Well, no. Let us start to the left. If you disconnect R6, C1 sees the parallel connection of R7 and R8 (13k). Now reconnect R6. The equivalent impedance seen through R6 depends on the transistor (the β of the transistor at the working point) and the emitter resistor. This is a somewhat cumbersome calculation, but to get a ballpark value, assume β=100. Then the impedance seen through R6 is approximately R6 + 100*R3. This impedance appears in parallel with the 13k calculated above.Is this ok?
There is no need to guess the transistor β. Both the Base current and Collector current are given.assume β=100.
Yes. But - we are not talking about the DC working point, but the AC current gain.There is no need to guess the transistor β. Both the Base current and Collector current are given.
No.Is this right?
Here is a pretty good tutorial on how to calculate the input impedance of a CE amplifier:Yes. But - we are not talking about the DC working point, but the AC current gain.