Part IB Paper 6: Information Engineering LINEAR SYSTEMS AND CONTROL Glenn Vinnicombe HANDOUT 2 Impulse responses, step responses and transfer functions. transfer function ū(s) u(t) Laplace transform pair G(s) g(t) ȳ(s)= G(s)ū(s) t y(t)= impulse response 0 u(τ)g(t τ)dτ =u(t) g(t) =g(t) u(t)
Summary The impulse response, step response and transfer function of a Linear, Time Invariant and causal (LTI) system each completely characterize the input output properties of that system. Given the input to an LTI system, the output can be deterermined: In the time domain: as the convolution of the impulse response and the input. In the Laplace domain: as the multiplication of the transfer function and the Laplace transform of the input. They are related as follows: The step response is the integral of the impulse response. The transfer function is the Laplace transform of the impulse response. 2
Contents 2 Impulse responses, step responses and transfer functions. 2. Preliminaries........................... 4 2.. Definition of the impulse function.......... 4 2..2 Properties of the impulse function.......... 5 2..3 The Impulse and Step Responses........... 6 2.2 The convolution integral.................... 7 2.2. Direct derivation of the convolution integral..... 7 2.2.2 Alternative statements of the convolution integral. 8 2.3 The Transfer Function (for ODE systems)........... 9 2.3. Laplace transform of the convolution integral.... 0 2.4 The transfer function for any linear system......... 2.5 Example: DC motor....................... 3 2.5. Impulse Response of the DC motor.......... 5 2.5.2 Step Response of the DC motor............ 6 2.5.3 Deriving the step response from the impulse response 7 2.6 Transforms of signals vs Transfer functions of systems.. 7 2.7 Interconnections of LTI systems................ 8 2.7. Simplification of block diagrams........... 9 2.8 More transfer function examples............... 20 2.9 Key Points............................. 22 3
2. Preliminaries 2.. Definition of the impulse function The impulse can be defined in many different ways, for example: a) T t or b) T 2 t as 0. (area= in each case.) taking limits, as 0 =δ(t T) T t impulse att=t (impulse occurs when argument= 0) Note: The convergence, as 0, occurs in the sense that both functions a) and b) have the same properties in the limit: 4
2..2 Properties of the impulse function Consider a continuous functionf(t), and leth (t) denote the pulse approximation to the impulse ( (a) on previous page): / if /2<t< /2 h (t)= 0 otherwise f(t) h (t T) Let T t T+ 2 I = f(t)h (t T)dt= T 2 f(t) dt = T+ 2 T 2 f(t)dt Now, I = max T /2 t T+ /2 f(t) and also T t I = min T /2 t T+ /2 f(t) T So, it must be the case that lim 0 I =f(t), that is t lim f(t)h (t T)dt=f(T) 0 5
A similar argument leads to the same result for the triangular approximation to the impulse ( defined above as (b) ). Formally, the unit impulse is defined as any function δ(t) which has the property f(t)δ(t T)dt=f(T) (Strictly speaking,δ(t) is a distribution rather than a function.) Laplace transform ofδ(t): L { δ(t)}= 0 δ(t)e st dt=e s 0 = 2..3 The Impulse and Step Responses Definition: The impulse response of a system is the output of the system when the input is an impulse, δ(t), and all initial conditions are zero. Definition: The step response of a system is the output of the system when the input is a step,h(t), and all initial conditions are zero. whereh(t) is the unit step function ift 0 H(t)= 0 ift< 0 If you know the impulse response of a system, then the response of that system to any input can be determined using convolution, as we shall now show: 6
2.2 The convolution integral 2.2. Direct derivation of the convolution integral u(t) LTI system (impulse response =g(t)) y(t) INPUT OUTPUT δ(t) impulse response g(t) δ(t τ) g(t τ) u(τ)δ(t τ) u(τ)δ(t τ)dτ }{{} u(t) u(τ)g(t τ) u(τ)g(t τ)dτ Hence, the response to the inputu(t) is given by: y(t)= u(τ)g(t τ)dτ 7
2.2.2 Alternative statements of the convolution integral The convolution integral is abbreviated as y(t)= u(τ)g(t τ)dτ y(t)=u(t) g(t) LetT =t τ, soτ=t T anddτ= dt It follows that y(t)= u(t T)g(T)dT =g(t) u(t) NOTE: in either form of the convolution integral, the arguments ofu( ) andg( ) add up tot If, in addition, g(t)=0fort< 0 (CAUSALITY) u(t)=0 fort< 0 (standing assumption), then and also asg(t τ)=0 fort τ< 0 (i.e.τ>t) t y(t)= u(τ)g(t τ)dτ 0 asu(τ)=0 forτ< 0 asu(t τ)=0 fort τ< 0 (i.e.τ>t) t y(t)= u(t τ)g(τ)dτ 0 asg(τ)=0forτ< 0 8
2.3 The Transfer Function (for ODE systems) As an alternative to convolution, the response of a linear system to arbitrary inputs can be determined using Laplace transforms. This is clear when the system is described as a linear ODE: For example, if a linear system has inputuand outputy satisfying the ODE d 2 y dt 2 +αdy dt +βy=adu dt +bu and if all initial conditions are zero, i.e. dy dt =y(0)=u(0)=0, t=0 then taking Laplace transforms gives s 2 ȳ(s)+αsȳ(s)+βȳ(s)=asū(s)+bū(s) or ( s 2 ) ( ) +αs+β ȳ(s)= as+b ū(s) and so ȳ(s)= as+b s 2 +αs+β }{{} Transfer function ū(s) The function as+b s 2 is called the transfer function fromū(s) (the +αs+β input) toȳ(s) (the output). Clearly the same technique will work for higher order linear ordinary differential equations (with constant coefficients). For such systems, the transfer function can be regarded as a placeholder for the coefficients of the differential equation. 9
2.3. Laplace transform of the convolution integral We have seen that both convolution with the impulse reponse (in the time domain) and multiplication by the transfer function (in the Laplace domain) can both be used to determine the output of a linear system. What is the relationship between these techniques? Assume thatg(t)=u(t)=0 fort< 0, ( ) L g(t) u(t) ( ) =L g(τ)u(t τ)dτ = 0 e st g(τ)u(t τ)dτdt = 0 e st g(τ)u(t τ)dtdτ Note thatτ is constant for the inner integration, and let T =t τ, which impliest=t+τ anddt=dt, giving (soe st =e st e sτ ) ( ) L g(t) u(t) = 0 0 e sτ e st g(τ)u(t)dtdτ τ = 0 e sτ g(τ)dτ 0 e st u(t)dt ( ) = L g(t) ( L u(t)) =ḡ(s)ū(s) In words, a Laplace transform (easy when you get used to them) turns a convolution (always hard!) into a multiplication (very easy). 0
2.4 The transfer function for any linear system It follows from the result of the previous section that, if y(t)=g(t) u(t) is the response of an LTI system with impulse responseg(t) to the inputu(t), then we can also write ȳ(s)=g(s)ū(s) whereg(s)=lg(t) is called the transfer function of the system. It follows that all LTI systems have transfer functions (given by the Laplace transform of their impulse response). transfer function ū(s) u(t) G(s) g(t) ȳ(s)= G(s)ū(s) y(t)=g(t) u(t) impulse response We shall use the notation x(s) to represent the Laplace transform of a signalx(t), and uppercase characters to represent transfer functions (e.g.g(s)).
In general a system may have more than one input. In this case, the transfer function from a particular input to a particular output is defined as the Laplace transform of that output when an impulse is applied to the given input, all other inputs are zero and all initial conditions are zero. This is most easily seen in the Laplace domain: If an LTI system has an inputuand an outputy then we can always write ȳ(s)=g(s)ū(s)+other terms independent ofu. G(s) is then called the transfer function fromū(s) toȳ(s). (Or, the transfer function relatingȳ(s) andū(s)) Here the other terms could be a result of non zero initial conditions or of other non zero inputs (disturbances, for example). NOTE: Remember that although the transfer function is defined in terms of the impulse response, it is usually most easily calculated directly from the system s differential equations. 2
2.5 Example: DC motor The following example will illustrate the impulse and step responses and the transfer function. We take the input to the motor to be the applied voltagee(t), and the output to be the shaft angular velocity R i L ω := θ(t). e(t) J B Kω(t) τ θ ) τ(t)=ki(t) Linearized motor equation 2) τ(t) Bω(t)=J ω(t) Newton 3) e(t)=ri(t)+l di(t) dt + Kω(t) Kirchoff Find the effect ofe(t) onω(t): Take Laplace Transforms, assuming thatω(0)=i(0)=0: ) τ(s)=kī(s) 2) τ(s) B ω(s)=j ( s ω(s) ω(0) ) 3) ē(s)=rī(s)+l ( s ī(s) i(0) ) +K ω(s) 3
We can now eliminateī(s) and τ(s) to leave one equation relatingē(s) and ω(s): First ) and 2) give: Kī(s)=(Js+B) ω(s) and rearranging 3) gives: ē(s)=(ls+r)ī(s)+k ω(s) putting these together leads to ( ē(s)= (Ls+R) (Js+B) ) +K ω(s) K or, in the standard form (outputs on the left) K ω(s) = (Ls+R)(Js+B)+K 2ē(s) k = (st + )(st 2 + )ē(s) for suitable definitions ofk,t andt 2 (assuming real roots). That is, output ω(s) = k We callg(s)= (st +)(st 2 +) (or, relating ω(s) andē(s)). The diagram k ē(s) (T s+ )(T 2 s+ ) }{{} Transfer Function input the transfer function fromē(s) to ω(s) ē(s) G(s) ω(s) represents this relationship between the input and the output: By using Laplace transforms, all LTI blocks can be treated as multiplication by a transfer function. For the lego NXT motor we havek 2.2,T 54ms andt 2 ms, as demonstrated in lectures. 4
2.5. Impulse Response of the DC motor To find the impulse response, we let e(t) = δ(t) unit impulse which implies and put ē(s)= ω(0)=i(0)=0. So, ω(s) = k (T 2 s+ )(T s+ ) Split into partial fractions: [ ] ω(s)=k T st 2 + + T 2 st + T 2 T [ ] k = + T T 2 s+ /T 2 s+ /T Hence, ω(t)= k T T 2 e.g. [ ] e t/t 2 }{{} + e t/t }{{} Fast Transient Slow Transient T 2 ms T 54 ms k (T T 2 ) 50 40 30 slow transient total impulse response ω (rad/sec) 20 0 0 0 20 initial value= 0 (= lim s 0 s ω(s)) fast transient k (T T 2 ) 30 40 50 0 0.05 0. 0.5 0.2 0.25 0.3 t (sec) 5
2.5.2 Step Response of the DC motor To find the step response, we let which implies that e(t)=h(t) ē(s)= s and put ω(0)=i(0)=0 (as for the impulse response.) So, Split into partial fractions: ω(s)=k ω(s) = k (T 2 s+ )(T s+ ) s s + T 2 T T 2 s+ T 2 T T T 2 s+ T Hence, [ ω(t)=k H(t)+ T 2 T T 2 e t/t 2 } {{ } Fast Transient T ] e t/t T T } 2 {{} Slow Transient k 2.5 2 step.5 total step response kt 2 (T T 2 ) ω (rad/sec) 0.5 0 0.5 fast transient initial slope= 0 kt (T T 2 ).5 2 slow transient 2.5 0 0.05 0. 0.5 0.2 0.25 0.3 t (sec) 6
2.5.3 Deriving the step response from the impulse response For a system with impulse responseg(t), the step response is given by i.e. y(t)= t t g(τ)h(t τ)dτ= g(τ)dτ 0 0 step response = integral of impulse response Check this on the example. Make sure you understand where the termh(t) in the step response comes from. 2.6 Transforms of signals vs Transfer functions of systems Mathematically there is no distinction, but in practice they have different interpretations G(s) Signals L G(s) Systems ȳ(s)=g(s) ū(s) δ(t) y(t) = u(t) s H(t) y(t) = t 0 u(τ)dτ (integrator) s+a e at ẏ(t)+ay(t)=u(t) (lag) ω s 2 +ω 2 sin(ωt) ÿ(t)+ω 2 y(t)=ωu(t) e st δ(t T) y(t) = u(t T) (delay) In each case, the signal is the impulse response of the system. 7
2.7 Interconnections of LTI systems x(s) a) F(s) Σ + v(s) H(s) z(s) ȳ(s) G(s) + represents the equations: v(s)=f(s) x(s)+g(s)ȳ(s) and z(s)=h(s) v(s) ( ) = z(s)=h(s) F(s) x(s)+g(s)ȳ(s) = H(s)F(s) x(s) + H(s)G(s) ȳ(s) }{{}}{{} transfer function transfer function from x(s) to z(s) from ȳ(s) to z(s) r(s) + ē(s) ȳ(s) b) Σ G(s) K(s) represents the simultaneous equations: ē(s)= r(s) K(s)ȳ(s) and ȳ(s) = G(s)ē(s) = ȳ(s)=g(s) r(s) G(s)K(s)ȳ(s) = (+G(s)K(s))ȳ(s)=G(s) r(s) = ȳ(s)= G(s) +G(s)K(s) r(s) 8
2.7. Simplification of block diagrams Recognizing that blocks represent multiplications, and using the above formulae, it is often easier to rearrange block diagrams to determine overall transfer functions, e.g. from ū(s) to ȳ(s) below. ū(s)+ + ȳ(s) Σ G (s) G 2 (s) Σ G 3 (s) + K (s) ū(s)+ Σ + G (s)g 2 (s) G 3 (s) +G 3 (s)k (s) ȳ(s) ū(s) + Σ G (s)g 2 (s)g 3 (s) +G 3 (s)k (s) ȳ(s) or y(s)= ȳ(s)= G (s)g 2 (s)g 3 (s) +G 3 (s)k (s) G ū(s) (s)g 2 (s)g 3 (s) +G 3 (s)k (s) G (s)g 2 (s)g 3 (s) +G 3 (s)k (s) G (s)g 2 (s)g 3 (s)ū(s) 9
2.8 More transfer function examples To obtain the transfer function, in each case, we take all initial conditions to be zero. The following three systems all have the same transfer function (a st order lag) ) Spring/damper system λ k y(t) x(t) = λẏ=k(x y) λ = kẏ+y=x λ sȳ(s)+ȳ(s)= x(s) k = ( ) λ k s+ ȳ(s)= x(s) = ȳ(s) = Ts+ x(s) T =λ/k Step Response: x(s)= s = ȳ(s)= s(st+ ) = s T st+ y(t) = y(t)=h(t) e t/t T t 20
2) RC network x(t) R C y(t) i=c dy dt =x y R = RCẏ+y=x = (RCs+ )ȳ(s)= x(s) = ȳ(s) = x(s) T =RC Ts+ 3) Water Heater q(t) (J/s) (assuming perfect mixing i.e. all the water in the tank is at the θ i M(kg) θ o same temperature,θ o ) ṁ (kg/s) specific heat capacity q=mc θ 0 +ṁc(θ o θ i ) = (Mcs+ṁc) θ o (s)= q(s)+ṁc θ i (s) = θo (s)= ṁc Ts+ q(s) input + Ts+ θ i (s) disturbance T =M/ṁ 2
2.9 Key Points The step response is the integral (w.r.t time) of the impulse response. The transfer function is the Laplace transform of the impulse response. By using transfer functions, block diagrams represent simple algebraic relationships (all blocks become multiplications). 22