Transcript Chapter 3
Network Analysis and Synthesis Chapter 3 Elements of Realizability Theory Introduction • In the last chapter we were concerned with the problem of identifying the response given the excitation and network. • When we discuss about synthesis we are concerned with the problem of constructing a network given the excitation and response. • The starting point for any synthesis is the system function H (s) R (s) E (s) • The first step in synthesis procedure is determining whether the system function can be realized with a physical passive network. • There are two considerations – Causality and – Stability 1. Causality • By causality we mean that a voltage doesn’t appear between any terminals in the network before a current/voltage is applied. • In other words, the impulse response of the network must be zero for t<0. h ( t ) 0 for t0 Example 1. h(t)=etu(t) • Is causal because for t<0 u(t)=0, hence, h(t)=0. 2. h(t)=e|t| • Is not causal because for t<0, h(t) is not zero. • In certain cases, the network can be made causal by delaying the impulse response by a certain time delay. • If we delay h(t) by T, then h(t-T) will be causal. • In the frequency domain causality is implied when the Paley-Wiener criterion is satisfied. • The Paley-Wiener criterion states that a necessary and sufficient condition for causality is log H ( jw ) w 1 2 dw • The physical implication of the Paley-Wiener criterion is that the amplitude response of a causal network can’t be zero over a finite band of frequency. Example 1. The ideal low pass filter – Is not causal because it is zero for all frequencies w>wc. 2. The Gaussian filter H ( jw ) e w 2 – Is not causal because log H ( jw ) w 1 2 is not finite. dw w w 2 2 1 dw 3. The amplitude function 1 H ( jw ) w 1 2 – Is causal because log H ( jw ) w 1 2 dw w 1 2 1 2 dw 2. Stability • If a network is stable, then for a bounded excitation e(t) the response will also be bounded. | e ( t ) | C 1 0t | r ( t ) | C 2 0t • Where C1 and C2 are real, positive and finite numbers. • If a linear system is stable, then we get from the convolution integral and the above definition of stability | r ( t ) | C h ( ) d C 1 2 0 • The above equation implies that the impulse response be absolutely integrable. h ( ) d 0 • One important requirement for h(t) to be absolutely integrable is h(t) approach zero as t increases to infinity. • Note that our definition of stability implies systems with sinwt impulse response are not stable because sin wt dt • However, a simple L-C network has such an impulse response. • Since we don't want to call these systems unstable, we call them marginally stable if they satisfy the following criterion. 0 lim t h ( t ) 0 and h (t ) C all t • Stability in the frequency domain implies that the system function should only have poles on the left had side of the ‘s’ plane or simple poles on the jw axis. • This is because if we have a pole on the right hand side, then the impulse response will have an exponentially increasing term, eαt. • Hence, our response will not be bounded. • If there is a double pole on the jw axis, then the impulse response of the network will have a term tsin(wt). • This will not be bounded. • If H(s) is given as a n s a n 1 s n H (s) n 1 b m s b m 1 s m m 1 ... a1 s a 0 ... b1 s b0 • Due to the requirement of simple poles on the jw axis, the order of the numerator shouldn’t exceed the order of the denominator by more than 1. That is n m 1 • If n m 1 then there would be multiple poles on the s=jw=infinity. • To summarize, for a network to be stabile the following three conditions must be satisfied 1. H(s) can’t have poles on the right side of the ‘s’ plane. 2. H(s) can’t have multiple poles on the jw axis. 3. The degree of the numerator of H(s) can’t exceed that of the denominator by more than 1. 3. Hurwitz polynomials • We mentioned in the previous section that in order for a system to be stable, its poles must lie in the left side of the ‘s’ plane; moreover the poles on the jw axis must be simple. • The denominator polynomial of a system function H(s) that satisfies these criteria belongs to a class of polynomials called Hurwitz polynomials. • In these section, we will discuss the properties of these types of polynomials. • A polynomial P(s) is said to be Hurwitz if it satisfies 1. P(s) must be real if s is real. 2. The real part of its roots must be negative or zero. • As a result of these conditions, if P(s) is a Hurwitz polynomial given by P ( s ) a n s a n 1 s n n 1 ... a1 s a 0 • Then all coefficients an must be real and if si=α+jβ is root of P(s), then α must be negative. Example 1. The polynomial Hurwitz because • • is For real s P(s) is real, P(s)=(s+1)(s2+3s+2) None of the roots lie on the right hand side of the ‘s’ plane. 2. The polynomial Hurwitz • P ( s ) ( s 1) s 1 j 2 s 1 j 2 G ( s ) ( s 1)( s 2 )( s 3 ) is not The root s=1 lies on the positive ‘s’ plane. Properties of Hurwitz polynomial 1. All the coefficients of the polynomial are non negative. • This is readily seen by examining the types of terms P(s) can have P ( s ) s i s i Simple real pole 2 2 s Simple pole on the jw axis i 2 i 2 Complex conjugate roots • The multiplication of these non negative coefficients can only give non negative coefficients. 2. The even and odd parts of P(s) have roots on the jw axis only. • If we denote the even and odd parts of P(s) as n(s) and m(s) P (s) n(s) m (s) • Then both n(s) and m(s) have roots on the jw axis only. 3. The continued fraction expansion of n(s)/m(s) or m(s)/n(s) of a Hurwitz polynomial yields only positive quotient terms. (s) n(s) m (s) 1 q1 s 1 q2s q3s 1 q4s 1 ..... 1 qns • All the q’s are positive. Example • Obtain the continued fraction expansion of F ( s ) s s 5s 3s 4 4 3 2 • Solution: n ( s ) s 5 s 4 and 4 2 m ( s ) s 3s 3 – Since the order of n(s) is higher than m(s), we start with n(s)/m(s). n(s) s 5s 4 4 m (s) 2 s 3s 3 2s 4 2 s s s 3s 3 1 s 3s 3 2s 4 2 s 1 1 s 2 s s 2s 4 2 1 1 s 2 1 2s 4 2 s s 1 1 1 s 2 2s 4 s s 1 1 2 s 1 2s 1 s 4 Note that all the coefficients of the quotients are positive 4. Positive Real Functions • These functions are important because they represent physically realizable passive driving point immitances. • A function is positive real if – F(s) is real for real s, that is F(σ) is real. – The real part of F(s) is greater or equal to zero when the real part of s is greater than or equal to zero. That is • In other words, the right half of the ‘s’ plane maps with the right half of F(s) plane. • In addition, the real axis of ‘s’ plane maps with real axis of F(s) plane. • A further restriction is that F(s) be rational. Example 1. F(s)=Ls (where L is positive real number), is positive real by definition. Inductor 2. F(s)=R (where R is positive real number), is positive real by definition. Resistor 3. F(s)=K/s (K real and positive) is positive real because when s is real F(s) is real and when the real part of s is positive the real part of F(s) is also positive. Capacitor F (s) 1 j 1 j * j j j 2 2 • The necessary and sufficient condition for F(s) to be a positive real function is – F(s) must have no poles on the right side of s plane. – F(s) may have only simple poles on the jw axis with real and positive residues. – Re (F(jw)) 0 for all w. Example 1. Is F (s) s2 s 3s 2 2 F (s) positive real function? s2 s 1( s 2 ) 1 s 1 – Its pole s=-1 lies on the left of s plane – No multiple poles on the jw axis – Its real part is 1 Re F ( jw ) Re 1 jw is always positive. 1 1 jw Re 2 2 1 w 1 w 2. Is F ( s ) ss 12 Solution: 2 F (s) positive real? s2 s j 2 s j 2 – No poles on the right hand side. – No multiple poles on jw axis. – The real part of F(jw) is 1 jw 1 Re F ( jw ) Re 2 2 w 2 2w this can be a negative number. For example for w=2. Exercise 3. Is F (s) s4 s 2s 1 2 positive real?