Capacitors in Hybrid-pi Model and Frequency Response

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MyNameGoesHere
Circuit schematic of a common-emitter BJT amplifier with a 2N3904 transistor, 15 VDC supply, R1 100k, R2 10k, R3 10k, R4 1k, capacitors C1, C2 and C3 at 0.1 uF, and a function generator input labelled Vin and output Vout.

This is what I have and I was wondering where C2 might go if I make apply the hybrid pi model on this circuit. I know that all the DC Voltages should be shorted if I were to make use of an equivalent circuit, so that in the hybrid pi model R1 should be in the input and R3 in the output. But what about C2? It was already grounded from the beginning. Does it go in the ground horizontally? If so, on the input or output?
I would normally ignore the capacitive reactances if the frequency was high enough that they would all go to zero. But I was told to do a frequency response and find -3 dB loss points, and they exist in both ends, in the low region and the high region. That means this circuit acts as a band pass filter which is a composite circuit of a low pass filter of RC such that I have R in series and C in parallel to the output and high pass filter of CR, C in series and R in parallel. Well, I can see that C1 and C3 are in series with the input and output. If C2 goes to the ground, I have no way of explaining decrease in the output in the high frequency range except arguing that the circuit is not perfect that it has capacitors and inductors everywhere.
This is as far as I was taught in my undergrad electronics physics class. I'm not an engineering student, so don't know a whole lot about circuits. It would be grateful if someone could give a hint. Thanks.
 
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  • Circuit schematic of a common-emitter transistor amplifier with 2N3904 NPN transistor, +15VDC supply, R1 100k, R2 10k, R3 10k, R4 1k, capacitors C1, C2 (0.1 µF) and C3 (0.1 µF), labelled Fig. 1.
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  • Schematic of a common-emitter BJT amplifier with a 2N3904 transistor, +15VDC supply, resistors R1 100k, R2 10k, R3 10k, R4 1k, capacitors C1, C2 and C3 at 0.1 µF, and a function generator input labelled Vin and output labelled Vout.
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  • Schematic of a common-emitter BJT amplifier with a 2N3904 transistor, +15VDC supply, R1 100kΩ, R2 10kΩ, R3 10kΩ, R4 1kΩ, capacitors C1, C2 and C3 at 0.1 µF, and a function generator input at Vin.
    upload_2017-10-26_9-45-33.png
    9.8 KB · Views: 976
  • Circuit schematic of a common-emitter BJT amplifier with a 2N3904 transistor, 15 VDC supply, R1 100k, R2 10k, R3 10k, R4 1k, capacitors C1, C2 and C3 at 0.1 uF, and a function generator input labelled Vin and output Vout.
    upload_2017-10-26_9-46-19.png
    9.8 KB · Views: 2,033
C2 is a power supply bypass capacitor. In the real world it is often required because, as you state,
MyNameGoesHere said:
the circuit is not perfect that it has capacitors and inductors everywhere.
That is, there may be long leads back to the power supply and/or the power supply may use low quality electrolytic capacitors that do not work well at high frequencies. C2 works to ensure a more ideal (low impedance) voltage source for the circuit. For the usual, non rigorous, simulations just ignore C2 and leave it out. That leaves you assuming a perfect voltage source for the +15V

For the low-pass situation, take account of the frequency response/delay time of the transistor.