Thanks all for the reply!
In this circuit you have your AC signals operating on top of a DC-bias (which makes the transistor influence the AC the way you want it to). The statement has to do with the difference between AC and DC analysis. You know what those are right?
When you are biasing the transistor you ignore the AC signals and only pay attention to the DC. You redraw the circuit the way that DC sees the circuit (inductors become short-circuits and capacitors become open-circuits, all AC sources are turned off). This sets the way the transistor behaves. THen you go to the AC analysis. You redraw the circuit the way AC sees the circuit (inductors and capacitors remain, all DC sources are turned off). You ignore the DC by "turning off the DC supply" so all the AC voltages are centered around 0V ...BUT you pretend the transistor still behaves as if it was still biased. You then add up the DC and AC voltages to get the actual voltage afterwards.
What happens when you turn off a DC current source? It goes to 0A flowing through both terminals- just like an open-circuit.
What happens when you turn off a DC voltage source? It goes to 0V across both terminals- just like a short-circuit.
SO in the AC analysis, it looks like +V is connected to ground via that short-circuit formed by turning off the DC voltage. In circuits where there is a current source, this effectively acts as an AC open-circuit.
Thanks, this is exactly the background information (DC/AC analysis and their respective treatment of power supply) I was not clear, though I saw similar phrases several times in books/webpages. I guess I need to grasp a text book to read related sections (I am sorry that I do not have the opportunity to do a systematical study on this field, so the information I obtained so far is piece by piece and not fit together yet).
I have been reading articles talking about linearity of circuit/elements, and/or linear region of circuit elements (e.g., it seems diode/transistors are not linear elements but have linear regions). So I guess why we can divide circuit analysis into DC/AC parts and "add" them together, is because the circuit in analysis is either linear, or it's assumed to be working in its linear region.
In terms of treatment of DC power supply in AC analysis, as you pointed out, they are treated as "short-circuit". My guessing is that, by "turn off" the DC power, it only means to remove the DC voltage from the ciruit, but not necessary means short-circuit the DC power terminals, especially considering the internal impedance of the power supply (which I believe is not going to disappear in AC analysis). Is this correct? If so, the following question is, why we can still consider the DC power terminals are short-circuited? Is this because the internal impedance of the DC power supply can be ignored (similar to what crutschow mentioned), or anything else?
Actually I am also a bit confused by the phrase like "power supply impedance to ground" as in crutschow's reply (as well as in other contexts talking about impedance "from power to ground"). My understanding of the DC power supply, take battery as an example, is that the battery is the "power supply", and it has two terminals, cathode and anode, and the cathode terminal is usually regarded as GND. So the impedance of the power supply (the battery) is the impedance from the cathode to the anode (or v.v.). Can I interprete the phrase "from power to ground" as a simplified form of "from cathode to anode", in this case?
Thanks again!