EECE 301 NS_10 Summary _ Solutions rev1

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NS-10 Periodic Inputs to LTI – Summary • What is the point of Fourier Series? – Breaks a periodic signal into a sum of sinusoids – Advantage… can analyze LTI system for each of those sinusoids individually using generalized “Phasor Method” • Frequency Response H (  ) links input and output FS coefficients – Output FS Coefficient d k = H ( k  o ) c k – This is just like Phasor Analysis… just “formalized”! • Frequency Response links plots of Input & Output Spectra – See Plot on next slide (Repeated from NS_10) • Find Frequency Response by doing Arbitrary Phasor Analysis – Analyze circuit with input being arbitrary phasor for arbitrary frequency 𝜔 – Find output phasor: ? = [?𝑢?? ?? 𝑅, 𝐿, 𝐶 𝜔] Ԧ ? – Frequency Response 𝐻(𝜔) is thing that multiplies input Phasor 1/18

Big Idea: “Frequency Response” Linear Circuit   − = = k t jk k e c t x 0 ) (  0 ( ) jk t k k y t d e   =− =  Input Output Input’s FS Coefficients Output’s FS Coefficients ( ) k o k d H k c  = H (  ) is the “Frequency Response” of the Circuit How to find the Frequency Response of a Circuit… • Assume arbitrary phasor X with frequency  • Analyze circuit to find output phasor Y − It will always take this multiplicative form: Y = H (  ) X − All impedances are evaluated at the arbitrary frequency  • The frequency response function H (  ) is the thing that multiplies X 2/18

Real-World Problem If the signal below is applied as the input to the circuit shown below, find the output signal y ( t ). R C y ( t ) x ( t ) We’ll break this down into several steps in the following problems!!! 3/18 2 t 1 – 1 3 5 – 3 – 5 ( ) x t (msec)

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1. For the circuit shown below, find its frequency response H ( ω ) for arbitrary RC time constant . R C y ( t ) x ( t ) 2. Given that the input signal has FS coefficients given by: 3. Plot the output’s FS magnitude spectrum vs k ω o for k = -4, -3, -2, -1, 0, 1 , 2, 3, 4 if RC = 0.3 msec 4. Use MATLAB to compute the resulting y ( t ) for RC = 0.3 msec and various other values. Really… we’d need to FIND these FS coefficients too but we’ve already done that last time! 4/18 0 1 c = ( 1) 0 k k j c k k  − =  Find the FS coefficients for the output signal (call them d k ) Then find the expression for | d k | 𝑇 = 2 × 10 −3 𝜔 𝑜 = 1000𝜋 ?𝑎?/???

R C y ( t ) x ( t ) Solution : Use Phasor analysis for arbitrary sinusoidal input. 1. For the circuit shown below, find its frequency response H ( ω ) for arbitrary RC time constant . R X Y 1 j C  1 R Y X R j C    =   +   ( ) H  ( ) 1 R H R j C   = + ( ) 1 j RC H j RC    = + Don’t want to leave “1 overs” here!! “Standard” form: rect. form over rect. form 5/18

2. Find the FS coefficients (call them d k ) for the output signal (again… in terms of arbitrary RC time constant ) given that the input signal has FS coefficients Solution : Since we know we have a periodic signal and we are given its FS coefficients (the c k values) we know that the given signal x ( t ) can be written as   − = = k t jk k e c t x 0 ) (  • We know the c k values • We can find the 𝜔 𝑜 from the given plot 3 2 2 10 1000 o T T    − =   = = 6/18 0 1 c = ( 1) 0 k k j c k k  − = 

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( ) 1 j RC H j RC    = + 1000 ( ) 1 1 1000 o o o jk RC jk RC H k jk RC jk RC      = = + + Linear Circuit   − = = k t jk k e c t x 0 ) (  0 ( ) jk t k k y t d e   =− =  Input Output Input’s FS Coefficients Output’s FS Coefficients ( ) k o k d H k c  = Now…Use this idea: So 𝐻(𝜔) is the Frequency Response over all values of 𝜔 1000 o   = We need it evaluated only at certain discrete values of 𝝎 : 𝒌𝝎 𝒐 integer k 7/18

8/18 0 1000 ( 1) , 0 1 1000 0 1000 1 0 1 0 1000 k k jk RC j d k jk RC k j RC d j RC        − =    +      = =   +    ( ) k o k d H k c  = Now use this in: But sometimes we want to plot the magnitudes | d k | and then it might make more sense to do the above computation differently using ( ) k o k d H k c  = ( ) k o k d H k c  = ( ) k o k d H k c  = See Next Problem! If all we want are the d k values then this form is perfectly fine… 0 1 c = ( 1) 0 k k j c k k  − =  with

3. Plot the output’s FS magnitude spectrum vs k ω o for k = -4, -3, -2, -1, 0, 1 , 2, 3, 4 if RC = 0.3 msec 0 1000 ( 1) , 0 1 1000 0 1000 1 0 1 0 1000 k k jk RC j d k jk RC k j RC d j RC        − =    +      = =   +    Solution : In problem #2 we found: We could work directly with this But for the magnitude spectrum it is better to use this approach: ( ) k o k d H k c  = ( ) k o k d H k c  = ( ) k o k d H k c  = Find each of these Find each of these 9/18

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( ) 1 j RC H j RC    = + 1000 ( ) 1 1 1000 o o o jk RC jk RC H k jk RC jk RC      = = + + Recall: Best to work directly with this first ( ) H  = 1 j RC j RC   = + ( ) 2 2 1 RC RC   + Now for insight we can sketch this: What does this look like for small 𝜔 ? …… …… …… …… …… for large 𝜔 ? For small 𝜔 : 1 + 𝜔 2 𝑅𝐶 2 ≈ 1 ( ) 1 RC H RC     = For large 𝜔 : 1 + 𝜔 2 𝑅𝐶 2 ≈ 𝜔 2 𝑅𝐶 2 ( ) 1 RC H RC     = A line! Constant! ( ) H   1 c RC  = 1 10/18 0.707 “Cutoff Freq ” “Cutoff Frequency” is the 𝝎 value where |𝑯 𝝎 | drops to 0.707 of value in “passband”

( ) k o k d H k c  = Returning to: Now let’s look at the ? 𝑘 0 1 0 1 k c k k c  =  = 0 ( 1) 0 1 k k k j c k c k c   − =   =   =  11/18 (We did this last time!)  o = 1000 π rad/sec 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 10,000  20,000  30,000  10,000  − 20,000  − 30,000  − (rad/sec) o k  |? 𝑘 |

Now put the pieces together… at least in principle: Note Even Symmetry! ( ) k o k d H k c  = 12/18 Zoomed in to [-20,000, +20,000] rad/sec We use this equation… … to get this plot from the two plots above it! Circles show the points on |𝐻 𝜔 | that are used to find the ? 𝑘 values!

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13/18 Here’s how we compute these values by hand: ( ) k o k d H k c  = ( ) ( ) ( ) 3 2 2 2 3 1000 0.3 10 0.3 ( ) 1 0.3 1 1000 0.3 10 o k k H k k k      − −   = = + +  0 1 0 1 k c k k c  =  = ( ) ( ) 2 2 0.3 1 ( 1000 ) 1 0.3 0.3 , 0 1 0.3 k k k d H k c k k k k            = =       +   =  + 0 0 (0) 0 1 0 d H c = =  =

14/18 ( ) 2 0.3 , 0 1 0.3 k d k k  =  + Now evaluate at k positive values (use even symmetry for neg k )…. 𝒌 ? 𝒌 0 0 1 0.218318 2 0.140595 3 0.100031 4 0.076917 These match the values shown on the bottom plot for |? 𝑘 | shown two slides back

4. Use MATLAB to compute the resulting y ( t ) for RC = 0.3 msec and various other values. See posted EECE_301_NS_10_Problem.m 15/18 Input Signal Output Signal This RC value is such that this term gets attenuated FS Mag. Spectrum of Input Signal FS Mag. Spectrum of Output Signal Freq. Resp. Mag. of Circuit ?(𝒕) ?(𝒕) |? 𝑘 | |? 𝑘 | |𝑯 𝝎 | Circuit makes ? 𝟎 = 𝟎 1 3333 rad/sec c RC  = = 𝒌𝝎 𝒐 (rad/sec) 𝝎 (rad/sec) 𝒌𝝎 𝒐 (rad/sec)

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RC = 6 msec 16/18 Re- run w/ larger RC value… so lower cutoff frequency! ?(𝒕) ?(𝒕) |? 𝑘 | |? 𝑘 | |𝑯 𝝎 | 1 167 rad/sec c RC  = = This new RC is such that NO terms get attenuated (except k=0 term!) But… still makes ? 𝟎 = 𝟎 So ?(𝒕) is now at 0 DC level!

T=2e-3; % set period to its value wo=2*pi/T; % compute the fundamental frequency t=linspace(-3*T,3*T,10000); % create vector of 10,000 time samples over range of -3T to 3T RC=0.3e-3; % set the RC value for the circuit k=-200:200; % create vector holding FS coefficient k index values (-200 to 200 rather than -inf to inf!!!) ck=j*((-1).^k)./(pi*k); % compute ck using formula for all k even though it only holds for odd!!! ck(k==0)=1; % index into k=0 position and set to value of 2 x=0; % initialize to 0 to allow later adding of terms for the FS summation for n=1:length(k) % loop that adds terms to FS summation x = x + ck(n)*exp(j*k(n)*wo*t); % each time through add a new term to the summation end x=real(x); % theoretically imaginary parts shoudl cancel... but due to rounding they don't so we keep only the real part subplot(3,2,1) % create a 3x2 grid of subplots and point to the 1st element plot(t/(1e-3),x) % plot the signal computed via FS summation set(gca,'Xtick',-6:6); xlim([-6 6]) xlabel('time (msec)') % ALWAYS lable axes!!!! ylabel('x(t)') title('Input Signal'); grid subplot(3,2,2) % point to 2nd plot in subplot grid stem(k.*wo,abs(ck)) % stem plot of FS coefficients of input signal xlabel('k\omega_o (rad/sec)'); ylabel('|c_k|') axis([-30*wo 30*wo 0 1.1]) % set the range of the axes... title('FS Magnitude Spectrum of Input Signal'); grid w=(-30*wo):10:(30*wo); % create a set of frequency points for plotting H(w) H_w=j*w*RC./(1 + j*w*RC); % compute H(w) at these w values (finer grid than needed for computing d_k) subplot(3,2,4) plot(w,abs(H_w)) % plot magnitude of frequency response xlabel('\omega (rad/sec)'); ylabel('|H(\omega)|') axis([-30*wo 30*wo 0 1.2]) %Set range of axes title('Magnitude of Frequency Response of Circuit'); grid 17/18 See posted EECE_301_NS_10_Problem.m

kwo=k*wo; %define convenient variable H_kwo=j*kwo*RC./(1 + j*kwo*RC); % compute H(k*wo)... dk=H_kwo.*ck; % compute output signal's FS coefficients y=0; % setup for computing FS summation for output signal for n=1:length(k) y = y + dk(n)*exp(j*k(n)*wo*t); end y=real(y); % suppress small imaginary part that is due to numerical precision subplot(3,2,5) plot(t/(1e-3),y) % plot output signal set(gca,'Xtick',-6:6) xlim([-6 6]) xlabel('time (msec)') ylabel('y(t)') title('Output Signal') grid subplot(3,2,6) stem(k.*wo,abs(dk)) % plot FS magnitude spectrum for output signal xlabel('k\omega_o (rad/sec)') ylabel('|d_k|') axis([-30*wo 30*wo 0 1.1]) title('FS Magnitude Spectrum of Input Signal') grid 18/18

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