EECE 301 Signals Systems Prof Mark Fowler Note Set 19 C T Systems Frequency Domain Analysis of Systems Reading Assignment Section 5 2 of Kamen and Heck 1 17 Course Flow Diagram The arrows here show conceptual flow between ideas Note the parallel structure between the pink blocks C T Freq Analysis and the blue blocks D T Freq Analysis New Signal Models Ch 1 Intro C T Signal Model Functions on Real Line System Properties LTI Causal Etc D T Signal Model Functions on Integers New Signal Model Powerful Analysis Tool Ch 3 CT Fourier Signal Models Ch 5 CT Fourier System Models Ch 6 8 Laplace Models for CT Signals Systems Fourier Series Periodic Signals Fourier Transform CTFT Non Periodic Signals Frequency Response Based on Fourier Transform Transfer Function New System Model New System Model Ch 2 Diff Eqs C T System Model Differential Equations D T Signal Model Difference Equations Ch 2 Convolution Zero State Response C T System Model Convolution Integral Zero Input Response Characteristic Eq D T System Model Convolution Sum Ch 4 DT Fourier Signal Models DTFT for Hand Analysis DFT FFT for Computer Analysis Ch 5 DT Fourier System Models Freq Response for DT Based on DTFT New System Model New System Model Ch 7 Z Trans Models for DT Signals Systems Transfer Function New System Model2 17 5 2 Response to Aperiodic Signals Impulse Response h t is a time domain description of the system Frequency Response H is a frequency domain description of the system Recall that Because h t and H form a FT pair one completely defines the other h t and convolution completely describe the zero state response of an LTI to an input i e h t completely describes the system Thus H must also completely describes the LTI system HOW 3 17 Conv Property from chapter 4 Proof Step 1 Think of the input as a sum of complex sinusoids Each component F e j t Step 2 We know how each component passes through an LTI F e j t This is the idea of frequency response H F e j t is the out component that is due to the input component Step 3 Exploit System Linearity again Step 2 was the first time 1 Total output is a sum of output components y t 2 H F e d 4 17 j t Input Output Relationship Characterized Two Ways 1 Time Domain y t h t f t 2 Freq Domain Y H F Given input f t and impulse response h t to analyze the system we could either 1 Compute the convolution h t f t or 2 Do the following a Compute H compute F b Compute the product Y H F c Compute the IFT y t F 1 H F Method 2 Freq Domain Method may not be necessarily easier but it usually provides a lot more insight than Method 1 From the Freq Domain view we can see how H boosts or cuts the amounts of the various frequency components 5 17 Relationships between various modeling methods Recall we are trying to find ways to model CT Linear Time Invariant Systems in Zero State Since these are all equivalent we can use any or all of them to solve a given problem 6 17 Example Scenario You need to send a pulse signal into a computer s interface circuit to initiate an event e g next PTT slide Q What kind of signal should you use Possibility A rectangular pulse Ap t A t 2 2 Q Will this work It depends on the interface circuitry already in the computer Suppose the interface circuitry consists of an AC Coupled transistor amplifier as shown below 7 17 We ll ignore the effects of this capacitor in our analysis AC coupled Output signal Input signal Model this as an equivalent Input Impedance simplify here Req Equivalent Circuit Model x t eq y t Now we need to find the System Model viewpoint 8 17 Equivalent System Model x t y t X H Actually one LIKE it Y What is H Use Sinusoidal Analysis to find it we did that once already for this circuit Use Phasors Impedances and Voltage Divider K K Req V0 Vi 1 Req j C x t eq y t H j Req C 1 j Req C 9 17 Now what does the input pulse look like in the frequency domain From FT table p t sinc 2 X j ReqC So the output FT looks like Y H X sinc 2 1 j R C eq Now how do we find y t y t F 1 Y So find IFT of this YUCK HARD 10 17 Well do we need to go back to the time domain NO Just look at Y and see what it tells Think Parseval s theorem The plots below show that very little energy gets through the system System s Freq Resp Input FT Output FT 0 1 0 08 0 8 0 08 0 06 0 6 0 06 H f 0 04 0 02 0 1000 01 Y 1 f 1 X 1 f 0 1 0 4 0 04 0 2 500 0 f Hz 500 1000 0 1000 1 0 02 500 0 f Hz 500 1000 0 1000 500 0 f Hz 500 1000 01 So this pulse signal is not usable here because very little of its energy gets through the interface circuitry 11 17 The problem lies in that H is small where X is big and vice versa Pick an X that does not do that Use a pulse that is Modulated Up to where H allows it to pass x2 t p t cos 0t A modulated pulse 2 2 t sinc shifted up sinc shifted up X 2 0 0 See actual plots on next page 12 17 System s Freq Resp Original Input FT Original Output FT 0 1 0 08 0 8 0 08 0 06 0 6 0 06 H f 0 04 0 02 0 1000 Y 1 f 1 X 1 f 0 1 0 4 0 04 0 2 500 0 f Hz 500 0 02 0 1000 1000 500 0 f Hz 500 0 1000 1000 0 1 0 08 0 8 0 08 0 06 0 6 0 06 H f 0 04 0 02 0 1000 0 4 0 f Hz 500 1000 Alternate Input FT 500 1000 500 0 f Hz 500 1000 0 04 0 2 500 0 f Hz Y2 f 1 X 2 f 0 1 500 0 02 0 1000 500 0 f Hz 500 1000 Same Freq Resp 0 1000 Alt Output FT Output FT is not changed much from Input FT this is a viable pulse 13 17 Example Attenuation of high frequency Disturbance X Desired Part of Signal X degrees Frequency rad sec This scenario could occur in an audio setting a high pitched interference We ve also seen it occur in the example of a radio receiver the de modulator created the desired lowfreq signal but it also …