EECE 301 NS_07 & NS_08 Summary & Solutions1

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Electrical Engineering

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Oct 30, 2023

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NS-07 & NS-08 Fourier Series – Summary • What is the usefulness of Fourier Series? – Breaks a periodic signal into a sum of sinusoids – Advantage… can analyze LINEAR system for each sinusoid individually – This leads to nothing more than a generalization of “Phasor Method” • Fundamental Frequency of FS Components – Determined from period of signal – All sinusoids in FS have frequencies an integer multiple of Fund Freq • There are Three Equivalent Forms of FS based on term type: – cosine with amplitude and phase – complex exponential with complex # holding amplitude & phase – sine & cosine, each with amplitude • But if we are given a periodic signal… how do we “find the Fourier Series”? – Use integral formula for Fourier Series Coefficients: c k = ???? – Note that the c k are essentially like phasors for the FS components! 1/13 The form we will exclusively use!

NS-07 Problems 1. For the periodic signal shown below… a) What is its fundamental frequency in rad/sec? … in Hz? b) What are the frequencies of the terms in its Fourier Series? ) ( t x 1 1 2 -2 -1 … … t 2/13

Solution #1 ) ( t x 1 1 2 -2 -1 … … t So… period is 2 seconds. 2 rad/sec o T    = = 2 o o f   = 0.5 o f Hz = The Fourier series consists of a sum sinusoids with frequencies that are integer multiples of the fundamental: π rad/sec, 2 π rad/sec, 3 π rad/sec, 4 π rad/sec, 5 π rad/sec, …. Each sinusoidal component has two exponential terms: + & – frequencies!!! 3/13

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2. Suppose you apply this periodic signal as input to the RL circuit shown below. Find the output signal y ( t ) 30  100mH y ( t ) x ( t ) 0 0.05 0.1 0.15 0.2 -2 -1 0 1 2 t (sec) x ( t ) You can’t really solve this as stated!! At least not until we learn the next step of determining the FS coefficients from a given periodic signal! 4/13

2. ALTERNATE VERSION … Suppose you have a periodic signal with fundamental frequency  o = 100 rad/sec and the Amplitude-Phase form of the FS has only three terms defined by If this signal x ( t ) goes through the following RL circuit, what is the output signal y ( t )? A 1 = 1.0  1 = 0 A 2 = 0.5  2 =  /4 A 3 = 0.5  3 =  /2 30  100mH y ( t ) x ( t ) NOW you can solve it!! 5/13 Hint : The info given allows you to write 𝑥(𝑡) as a sum of 3 cosines Superposition says you can handle each cosine by itself Use phasor analysis for each of those terms!

Solution #2 L Z j L  = ( ) cos(100 ) 0.5cos(200 / 4) 0.5cos(300 / 2) x t t t t   = + + + + 0 1 1 1 j X e = = /4 2 0.5 j X e  = /2 3 0.5 j X e  = 300 k k k k j L j j Y X X X R j L R L j j             = = =       + + +       1.2490 1 100 1 0.3162 300 100 j j Y e j   = =   +   /4 1.7682 2 200 0.5 0.2774 300 200 j j j Y e e j    = =   +   /2 2.3562 3 300 0.5 0.3535 300 300 j j j Y e e j    = =   +   ( ) 0.3162cos(100 1.2490) 0.2774cos(200 1.7682) 0.3536cos(300 2.3562) y t t t t = + + + + + A 1 = 1.0  1 = 0 A 2 = 0.5  2 =  /4 A 3 = 0.5  3 =  /2 These tell how to express the periodic signal in terms of FS components Convert each cosine into its corresponding static phasor Find each output phasor using voltage divider (or other method): Convert to polar form Convert each output phasor to cosine & add 6/13

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0 0.05 0.1 0.15 0.2 -2 -1 0 1 2 t (sec) 30  100mH y ( t ) x ( t ) x ( t ) y ( t ) ( ) cos(100 ) 0.5cos(200 / 4) 0.5cos(300 / 2) x t t t t   = + + + + 0 0.05 0.1 0.15 0.2 -1 -0.5 0 0.5 1 t (sec) ( ) 0.3162cos(100 1.2490) 0.2774cos(200 1.7682) 0.3536cos(300 2.3562) y t t t t = + + + + + I used MATLAB to add these three sinusoids to get this input form I used MATLAB to add these three sinusoids to get this output form 7/13

NS-08 – Problems 2. Evaluate the integral found in #1 8/13 1. Write the integral needed to find the complex-exponential form FS coefficients for the periodic signal below 2 t 1 – 1 3 5 – 3 – 5 ( ) x t

NS-08 – Solutions 9/13 T 0 2 T   =  = 0 1 pick t = − 1 1 2 1 ( 1) jk t k c t e dt  − − = +  1 ( ) o o o t T jk t k T t c x t e dt  + − =  1. Write the integral needed to find the complex-exponential form FS coefficients for the periodic signal below 2 t 1 – 1 3 5 – 3 – 5 ( ) x t

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NS-08 – Solutions 2. Now evaluate the integral: 1 1 2 1 ( 1) jk t k c t e dt  − − = +  2 From Integral Table ( 1) 0 ax ax e xe dx ax a a = −   1 1 1 1 2 2 1 1 jk t jk t te dt e dt   − − − − = +   One Way Another way: change of variables… t + 1 = τ 1 0 1 2 t t   = −  = =  = 2 2 ( 1) 1 1 2 2 0 0 jk jk jk k c e d e e d         − − − = =   dt d  = 2 1 2 2 0 ( 1) 0 ( ) jk t jk e e jk t k jk     − −   = − −    −   1 t  = − 10/13

2 1 2 2 0 ( 1) 0 ( ) jk t jk e e jk t k jk     − −   = − −    −   2 2 ( ) 1 j j − = = − ( ) ( 1) k jk j k e e     = = − ( ) 2 1 2 2 2 0 1 1 ( 1) 0 k jk t e jk t k k    −     = − − −      −   ( ) 1 2 2 2 2 0 1 ( 1) 0 k jk t e jk t k k    − − −   = − −    ( ) 1 2 0 2 2 2 1 ( 2 1) ( 0 1) 0 k jk jk e jk e jk k k      − − − −   = − − − − −    2 t = 0 t = Carry over result from previous slide: 11/13

( ) 1 2 0 2 2 2 1 ( 2 1) ( 0 1) 0 k jk jk e jk e jk k k      − − − −   = − − − − −    0 1 e = Carry over result from previous slide: ( ) 2 1 1 k j k e  − = = =0 ( )   1 2 2 2 1 ( 2 1) ( 1) 0 k j k k k   − − = − − − −  cancel ( )   1 2 2 2 1 2 0 k j k k k   − − = −  ( 1) 0 k k j c k k  − =  cancel cancel 12/13

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0 1 c = For k = 0: 1 1 0 1 1 0 2 2 1 1 ( 1) ( 1) 1 j t c t e dt t dt  − − − = + = + =   2 t 1 – 1 3 5 – 3 – 5 ( ) x t So… the final result is: 0 1 c = ( 1) 0 k k j c k k  − =  c 0 is the “DC Level” or “DC Offset” 13/13 Big Idea: • We can use these like phasors!!! • They are not in polar form, but we could put them in that form • We can now use them to figure out how this 𝑥(𝑡) goes through some LTI system (e.g., an RLC circuit)!!!!