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Question:Problem 4.8 Sketch the FT representation X6(ja) of the discrete-time signal x(n) = sin(3mm/8) ass...
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Problem 4.8 Sketch the FT representation X6(ja) of the discrete-time signal x(n) = sin(3mm/8) ass...
Problem 4.8 Sketch the FT representation X6(ja) of the discrete-time signal x(n) = sin(3mm/8) assuming that (a) T- 1/2, (b) T,-3/2. See Fia 4 19
Problem 4.8 Sketch the FT representation X6(ja) of the discrete-time signal x(n) = sin(3mm/8) assuming that (a) T- 1/2, (b) T,-3/2. See Fia 4 19
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Dont copié formé thé book oh ya dont copié formé thé book cause you Oiil inde up being triste soi remembré not toi copié frome thé book oh ya
Problem 4.8 Sketch the FT representation X6(ja) of the discrete-time signal x(n) = sin(3mm/8) ass...
find all solutions in the interval [0,2 pi) sin(x+(3.14/3) + sin(x- 3.14/3) =1 sin^4 x cos^2 x Since sin (a+b) = sina cosb + cosb sina and sin (a-b) = sina cosb - cosb sina, the first problem can be written 2 sin x cos (pi/3)= sin x The soluti
2) Use the sum and difference identites sin[x + pi/4] + sin[x-pi/4] = -1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4 - cosx sin pi/4 = -1 2 sin x cos pi/4 =-1 cos pi/4 = sqr2/2 2sin^x(sqrt2/2) = -1 sin x = -sqrt2 x = 7pi/4 and 5pi/4 Am I co
the original problem was: (sin x + cos x)^2 + (sin x - cos x)^2 = 2 steps too please I got 1 for (sin x + cos x)^2 but then what does (sin x - cos x)^2 become since it's minus
the original problem was: Solve: sin(3x)-sin(x)=cos(2x) so far i've gooten to: sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x) Where would I go from here
the original problem was: Solve: sin(3x)-sin(x)=cos(2x) so far i've gotten to: sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x) Where would I go from here
Q2.) Consider the sampling of the continuous-time signal x(t) to obtain a discrete-time signal x[...
prove these identies sin^2+tan^2=sec^2-cos^2 sin^2 sec^2 +sin^2=tan^2+sin^2
(sin/sin+1)-(sin/sin-1)=-2tan^2
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Dont copié formé thé book oh ya dont copié formé thé book cause you Oiil inde up being triste soi remembré not toi copié frome thé book oh ya
Problem 4.8 Sketch the FT representation X6(ja) of the discrete-time signal x(n) = sin(3mm/8) ass...
find all solutions in the interval [0,2 pi) sin(x+(3.14/3) + sin(x- 3.14/3) =1 sin^4 x cos^2 x Since sin (a+b) = sina cosb + cosb sina and sin (a-b) = sina cosb - cosb sina, the first problem can be written 2 sin x cos (pi/3)= sin x The soluti
2) Use the sum and difference identites sin[x + pi/4] + sin[x-pi/4] = -1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4 - cosx sin pi/4 = -1 2 sin x cos pi/4 =-1 cos pi/4 = sqr2/2 2sin^x(sqrt2/2) = -1 sin x = -sqrt2 x = 7pi/4 and 5pi/4 Am I co
the original problem was: (sin x + cos x)^2 + (sin x - cos x)^2 = 2 steps too please I got 1 for (sin x + cos x)^2 but then what does (sin x - cos x)^2 become since it's minus
the original problem was: Solve: sin(3x)-sin(x)=cos(2x) so far i've gooten to: sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x) Where would I go from here
the original problem was: Solve: sin(3x)-sin(x)=cos(2x) so far i've gotten to: sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x) Where would I go from here
Q2.) Consider the sampling of the continuous-time signal x(t) to obtain a discrete-time signal x[...
prove these identies sin^2+tan^2=sec^2-cos^2 sin^2 sec^2 +sin^2=tan^2+sin^2
(sin/sin+1)-(sin/sin-1)=-2tan^2
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