Amplifier with low frequency question?
For AC signal analysis, there are mainly two steps in order to obtain the the h-parameter model

Step 1: Short all power dc power sources and coupling capacitors.

Step 2: Replace BJT with its h-parameter model.

Following step1, we have to ground the Vcc power source (here, there is no coupling capacitors). Resulting circuit is shown below.







simulate this circuit &ndash; Schematic created using CircuitLab

Now we need to replace BJT with h-parameter model. So resulting figure is given below..







simulate this circuit

Note: Rb resistor is in between collector and base terminals. In your model, one of its terminal is connecting to ground and it is wrong. hoe is the outptut transconductance and it is also present in h-parameter model(we can avoid it in some cases, since COLLECTOR RESISTANCE is almost equal to Rc)

Now you need to use miller's theorem, to remove the resistance Rb From the current position. Presence of Rb make it difficult to analyse the circuit. In order to apply millers theorem, we need a voltage gain circuit. But unfortunately we have a current gain circuit. We can convert a current gain source to voltage  gain source as follows

\$i_b=\cfrac{V_b}{h_{ie}}\$

Hence, \${h_{fe}}{{i_b}}={h_{fe}}{\cfrac{V_b}{h_{ie}}}=(\cfrac{h_{fe}}{hie}).Vb=0.0454V_b\$

Now apply millers theorem,we will result the following circuit







simulate this circuit

Analysis

Note: since value of hoe is not given, we can neglect it from small signal equivalent circuit

According to millers theorem,

Effect of resistor Rb at input terminal is

\$R_{bi}=\cfrac{R_b}{(1-A)}; A = 0.0454;\$voltage gain of circuit

\$R_{bi}=\cfrac{R_b}{(1-0.045)} =1.048R_b=209.6k\Omega\$

Similarly,Effect of resistor Rb at input terminal is

\$R_{bo}=\cfrac{{A}{R_b}}{(1-A)}; A = 0.0454\$

\$R_{bo}=\cfrac{{0.0454}{R_b}}{(1-0.045)} ={0.048}{R_b} =9.6k\Omega\$

Impedance at transmitter terminal

\$Z_b=h_{ie}=1.1k\Omega\$

Circuit input impedance

\$Z_i=R_{bi} || h_{ie} = 1.094k\Omega       (approx. hie)\$

Circuit output impedance

\$Z_o=R_{bo} || R_c = 4.9k\Omega      (approx. half of Rc)\$

Voltage gain

\$A_v=\cfrac{V_o}{V_i}=\cfrac{({-}{i_c}{R_c})}{({i_b}{h_{ie}})}\$

\$A_v=\cfrac{({-}{h_{fe}}{i_b}.\cfrac{R_{bo}}{(R_{bo}+R_c)}).R_c}{i_b.h_{ie}}\$

\$A_v=\cfrac{{-}{h_{fe}}{R_{bo}}{R_c}}{({h_{ie}}(R_{bo}+R_c))}=-222.634\$

Current gain

\$A_i=\cfrac{i_c}{i_i}=\cfrac{{({h_{fe}}{i_b}{\cfrac{(R_{bo}}{(R_{bo}+R_{c})}}))}}{\cfrac{{i_b}{h_{ie}}}{Z_i}}=\cfrac{{h_{fe}}{R_{bo}}{Z_i}}{{h_{ie}}{(R_{bo}+R_c)}}=24.36\$