Ensayo Y Ecologia Medio Ambiente

SOLUCION TALLER

Solución punto 1

*Obtenga los siguientes parámetros: la tensión pico a pico (Vpp), la tensión mínima (Vmin), la tensión máxima (Vmax), el valor eficaz de la tensión (Vrms), el valor medio de la tensión (VAV), el valor medio absoluto de la tensión (VAAV), el valor eficaz de la componente fundamental, la distorsión armónica total, el factor de forma y el factor de cresta de las siguientes señales de tensión:

v(t)=150 2cos(wt)

V_pp= 2V_p= 2(150√2) = 300√2 [V_pp ]

V_min=-(150√2) [v]

V_max= (150√2) [v]

V_rms= (150√2)/√2 =150 [V_rms ]

V_AV= 0 [v]

V_(AV.abs)= 1/T ∫_0^T▒|150√2 cos⁡〖(wt)〗 | □(24&dt)

= 1/T [∫_((-Π)/2w)^(Π/2w)▒〖[150√2 cos⁡〖(wt)〗 ] □(24&dt)〗-∫_(Π/2w)^(3Π/2w)▒〖[150√2 cos⁡〖(wt)〗 ] □(24&dt)〗]

= 1/T [(150√2)/w ├ sin⁡〖(wt)〗 ┤| (Π/2w)¦((-Π)/2w)-(150√2)/w ├ sin⁡〖(wt)〗 ┤| (3Π/2w)¦(Π/2w)]

=1/T [(150√2)/w(4)]=(150√2)/2Π(4)= (150√2)/Π(2)=

= (300√2)/Π [v]

Vrms1=(150√2)/√2 =150 [V_rms ]

Vrmsh=√(150^2-150^2)=0

DATv=THDv=(Vrmsh/Vrmas1)*100=0

*F.F=V_rms/( V_(AV.abs) ) =150Π/(300√2) =Π/( 2√2) = 1.110

*F.C=V_màx/( V_rms ) = (150√2)/150 =√2 = 1.414

b) vt150√2cost

V_pp= 150√2 [V_pp ]

V_min=0[v]

V_max= 150√2 [v]

〖V_rms〗^21/T [∫_((-Π)/2w)^(Π/2w)▒〖[150√2 cos⁡〖(wt)〗 ]^2 □(24&dt)〗+∫_(Π/2w)^(3Π/2w)▒〖[150√2 cos⁡〖(wt)〗 ]^2 □(24&dt)〗]

( 1)/T 〖(150√2 )〗^2 [( 1t/2 ├ +sin⁡(2wt)/4w ) ┤| (Π/2w)¦((-Π)/2w)+( 1t/2+sin⁡(2wt)/4w ├ )┤| (3Π/2w)¦(Π/2w)]

〖V_rms〗^2( 1)/T 〖(150√2 )〗^2 [(Π/4w+Π/4w)+2Π/4w]

〖V_rms〗^2( 1)/T 〖(150√2 )〗^2 [Π/w]

〖V_rms〗^2( 1)/2 〖(150√2 )〗^2

〖 V〗_rms= (150√2)/√2 =150 [V_rms ]

V_(AV.abs)= 1/T ∫_0^T▒|150√2 cos⁡〖(wt)〗 | □(24&dt)

= 1/T [∫_((-Π)/2w)^(Π/2w)▒〖[150√2 cos⁡〖(wt)〗 ] □(24&dt)〗+∫_(Π/2w)^(3Π/2w)▒〖[150√2 cos⁡〖(wt)〗 ] □(24&dt)〗]

= 1/T [(150√2)/w ├ sin⁡〖(wt)〗 ┤| (Π/2w)¦((-Π)/2w)+(150√2)/w ├ sin⁡〖(wt)〗 ┤| (3Π/2w)¦(Π/2w)]

=1/T [(150√2)/w(4)]=(150√2)/2Π(4)= (150√2)/Π(2)=

= (300√2)/Π [v]

〖* V〗_AV =V_(AV.abs)= (300√2)/Π [v]

Vrms1=(150√2)/√2 =150 [V_rms ]

Vrmsh=√(150^2-150^2)=0

DATv=THDv=(Vrmsh/Vrmas1)*100=0

*F.F=V_rms/( V_(AV.abs) ) =150Π/(300√2) =Π/( 2√2) = 1.110

*F.C=V_màx/( V_rms ) = (150√2)/150 =√2 = 1.414

c)

v(t)={(150√2 cos⁡(wt) si v(t)≥0)¦( 0 si v(t)<0 )}

V_pp= 150√2 [V_pp ]

V_min=0 [v]

V_max= (150√2) [v]

〖V_rms〗^21/T [∫_((-Π)/2w)^(Π/2w)▒〖[150√2 cos⁡〖(wt)〗 ]^2 □(24&dt)〗+∫_(Π/2w)^(3Π/2w)▒〖0□(24&dt)〗]

V_rms= (150√2)/2 =75√2 [V_rms ]

V_(AV.abs)= 1/T ∫_0^T▒|150√2 cos⁡〖(wt)〗 | □(24&dt)

= 1/T [∫_((-Π)/2w)^(Π/2w)▒〖[150√2 cos⁡〖(wt)〗 ] □(24&dt)〗]

= (150√2)/Π [v]

〖* V〗_AV =V_(AV.abs)= (150√2)/Π [v]

Vrms1=(150√2)/2 =75√2 [V_rms ]

Vrmsh=√(75√2^2-75√2^2)=0

DATv=THDv=(Vrmsh/Vrmas1)*100=0

*F.F=V_rms/( V_(AV.abs) ) =(75√2 Π)/(150√2) =Π/( 2) = 1.571

*F.C= V_màx/( V_rms ) = (150√2)/(75√2) =2

d)

sea v(t)={(〖 v〗_p si 0<t<T/2)¦( 〖-v〗_(p ) si T/2<t<T )}

V_pp= 2V_p [V_pp ]

V_min=-V_p [v]

V_max= V_p [v]

V_AV= 0 [v]

〖V_rms〗^21/T [∫_0^(T/2)▒〖[V_p ]^2 □(24&dt)〗+∫_(T/2)^T▒〖[V_p ]^2 □(24&dt)〗]

( 1)/T [( V_P^2 t ├ )┤| (T/2)¦0+(V_P^2 t ├ )┤| T¦(T/2)]

〖V_rms〗^2( 1)/T [V_P^2 (T/2)+V_P^2 (T/2)]

〖V_rms〗^2( 1)/T [V_P^2 T]

〖 V〗_rms= V_p [V_rms ]

Vrms1=(4Vp/√2 Π)

Vrmsh=√(V_P^2-(16V_P^2/√2 Π))=0.435Vp

DATv=THDv=(Vrmsh/Vrmas1)*100=(0.435Vp/(4Vp/√2 Π))*100=0.483164*100=48.3164

V_(AV.abs)= 1/T ∫_0^T▒|V(t)| □(24&dt)

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