1/8 1/31/2011 ( ) ( ) Amplifiers lecture. out. Jim Stiles. Dept. of o EECS

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1 1/31/2011 Amplifiers lecure 1/8 Amplifiers An ideal amplifier is a wo-por circui ha akes an pu signal v and reproduces i exacly a is oupu, only wih a larger magniude! ( ) i ( ) v + ( ) A I v + ou ( ) ( ) = A v The real value A is he open-circui lage ga of hiss ideal amplifier, and has a magniude much larger han uniy ( A 1). Jim Siles The Univ. of Kansas Dep. of o EECS

2 1/31/2011 Amplifiers lecure 2/8 We acually can fd g()! Now, le s express his resul usg our knowledge of lear circui heory! Recall, he oupu u ( ) of a lear device can be deermed by conlvg is pu v( ) wih he device impulse response g ( ): v ( ) = g( ) v ( ) d ou Q: Yikes! Wha is he impulse response of his ideal amp? How can we deerme i? A: I s acually quie simple! Remember, he impulse response of lear circui is jus he oupu ha resuls when he pu is an impulse funcion ( ) δ. Jim Siles The Univ. of Kansas Dep. of EECS

3 1/31/2011 Amplifiers lecure 3/8 Every funcion an Eigen funcion Sce he oupu of an ideal amplifier is jus he pu muliplied by A, we conclude if v ( ) = δ ( ): ( ) = ( ) = A δ ( ) g v ou Thus: v ( ) = g( ) v ( ) d ou = A δ ( ) v ( ) d = A δ ( ) v ( ) d = Av( ) Any and every funcion v ( ) is an Eigen funcion of an ideal amplifier!! Jim Siles The Univ. of Kansas Dep. of EECS

4 1/31/2011 Amplifiers lecure 4/8 And now he Eigen value Now, we can deerme he Eigen value of his lear operaor relag pu o oupu: v ou { } ( ) = L v ( ) Recall his Eigen value is found from he Fourier ransform of he impulse response: jω G ω h e d ( ) = ( ) = jω A δ ( ) e d = A + j0 = Ae j 0 This resul, alhough simple, has an eresg erpreaion Jim Siles The Univ. of Kansas Dep. of EECS

5 1/31/2011 Amplifiers lecure 5/8 DC o dayligh i means ha he amplifier exhibis ga of A for susoidal signals of any and all frequencies! BUT, here is one big problem wih an ideal amplifier: They are impossible o build!! Jim Siles The Univ. of Kansas Dep. of EECS

6 1/31/2011 Amplifiers lecure 6/8 Real amplifier have fie bandwidhs The ideal amplifier has a frequency response of G ( ω ) A =. Noe his means ha he amplifier ga is A for all frequencies 0 (D.C. o dayligh!). The bandwidh of he ideal amplifier is herefore fie! < ω < * Sce every elecronic device will exhibi some amoun of ducance, capaciance, and resisance, every device will have a fie bandwidh. * In oher words, here will be frequencies ω where he device does no work! * From he sandpo of an amplifier, no workg means G ( ω ) A low ga). Amplifiers herefore have fie bandwidhs. (i.e., Jim Siles The Univ. of Kansas Dep. of EECS

7 1/31/2011 Amplifiers lecure 7/8 Amplifier bandwidh There is a range of frequencies ω beween ω and ω where he ga will (approximaely) be A. For frequencies ouside his range, he ga will ypically be small (i.e. ( ) A G ω ): ( ) G ω A ω < ω < ω L H = A ω < ω, ω > ω L H L H The widh of his frequency range is called he amplifier bandwidh: Bandwidh ω ω L H f f H L (radians/sec) (cycles/sec) Jim Siles The Univ. of Kansas Dep. of EECS

8 1/31/2011 Amplifiers lecure 8/8 Wideband is desirable One resul of a fie bandwidh is ha he amplifier impulse response is no an impulse funcion! + jω h ( ) = Hω ( ) e d A δ ( ) herefore generally speakg: ( ) ( ) v A v!! ou However, if an pu signal specrum V ( ) ω lies compleely wih he amplifier bandwidh, hen we fd ha will (approximaely) behave like an ideal amplifier: ( ) ( ) if V ( ) v A v ou ω is wih he amplifier bandwidh As a resul, maximizg he bandwidh of an amplifier is a ypically and imporan design goal! Jim Siles The Univ. of Kansas Dep. of EECS