critical point shouldn t be encircled: 0 sau 0 sau 0 < > > < < > 3
Counterclocwise ω0 ω0 ω H s G s H( jω) G( jω) + + f. ( s ) ( ω) ( ω) H j G j ( jω ) 3 3 ( + ω ) arg 3 ω 0log H( jω) G( jω) 0log + { H( jω ) G( jω) } 3arctg ( ω). Counterclocwise sense or 0 or 8,8, 0 > < 8 < < < 3 4
Counterclocwise ω 3 Re-/8 ω ω0 ω g. homewor. s + h. H( s) G( s) s p, ± Re{ sp} > 0 s 4 The open loop system is unstable. jω + H( jω) G( jω) Re { H( jω) G( jω) }, ω 4 ω + 4 ω Im{ H( jω) G( jω) } ω + 4 Re H j G j Im H jω G jω { } { } ω ( ω) ( ω ) 0 --4 0 -/5 -/5 0 0 The system is stable for < < 0 > 4 and > 0 > 4. 4 5
Counter clocwise ω0 ω ω i. H( s) G( s) s p, j Re{ sp,} 0 s + s+ ± < It results the open loop system is stable. ( ω) ( ω) H j G j Re ω jω ω jω 4 ω + jω+ + 4ω 4+ ω ω 4 4 ω ( ω ) { } 4 { H( jω) G( jω) } ; Im H( jω) G( jω) + ω 0 { } { } Re H jω G jω / 0 0 Im H jω G jω 0 /6 0 < 0 sau > > 0 sau > (, ) ω 4 + ω 6
Counter-clocwise ω ω0 ω 7