ECE2101_HW5_PPSOL_SpS19.pdf - Spring Semester 2019 ECE 2101 — Electrical Circuit Analysis II A Gonzalez California State Polytechnic University Pomona

# ECE2101_HW5_PPSOL_SpS19.pdf - Spring Semester 2019 ECE 2101...

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Spring Semester 2019 ECE 2101 — Electrical Circuit Analysis II A. Gonzalez California State Polytechnic University, Pomona Department of Electrical and Computer Engineering Homework 5 Practice Problems Solutions Due : N/A 4. Find f ( t ) if (a) F ( s ) = 8 s 2 + 37 s + 32 ( s + 1)( s + 2)( s + 4) K 1 s + 1 + K 2 s + 2 + K 3 s + 4 %4(a): Ns=[8 37 32]; %N(s) polynomial Ds=poly([-1 -2 -4]); %D(s) polynomial [r,p]=residue(Ns,Ds) %Residue method Partial fractions: F ( s ) = 8 s 2 + 37 s + 32 ( s + 1)( s + 2)( s + 4) 1 s + 1 + 5 s + 2 + 2 s + 4 Inverse-transform: f ( t ) = L - 1 [ F ( s )] = [ e - t + 5 e - 2 t + 2 e - 4 t ] u ( t ) (b) F ( s ) = 13 s 3 + 134 s 2 + 392 s + 288 s ( s + 2)( s 2 + 10 s + 24) K 1 s + K 2 s + 2 + K 3 s + 4 + K 4 s + 6 %4(b): Ns=[13 134 392 288]; %N(s) polynomial Ds=poly([0 -2 roots([1 10 24]).']); %D(s) polynomial [r,p]=residue(Ns,Ds) %Residue method Partial fractions: F ( s ) = 13 s 3 + 134 s 2 + 392 s + 288 s ( s + 2)( s 2 + 10 s + 24) 6 s + 4 s + 2 + 2 s + 4 + 1 s + 6 Inverse-transform: f ( t ) = L - 1 [ F ( s )] = [6 + 4 e - 2 t + 2 e - 4 t + e - 6 t ] u ( t ) 1

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Spring Semester 2019 ECE 2101 — Electrical Circuit Analysis II A. Gonzalez (c) F ( s ) = 20 s 2 + 16 s + 12 ( s + 1)( s 2 + 2 s + 5) K 1 s + 1 + K 2 s + 1 - j 2 + K * 2 s + 1 + j 2 %4(c): Ns=[20 16 12]; %N(s) polynomial Ds=poly([-1 roots([1 2 5]).']); %D(s) polynomial [r,p]=residue(Ns,Ds) %Residue method Partial fractions: F ( s ) = 20 s 2 + 16 s + 12 ( s + 1)( s 2 + 2 s + 5) 4 s + 1 + 8 + j 6 s + 1 - j 2 + 8 - j 6 s + 1 + j 2 4 s + 1 + 10 36 . 87 s + 1 - j 2 + 10 - 36 . 87 s + 1 + j 2 Inverse-transform: f ( t ) = L - 1 [ F ( s )] = [4 e - t + 20 e - t cos(2 t + 36 . 87 )] u ( t ) (d) F ( s ) = 250( s + 7)( s + 14) s ( s 2 + 14 s + 50) K 1 s + K 2 s + 7 - j + K * 2 s + 7 + j %4(d): Ns=250 * poly([-7 -14]); %N(s) polynomial Ds=poly([0 roots([1 14 50]).']); %D(s) polynomial [r,p]=residue(Ns,Ds) %Residue method Partial fractions: F ( s ) = 250( s + 7)( s + 14) s ( s 2 + 14 s + 50) 490 s + - 120 - j 35 s + 7 - j + - 120 + j 35 s + 7 + j 490 s + 125 - 163 . 74 s + 7 - j + 125 163 . 74 s + 7 + j Inverse-transform: f ( t ) = L - 1 [ F (
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