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HKN ECE 210 Exam 3 Review Session
Corey Snyder Grant Greenberg
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Topics Fourier Transform Signal Energy and Bandwidth
LTI System Response with Fourier Transform Modulation, AM, Coherent Demodulation Impulse Response and Convolution Sampling and Analog Reconstruction LTIC and BIBO Stability
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Fourier Transform The Fourier Transform of a signal shows the frequency content of that signal In other words, we can see how much energy is contained at each frequency for that signal This is a big deal! This is the biggest deal! πΉ π = ββ β π π‘ π βπππ‘ ππ‘ π π‘ = 1 2π ββ β πΉ π π πππ‘ ππ
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Important Signals for Fourier Transform
ππππ‘ π‘ π = 1, πππ π‘ < π 2 0, πππ π‘ > π 2 π’ π‘ = 0, πππ π‘<0 1, πππ π‘>0 Ξ π‘ π = 1+ 2π‘ π , πππ βπ 2 <π‘<0 1 β 2π‘ π , πππ 0<π‘< π 2 π πππ π‘ = sin π‘ π‘ , πππ π‘β 0 1, πππ π‘=0
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Fourier Transform Tips
Convolution in the time domain is multiplication in the frequency domain π π‘ βπ π‘ β± πΉ π πΊ(π) Conversely, multiplication in the time domain is convolution in the frequency domain π π‘ π π‘ β± 1 2π πΉ π βπΊ(π) Scaling your signal can force properties to appear; typically time delay Ex: π β2 π‘+1 π’ π‘β1 β π β4 π β2 π‘β1 π’(π‘β1) The properties really do matter! Take the time to acquaint yourself with them. Remember that the Fourier Transform is linear, so you can express a spectrum as the sum of easier spectra Ex: Staircase function Magnitude Spectrum is even symmetric, Phase Spectrum is odd symmetric for real valued signals
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Signal Energy and Bandwidth
πΈπππππ¦=π= ββ β π π‘ 2 ππ‘= 1 2π ββ β πΉ π 2 ππ Energy signals can be either low-pass or band-pass signals Why not high-pass? Bandwidth for Low-pass Signals 3dB BW πΉ Ξ© πΉ = 1 2 r% BW 1 2π βΞ© Ξ© πΉ π 2 ππ=ππ Bandwidth for Band-pass signals 1 2π π π π π’ πΉ π 2 ππ= ππ 2 , Ξ©= π π’ β π π
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LTI System Response using Fourier Transform
Given the following LTI system: π π‘ β βπ¦(π‘) π π =πΉ π π»(π) Computationally speaking, and in future courses, we prefer to use Fourier Transforms instead of convolution to evaluate an LTI system Why? So much faster: π πππππ π£π . π( π 2 ) π»(π)
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Modulation, AM Radio, Coherent Demodulation
Modulation Property π π‘ βπΉ π , π π‘ cos π π π‘ β 1 2 [πΉ πβ π π +πΉ π+ π π ] In general, for Amplitude Modulation in communications, we modulate with cosine in order to shift our frequency spectrum into different frequency bands Coherent Demodulation refers to the process of modulating a signal in order to make it band-pass, then modulating it back to the original low-pass baseband before low-pass filtering in order to recover the original signal Envelope detection is the process of Full-wave Rectification (absolute value), then low-pass filtering in order to extract the signal
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Impulse Response and Convolution
π₯ π‘ βπ¦ π‘ = ββ β π₯ π π¦ π‘βπ ππ= ββ β π₯ π‘βπ π¦ π ππ We βflip and shiftβ one signal and evaluate the integral of the product of the two signals at any value of t (our delay for the shift) Representing LTI Systems π¦ π‘ =π₯ π‘ ββ π‘ , π€βπππ β π‘ ππ π‘βπ πππππππ ππππππππ ππ π‘βπ π π¦π π‘ππ Impulse Response is the system output to a πΏ π‘ input Graphical convolution helps to visualize the process of flipping and shifting
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Helpful Properties for Convolution
Derivative β π‘ βπ π‘ =π¦ π‘ β π ππ‘ β π‘ βπ π‘ =β π‘ β π ππ‘ π π‘ = π ππ‘ π¦(π‘) Use of Derivative property: Finding the impulse response from the unit-step response πΌπ π¦ π‘ =π’ π‘ ββ π‘ , π‘βππ π ππ‘ π¦ π‘ = π ππ‘ π’ π‘ ββ π‘ =πΏ π‘ ββ π‘ =β(π‘) Start Point If the two signals have start points at π‘ 1 πππ π‘ 2 , then the start point of their convolution will be at π‘ 1 + π‘ 2 End Point Similarly for the end points, if the two signals have end points at π‘ 1 πππ π‘ 2 , then the end point of their convolution will be at π‘ 1 + π‘ 2 Width From the above two properties, we can see that if the two signals have widths π 1 and π 2 , then the width of their convolution will be π 1 + π 2
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The Impulse Function πΉ(π)
The impulse function is the limit of 1 π ππππ‘ π‘ π ππ πβ0 Infinitesimal Width Infinite Height Of course, it integrates to 1. (0ββ=1) Sifting π π πΏ π‘β π‘ π π π‘ ππ‘= π π‘ 0 , π‘ 0 β[π,π] 0, πππ π Sampling π π‘ πΏ π‘β π‘ π =π π‘ π πΏ(π‘β π‘ π ) Unit-step derivative ππ’ ππ‘ =πΏ(π‘)
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Sampling and Analog Reconstruction
If we have an original analog signal π(π‘) Our digital samples of the signal are obtained through sampling property as: π π =π ππ where T is our sampling period; this is Analog to Digital (A/D) conversion We must make sure to satisfy Nyquist Criterion: π< 1 2π΅ ππ π π >2π΅, π΅= π΅ππππ€πππ‘β (in Hz) To learn more about Nyquist Criterion and Sampling take ECE 110!
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Sampling and Analog Reconstruction
If we have an original analog signal π(π‘) Our digital samples of the signal are obtained through sampling property as: π π =π ππ where T is our sampling period; this is Analog to Digital (A/D) conversion We must make sure to satisfy Nyquist Criterion: π< 1 2π΅ ππ π π >2π΅, π΅= π΅ππππ€πππ‘β (in Hz) To learn more about Nyquist Criterion and Sampling take ECE 110! Just kidding, take ECE 310
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Sampling and Analog Reconstruction
If we have an original analog signal π(π‘) Our digital samples of the signal are obtained through sampling property as: π π =π ππ where T is our sampling period; this is Analog to Digital (A/D) conversion This results in infinitely many copies of the original signalβs Fourier Transform spaced by 2π π We must make sure to satisfy Nyquist Criterion: π< 1 2π΅ ππ π π >2π΅, π΅= π΅ππππ€πππ‘β (in Hz) To learn more about Nyquist Criterion take ECE 110! Just kidding, take ECE 310 Following A/D conversion, we perform D/A conversion, then low-pass filter our signal in order to obtain our original signal according to the following relation π π‘ = π π π π πππ π π π‘βππ For a more complete explanation, take ECE 310!
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Sampling and Analog Reconstruction (contβd)
The complete mathematical relationship between the continuous frequency spectrum and the discrete time (sampled) spectrum in analog frequency πΉ π π = 1 π π=ββ β πΉ π+ 2ππ π The nuances of this representation will be explored and clarified in ECE 310!
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BIBO Stability A system is BIBO stable if for any bounded input, we obtain a bounded output Two ways to check BIBO stability By the definition: πΌπ π π‘ β€πΌ<β, π‘βππ π¦ π‘ β€π½<β, β π‘ By Absolute Integrability ββ β β π‘ ππ‘<β
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LTIC LTIC stands for Linearity Time-Invariance and Causality Linearity
Satisfy Homogeneity and Additivity Can be summarized by Superposition If π₯ 1 (π‘)β π¦ 1 (π‘) πππ π₯ 2 (π‘)β π¦ 2 (π‘), π‘βππ π π₯ 1 (π‘)+π π₯ 2 (π‘)βπ π¦ 1 (π‘)+π π¦ 2 (π‘) Shift Invariance If π₯(π‘)βπ¦(π‘) π‘βππ π₯(π‘β π‘ π )βπ¦(π‘β π‘ π ) β π‘ 0 πππ π₯(π‘) Causality Output cannot depend on future input values β π =0 for π<0.
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Problem 1 SP16 Consider a real-valued function π(π‘) with bandwidth Ξ© and let π π >Ξ©. Obtain the bandwidth of the following functions All answers may be left in terms of Ξ© and π π . (a) π 1 π‘ =π π‘ sin π π π‘ (b) π 2 π‘ =π π‘ + sin π π π‘ (c) π 3 π‘ =π π‘ sin 2 π π π‘ (d) π 4 π‘ = π 2 π‘ sin π π π‘ (e) π 5 π‘ =π π‘ β sin π π π‘
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Problem 3 FA16 Given the circuit on the right: Find the step response
Find the response to π₯ π‘ =ππππ‘ π‘β1 2
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Problem 3 SP14
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Problem 2 SP13
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Problem 2 FA15
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Problem 4 SP14 iii) Determine y(t)
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Problem 5 FA13 a)i. Determine h(t) ii. Is the system causal?
iii. Is the system BIBO stable? b)i. Determine h(t)
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