Transfer Functions Transfer Functions Assume zero initial conditions. • Transfer functions defined N • Examples • System stability dk y(t) dtk = ak sk Y (s) = ak k=0 • Pole-Zero Plots N • Sinusoidal steady-state analysis Y (s) • Transfer function synthesis Portland State University ak sk Transfer Functions Ver. 1.66 1 J. McNames Initial Conditions Assume zero initial conditions. N k=0 N dk y(t) dtk = ak sk Y (s) = ak k=0 M k=0 M bk bk sk X(s) = X(s) k=0 ECE 222 M dk x(t) dtk k=0 N Y (s) J. McNames bk k=0 k=0 • Linearity and time invariance defined M = M bk sk k=0 M k k=0 bk s N k k=0 ak s Portland State University ECE 222 X(s) = H(s)X(s) Transfer Functions Ver. 1.66 2 Transfer Functions Continued M k k=0 bk s X(s) = H(s)X(s) Y (s) = N k k=0 ak s dk x(t) dtk • In the time domain, the relationship can be complicated bk sk X(s) • In the s domain, the relationship of Y (s) to X(s) of LTI systems simplifies to a rational function of s k=0 • All voltages and currents are due to independent sources (superposition) • H(s) is usually a rational ratio of two polynomials • Energy stored in capacitors and inductors also act like independent sources • Specifically, the transfer function of an LTI system can be defined as the ratio of Y (s) to X(s) • H(s) is called the transfer function • We will now focus a specific class of circuits – Only one independent source (input) – No energy stored in capacitors or inductors • Usually denoted by H(s), sometimes G(s) • Without loss of generality, usually aN 1 • Greatly simplifies analysis J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 3 J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 4 Example 1: Transfer Function vs. Impulse Response Transfer Functions and the Impulse Response Fill in the missing parts to determine how the transfer function of an LTI system G(s) is related to the impulse response h(t) x(t) = δ(t) X(s) = y(t) = Y (s) = L {h(t)} = L−1 {G(s)} = x(t) h(t) y(t) H(s) x(t) y(t) • Because of their relationship, both H(s) and h(t) completely characterize the LTI system • If the LTI system is a circuit, once you know either H(s) or h(t), you have sufficient information to calculate the output • You now have three different approaches to solve for the output of an LTI circuit – y(t) = x(t) ∗ h(t) – Solve for H(s), X(s), and then y(t) = L−1 {H(s) X(s)} – Use Laplace transform circuit analysis to solve for the outputs of interest • All three have limitations, advantages, and disadvantages J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 5 J. McNames Portland State University Continous-Time Convolution Tradeoffs x(t) h(t) y(t) x(t) H(s) y(t) x(t) 6 Transfer Functions h(t) y(t) H(s) x(t) Ver. 1.66 8 y(t) • Advantages – Reduces differential equation to an algebra problem – Usually the easiest approach – Easy to find the output for different input signals • Disadvantages – Cannot account for non-zero initial conditions, requires complete x(t) and y(t) – Can be difficult to write and solve integrals – Can only be used for single-input single-output (SISO) systems ECE 222 Ver. 1.66 Transfer Function: y(t) = L−1 {H(s) X(s)} • Advantages – Can find solution for all t, not just t > 0 – Can be approximated using discrete-time convolution Portland State University Transfer Functions Transfer Function Analysis Tradeoffs Continuous-time Convolution: y(t) = x(t) ∗ h(t) J. McNames ECE 222 Ver. 1.66 • Disadvantages – Can only solve for y(t) for t > 0 – Requires zero initial conditions – Can only be used for SISO systems 7 J. McNames Portland State University ECE 222 Transfer Functions Laplace Transform Circuit Analysis Tradeoffs h(t) x(t) y(t) x(t) H(s) Example 6: Transfer Functions R y(t) + Laplace Transform Circuit Analysis vs(t) • Advantages – Elegant method of handling non-zero initial conditions – Can handle multiple sources (multiple inputs) & can solve for multiple outputs (any voltage or current) — MIMO systems Portland State University ECE 222 Transfer Functions Ver. 1.66 vo(t) - Find the transfer function for the circuit above. The input is the voltage source vs (t) and the output is labeled vo (t). • Disadvantages – Can only solve for y(t) for t > 0 – Cannot account for full history, x(t) for t < 0. Requires this effect to be captured in the initial conditions – Can be tedious – Specific to application (circuits), we did not discuss generalization to other types of systems J. McNames C 9 J. McNames Portland State University Example 7: Transfer Functions Ver. 1.66 10 C C R + + vs(t) vo(t) vo(t) - - Find the transfer function for the circuit above. Do you recognize this function? J. McNames Transfer Functions Example 8: Transfer Functions R vs(t) ECE 222 Portland State University ECE 222 Transfer Functions Ver. 1.66 Find the transfer function for the circuit above. Do you recognize this function? 11 J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 12 Example 9: Transfer Functions Example 9: Workspace CB CA RB RA + vs(t) vo(t) RL - Find the transfer function for the circuit above. J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 13 J. McNames Example 10: Transfer Functions + vR(t) R vs(t) - + vL(t) Portland State University ECE 222 Transfer Functions Ver. 1.66 14 Ver. 1.66 16 Example 10: Workspace - L + C vC(t) - Find the transfer function from the input voltage to an output voltage across each element of the three passive elements in a series RLC circuit. J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 15 J. McNames Portland State University ECE 222 Transfer Functions Poles and Stability Pole-Zero Plots Assume all of the poles in a transfer function H(s) are unique. Then H(s) can be written as follows using partial fraction expansion: N N (s) k H(s) = = D(s) s − p H(s) = • Zeros: roots of N (s) • Poles: roots of D(s) =1 L−1 {H(s)} = h(t) = N N (s) D(s) • Poles must be in the left half plane for the system to be stable k e+p t u(t) • As the poles get closer to the boundary, the system becomes less stable =1 • Note the expansion is in terms of the poles, rather than −p ∞ • If −∞ |h(t)| dt < ∞, the LTI system is bounded-input bounded-output (BIBO) stable • Pole-Zero Plot: plot of the zeros and poles on the complex s plane • You will use these throughout the junior sequence (ECE 32x) • That is |h(t)| < α < ∞ for all t • h(t) is bounded if Re{p } < 0 for all • The system is BIBO stable if and only if all the poles are in the left half of the complex plane J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 17 J. McNames Portland State University Example 11: Pole-Zero Plots s3 Ver. 1.66 18 Pole−Zero Plot 1.5 s2 − 1 + 4s2 + 6s + 4 1 Imaginary Axis Using the MATLAB, we can quickly find the roots H(s) = Transfer Functions Example 11: Pole-Zero Plot Use MATLAB to generate a Pole-Zero plot for a system with the following transfer function H(s) = ECE 222 (s + 1)(s − 1) (s + 2)(s + 1 − j)(s + 1 + j) Is the system stable? The pole-zero plot, impulse response, and step response are shown on the following slides. 0.5 0 −0.5 −1 −1.5 −3 J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 19 J. McNames −2.5 −2 −1.5 −1 Portland State University −0.5 0 Real Axis ECE 222 0.5 1 Transfer Functions 1.5 2 Ver. 1.66 20 Example 11: MATLAB Code for Pole-Zero Plot Example 11: Impulse Response sys = tf([1 0 -1],[1 4 6 4]); Impulse Response figure; [p,z] = pzmap(sys); h = plot(real(p),imag(p),’bx’,real(z),imag(z),’ro’); set(h,’LineWidth’,1.2); set(h,’MarkerSize’,5); hold on; plot([0 0],[-2 2],’k:’,[-3 2],[0 0],’k:’); hold off; xlabel(’Real Axis’); ylabel(’Imaginary Axis’); title(’Pole-Zero Plot’); axis([-3 2 -1.5 1.5]); 1 0.8 h(t) 0.6 0.4 0.2 0 −0.2 −0.4 0 J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 21 J. McNames 1 2 Portland State University Example 11: MATLAB Code for Impulse Response 3 4 Time (seconds) ECE 222 5 6 Transfer Functions 7 Ver. 1.66 22 Example 11: Step Response sys = tf([1 0 -1],[1 4 6 4]); Step Response figure; t = 0:0.01:7; [h,t] = impulse(sys,t); h = plot(t,h); set(h,’LineWidth’,1.5); hold on; plot([0 0],[-2 2],’k:’,[0 max(t)],[0 0],’k:’); hold off; axis([0 max(t) -0.5 1.2]); xlabel(’Time (seconds)’); ylabel(’h(t)’); title(’Impulse Response’); 0.1 0 y(t) −0.1 −0.2 −0.3 −0.4 −0.5 J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 23 J. McNames 0 1 2 Portland State University 3 4 Time (seconds) ECE 222 5 6 Transfer Functions 7 Ver. 1.66 24 Example 11: MATLAB Code for Step Response Steady-State Sinusoidal Analysis Assume a system H(s) is BIBO stable. Consider the case where the source is sinusoidal: sys = tf([1 0 -1],[1 4 6 4]); figure; t = 0:0.01:7; [h,t] = step(sys,t); h = plot(t,h); set(h,’LineWidth’,1.5); hold on; plot([0 0],[-2 2],’k:’,[0 max(t)],[0 0],’k:’); hold off; axis([0 max(t) -0.5 0.2]); xlabel(’Time (seconds)’); ylabel(’y(t)’); title(’Step Response’); J. McNames Portland State University ECE 222 x(t) = A cos φ cos ωt − A sin φ sin ωt A (s cos φ − ω sin φ) X(s) = s2 + ω 2 Y (s) = H(s)X(s) A (s cos φ − ω sin φ) Y (s) = H(s) s2 + ω 2 Transfer Functions Ver. 1.66 25 J. McNames Steady-State Sinusoidal Analysis Continued Y (s) = 2|k| cos(ωt + ∠k)u(t) + N Ver. 1.66 26 t→∞ • If the input to an LTI system is sinusoidal, – The steady-state output is sinusoidal at the same frequency – The amplitude and phase of y(t) differ from that of x(t) k e−pt u(t) • We applied this idea when we did phasor analysis • But how is k related to H(s), A, and φ? = yss (t) + ytr (t) = Transfer Functions yss (t) = lim y(t) = 2|k| cos(ωt + ∠k) =1 yss (t) ECE 222 If x(t) = A cos(ωt + φ), A (s cos φ − ω sin φ) s2 + ω 2 N k k k∗ + + s − jω s + jω s + p =1 y(t) Portland State University Steady-State Sinusoidal Analysis Comments = H(s) = = A cos(ωt + φ) lim y(t) t→∞ = 2|k| cos(ωt + ∠k) J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 27 J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 28 Solving for the Complex Residue Sinusoidal Steady-State Output Since H(jω) is complex, we can write it in polar form as A (s cos φ − ω sin φ) Y (s) = H(s) s2 + ω 2 N k k k∗ = + + s − jω s + jω s + p =1 A (s cos φ − ω sin φ) k = H(s) s + jω s=+jω = = = J. McNames H(jω) = |H(jω)| ej∠H(jω) Then using the results of the previous slide, we have k = 12 H(jω)A ejφ |k| = 12 |H(jω)|A A(jω cos φ − ω sin φ) 2jω A(cos φ + j sin φ) H(jω) 2 jφ 1 H(jω)A e 2 = 12 |H(jω)|A ej(φ+∠H(jω)) ∠k = φ + ∠H(jω) H(jω) Portland State University ECE 222 Transfer Functions yss (t) H(s) 2|k| cos(ωt + ∠k) = |H(jω)|A cos (ωt + φ + ∠H(jω)) Ver. 1.66 29 J. McNames Sinusoidal Steady-State Output x(t) = Portland State University ECE 222 Transfer Functions Ver. 1.66 30 Example 12: Steady-State Sinusoidal Analysis R y(t) + x(t) yss (t) vs(t) = A cos(ωt + φ) = |H(jω)|A cos (ωt + φ + ∠H(jω)) C vo(t) - • The input is sinusoidal Find the steady-state sinusoidal response to an input voltage of vs (t) = cos(ωt). • The steady-state signal yss (t) is also a sinusoid – Same frequency as x(t): ω – Amplitude is scaled by |H(jω)| – The phase is shifted by ∠H(jω) • If we know H(s), we can easily find the steady-state solution for any sinusoidal input signal J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 31 J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 32 Example 12: Workspace Steady-State Sinusoidal Analysis Comments • We will study this in depth shortly • There is analytical significance to how the magnitude and phase of H(s) vary with s = jω J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 33 LTI Systems x(t) h(t) y(t) ; ; J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 34 Example 13: Transfer Function Analysis H(s) x(t) y(t) b k m x(t) • If we know the transfer function, we have sufficient information to calculate the output for any input y(t) • This enables us to treat the circuit more abstractly as H(s) Find the transfer function for the linear system shown above. The external force x(t) is the input to the system and the displacement y(t) is the output. Find the transfer function. • The transfer function may be for another type of system: mechanical, chemical, hydraulic, etc. • Mathematically they are treated the same • Field-specific analysis is used only to find H(s) J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 35 J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 36 Transfer Function Synthesis x(t) h(t) y(t) H(s) x(t) Cascade Transfer Function Synthesis x(t) y(t) • Thus far we have talked only about circuit analysis H1(s) H(s) = H2(s) HP(s) y(t) N (s) = H1 (s) × H2 (s) × · · · × HP (s) D(s) • We now know several ways to solve for the output of a given system • There are many approaches to transfer function synthesis • If there are zero initial conditions, then we can find the transfer function H(s) of a given circuit • Will discuss how to specify H(s) to meet the requirements for a given application later this term • Now we will discuss how to design a circuit that implements a given H(s) • The most common (and perhaps easiest) approach to synthesis is to break H(s) up into 1st (real poles) or 2nd (complex poles) order components • This is called transfer function synthesis • Thus each component, Hi (s) has either a 1st or 2nd order polynomial in the numerator and denominator • There are many circuits that have the same transfer function J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 37 J. McNames Portland State University Cascade Transfer Function Synthesis Continued x(t) H1(s) H2(s) HP(s) ECE 222 Transfer Functions Ver. 1.66 Summary • Circuits with a single input (independent source) and zero initial conditions can be represented generically by their transfer functions y(t) • There are robust, standard circuits for implementing these low-order components • H(s) is the Laplace transform of the system impulse response • The output of the system is y(t) = L−1 {H(s)X(s)} for any causal input signal (x(t) = 0 for t < 0) • The output of each transfer function is generated by an operational amplifier • For sinusoidal inputs, the output is also sinusoidal at the same frequency but amplified by |H(jω)| and shifted in phase by ∠H(jω) • This is essential for the cascade synthesis to work (will explain later) • Some of these 1st and 2nd order components are discussed in the text (Chapter 15) • Thus, transfer functions make sinusoidal steady-state analysis easy • Others can be found in more advanced analog circuits texts • Transfer function analysis used for all types of LTI systems, not just circuits • Generalization of phasors • You will use cascade synthesis in the first lab for ECE 203 J. McNames Portland State University ECE 222 Transfer Functions 38 • Can synthesize a transfer function using a cascade of 1st and 2nd order active circuits Ver. 1.66 39 J. McNames Portland State University ECE 222 Transfer Functions Ver. 1.66 40