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# Signals and Systems ECE 350

BSU

GPA 3.91

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This 12 page Class Notes was uploaded by Judd Okuneva on Saturday October 3, 2015. The Class Notes belongs to ECE 350 at Boise State University taught by Elisa Smith in Fall. Since its upload, it has received 13 views. For similar materials see /class/217979/ece-350-boise-state-university in Engineering Electrical & Compu at Boise State University.

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Date Created: 10/03/15

Frequency Response of a Physical and Simulated Lowpass Filter an 0 Vin quot v253 Vuut Nate Stutzke Ap l92001 Introduction ller u rm w mew 2 M e E mew 40 Fig1 RC an39pm Filter Theory and Simulations th me use uforchu s Vunage and Currant Laws me luwpass RC mar m Fxgure 1 an be desmbed by me differmual equauun yltrgtRcmxltrgt m if x in m lavmm system H m1s rewunsble runhe salmg and sh hng urme uulpul Restated er x9 yt Emma Pluggmg 39hesewlues urm anuym mm the mrfaenual equauun 1 and sulvmg furthe 39equencyrespunse gves 1 HQ 47gt m 2 How by mm mama42gt 3 The mmmdes and phases urme mpuL uulpuL and frequmcy rewunse are related by mwexmmu H04 4 and 4mm agmmmm 5 quot L L 39 39 39 39 39 can easilybe a a p in circuitslab As observedby 39 seen in Figure l L and a 22m capacitor The lsim command was utilized to simulate the output ofthe RC circuit in Matlab The run we began by 39 10Vpp sinusoid At 5H2 010 Equation 2 gives Hj107z 02942 1 2 Equations 4 ands tell us that the output of the circuit should have a magnitude of0294 lovpp 294m andit shouldhave a phase of 71272 radians 39 39 39 39 Figure 2 By 39 39 39 39 it L 39 w a p t d To get an exact value we used Matlab max command This gave a value of292Vpp very close to the predicted value Visual 39 39 39 quot A ab mi 404 nu 39 26 radians angle 39 172 radians exactly as predi cted 5 One Second anpui m 9192 93 94 as 95 a7 98 991 t NS One Second uHslm Ouipui yt t n 4 9192 93 94 as 95 a7 98 991 t Fig 2 Simulated RC Circuit Response Physical Circuits 39 quotL prisian Rr lterinFigurelWebeganby examining the response to a 5H2 5Vpp sinusoid As in the simulation theory predicts the frequency response Hj107z 02942 1 272 The output should again have a phase ofel272 radians This time the magnitude of the output should be 0294 SVPP 147Vpp Figure 3 shows the input and output of this physical lowpass filter The yellow sinusoid represenm the input whereas the green represents the output Measurement tools on the oscilloscope gave an output magnitude of about 15Vpp This is within 20 of the predicted value Using the movable markers we estimated a phase of about 7126 radians This is within 09 of the predicted value We also examined the response of the lowpass filter to 10Hz and 20Hz 5Vpp sinusoids Their responses are shown in Figures 4 and 5 The data for these responses is summarized in Table l Fig 4 Response to 10Hz sinusoid Fig 5 Response to 20Hz sinusoid mm 0152 505003 0104 01 4410 137 34 20Hz 00707 505044 0007 130 4494 146 23 Fourier Transfo rm Analysis Another way to characterize the frequency response ofthis circuit is by examining the fast Fourier transformsFFT H 39 A T 39 39 39 FFTsshoulduuul 39 39 inputand output both quot 39 L39 W 39 A H 39 A 39 39 39 quot t yuan ct o theoutputpeak L 394 quot 39 39 39 pi Inthe 39L 39 l 39 weonly inspected the output ofthe simulated circuit Using the command gives the FFT shown in Figure 6 As expected the peak is located at 107tradsec 5H2 r39 quoto pasnne 39 AL39 A quotquotquot Thi A 39 L builtin FFT function on the oscilloscope Figure 7 shows the FFT ofthe 5H2 sinusoidal input The largest peak was at 5H2 as expected This peak had an amplitude of1786dB Figure 8 shows the FFT ofthe output ofthe physical lowpass lter The largestpeak was again at 5H2 as expected The largest peak had an amplitude of 714dB NS Mannlude utFFT utOutput yl Fig 6 FFT of Simulated Output m m mu m mu Table 2 n A 1 r t t w W mm 4b Ts 39 me w t L thew er uuiu mi N 39 L 39 y Werepeatedthis Value 0f0291 39 pmcess for the 10Hz and onz inputs This data for the FPT analysis at all three frequencies is locatedin Table 2 FFT Results for Physical Lowpass Filler Theor Hiw FFTiquot FFTm VFFTnuFFTiquot quotA Diff 5HzI 0294 1786 I 714 I 0291 I 10 I 10Hzl 0152 I 1788 I 115 I 0146 I 39 I 20HzI 00767 I 179 I 596 I 00641 I164I 39r L L r c L I wel L L IL 1 function generator quotquot 39 39 39 39 39 39 39 ofthe input rl 1 L mi up 39I L r L 1 1 increases This is expected behavior since the circuit is alowpass filter Conclusion We have successfully shown that I Yjw II Xjw H H 60 I rst by simulating the circuit with Matlab then by alab W 39 A q r I r r rhesystenr 39 inmae 39 l l w 1 increases This iswhyL 39 quot quot4 lnwna Filter met Li frequencies one interesting trend that ms observed was adecrease in accuracy as frequency increased In general IHGmI ms larger than predicted As frequency increased the experimental value got further on hi h 391 l 39 This is likely 39 39 39 39 39 i not iuear It is not able to filter out higher frequencies as well as theory suggests This is the case in rnany situations when going from theory to reality Useful Identities Euler e cosx j sinx efr e 2 Jr eijr 2 j Trigonometric Identities cos x cos x sinx sin x cosa b cosa cosb sina sinb sina b sina cosb cosa sinb l cos2x 2 cosx sinx 8 cos2 x 5111200 l cc2s2x Note all of these can be derived using Euler s formulas above Complex Numbers A complex number can be written in rectangular or polar form Rectangular 0r Cartesian form 2 a jb a Rez a is the real portion ofz b lmz b is the imaginary portion ofz Polar form 2 m m zl m is the magnitude of 2 distance from origin 9 AZ 9 is the angle or phase of 2 measured from the positive real axis To convert between forms use trigonometric formulas amp Pythagorus m Val 12 9 atanb a 6 Note you need to consider in which quadrant the point lies and adjust accordingly a m cos9 b m sin9 EE350 Signals amp Systems Boise State University Notes on Stochastic Processes Autocorrelation EXt1Xtz Rxxt1tz 13320 Rxxm If Xt is compleX EXI1X 12 RXXI1I2 Rxxt1tz Z 0 Crosscorrelation nyt1tz Ext1yquottz Rquotyxtzt1 Autocovariance Cxxt1tz Rxxt1tz Hxt1Hx t2 Crosscovariance nyObtz nyt1tz MODW 02 Wide sense stationa EXt ux a constant EXt1xt139c Rxxt1t139c Rxx39c Power Spectrum 30 IRre39 dr forward Fourier Transform R39c 2L Isaak4 da Inverse Fourier Transform 7 Note 80 will always be a real function since R39c R39c In general books use many different letters to denote the Power Spectrum Gn m p00 Systems with Stochastic Inputs yt TXt T any transformation from X to y not necessarily LTI if T is memoryless then if the input Xt is stationary then the output will also be stationary If the system is linear yt LXt Cross and autocorrelations with yt will be related to the autocorrelations for Xt nyt1tz L2Rxxt1tz the L2 indicates action wrt time t2 nyt1tz TR 1112 00MBth Rxxt1t2ht2 Ryyt1tz nyt1tz the L2 indicates action wrt time t1 Ryyalatz 321 Ii or Izho d0 nyt1tzht1 When comhiiied Ryy 512 2 32R 11 Dhlz hah d0d Rxxt1t2ht1ht2 although it coonceptually and computationally easier to nd nyt1tz rst If the processes are complex then compleX conjugates are needed mam Rxxt1tzhquottz Ryyt1tz nyt1tzh 1 Ryyt1tz Rxxt1t2h t2ht1 If the random processes are stationary Ryy Rxx1h1 1 Rxxt1tz 321 Syy n SxxwHTH 03 SxxTlHwl White Noise values Xti and XIj are uncorrelated for every ti 7E tj 9 Cxxtitj 0 ti 7E tj has power q SO CXXI1I2 150142 if mean is zero and process is stationary then Rxxt1tz q5t1tz q539c and Sxxoa jq5t q1 a constant over all frequency hence white Normal Process The random variables Xt1 Xtz Xtn are jointly normal for any n and t1 t2 tn Xti is a normal random variable with mean uxti EXIi and variance 6x2ti Ctiti C ll 12 JCOIJOCOZJZ and correlation coef cient rt1tz Partial Fractions Goal Take a polynomial fraction and convert it to a sum of polynomial fractions based on the roots of the denominator nzx2 nlxn0 nzx2 nlxn0 A B C d3x3d2x2 d1xd0 xrxrxr3 ml 39 ml 39 mg nzx2 nlxn0 nzx2 nlxn0 A B C dsxs39l39dzx2 d1x l39do x71xrzxrz xV1lxVzlxrz2 nzx2nlxn0 nzx2 nlxn0 A BxC d3x3 d2x2 d1xd0 xr1wc2 bxc xr1 39 11x2 bxc For non repeated roots it is easiest to use method 2 For repeated roots it is easiest to use a hybrid of the rst two methods Use Method 2 to nd the value of the coef cients for the nonrepeated roots and highest order repeated roots and then use Method 1 to nd the remaining variables You will have less complicated equations to solve in Method 1 and won t need to take derivatives as in Method 2 for repeated roots Method 1 A No repeated roots r1 r2 r3 0 Get common denominator on right side of equation nix2 n1x n0 Axr2 xr3Bxqxr3Cx13xr2 xrlxrzxrs xrlxrzxrs 0 Expand numerator sz Ar2xAr3xAr2r3 Bx2 Br1xw x71xrzx73 0 Set numerator constant terms equal and same for each power of x Note n0 n1 or 11 may be zero n0 Arzr3 Br1r3 Cr1r n1 Ar2 Ar3 Br1 393 r3 Cr1 393 r2 n2 A B C 0 Now solve the 3 equations for the 3 unknowns B Repeated roots 0 Set right side such that you have an increasing order term for each repeated root nix2 nlxn0 A I B I C xrxrxrz ml 39 ow 39 xr22 0 Proceed as above to get 3 equations and solve for the 3 unknowns C Complex roots 0 If a pair of roots are complex leave the pair together so that the form of the express1on 1s nzxznlxn0 nix2 nlxn0 A I BxC d3x3 d2x2 d1xd0 xr1wc2 bxc xr1 39 11x2 bxc 0 Proceed as above to get 3 equations and solve for the 3 unknowns Method 2 A No repeated roots rlqtrzqtrg nix2 nlx n0 A B C xrxrxr3 ml ow mg 0 Multiply both sides by rst factored denominator term 7sz2 n1xn0xr1 Axr1 I Bxr1 I Cxr1 xrxrxr3 ml 39 ow 39 mg 0 Cancel where term appears in both numerator and denominator one place on both sides of equation 7sz2 n1xn0 A Bxr1 Cxr1 xrzx73 x l39rz x l39rs 0 Set your variable to the value of the root ie x r1 quotA 702 n1r1n0 A 30 C01 r1r2r1r3 1rz 71 73 n2r12 n1r1n0 A 71 rzri 73 0 Repeat this for every other term B Repeated roots 0 Set right side such that you have an increasing order term for each repeated root nix2 nlxn0 A B C xrxrxrz ml 39 ow 39 mm For the non repeated roots proceed as above For the repeated roots multiply both sides by highest power of denominator term nix2 nlxnoxr22 Axr22 Bxr22 Cxr22 Axr22 xrxrz2 W1 39 W2 39 my m1 Set your variable to the value of the root ie x V2 and solve n2rz2 nir2 no A02 BOC rz r1 r2 H For the coefficients of the lower order terms of the repeated roots B here take the derivative of both sides of the equation d nzx2 n1xn0xr221 d lAxr22 Bxr Mclzi Axr2ZLB0 dxl xr1xr22 l dxl xr1 2 dx xr1 Bxr2C Set your variable to the value of the root ie x r2 and solve The A term on the right will go to zero when your variable equals the root value For roots repeated more than twice repeat the procedure taking the second then the third derivative etc until all variable values are found Meth0d3 Use Maple gt convert n2x2nlxn0 d2x2dlxd0 parfrac x

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