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)

= 1/T [∫_0^(T/2)▒〖[V_p ] □(24&dt)〗-∫_(T/2)^T▒〖[-V_p ] □(24&dt)〗]

= 1/T [├ V_p t┤| (T/2)¦0+├ V_p t┤| T¦(T/2)]

=1/T [V_p (T/2)+V_p (T/2)]=

= V_p [v]

*F.F=( V_p)/(〖 V〗_p ) =1

*F.C=( V_p)/(〖 V〗_p ) =1

e)

sea v(t)={(□((〖 V〗_p )/t_0 ) t si 0<t<t_0)¦( 0 si 〖 t〗_0<t<T )}

V_pp= V_p [V_pp ]

V_min=0 [v]

V_max= V_p [v]

〖V_rms〗^21/T [∫_0^(t_0)▒〖[□((〖 V〗_p )/t_0 ) t]^2 □(24&dt)〗+∫_(t_0)^T▒〖[0]^2 □(24&dt)〗]

( 1)/T [( □((v_p^2 )/(〖3t〗_0^2 ))t^3 ├ )┤| t_0¦0]

〖V_rms〗^2( 1)/T [( □((v_p^2 t_0 )/3) )] [V_rms ]

〖 V〗_rms=V_p/√3 (√(t_0/T) ) [V_rms ] si t_0= T entonces 〖 V〗_rms=V_p/√3 [V_rms ]

Ahora

*V_AV= V_p/2 [v]

〖* V〗_AV =V_(AV.abs)=V_p/2 [v]

Vrms1=(Vp/2√2 Π)

Vrmsh=√(V_P^2-(V_P^2/8 Π* Π)=0.698094Vp

DATv=THDv=(Vrmsh/Vrmas1)*100=0.698094Vp/(Vp/2√2 Π)=6.2031*100=620.31

*F.C=( √3 V_p)/(〖 V〗_p ) =√3

*F.F=( 2 V_p/√3)/(〖 V〗_p ) =2/√3

f)

sea v(t)={(V_p/(T/2) t- V_p si 0<t<T/2)¦( -(v_(pt-2v_(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)▒〖[□((〖 2V〗_p )/T) t-V_p ]^2 □(24&dt)〗+∫_(T/2)^T▒〖[□((〖

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