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In the following problem, I know how to solve the whole problem but I never understood what this statement means: " All capacitors are assumed to be large enough at the operating frequency of Vsig" enter image description here

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    \$\begingroup\$ It just means that the operating point voltage across those capacitors can be considered "fixed" and that applying the signal won't change that voltage. Or, put another way, that the capacitor's impedance is ignorably low at frequencies of interest. \$\endgroup\$ – jonk Apr 12 '18 at 21:47
  • \$\begingroup\$ Depending on your criteria for phase shift or attenuation, {Rc+RL}Cc =T, determines the breakpoint for f{-3dB} and 45 deg. Phase shift. Same for RbCb=T and that T>>0.16/f with choice of C \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 12 '18 at 22:55
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Agreed with Jonk. The impedance of a capacitor is 1 / (jwC) so very large wC make this 1 / BigNumber which is approximately 0

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This means that for the operating frequency of your circuit (or mid-band as its sometimes called), the capacitors can be assumed to act as short cicuits, and in your small signal model - safely ignored.

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