ECE 352 Systems II Manish K. Gupta, PhD Office: Caldwell Lab ece. osu. ece. osu. edu Home Page: TA: Zengshi Chen osu. eduZengshi Chen Office Hours for TA : in CL 391: Tu & Th 1:00-2:30 pm Home Page:

Acknowledgements Various graphics used here has been taken from public resources instead of redrawing it. Thanks to those who have created it. Thanks to Brian L. Evans and Mr. Dogu Arifler Thanks to Randy Moses and Bradley Clymer

ECE 352 Slides edited from: Prof. Brian L. Evans and Mr. Dogu Arifler Dept. of Electrical and Computer Engineering The University of Texas at Austin course: EE 313 Linear Systems and Signals Fall 2003

Continuous-Time Domain Analysis Example: Differential systems –There are n derivatives of y(t) and m derivatives of f(t) –Constants a 0, a 1, …, a n-1 and b 0, b 1, …, b m –Linear constant-coefficient differential equation Using short-hand notation, above equation becomes

Continuous-Time Domain Analysis Polynomials Q(D) and P(D) where a n = 1: –Recall n derivatives of y(t) and m derivatives of f(t) –We will see that this differential system behaves as an (m-n)th-order differentiator of high frequency signals if m > n –Noise occupies both low and high frequencies –We will see that a differentiator amplifies high frequencies. To avoid amplification of noise in high frequencies, we assume that m  n

Continuous-Time Domain Analysis Linearity: for any complex constants c 1 and c 2,

Continuous-Time Domain Analysis For a linear system, –The two components are independent of each other –Each component can be computed independently of the other T[·] y(t)y(t)f(t)f(t)

Continuous-Time Domain Analysis Zero-input response –Response when f(t)=0 –Results from internal system conditions only –Independent of f(t) –For most filtering applications (e.g. your stereo system), we want no zero-input response. Zero-state response –Response to non-zero f(t) when system is relaxed –A system in zero state cannot generate any response for zero input. –Zero state corresponds to initial conditions being zero.

Zero-Input Response Simplest case Solution: For arbitrary constant C –Could C be complex? –How is C determined?

Zero-Input Response General case: The linear combination of y 0 (t) and its n successive derivatives are zero. Assume that y 0 (t) = C e t

Zero-Input Response Substituting into the differential equation y 0 (t) = C e  t is a solution provided that Q( ) = 0. Factor Q( ) to obtain n solutions: Assuming that no two i terms are equal

Zero-Input Response Could i be complex? If complex, For conjugate symmetric roots, and conjugate symmetric constants,

Zero-Input Response For repeated roots, the solution changes. Simplest case of a root repeated twice: With r repeated roots

System Response Characteristic equation –Q(D)[y(t)] = 0 –The polynomial Q( ) Characteristic of system Independent of the input –Q( ) roots 1, 2, …, n Characteristic roots a.k.a. characteristic values, eigenvalues, natural frequencies Characteristic modes (or natural modes) are the time-domain responses corresponding to the characteristic roots –Determine zero-input response –Influence zero-state response

RLC Circuit Component values L = 1 H, R = 4 , C = 1/40 F Realistic breadboard components? Loop equations (D D + 40) [y 0 (t)] = 0 Characteristic polynomial = ( j 6)( j 6) Initial conditions y(0) = 2 A ý(0) = A/s  LR C y(t)y(t) f(t)f(t) Envelope y 0 (t) = 4 e -2t cos(6t -  /3) A

ECE 352 Linear Time-Invariant System Any linear time-invariant system (LTI) system, continuous-time or discrete-time, can be uniquely characterized by its –Impulse response: response of system to an impulse –Frequency response: response of system to a complex exponential e j 2  f for all possible frequencies f –Transfer function: Laplace transform of impulse response Given one of the three, we can find other two provided that they exist

ECE 352 Impulse response Impulse response of a system is response of the system to an input that is a unit impulse (i.e., a Dirac delta functional in continuous time)

ECE 352 Example Frequency Response System response to complex exponential e j  for all possible frequencies  where  = 2  f Passes low frequencies, a.k.a. lowpass filter  |H(  )| pp ss  s  p passband stopband   H() H()

ECE 352 Kronecker Impulse Let  [k] be a discrete-time impulse function, a.k.a. the Kronecker delta function: Impulse response h[k]: response of a discrete-time LTI system to a discrete impulse function k [k][k] 1

Transfer Functions

Zero-State Response Q(D) y(t) = P(D) f(t) All initial conditions are 0 in zero-state response Laplace transform of differential equation, zero- state component

Transfer Function H(s) is called the transfer function because it describes how input is transferred to the output in a transform domain (s-domain in this case) Y(s) = H(s) F(s) y(t) = L -1 {H(s) F(s)} = h(t) * f(t)  H(s) = L{h(t)} The transfer function is the Laplace transform of the impulse response

Transfer Function Stability conditions for an LTIC system –Asymptotically stable if and only if all the poles of H(s) are in left-hand plane (LHP). The poles may be repeated or non-repeated. –Unstable if and only if either one or both of these conditions hold: (i) at least one pole of H(s) is in right- hand plane (RHP); (ii) repeated poles of H(s) are on the imaginary axis. –Marginally stable if and only if there are no poles of H(s) in RHP, and some non-repeated poles are on the imaginary axis.

Examples Laplace transform Assume input f(t) & output y(t) are causal Ideal delay of T seconds

Examples Ideal integrator with y(0 - ) = 0 Ideal differentiator with f(0 - ) = 0

Cascaded Systems Assume input f(t) and output y(t) are causal Integrator first, then differentiator Differentiator first, then integrator Common transfer functions –A constant (finite impulse response) –A polynomial (finite impulse response) –Ratio of two polynomials (infinite impulse response) 1/ss F(s)/s f(t)f(t) F(s) f(t)f(t) s1/s s F(s) f(t)f(t) F(s) f(t)f(t)

Frequency-Domain Interpretation y(t) = H(s) e s t for a particular value of s Recall definition of frequency response: h(t)h(t) y(t)y(t)e st h(t)h(t) y(t)y(t)e j 2  f t

Frequency-Domain Interpretation s is generalized frequency: s =  + j 2  f We may convert transfer function into frequency response by if and only if region of convergence of H(s) includes the imaginary axis What about h(t) = u(t)? We cannot convert this to a frequency response However, this system has a frequency response What about h(t) =  (t)?

Unilateral Laplace Transform Differentiation in time property f(t) = u(t) What is f ’(0)? f’(t) =  (t). f ’(0) is undefined. By definition of differentiation Right-hand limit, h =   h = 0, f ’(0 + ) = 0 Left-hand limit, h = -   h = 0, f ’(0 - ) does not exist t f(t)f(t) 1

Block Diagrams H(s)H(s) F(s)F(s)Y(s)Y(s) H 1 (s) + H 2 (s) F(s)F(s)Y(s)Y(s) H1(s)H1(s) F(s)F(s)Y(s)Y(s) H2(s)H2(s)  = H1(s)H1(s) F(s)F(s)Y(s)Y(s) H2(s)H2(s)H1(s)H2(s)H1(s)H2(s) F(s)F(s)Y(s)Y(s) = W(s)W(s) G(s) 1 + G(s)H(s) F(s)F(s)Y(s)Y(s) G(s)G(s) F(s)F(s)Y(s)Y(s) H(s)H(s)  - = E(s)E(s)

Derivations Cascade W(s) = H 1 (s)F(s) Y(s) = H 2 (s)W(s) Y(s) = H 1 (s)H 2 (s)F(s)  Y(s)/F(s) = H 1 (s)H 2 (s) One can switch the order of the cascade of two LTI systems if both LTI systems compute to exact precision Parallel Combination Y(s) = H 1 (s)F(s) + H 2 (s)F(s) Y(s)/F(s) = H 1 (s) + H 2 (s) H1(s)H1(s) F(s)F(s)Y(s)Y(s) H2(s)H2(s)H2(s)H2(s) F(s)F(s)Y(s)Y(s) H1(s)H1(s) 

Derivations Feedback System Combining these two equations What happens if H(s) is a constant K? –Choice of K controls all poles in the transfer function –This will be a common LTI system in Intro. to Automatic Control Class (required for EE majors)

Stability

Many possible definitions Two key issues for practical systems –System response to zero input –System response to non- zero but finite amplitude (bounded) input For zero-input response –If a system remains in a particular state (or condition) indefinitely, then state is an equilibrium state of system –System’s output due to nonzero initial conditions should approach 0 as t  –System’s output generated by initial conditions is made up of characteristic modes

Stability Three cases for zero-input response –A system is stable if and only if all characteristic modes  0 as t   –A system is unstable if and only if at least one of the characteristic modes grows without bound as t   –A system is marginally stable if and only if the zero- input response remains bounded (e.g. oscillates between lower and upper bounds) as t  

Characteristic Modes Distinct characteristic roots 1, 2, …, n –Where =  + j  in Cartesian form –Units of  are in radians/second StableUnstable Im{ } Re{ } Marginally Stable Left-hand plane (LHP) Right-hand plane (RHP)

Characteristic Modes Repeated roots –For r repeated roots of value. –For positive k Decaying exponential decays faster than t k increases for any value of k –One can see this by using the Taylor Series approximation for e t about t = 0:

Stability Conditions An LTIC system is asymptotically stable if and only if all characteristic roots are in LHP. The roots may be simple (not repeated) or repeated. An LTIC system is unstable if and only if either one or both of the following conditions exist: (i) at least one root is in the right-hand plane (RHP) (ii) there are repeated roots on the imaginary axis. An LTIC system is marginally stable if and only if there are no roots in the RHP, and there are no repeated roots on imaginary axis.

Response to Bounded Inputs Stable system: a bounded input (in amplitude) should give a bounded response (in amplitude) Linear-time-invariant (LTI) system Bounded-Input Bounded-Output (BIBO) stable h(t)h(t) y(t)y(t)f(t)f(t)

Impact of Characteristic Modes Zero-input response consists of the system’s characteristic modes Stable system  characteristic modes decay exponentially and eventually vanish If input has the form of a characteristic mode, then the system will respond strongly If input is very different from the characteristic modes, then the response will be weak

Impact of Characteristic Modes Example: First-order system with characteristic mode e  t Three cases

 e -1 /RC 1/RC t h(t)h(t) System Time Constant When an input is applied to a system, a certain amount of time elapses before the system fully responds to that input –Time lag or response time is the system time constant –No single mathematical definition for all cases Special case: RC filter –Time constant is  = RC –Instant of time at which h(t) decays to e -1  of its maximum value

t0t0 thth t h(t)h(t) h(t0)h(t0) h(t)h(t) ĥ(t)ĥ(t) System Time Constant General case: –Effective duration is t h seconds where area under ĥ(t) –C is an arbitrary constant between 0 and 1 –Choose t h to satisfy this inequality General case applied to RC time constant:

h(t)h(t) y(t)y(t)u(t)u(t) t u(t)u(t) 1 t h(t)h(t) A t y(t)y(t) A t h trtr Step Response y(t) = h(t) * u(t) Here, t r is the rise time of the system How does the rise time t r relate to the system time constant of the impulse response? A system generally does not respond to an input instantaneously trtr

Filtering A system cannot effectively respond to periodic signals with periods shorter than t h This is equivalent to a filter that passes frequencies from 0 to 1/t h Hz and attenuates frequencies greater than 1/t h Hz (lowpass filter) –1/t h is called the cutoff frequency –1/t r is called the system’s bandwidth (t r = t h ) Bandwidth is the width of the band of positive frequencies that are passed “unchanged” from input to output

Transmission of Pulses Transmission of pulses through a system (e.g. communication channel) increases the pulse duration (a.k.a. spreading or dispersion) If the impulse response of the system has duration t h and pulse had duration t p seconds, then the output will have duration t h + t p

System Realization

Passive Circuit Elements Laplace transforms with zero-valued initial conditions Capacitor Inductor Resistor + – v(t)v(t) + – v(t)v(t) + – v(t)v(t)

First-Order RC Lowpass Filter x(t)x(t)y(t)y(t) ++ C R X(s)Y(s) ++ R Time domain Laplace domain i(t)i(t) I(s)

Passive Circuit Elements Laplace transforms with non-zero initial conditions Capacitor Inductor

Operational Amplifier Ideal case: model this nonlinear circuit as linear and time-invariant Input impedance is extremely high (considered infinite) v x (t) is very small (considered zero) + _ y(t)y(t) + _ + _ vx(t)vx(t)

Operational Amplifier Circuit Assuming that V x (s) = 0, How to realize gain of –1? How to realize gain of 10? + _ Y(s)Y(s) + _ + _ Vx(s)Vx(s) Zf(s)Zf(s) Z(s)Z(s) +_ F(s)F(s) I(s)I(s)H(s)H(s)

Differentiator A differentiator amplifies high frequencies, e.g. high-frequency components of noise: H(s) = s where s =  + j 2  f Frequency response is H(f) = j 2  f  | H( f ) |= 2  f | Noise has equal amounts of low and high frequencies up to a physical limit A differentiator may amplify noise to drown out a signal of interest In analog circuit design, one would use integrators instead of differentiators

Initial and Final Values Values of f(t) as t  0 and t   may be computed from its Laplace transform F(s) Initial value theorem If f(t) and its derivative df/dt have Laplace transforms, then provided that the limit on the right- hand side of the equation exists. Final value theorem If both f(t) and df/dt have Laplace transforms, then provided that s F(s) has no poles in the RHP or on the imaginary axis.

Final and Initial Values Example Transfer function Poles at s = 0, s = -1  j2 Zero at s = -3/2