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(e) Let y(0)=0 and u(t)=cos2t, when t>=0. Find y_(ss)(t), the output of the system at steady state using the approach described in Corollary 2.33. (20%) (e) Let y(0) = 0 and u(t)= cos 2t, when t 0. Find y,s(t), the output of the system at steady state using the approach described in Corollary 2.33. (20%)

          (e) Let y(0)=0 and u(t)=cos2t, when t>=0. Find y_(ss)(t), the output of the system at steady state using the
approach described in Corollary 2.33. (20%)
(e) Let y(0) = 0 and u(t)= cos 2t, when t 0. Find y,s(t), the output of the system at steady state using the approach described in Corollary 2.33. (20%)
        
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e let y00 and utcos2t when t0 find ysst the output of the system at steady state using the approach described in corollary 233 20 e let y0 0 and ut cos 2t when t 0 find yst the output of th 78044

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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(e) Let y(0)=0 and u(t)=cos2t, when t>=0. Find y_(ss)(t), the output of the system at steady state using the approach described in Corollary 2.33. (20%) (e) Let y(0) = 0 and u(t)= cos 2t, when t 0. Find y,s(t), the output of the system at steady state using the approach described in Corollary 2.33. (20%)
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Transcript

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00:01 Hi, from the question given that a system is modeled as dy by dt plus a of y of t which is equal to b times of dx by dt plus cx of t.
00:17 Here a, b, c are all unknowns that is the thing we need to find here and the response to input x of t is equal to 9 plus 15 cos 12t and y of t is equal to 5 plus 13 cos 12t plus 0 .2487.
00:39 Now apply the laplace transform so sy of s is equal to plus not equal to plus a y of s is equal to bs x of s plus cx of s.
00:56 Now the system transfer function y of s by y of s by x of s is equal to bs plus c divided by s plus a.
01:10 So here so let s is equal to j omega so that b j omega plus c divided by j omega plus a.
01:24 So system gain absolute value of h of j omega is equal to square root of c square plus b square omega square divided by square root of a square plus omega square.
01:39 Therefore angle h of j omega is equal to tan inverse of omega b by c minus tan inverse of omega by a.
01:59 So gain of system here it is omega is equal to 0 so this implies absolute value of h of j of 0 is equal to c by a and c by a by comparing the given expression 5 by 9.
02:17 So this implies c is equal to 5 and a is equal to 9 and also gain of system omega is equal to 12.
02:26 Then absolute value of h of j omega is equal to square root of c square plus b square times of 144 divided by square root of a square plus 144.
02:43 So which is equal to 13 over 15...
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