[PPT] - Transfer Functions Transfer Functions Assume zero initial PowerPoint Presentation

SLIDE 1

Initial Conditions Assume zero initial conditions.

ak dky(t) dtk =

bk dkx(t) dtk

akskY (s) =

bkskX(s)

(superposition)

sources

– Only one independent source (input) – No energy stored in capacitors or inductors

Transfer Functions

Transfer Functions Continued Y (s) = M

N

simplifies to a rational function of s

as the ratio of Y (s) to X(s)

Transfer Functions Assume zero initial conditions.

ak dky(t) dtk =

bk dkx(t) dtk

aksk Y (s) =

bksk X(s) Y (s)

aksk = X(s)

bksk Y (s) = M

N

SLIDE 2

Continous-Time Convolution Tradeoffs H(s)

x(t) y(t)

h(t)

x(t) y(t)

Continuous-time Convolution: y(t) = x(t) ∗ h(t)

– Can find solution for all t, not just t > 0 – Can be approximated using discrete-time convolution

– 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

Example 1: Transfer Function vs. 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)} =

Transfer Function Analysis Tradeoffs H(s)

x(t) y(t)

h(t)

x(t) y(t)

Transfer Function: y(t) = L−1 {H(s) X(s)}

– Reduces differential equation to an algebra problem – Usually the easiest approach – Easy to find the output for different input signals

– Can only solve for y(t) for t > 0 – Requires zero initial conditions – Can only be used for SISO systems

Transfer Functions and the Impulse Response H(s)

x(t) y(t)

h(t)

x(t) y(t)

characterize the LTI system

you have sufficient information to calculate the output

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

SLIDE 3

Example 7: Transfer Functions

vs(t) R vo(t)

C

Find the transfer function for the circuit above. Do you recognize this function?

Laplace Transform Circuit Analysis Tradeoffs H(s)

x(t) y(t)

h(t)

x(t) y(t)

Laplace Transform Circuit Analysis

– 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

– 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

Example 8: Transfer Functions

vs(t) vo(t)

R C

Find the transfer function for the circuit above. Do you recognize this function?

Example 6: Transfer Functions

R C vs(t) 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).

SLIDE 4

Example 10: Transfer Functions

R L C vs(t) vR(t)

vL(t)

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.

Example 9: Transfer Functions

vo(t)

vs(t) RL RA CA RB CB

Find the transfer function for the circuit above.

Example 10: Workspace

Example 9: Workspace

SLIDE 5

Example 11: Pole-Zero Plots Use MATLAB to generate a Pole-Zero plot for a system with the following transfer function H(s) = s2 − 1 s3 + 4s2 + 6s + 4 Using the MATLAB, we can quickly find the roots H(s) = (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.

Poles and Stability 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: H(s) = N(s) D(s) =

kℓ s − pℓ L−1 {H(s)} = h(t) =

kℓ e+pℓtu(t)

∞

bounded-output (BIBO) stable

left half of the complex plane

Example 11: Pole-Zero Plot

−3 −2.5 −2 −1.5 −1 −0.5 0.5 1 1.5 2 −1.5 −1 −0.5 0.5 1 1.5 Real Axis Imaginary Axis Pole−Zero Plot

Pole-Zero Plots H(s) = N(s) D(s)

stable

plane

SLIDE 6

Example 11: MATLAB Code for Impulse Response

Example 11: MATLAB Code for Pole-Zero Plot

Example 11: Step Response

1 2 3 4 5 6 7 −0.5 −0.4 −0.3 −0.2 −0.1 0.1 Time (seconds) y(t) Step Response

Example 11: Impulse Response

1 2 3 4 5 6 7 −0.4 −0.2 0.2 0.4 0.6 0.8 1 Time (seconds) h(t) Impulse Response

SLIDE 7

Steady-State Sinusoidal Analysis Continued Y (s) = H(s)A (s cos φ − ω sin φ) s2 + ω2 = k s − jω + k∗ s + jω +

kℓ s + pℓ y(t) = 2|k| cos(ωt + ∠k)u(t) +

kℓ e−ptu(t) = yss(t) + ytr(t) yss(t) = lim

= 2|k| cos(ωt + ∠k)

Example 11: MATLAB Code for Step Response

Steady-State Sinusoidal Analysis Comments If x(t) = A cos(ωt + φ), yss(t) = lim

– The steady-state output is sinusoidal at the same frequency – The amplitude and phase of y(t) differ from that of x(t)

Steady-State Sinusoidal Analysis Assume a system H(s) is BIBO stable. Consider the case where the source is sinusoidal: x(t) = A cos(ωt + φ) = A cos φ cos ωt − A sin φ sin ωt X(s) = A (s cos φ − ω sin φ) s2 + ω2 Y (s) = H(s)X(s) Y (s) = H(s)A (s cos φ − ω sin φ) s2 + ω2

SLIDE 8

Sinusoidal Steady-State Output H(s)

x(t) y(t)

x(t) = A cos(ωt + φ) yss(t) = |H(jω)|A cos (ωt + φ + ∠H(jω))

– Same frequency as x(t): ω – Amplitude is scaled by |H(jω)| – The phase is shifted by ∠H(jω)

any sinusoidal input signal

Solving for the Complex Residue Y (s) = H(s)A (s cos φ − ω sin φ) s2 + ω2 = k s − jω + k∗ s + jω +

kℓ s + pℓ k = H(s)A (s cos φ − ω sin φ) s + jω

= H(jω)A(jω cos φ − ω sin φ) 2jω = H(jω)A(cos φ + j sin φ) 2 =

Example 12: Steady-State Sinusoidal Analysis

R C vs(t) vo(t)

Find the steady-state sinusoidal response to an input voltage of vs(t) = cos(ωt).

Sinusoidal Steady-State Output Since H(jω) is complex, we can write it in polar form as H(jω) = |H(jω)| ej∠H(jω) Then using the results of the previous slide, we have k = 1

= 1

|k| = 1

∠k = φ + ∠H(jω) yss(t) = 2|k| cos(ωt + ∠k) = |H(jω)|A cos (ωt + φ + ∠H(jω))

SLIDE 9

LTI Systems H(s)

x(t) y(t)

h(t)

x(t) y(t)

calculate the output for any input

mechanical, chemical, hydraulic, etc.

Example 12: Workspace

Example 13: Transfer Function Analysis

b

m

x(t) y(t)

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.

Steady-State Sinusoidal Analysis Comments

H(s) vary with s = jω

SLIDE 10

Cascade Transfer Function Synthesis Continued H1(s)

x(t)

H2(s) HP(s)

y(t)

low-order components

later)

text (Chapter 15)

Transfer Function Synthesis H(s)

x(t) y(t)

h(t)

x(t) y(t)

system

function H(s) of a given circuit

given H(s)

Summary

conditions can be represented generically by their transfer functions

causal input signal (x(t) = 0 for t < 0)

frequency but amplified by |H(jω)| and shifted in phase by ∠H(jω)

just circuits

Cascade Transfer Function Synthesis H1(s)

x(t)

H2(s) HP(s)

y(t)

H(s) = N(s) D(s) = H1(s) × H2(s) × · · · × HP (s)

given application later this term

to break H(s) up into 1st (real poles) or 2nd (complex poles)

polynomial in the numerator and denominator