EECE 301 Signals & Systems Prof. Mark Fowler

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1 EECE 3 Signals & Systems Prof. Mark Fowler Note Set #9 C-T Systems: Laplace Transform Transfer Function Reading Assignment: Section 6.5 of Kamen and Heck /7

2 Course Flow Diagram The arrows here show conceptual flow between ideas. Note the parallel structure between the pink blocks (C-T Freq. Analysi and the blue blocks (D-T Freq. Analysi. New Signal Models Ch. Intro C-T Signal Model Functions on Real Line System Properties LTI Causal Etc Ch. 3: CT Fourier Signal Models Fourier Series Periodic Signals Fourier Transform (CTFT) Non-Periodic Signals Ch. Diff Eqs C-T System Model Differential Equations D-T Signal Model Difference Equations Zero-State Response Ch. 5: CT Fourier System Models Frequency Response Based on Fourier Transform New System Model Ch. Convolution C-T System Model Convolution Integral Ch. 6 & 8: Laplace Models for CT Signals & Systems Transfer Function New System Model New System Model D-T Signal Model Functions on Integers Zero-Input Response Characteristic Eq. D-T System Model Convolution Sum New Signal Model Powerful Analysis Tool Ch. 4: DT Fourier Signal Models DTFT (for Hand Analysi DFT & FFT (for Computer Analysi Ch. 5: DT Fourier System Models Freq. Response for DT Based on DTFT New System Model Ch. 7: Z Trans. Models for DT Signals & Systems Transfer Function New System Model /7

3 6.5 Transfer Function We ve seen that the system output s LT is: C( B( Y ( X ( A( A( Part due to IC s Part due to Input So, if the system is in zero-state then we only get the second term: B( Y ( X ( A( Y ( H ( X ( Define: H ( Δ B( A( H( transfer function System effect in zero-state case is completely set by the transfer function 3/7

4 Note: If the system is described by a linear, constant coefficient differential equation, we can get H( by inspection!! Let s see how Recall: x ( s X ( s x() s x () x ( n) n n n ( n ) () With zero ICs we have that each higher derivative corresponds to just another power of s. We can then apply this idea to get the Transfer Function The condition under which we get the TF To illustrate Take the LT of a Diff. Eq. under the zero-state case: d y( a dy( a y( b dx( b x( s Y ( asy ( ay ( bsx ( b X ( Solve for Y( and identify the H(: Y ( b s b X ( s ) s as a H ( 4/7

5 5/7 ) ( ) ( ) ( ) ( ) ( t x b t dx b t y a t dy a t y d ) ( a s a s b b s s H So now it is possible to directly identify the TF H( from the Diff. Eq.:

6 But, we have also seen that for the zero-state case the system output is: t y( h( λ) x( t λ) dλ h( * x( These limits arise by assuming h( and x( are causal But We have an LT property for convolution that says: Y ( H ( X ( y( h( * x( where: H ( L{ h( } Transfer Function L { Impulse Response} So, we have two ways to get H( - Inspect the Diff. Eq. and identify the transfer function H( - Take the LT of the impulse response h( This gives an easy way to get the impulse response from a Diff. Eq.: - Identify the H( from he Diff. Eq. and then find the ILT of that 6/7

7 Note that the transfer function does essentially the same thing that the frequency response does Y ( H ( X ( y( h( * x( Y ( ω) H ( ω) X ( ω) y( h( * x( Recall: If the ROC of H( includes the jω axis, then H ( ω) H ( s jω This is the connection between The transfer function and frequency response. Recall that the LT is a generalized, more-powerful version of the FT this result just says that we can do the same thing with H( that we did with H(ω), but we can do it for a larger class of systems There are some systems for which we can use either method those are the ones for which the ROC of H( includes the jω axis. 7/7

8 So we know that H( is completely described by the Diff. Eq. Therefore we should expect that we can tell a lot about a system by looking at the structure of the transfer function H( This structure is captured in the idea of Poles and Zeros Poles and Zeros of a system Given a system with Transfer Function: H ( M M bm s bm s... bs b B( N N s an s... as a We can factor B( and A(: (Recall: A( characteristic polynomial) A( H ( bm ( s z)( s z)...( s z ( s p )( s p )...( s p N ) ) M Assume any common factors in B( and A( have been cancelled out Note: H ( i,,..., M s z i { } are called "zeros of H ( " z i { p } are called "poles of H ( " i H ( i,,..., N s p i Note: p i are the roots of the char. polynomial 8/7

9 Note that knowing the sets { } M { } N i p i i i z & tells us what H( is: (up to the multiplicative scale factor b M ) -b M is like a gain (i.e. amplification) Pole-Zero Plot This gives us a graphical view of the system s behavior Example: s s H ( 3 s 6s s 8 ( s 3 j)( s 3 j) ( s 4)( s j)( s j) Real coefficients complex conjugate pairs Pole-Zero Plot for this H( jω s - plane σ x denotes a pole o denotes a zero 9/7

10 From the Pole-Zero Plots we can Visualize the TF function on the s-plane: /7

11 From our Visualization of the TF function on the s-plane we can see the Freq. Resp.: /7

12 Can also look at a pole-zero plot and see the effects on Freq. Resp. As the pole moves closer to the jω axis it has a stronger effect on the frequency response H(ω). Poles close to the jω axis will yield sharper and taller bumps in the frequency response. By being able to visualize what H( will look like based on where the poles and zeros are, an engineer gains the ability to know what kind of transfer function is needed to achieve a desired frequency response then through accumulated knowledge of electronic circuits (requires experience accumulated AFTER graduation) the engineer can devise a circuit that will achieve the desired effect. /7

13 Continuous-Time System Relationships Time Domain Freq Domain N d y( a N b M N M d N d x( b M y( a N M d M dy( a x( b M y( dx( b x( H ( Differential Inspection in Practice Equation LT in Theory b M M s N s a M bm s N N s Transfer Function b s b a s a Pole-Zero Diagram Char. Modes Impulse Impulse Response h( Laplace Transform Fourier Transform s jω Frequency Response H ( ω) H ( s jω This Chart provides a Roadmap to the CT System Relationships!!! 3/7

14 In practice you may need to start your work in any spot on this diagram. From the differential equation you can get: a. Transfer function, then the impulse response, the pole-zero plot, and if allowable you can get the frequency response. From the impulse response you can get: a. Transfer function, then the Diff. Eq., the pole-zero plot, and if allowable you can get the frequency response 3. From the Transfer Function you can get: a. Diff. Eq., the impulse response, the pole-zero plot, and if allowable you can get the frequency response 4. From the Frequency Response you can get: a. Transfer function, then the Diff. Eq., the pole-zero plot, and the impulse response 5. From the Pole-Zero Plot you can get: a. (up to a scaling factor) Transfer function, then the Diff. Eq., the impulse response, and possible the Frequency Response 4/7

15 In Practice we often get the transfer function from a circuit Here s an example: Recall: We get the frequency response from a circuit by using frequency dependent impedances Z R ( ω ) R ZC ( ω) Z L ( ω) jωc and then doing circuit analysis. jωl Similarly, we can get the transfer function using the s-domain impedances: ZR ( R ZC ( ZL( sc sl 5/7

16 Example 6.37 (Assume zero-state) We must assume this to find the transfer function x ( y( X ( Y ( Now we treat everything here as if we have a DC Circuit which leads to simple algebraic manipulation rather than differential equations! For this circuit the easiest approach is to use the Voltage Divider / Cs Y ( X ( R Ls (/ C A standard form: a ratio of two polynomials in s unity coefficient on the highest power in the denominator / LC X ( s ( R / L) s (/ LC) H ( 6/7

17 Some comments: For RLC circuits, the ROC always includes jω Once you start including linear amplifiers with gain > this may not be true If you include non-linear devices the system becomes non-linear But, you may be able to linearize the system over a small operating range E.g. A transistor can be used to build a (nearly) linear amplifier even though the transistor is itself a non-linear device 7/7

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